Aschbacher M. Finite Group Theory (2Ed., CUP 2000)(ISBN

Home , Finite group

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Library of Congress Cataloging in Publication DataData Aschbacher, Michael,Michael, 1944- Finite group theory / M. Aschbacher. -- 2nd ed. p. cm. - (Cambridge (Cambridge studiesstudies inin advancedadvanced mathematicsmathematics ;; 10) 10) Includes bibliographical referencesreferences andand index.index. ISBN 0-521-78145-0 (hb) (hb) - ISBN ISBN 0-521-78675-40-521-78675-4 (pbk.)(pbk.) 1. Finite groups. 1.I. Title. 11.II, Series. QA177.A82QA177 .A82 2000 512'.2 -- dc2ldc21 99-055693

ISBN 0 521 78145 0 hardback ISBN 0 521 78675 4 paperback To PamPam

ContentsContents

PrefacePreface 11 PreliminaryPreliminary results results 11 Elementary Elementary groupgroup theorytheory 22 Categories Categories 33 GraphsGraphs andand geometriesgeometries 44 Abstract Abstract representations representations 22 PermutationPermutation representations representations 55 PermutationPermutation representations representations 66 Sylow's Sylow's TheoremTheorem 33 RepresentationsRepresentations of of groupsgroups onon groupsgroups 77 Normal seriesseries 88 Characteristic Characteristic subgroupssubgroups andand commutatorscommutators 99 Solvable Solvable andand nilpotentnilpotent groups groups 1010 SemidirectSemidirect productsproducts 1111 CentralCentral productsproducts andand wreathwreath productsproducts 44 LinearLinear representationsrepresentations 1212 ModulesModules overover thethe groupgroup ringring 1313 TheThe generalgeneral linearlinear groupgroup andand specialspecial linearlinear groupgroup 1414 The The dualdual representationrepresentation 55 PermutationPermutation groups groups 1515 The symmetricsymmetric andand alternatingalternating groupsgroups 1616 RankRank 3 3permutation permutation groups groups 66 Extensions ofof groups and modulesmodules 1717 1-cohomology1-cohomology 1818 Coprime Coprime actionaction 7 SpacesSpaces withwith formsforms 1919 Bilinear,Bilinear, sesquilinear,sesquilinear, and quadratic forms 2020 Witt's Witt's LemmaLemma 2121 Spaces overover finitefinite fields fields 22 The The classical classical groups groups 8 p-groups 2323 Extremal p-groupsp-groups 2424 Coprime action on p-groups viii Contents

9 Change of field ofof a a linearlinear representationrepresentation 117 25 Tensor products 117 26 Representations overover finitefinite fieldsfields 123 2727 MinimalMinimal polynomialspolynomials 127 1010 PresentationsPresentations of of groupsgroups 138 138 2828 FreeFree groups 2929 Coxeter groups 141 3030 Root systems 148 11 TheThe generalized generalized Fitting Fitting subgroup subgroup 156 313 1 The generalized FittingFitting subgroupsubgroup 157 32 ThompsonThompson factorization factorization 162 3333 CentralCentral extensionsextensions 166 12LinearLinear representationsrepresentations of of finite finite groupsgroups 177 34CharactersCharacters inin coprimecoprime characteristic characteristic 178 35CharactersCharacters inin characteristiccharacteristic 00 181 36 SomeSome special special actions actions 192 13Transfer andand fusion fusion 197 37 Transfer 197 38Alperin's FusionFusion TheoremTheorem 200 39NormalNormal p-complementsp-complements 202 40SemiregularSemiregular actionaction 205 14 TheThe geometry geometry of of groups groups of of Lie Lie type type 209 41ComplexesComplexes 209 42BuildingsBuildings 215 43BN-pairsBN-pairs andand Tits Tits systemssystems 218 15SignalizerSignalizer functors functors 229 44SolvableSolvable signalizer signalizer functorsfunctors 229 16FiniteFinite simplesimple groups 242 45InvolutionsInvolutions inin finitefinite groupsgroups 243 46ConnectedConnected groupsgroups 245 47TheThe finitefinite simplesimple groupsgroups 249 48 AnAn outline outline of of the the Classification Classification Theorem Theorem 260 AppendixAppendix 269 ReferencesReferences 297 ListList of of SymbolsSymbols 299 IndexIndex 301 Preface

Finite Group TheoryTheory is intended to serve both as a text and as a basic reference on finite groups. In neither role do II wishwish thethe bookbook toto bebe encyclopedic, encyclopedic, so so I've includedincluded onlyonly thethe materialmaterial II regardregard asas mostmost fundamental.fundamental. WhileWhile suchsuch judgments areare subjective,subjective, I've been guided by a fewfew basic principles which I feel are important andand shouldshould bebe mademade explicit.explicit. One unifying notion is that of a group representation. The term representation is used here in a muchmuch broader sense than usual. Namely inin this bookbook a representation of a groupgroup GG inin aa categorycategory +?is6 is a homomorphism of G into the automorphism group of some object of -C6 Among these representations, the permutation representations, the linear representations, and and the representations of groups on groups seem to be the most fundamental. As a result much of the book is devoted to these three classes ofof representations. The first step in investigatinginvestigating representations of finitefinite groups oror finitefinite di-dimensional groups is to break up thethe representationrepresentation intointo indecomposableindecomposable or irreducible representations.representations. This This processprocess focusesfocuses attentionattention onon twotwo areasareas of study: first on thethe irreducibleirreducible andand indecomposableindecomposable representationsrepresentations themselves, and second on the recovery of the general representation from its irreducible constituents. Both areas receive attention here. The irreducible objects in the category of groups are the simplesimple groups. I regard the the finitefinite simplesimple groupsgroups and and their their irreducibleirreducible linear linear andand permutationpermutation representations as the centercenter of interestinterest inin finitefinite groupgroup theory.theory. This pointpoint ofof view above all others has dictated the choice of material. In particularparticular I feelfeel many of the deeperdeeper questionsquestions about finite groupsgroups areare bestbest answeredanswered throughthrough thethe following process. First reduce the question to aa question about somesome class of irreducible representations of simple groups or almost simple groups. Second appeal to the classification of the finite simple groups to conclude the group is an alternating group, a groupgroup of LieLie type,type, oror oneone ofof thethe 2626 sporadicsporadic simple simple groups. Finally invoke the irreducible representation theory of these groups. The book serves as a foundation for the proof of the Classification Theorem. Almost all material covered plays a role in the classification, but as itit turnsturns outout almost all is of interestinterest outside that frameworkframework too. The only major result treatedtreated here which has not found application outside of simple group theory is the Sig- nalizer Functor Theorem. Signalizer functors areare discussed near the end of the book. The last section of the book discusses the classification inin general terms. x Preface

The first edition of the book included a new proof of the Solvable Signalizer Functor Theorem, basedbased onon earlier work of Helmut Bender. Bender'sBender's proof was valid only for the prime 2, but it is very short and elegant. I've come to believe that my extension to arbitrary primes inin thethe firstfirst editionedition isis soso compli- compli- cated that it obscuresobscures the proof, so this editionedition includes only a proof of thethe Solvable 2-Signalizer Functor Theorem, which is closer to Bender's originaloriginal proof. Because of this change, section 36 has also been truncated. In some sense most of thethe finitefinite simplesimple groupsgroups areare classicalclassical linearlinear groups.groups. Thus the classical groups serve as the best exampleexample of finitefinite simple groups. They are also representative of the groups of Lie type, both classical andand ex-exceptional, finite or infinite. A significant fraction of the book is devoted to the classical groups. The discussion is not restricted to groupsgroups overover finitefinite fields. The classical groups are examined via theirtheir representationrepresentation as the automorphismautomorphism groups of spaces of formsforms and theirtheir representationrepresentation as thethe automorphismautomorphism groups of buildings. The Lie theoretic point of view entersenters intointo thethe latterlatter representationrepresentation and into a discussion of Coxeter groups andand rootroot systems.systems. I assumeassume the the readerreader hashas beenbeen exposed exposed to aa firstfirst coursecourse inin algebraalgebra oror its its equivalent; Herstein'sHerstein's Topics inin AlgebraAlgebra wouldwould bebe a representative text for such a course.course. Occasionally Occasionally some deeper algebraicalgebraic results areare alsoalso needed;needed; in such instances thethe result is quoted and a referencereference is given for itsits proof.proof. Lang's Algebra is one reference for such results. The group theory I assume is listed explicitly in section 1. ThereThere isn'tisn't much; for example Sylow'sSylow's Theorem is proved in chapter 2. As indicated earlier,earlier, thethe bookbook isis intendedintended toto serveserve bothboth asas aa texttext andand as a basic reference. Often these objectives are compatible, but when compromise is necessary it is usually in favorfavor of the rolerole asas aa reference.reference. Proofs are more terse thanthan inin mostmost texts.texts. TheoremsTheorems areare usuallyusually notnot motivatedmotivated oror illustratedillustrated with examples, but there are exercises. Many of the results in the exercisesexercises are interesting in their own right; often there is anan appealappeal toto thethe exercisesexercises inin thethe book proper. In this second edition I've added an appendix containing solutions to some of the most difficult and/orandlor important exercises. If the bookisbook is used as a texttext thethe instructorinstructor will probablyprobably wish to expand many proofs in lecture and omit some of the more difficult sections. HereHere are some suggestions about which sections toto skipskip oror postpone.postpone. A good basic course in finite group theory would consist of the first eight chapters, omitting sections 14, 16, and 17 and chapter 7, and adding sections 28, 31, 34, 35, and 37. Time permitting, sections 32, 33,33,38, 38, and 39 could be added. The classical groups and some associated Lie theory are treated in chapter 7, sections 29 and 30, chapter 14, and the latter part of section 47. A different sort of course could be built around this material. Preface xi

Chapter 9 deals with various concepts in the theory of linear representations which are somewhat less basic than most of thosethose inin chapterschapters 44 andand 12.12. MuchMuch ofof the material in chapter 9 is of principal interest for representations over fields of primeprime characteristic. characteristic. AA coursecourse emphasizingemphasizing representationrepresentation theory would probably include chapter 9. Chapter 15 is the the most most technicaltechnical and specialized. ItIt is probably only of interest to potential simple groups theorists. Chapter 1616 discussesdiscusses the finitefinite simplesimple groupsgroups andand thethe classification.classification. The latter part of section 47 builds on chapter 14,14, but the rest of chapter 1616 is pretty easy reading.reading. Section 48 consists of a very brief outline of thethe proofproof ofof thethe finitefinite simple groups makes use of results from earlier inin the book and thus motivates those results by exhibiting applications of thethe results.results. Each chapter begins with a short introduction describing the major results in the the chapter. chapter. MostMost chapters close with a fewfew remarks.remarks. SomeSome remarksremarks ac- knowledge sources for material covered in the chapter or suggestsuggest references for further reading. Similarly,Similarly, some ofof thethe remarks place certain resultsresults inin con-con- text and and hencehence motivatemotivate those results. Still others warn that some some sectionsection inin the chapter is technicaltechnical or specializedspecialized andand suggestssuggests thethe casualcasual readerreader skipskip oror postpone the section. In addition to thethe introductionintroduction and the remarks, there is another good way to decide which results inin aa chapterchapter areare ofof mostmost interest:interest: thosethose resultsresults whichwhich bear some sortsort of descriptivedescriptive label (e.g.(e.g. Modular Property of Groups,Groups, FrattiniFrattini Argument) are oftenoften ofof most importance.importance.

Preliminary results

I assume familiarityfamiliarity with material from aa standardstandard coursecourse onon elementaryelementary alge- alge- bra. A typical text for such a course is Herstein [He]. A few deeper algebraic results areare alsoalso needed;needed; theythey cancan bebe foundfound forfor example example in in LangLang [La]. [La]. Section Section 1 1 lists the elementary group theoretic results assumed and also contains a list of basic notation. Later sectionssections inin chapter 11 introduce some some terminology andand no-notation from a few other areas ofof algebra.algebra. DeeperDeeper algebraicalgebraic resultsresults are introducedintroduced when they are needed.needed. The last section of chapter 11 contains a brief discussion of group representations. The termterm representationrepresentation isis used here in aa more general sensesense thanthan usual. Namely a representation of a group GG willwill bebe understoodunderstood toto bebe aa group homomorphism of G into the group of automorphisms of an objectobject X. Standard useuse of the term representation requires X to be a vector space.

1 ElementaryElementary groupgroup theorytheory Recall that a binary operation on aa setset GG isis aa function function from. from the set product GGxG x G into G. Multiplicative notation will usually be used. Thus the image of a pair (x, y) underunder thethe binarybinary operation will be written xy. The operation is associative if (xy)z = x(yz) x(yz) for for all all x, x, y, y, z z in in G. G. TheThe operation operation is is commutativecommutative if xy = yx yx for for all all x, x, y y in in G. G. AnAn identity identity for for thethe operationoperation is anan element 1 inin G suchsuch that xlx 1 == lxlx == x x for for all all x x in in G. G. An An operation operation possesses possesses at at most most oneone identity. Given an operation onon G possessing an identity 1, an inverse for an element x of G is an element yy inin GG suchsuch thatthat xy x y = = yxyx = 1. 1. If If ourour operationoperation is associative and x possesses anan inverse then that inverse isis uniqueunique and isis denoteddenoted by x-'x-1 inin multiplicativemultiplicative notation.notation. A group isis aa setset G G together together with with an an associative associative binary binary operation operation which which possesses an identity and such that each element of G possessespossesses anan inverse.inverse. The group is abelian ifif itsits operationoperation isis commutative.commutative. InIn thethe remainderremainder ofof thisthis section G isis aa groupgroup writtenwritten multiplicatively.multiplicatively. Let xX E G and n a positivepositive integer. xnx" denotes the product of x withwith itselfitself x_n n times. Associativity insures xnx" is aa well-definedwell-defined element of G. DefineDefine x-" to be (x-')"(x-1)" andand x°x0 to to bebe 1.1. TheThe usualusual rulesrules ofof exponentsexponents can be derived from 2 Preliminary resultsresults

this definition:

(1.1) Let G be a group, x EE G,G, andand nn andand m integers. Then (1) (xn)(xm)(xn)(xm) = xn+m = (xm)(xn). (xrn)(x"). (2) (xn)m == xnm.xnm

A subgroup of G is a nonemptynonempty subsetsubset HH of G such thatthat forfor eacheach x, x, yy E H, xy and x-1x-' areare inin H.H. This This insures insures that that thethe binarybinary operationoperation on G restricts to a binary operation on H whichwhich makesmakes HH intointo a a groupgroup with with thethe samesame identityidentity asas G and the same inverses. II writewrite HH (< GG toto indicateindicate that H isis a subgroup of G.

(1.2) TheThe intersectionintersection of of anyany setset ofof subgroupssubgroups of of GG isis alsoalso aa subgroupsubgroup of of G.G.

Let SS CE GG and and define define (S) = n H scH

(1.3) Let S Cg G.G. Then (S) = {(sl)E1 {(s~)~'...(s~)~":s~ ... (S,)"': Si EE S,S, e&=+I = +1 oror-1). -1}.

(1.4) Let xX EE G.G. ThenThen (x)(x) = {xn:{xn: nn Ec 1].Z].

Of course 1.41.4 is a special case of 1.3.1.3. A group G isis cycliccyclic if it is generated by some element x.x. In that case x is said to be a generator ofof GG andand byby 1.4,1.4, GG consists of the powers of x. The orderorder ofof aa groupgroup GG isis the the cardinality cardinality ofof thethe associatedassociated setset G.G. Write Write IGI IGl forfor thethe orderorder of of aa set set G G or or a a group group G. G. For For X x E E G, G, Ix Ix I Idenotes denotes I I(x) (x) I1 and is called the order of x. AA group homomorphism from a group G into a groupgroup H is a functionfunction a:a : GG 4-* H ofof thethe setset GG into into the the set set H H which which preserves preserves the the group group operations:operations: that isis forfor allall x, y inin G,G, (xy)a(xy)cr == xaya. xaya. Notice Notice that that I Iusually usually writewrite mymy mapsmaps onon thethe right,right, particularlyparticularly thosethose thatthat are are homomorphisms. homomorphisms. TheThe homomorphism homomorphism a isis an an isomorphismisomorphism if acr isis a a bijection. bijection. InIn thatthat casecase aa possesses possesses an an inverse inverse functionfunction a-':a`1: HH -*+ G G and and it it turns turns out out a-1 a-' is is also also a agroup group homomorphism. homomorphism. GG isis isomorphic toto HH if therethere existsexists anan isomorphism isomorphism of of G G and and H. H. Write Write G G Z = H to indicateindicate that G is isomorphic to H. IsomorphismIsomorphism is an equivalence relation.relation. HH isis said to be a homomorphic imageimage of G ifif therethere is a surjective homomorphism of G onto H. Elementary group theory 3

A subgroup HH of G is normal ifif g-'hgg-lhg c-E H for for eacheach g EE GG andand h c-E H. Write HHI! a GG toto indicateindicate HH is is a a normal normal subgroupsubgroup of G. IfIf a:a: G G + X is aa group homomorphismhomomorphism thenthen the the kernel kernel of of a ais is ker(a) ker(a) = _ {g E G: gaga == 1} 11 and and it turns out that ker(a) is a normal subgroup ofof G. Also write Ga forfor thethe imageimage {ga: g E G}GI of GGin in X.X. GaGa is is a a subgroup subgroup ofof X.X. Let H <5 G. G. For Forx X EE GG writewrite Hx = {hx:{hx: h EE H}HI andandxH xH = {xh:{xh: h EE H}.HI. Hx and xH are cosets of H inin G.G. HxHx isis aa rightright cosetcoset andand xH a left coset. To bebe consistentconsistent I'llI'll work with right cosets HxHx inin thisthis section.section. GIHG/H denotesdenotes the set ofof allall (right)(right) cosetscosets of of H H inin G. G. GIHG/H is the coset spacespace ofof HH in G. Denote by IIG G :: HIH Ithe the order order of of the the coset coset space space G/H. G/H. As As the the map map h h t+ H hxhx isis aa bijectionbijection of H withwith Hx,Hx, allall cosets have the same order, so

(1.5) (Lagrange's Theorem)Theorem) LetLet GG bebe aa group group and and H H i < G.G. ThenThen IGIG I= _ HIHJ I I(G G:: H 1. In In particularparticular ifif GG is is finite finite then then I HIH I ( dividesdivides I \GI.G I.

If H a5 G G the the coset coset space space G/H GIH is is made made into into a a group group by by definingdefining multiplicationmultiplication via (Hx)(Hy) = Hxyx, y E G Moreover there is a natural surjective homomorphismhomomorphism n: rr: G G + -* GIHG/H defineddefined by n:rr: x x t+H Hx.Hx. NoticeNotice ker(n)ker(rr) = H.H. Conversely Conversely if a: G -*+ L L isis aa surjective surjective homomorphism with ker(a) == H H then then the the map map ,B: /3: Hx t+i-+ xaxa is an isomorphism of G/HGIH with with LL suchsuch thatthat ir,Bnp == a. The The group G/HGIH is is calledcalled thethe factor group of G by H. ThereforeTherefore thethe factorfactor groupsgroups of GG overover itsits variousvarious normal subgroups are, up to isomorphism, precisely the homomorphic images of G.

(1.6) Let H aI! G. G. Then Then the the mapmap LL Ht+ L/HLIH is isa bijectiona bijection between between the the set set of of allall subgroups of G containing HH and the setset of allall subgroupssubgroups of G/H.GIH. NormalNormal subgroups correspond to normal subgroups under this bijection.

For x,x, yy E G, set XYxy = y-lxy.y-lxy. ForFor X X CG g Gset set Xy XY _ ={xY: {xY:x x E E X). X). XY XY isis thethe conjugate of X under y. Write XGxG for the set {X8:{Xg: gg Ec G}G) of conjugates of X under G. Define

NG(X)NG(X)={gEG:XB=X). = {g E G:Xg = X). NG(X) is the normalizer inin GG ofof XX and isis aa subgroupsubgroup ofof G.G. IndeedIndeed if if XX 5< G then NG(X) is the largest subgroup of G in which X isis normal.normal. DefineDefine CG(X) =_ {g E G: xgxg == gx for all x EE X}.X). CG(X) is the centralizer inin GG of X. CG(X) is is alsoalso aa subgroup of G. 4 Preliminary resultsresults

ForFor X, Y C GG definedefine XYXY == {xy: {xy: xx EE X,X, y y E E Y).Y}. TheThe set set XY XY isis thethe productproduct of X withwith Y.Y.

(1.7) Let X, Y 5< G.G. Then Then (1)(1) XYis XY is a a subgroup subgroup of of G G if if and and only only if if XY XY == YX. YX. (2)(2) If If Y Y < 5 NG(X) NG(X) thenthen XY XY isis aa subgroupsubgroup of G and XY/X 2= Y/(Y n X). (3)(3) IXYIIXYt == )XHHYl/IX tXIIYIIIX nn YI.Y.

(1.8)(1.8) LetLet H andand K be normal subgroupssubgroups of of G G with with K K 5 < H.H. ThenThen G/K/H/K G/K/H/K G G/H.G/H.

LetLet G1, G1,...... , ,G. G, be be a a finite finite set set ofof groups. TheThe directproductdirectproduct GIG1x x - .... . x GnG, = lny=, 1n=1GiGi ofof thethe groupsgroups G1,GI, ...,. . . Gn, G, is is the the group group defined defined on on the the set set product product G1G1 x ...... xx GnG, byby thethe operationoperation (x1,...,Xn)(y1,...,yn)=(xiyi,...,xnyn)xi,yi c Gi

(1.9)(1.9) LetLet GG bebe aa groupgroup and (Gi: 1 1 <5 ii <5 n) n) a afamily family of of subgroups subgroups of of G.G. Then Then thethe followingfollowing areare equivalent:equivalent:

(1)(1) The The map map (xi,(xl,...,. . . xn) , x,) HI+ x1xl ... . x,. . x,is anis an isomorphism isomorphism of of G G with with G1GI xx .. . xx G.G, . (2)(2) G G == (Gi: (Gi: 11 <5 i i

(1.10)(1.10) LetLet GG = = (g) (g) be be a acyclic cyclic group group and and Z Z thethe group group of of integersintegers under under addi-addition.tion. ThenThen (1)(I) IfIf HH is is a anontrivial nontrivial subgroup subgroup ofof ZZ thenthen HH == (n), (n), wherewhere n n is is the the least least positivepositive integer in H. (2)(2) The The map map a: a: 71Z -±-+ GG defined defined by mama = g' g" is is a asurjective surjective homomorphismhomomorphism withwith kernelkernel (n),(n), wherewhere n == 0 0 if if gg is is of of infinite infinite order order and and nn = = min min {m {rn >> 0: 0: gmgm == 11 1) if if gg hashas finitefinite order. order. (3)(3) Iflf gg hashas finitefinite order n n then then G G = = {gi:0(g" 0 <5 ii << n} n) and and n n is is the the least least positivepositive integer rnm withwith gm g' == 1. 1. (4)(4) Up Up to to isomorphism isomorphism 71 Z isis thethe uniqueunique infiniteinfinite cycliccyclic groupgroup andand forfor eacheach positivepositive integer integer n, n, the the group group 71n Z, of integers modulo nn isis thethe uniqueunique cycliccyclic group group ofof orderorder n.n. Elementary group theory 5

(5) Let Let IgI Ig ( == n. n. Then Then for for each each divisor divisor m m of n, (g"/')(gn/m) isis the the unique unique subgroupsubgroup of G of order m. In particular subgroups of cyclic groups are cyclic.

(1.11) Each finitelyfinitely generated abelian group is thethe directdirect productproduct ofof cycliccyclic groups.

Let p be aa prime.prime. AA p-group is a group whose order is a power ofof p.p. More generally if nTr isis aa setset ofof primes thenthen a n-grouprr-group is a group G ofof finitefinite order such that 7rn(G) (G) gC n,jr, where where 7rn(G) (G) denotesdenotes the set of prime divisors ofof I/GI. GI. p' denotes the set of all primesprimes distinctdistinct fromfrom p.p. AnAn elementelement xx inin a groupgroup GG is a n-element7r -element if if (x)(x) is a n-group.7r -group. An An involution involution is is anan elementelement of order 2.

(1.12) LetLet 1 #: G G be be an an abelian abelian p-group.p-group. ThenThen GG isis the the direct direct product product ofof cyclic subgroupssubgroups Gi G, 2 = Zpe,,Z pe; 1 ,1_( ez e2 2 > ...... L.> een > > 1. 1.Moreover Moreover the the integers n and (e1:(ei: 11 5< ii <5 n)n) areare uniquely uniquely determineddetermined by by G.G.

The exponent of a finite group G is thethe leastleast commoncommon multiple of the ordersorders of the elements of G. An elementaryelementary abelian p-group is an abelian p-group of exponent p. Notice that by 1.12,1.12, G is anan elementaryelementary abelian p-groupp-group of order p"pn if if andand onlyonly ifif GG is is the the direct direct product product of of n n copies copies of of TLp. Z,. In particular up to isomorphism there isis aa uniqueunique elementaryelementary abelianabelian p-group p-group ofof orderorder pn,p", which will be denoted by Ep.Ep,. . The integer n is the p-rank of Ep».Epa. The p-rankp-rank of a general finite group G is the maximum p-rank ofof anan elementaryelementary abelianabelian p-subgroup ofof G,G, and and isis denoteddenoted by mp(G).

(1.13) EachEach groupgroup ofof exponentexponent 22 isis abelian.abelian.

If nTr isis a a set set of of primes primes and and G G aa finite finite group, group, write write 0, 0, (G) for the largest normal n-subgroup7r-subgroup ofof G,G, and OK(G)O" (G) for for thethe smallestsmallest normalnormal subgroupsubgroup HH of GG suchsuch that G/H isis aa 7r-group.n-group. O,(G)0, (G) andand 0'OK(G) (G) areare well well defined defined by by ExerciseExercise 1.1. 1.1. Define Z(G) == CG CG(G) (G) and call Z(G) the center of G. If G is aa p-groupp-group thenthen define

Q,(G)=(xEG:x"Q,,(G) = (x E G: xp" _1) = 1)

Un(G)Y(G)=(x'':xEG). = (xp":x E G).

For X <( G G definedefine AUtG(X)AutG(X) == NG(X)/CG(X)NI(X)/CG(X) to be the automizer inin G of X. Notice that by ExerciseExercise 1.3,1.3, AutG(X) AutG(X) _( < Aut(X)Aut(X) and indeedindeed Autc(X)Autc(X) is the group of automorphisms induced on X in G. A maximal subgroupsubgroup ofof aa groupgroup G is a properproper subgroupsubgroup of G whichwhich is is properly containedcontained in no proper subgroup of G. That isis aa maximal subgroup is 6 Preliminary resultsresults

a maximal member of the set of proper subgroups ofof G, partially ordered byby inclusion.inclusion. IfIf a: S S -++ T T is is a a function function and and RR CG S S then then a)R ~IR denotesdenotes thethe restrictionrestriction of a toto R.R. ThatThat isis aalR: 4R:R R + - * TT isis thethe functionfunction fromfrom R into T agreeingagreeing with a. Here's aa littlelittle result that'sthat's easy easy toto prove prove butbut useful.useful.

(1.14)(1.14) (Modular(Modular Property Property of of Groups)Groups) Let Let A,A, B,B, andand CC bebe subgroupssubgroups of of a group GGwithA with A <5 C. C.ThenABnC Then AB F) C ==A(BnC). A(B F) C).

IfIf GG isis aa groupgroup write G#G' for the set G - (1) (1) of of nonidentity nonidentity elementselements ofof G.G. OnOn thethe otherother hand if R is a ring define R'R# == R -- (O). (0). DenoteDenote by C,C, R,R, andand 0Q the the complex complex numbers, numbers, the the reals, reals, and and thethe rationals, rationals, respectively.respectively. Often Z1 willwill denotedenote thethe integers.integers. GivenGiven a group G, a subgroup H ofof G,G, and and aa collectioncollection C C ofof subgroupssubgroups of of G,G, I'llI'll oftenoften writewrite cC n n H H for for the the set set of of members members of of CC which which are are subgroups subgroups of of H.H. I'llI'll useuse thethe barbar convention. convention. ThatThat isis I'llI'll oftenoften denote denote aa homomorphichomomorphic image GaGa of of aa group group GG byby GG (or(or G* G* or or G)G) andand write write gg (or(or g*g* or g)8) for ga. This This willwill be donedone withoutwithout comment.comment. OtherOther notation and terminology areare introduced in later chapters. The List of Symbols gives gives the page number wherewhere aa notation is first introduced and defined.

22 CategoriesCategories ItIt willwill bebe convenientconvenient toto havehave availableavailable somesome ofof thethe elementaryelementary conceptsconcepts andand languagelanguage of of categories.categories. For For a a somewhatsomewhat more more detailed detailed discussion, discussion, see see chapter chapter 1 1 ofof LangLang [La].[La]. AA categorycategory fi?i' consistsconsists ofof (1)(1) AA collectioncollection Ob(&)Ob(i) of of objects. objects. (2)(2) ForFor each each pair pair A,B A,B ofof objects,objects, a a set set Mor(A,B) Mor(A,B) ofof morphismsmorphisms fromfrom AA toto B.B. (3)(3) ForFor each each triple triple A,A, B,B, C C of of objectsobjects a a mapmap Mor(A, B) x Mor(B, C) Mor(A, C) calledcalled composition.composition. WriteWrite fgf g for the image of the pair (f, g)g) under under thethe compositioncomposition map. map. MoreoverMoreover thethe followingfollowing three three axiomsaxioms are are required required to to hold: hold: CatCat (1)(1) ForFor each each quadruple quadruple A, A, B, B, C, C, D D of of objects, objects, Mor(A, Mor(A, B) B) f1 n Mor(C, Mor(C, D) D) is is emptyempty unless AA == C and B = D. D. CatCat (2)(2) Composition Composition is is associative. associative. CatCat (3) (3) For For each each object object A, A, Mor(A, Mor(A, A) A) possesses possesses an an identity identity morphism morphism 1A 1.4 such such thatthat forfor allall objectsobjects B andand allall ff in in Mor(A, Mor(A, B) B) and and g g in in Mor(B, Mor(B, A), A), lAf=fandglA=g.1.4 f = f and g1.4 = g. Graphs and geometries 7

Almost allall categories considered here will be categories of sets withwith structure.structure. That is the objectsobjects of the category are setssets together with some extra structure, Mor(A, B) consists of all functions from thethe set associated toto A to the set associated to B which preserve the extra structure, and compositioncomposition is ordinary composition of of functions. The identity morphism IAlA is forced to be the identityidentity map on A.A. ThusThus wewe needneed toto knowknow the the identity identity mapmap preserves preserves structure. structure. WeWe also needneed toto knowknow thethe compositioncomposition ofof mapsmaps whichwhich preservepreserve structure structure also also preserves structure. These facts will usually be obvious in the examplesexamples wewe consider. We'll bebe mostmost interestedinterested inin thethe followingfollowing threethree categories,categories, which are all all categories ofof setssets withwith structure.structure.

(1) The categorycategory of setssets andand functions:functions: Here the objects are the sets sets andand Mor(A, B)B) isis thethe setset ofof allall functionsfunctions fromfrom thethe setset AA intointo thethe setset B.B. (2) TheThe categorycategory ofof groupsgroups andand groupgroup homomorphisms:homomorphisms: The objects are the groups and morphisms areare thethe groupgroup homomorphisms.homomorphisms. (3) TheThe category category of of vectorvector spaces and linear transformations: Fix aa fieldfield F.F. TheThe objects are the vector spaces overover FF and the morphisms are the F-linear transformations.

Let f bebe a a morphism morphism from an object A toto anan object object B. B. An An inverse inverse for for f fin in ff? -' is a morphismmorphism gg Ee Mor(B, A)A) such that lA1A = = f g and 1lB B= = ggf.f. The morphism f isis an an isomorphism isomorphism if if itit possessespossesses an an inverseinverse inin w.ff?. AnAn automorphismautomorphism of of AA isis an isomorphism from A to A. Denote by Aut(A) the set of all automorphisms of A and observe Aut(A)Aut(A) formsforms aa group group under under the the composition composition in in 6'. i. If a: A + Bisanisomorphismdefinea*:Mor(A,B is an isomorphismdefine a*:Mor(A , A) A) + Mor(B, B)B) by B + a-l~aa`1 fla andand observe observe a* a* restrictsrestricts to to a group isomorphism of Aut(A) with Aut(B).Aut(B). Let (Ai: i Ee I)I) be aa familyfamily of objects in aa categorycategory -i'.ff?. AA coproductcoproduct ofof thethe family isis anan objectobject C C together together with with morphisms morphisms ci: c, Ai: A; + -- C,C, i i E I, satisfying satisfying the the universal property: whenever X is an object and at:ai: AlAi --+ -+ X areare morphisms,morphisms, there existsexists aa uniqueunique morphism morphism a: a: C C + -+ X X with with cia cia = = aiai for each ii eE I.I. As As aa consequence of of thethe universal property, the coproduct ofof a family is determined up to isomorphism, if it exists. The product of thethe familyfamily isis defineddefined dually.dually. That is toto obtainobtain thethe definitiondefinition of the product, take the definition of the coproduct and reverse the direction of all arrows. Exercise 1.2 gives a descriptiondescription of coproducts and products in the threethree categories listed above.

3 GraphsGraphs and and geometries geometries This section contains a briefbrief discussiondiscussion ofof twotwo moremore categoriescategories whichwhich willwill make occasional appearances in these notes. 8 Preliminary results

A graph fi=B = (V, (V, *) *) consists consists of of a aset set V V of of vertices vertices (or (or objects objects or or points)points) to-together with a symmetricsymmetric relation * called adjacency (or incidence or something else). The ordered pairs in the relation are called the edges of the graph. I write u * vv toto indicateindicate twotwo verticesvertices areare relatedrelated viavia ** andand saysay uu isis adjacentadjacent to v. ApathA path of length n from u to vv isis aa sequencesequence ofof verticesvertices u == uo, ug, u u 1, 1, ...... , , u, U, == v v such such that ui.ui+lu,*u,+ forfor eacheach i.i. Denote Denote byby d(u,d(u, v) v) thethe minimal minimal length length ofof aa pathpath fromfrom uu to v.v . If nono suchsuch path path exists exists set set d d(u, (u , v) v) == no.oo. d(u,d (u ,v) v) is is the the distancedistance fromfrom u to v. The relation - - onon VV defineddefined byby uu -- v v ifif andand onlyonly if d(u, v) < nooo is an equivalence relation on V. The equivalence classes of this relation are called the connected components ofof thethe graph. TheThe graphgraph isis connected ifif itit hashas justjust one connected component. Equivalently therethere isis aa path between any pair of vertices. A morphism a: a: B9 -, i'g' of of graphs graphs is is a a function function a: a: V V -+ + V' V' fromfrom the the vertex vertex set V of $?to9 to the vertex setset V'V' ofof @''" whichwhich preserves preserves adjacency; adjacency; thatthat is if u and v are vertices adjacent inin @'then07thenua ua isis adjacentadjacent to to vava inin i?'.9'. So much for graphs; on to geometries. In this book I adopt a notion of ge-geometry due to Tits.Tits. LetLet II be a finite set. AA geometrygeometry over over I I isis aa tripletriple (r,(F, t.,r, *) where F is a set ofof objects,objects, t.:r: rr -++ I I is is a atype type function, function, and and ** isis aa symmetricsymmetric incidenceincidence relation on rl? suchsuch that that objects objects u u and and vv ofof thethe samesame type type areare incidentincident if andand only if u = v. v. r(u)t.(u) is is the the type type of of the the object object u.u. NoticeNotice (F,(r, *) *) isis aa graph.graph. I'll usuallyusually write r forfor thethe geometrygeometry (F,(r, r,t., *). *). A morphism a: a: rF +-+ F'r' of of geometriesgeometries is a function a:a: rF -++ F'r' of of thethe asso-associated object sets whichwhich preservespreserves typetype and and incidence; incidence; that that is is if if u, u, v v E c rF with u * v thenthen t.(u) r(u) == ~'(ua)r'(ua) andand ua *'*' va. AA$ag flag of the geometrygeometry rr isis aa setset TT ofof objectsobjects such such thatthat eacheach pairpair ofof objectsobjects in TT isis incident.incident. NoticeNotice ourour oneone (weak)(weak) axiomaxiom insuresinsures thatthat aa flagflag TT possessespossesses at most one object of each type, so that the typetype functionfunction t.r induces anan injection of T intointo I. The The imageimage rt.(T) (T) isis calledcalled thethe typetype of T. The rank and corank of T are the order of t.(T)r (T) and I - r t.(T),(T ), respectively.respectively. The residueresidue rT ofof thethe flagflag TT is (v cE Fr -- T: T: v v * * t tfor for all all t t cE T T] J regarded regarded as aa geometrygeometry over over I I- - t.(T).r(T). The geometry r is is connected connected if if itsits graph graph (I',(r, *)*) isis connected.connected. Fr isis residually residually connected if the residue of every flag of corank at least 2 is connected andand the residueresidue ofof everyevery flagflag of corank 11 is is nonempty.nonempty. Here's aa wayway toto associate associate geometriesgeometries to groups. Let G bebe aa groupgroup andand 9= (Gi: (G,: ii EE I) I) a afamily family of of subgroupssubgroups of of G.G. Define Define F(G,r(G, F)) to be the geo- metrymetry whose set of objects of type i isis thethe coset space G/GiGIG, andand withwith objects objects GixG,x andand GayG,y incidentincident ifif GixG,x fln GayG, y is nonempty.nonempty. For For JJ Gc_ I I write J' forfor thethe complement I I- - JJ ofof JJ in I and definedefine G G J =j = nJEJ flG3.G,. Observe that forfor xEG,Sj,X=(Gjx:jEJ}isaflagx E G, Sj,, = (GJx:j E J) is a flag of r(G, g)of type J. A group H of automorphisms of a geometry r isis saidsaid toto bebeflag flag transitive if HH isis transitivetransitive on flags ofof typetype JJ for each subsetsubset JJ of I. Abstract representations 9

4 AbstractAbstract representations representations Let 6'B bebe aa category.category. A -2-representation&-representation of a group G is a groupgroup homomorphism nn: : GG -f+ Aut(X) Aut(X) of of G G into into the the group group Aut(X) Aut(X) of of automorphisms automorphisms of of somesome object X in 6.6'. (Recall(Recall the the definition definition of of Aut(X)Aut(X) inin sectionsection 2.)2.) WeWe will be most concerned with the following three classes of representations. A permutationpermutation representation is aa representation in the category of sets and functions. TheThe group Aut(X) of automorphisms ofof setset X isis thethe symmetricsymmetric group Sym(X) of X. ThatThat is Sym(X)Sym(X) isis thethe groupgroup ofof allall permutationspermutations of X under composition. A linear representationrepresentation is is aa representationrepresentation in in thethe categorycategory ofof vectorvector spacesspaces and linear transformations. Aut(X)Aut(X) isis thethe generalgeneral linearlinear group GL(X) of the vector space X. That is GL(X) is the group of all invertible linear transformations of X. Finally we will of course be interested in the category of groups andand groupgroup homomorphisms. Of particular interest is the representation of G viavia conjugation on itself (cf.(cf. Exercise 1.3).1.3). Two '-representationsB-representations nl ni: : G -++ Aut(X1 Aut(Xi), ), i i == 1,1,2, 2, are said to be equivalent if there exists an isomorphism a:a: X1 +-+X2 X2 such such thatthat n2n2 == Trla*, nla*, where where a*: Aut(X1) -++ Aut(X2) Aut(X2) is is thethe isomorphism described in section 2. The map a is said toto bebe anan equivalenceequivalence of of the the representations. representations. &-representations 6-representations n,ni: : Gi +-+ Aut(Xi)Aut(Xl) i == 1, 1, 2, 2, are are said said to to be be quasiequivalentquasiequivalent if if therethere existsexists aa groupgroup isomor-isomor- phismphismQQ,B: /3: G2 G2-++ G1 G1 of ofgroups groups and and a a&-isomorphism 6-isomorphism a: a: X1 X1 +-+ X2 such that n2 = N7rla*/3n1a*. Equivalent representationsrepresentations of of a a groupgroup GG are the same for ourour purposes. purposes. Quasiequivalent representations are almost the same, differing only by an automorphism of G. A representation 7r n of G isfaithful faithful if n7r is is an an injection. injection. InIn that that event event n n inducesinduces an isomorphism of G with the subgroup G7rGn of Aut(X), so G may be regarded as a group ofof automorphismsautomorphisms of of X X via via 7T n. . Let ni:ni: G -++ Aut(XZ Aut(Xi), ), ii == 1,1,2, 2, be 6-representations.&-representations. Define Define a a G-morphism a: X1 X1 -++ X2 X2 to to be be aa morphism morphism a a Of of X1X1 to X2 which commutes with the action of G in the sense that (gnl)a(gal)a = = a(g7r2) a(gnz) for for each each g g E E G. G. Write Write MorG(X1, Morc(X1, X2)X2) for the set of G-morphismsG-morphisms of X1 to X2. Notice that the composition ofof G- morphisms is a G-morphismG-morphism and thethe identity identity morphism morphism is aa G-morphism.G-morphism. Similarly define a G-isomorphism toto bebe a G-morphism which is also an isomorphism. Notice the G-isomorphisms are the equivalences ofof representationsrepresentations of G. One focus of this book is the decompositiondecomposition of of a representation 7r n into smaller representations. Under suitable finiteness conditions (which are always present in thethe representationsrepresentations considered here) this process of decompositiondecomposition must terminate, at which point wewe havehave associated associated toto Trn certaincertain indecomposableindecomposable 1010 Preliminary resultsresults or irreducibleirreducible representationsrepresentations which cannot be brokenbroken downdown further.further. ItIt willwill develop that the indecomposables associated associated toto rrn areare determineddetermined up up toto equiv-equivalence.alence. Thus we are reduced to a consideration of indecomposable representations.tions. In general indecomposables are not irreducible, so an indecomposable rep- resentationresentation itn cancan bebe brokenbroken downdown further,further, and we can associate to rrn a set of irreducibleirreducible constituents. SometimesSometimes these irreducibleirreducible constituentsconstituents areare deter-deter- minedmined upup toto equivalence,equivalence, and and sometimessometimes not. not. EvenEven whenwhen thethe irreducibleirreducible con- con- stituentsstituents are determined, they they usually do not determine jr.n. Thus we will also be concernedconcerned withwith thethe extensionextension problem: Given a set S of irreducible representations,tations, whichwhich representationsrepresentations havehave SS as as their their setset of of irreducibleirreducible constituents? constituents? . ThereThere isis alsoalso thethe problemproblem ofof determiningdetermining the the irreducibleirreducible and and indecomposable indecomposable representationsrepresentations of of thethe group.group. ItIt isis possiblepossible toto givegive aa categorical categorical definitiondefinition ofof indecomposabilityindecomposability (cf.(cf. Ex-Ex- erciseercise 1.5).1.5). There There isis alsoalso aa uniformuniform definition definition ofof irreducibilityirreducibility forfor thethe classesclasses ofof representationsrepresentations considered considered mostmost frequentlyfrequently (cf. (cf. ExerciseExercise 1.6). 1.6). I I have have cho- cho- sensen howeverhowever to relegaterelegate thesethese definitionsdefinitions to thethe exercisesexercises andand toto makemake thethe appropriateappropriate definitions of indecomposabilityindecomposability and irreducibilityirreducibility forfor eacheach cat-cat- egoryegory in the the chapterchapter discussingdiscussing thethe elementaryelementary representation theory of the category.category. This process begins in the next chapter, which discusses permutation representations.representations. HoweverHowever one case is of particular interest. A representation of a group G on itselfitself viavia conjugationconjugation (in(in thethe categorycategory ofof groupsgroups andand groupgroup homomorphisms)homomorphisms) isis irreducibleirreducible ifif GG possessespossesses nono nonidentitynonidentity properproper normalnormal subgroups.subgroups. InIn thisthis casecase GG isis saidsaid toto bebe simple. To my mind the simple groups and their irreducibleirreducible linearlinear andand permutation representationsrepresentations are are thethe centercenter ofof interestinterest inin finitefinite group theory.theory.

ExercisesExercises for chapter 1 1 1.1. LetLet G G be be a a finite finite group, group, jr n a aset set of of primes, primes, 0 52 the the set set of of normal normal 7r n-subgroups -subgroups ofof G,G, and r the the set set of of normalnormal subgroupssubgroups X of G with G/X aa rr-group.n-group. Prove (1)(1) IfIf H,H, K K E E 0 thenthen HK HK E 0.E a.HenceHence (S2) (a) isis the the unique unique maximal maximal member member of0.of a. (2)(2) IfIf H,H, K K EE Pr thenthen HH fl n K K EE P. r. Hence Hence HH is is the the unique unique minimal minimal InH,,IHEP membermember of P.r. 2.2. LetLet e,8, bebe the the category category of of sets sets and and functions, functions, 82'2 thethe categorycategory ofof vec-vec- tortor spacesspaces andand linearlinear transformations,transformations, and 03ti?, thethe category category ofof groupsgroups and and homomorphisms.homomorphisms. Let Let FF == (A,:(Ai: 1 _(< i <_( n)n) bebe aa familyfamily ofof objectsobjects in 1k.gk. ProveProve (1)(1) Let Let kk == 1. 1. Then Then thethe coproduct coproduct CC ofof FF is is the the disjoint disjoint union union of of thethe setssets A,Ai with with c1:ci: AiA, + -). C the inclusion map.map. TheThe productproduct PP of F isis Abstract representationsrepresentations 1111

the setset product Al x .... . -x x An A, withwith pi:pi: P P -+ + Ai Ai the the projection projection map map Pi:(a1,...,an)Hai.pi :(al, .. .,a,) H ai. (2)(2) LetLet k = 2. 2. ThenThen C = PP == i @:='=,Ai1 Ai is the is the direct direct sum sum of of the the subspaces subspaces

Ai, withwith cici: : AiAi +-). C defined by aiciai ci == (0,(0,. . . ., , ai, . . . .., ,0) 0) and pi: P + Ai thethe projectionprojection map.map. (3)(3) LetLet k k = = 3. 3. Then Then the the product product P P of of F F is is the the direct direct product product A Al 1 x .. . . x An.A,. with pi: P + AiAi thethe projectionprojection map.map. (The(The coproductcoproduct turnsturns outout toto bebe the so-calledso-called freefree productproduct ofof thethe family.)family.) 3. Let G G bebe a agroup, group, H H < II G, G, and and for for g gE EG G define define g7r:gn: H -).+ H byby x(gir)x(gn) = = xg, xg, x x E E H. H. Let Let -' &be be the the category category of of groups groups and and homomorphisms. homomorphisms. Prove nit is is aa -'-representation&-representation ofof G G withwith kernelkernel CG(H).CG(H). itn isis thethe repre-representation by conjugation of of GG onon H.H. If H == G, G, the the image image ofof GG underunder itn isis thethe inner inner automorphismautomorphism group of GG andand isis denoteddenoted byby Inn(G).Inn(G). ProveProve Inn(G) Aut(X) be a gk-representation,-'k-representation, where where X X = = C C if if k k = = 1 oror 2,2, and and X X = = PP if k = 3.3. Prove the following are equivalent: (a) There There exist exist -1'k-representations &k-representations r1: ni: GG + Aut(AiAut(Ai), ), 11 5< i <5 n,n, suchsuch that 7rn = $4, where 40 is the injection ofof Exercise 1.41.4 and n $ : G + F1n Aut(Ai) is defined byby g$ g* = (girl (gnl, , ...... , ,gJrn).gn,). i-1 (b) jrn isis decomposable. decomposable. (If k == 1, 1, transitivity transitivity is is the the same same as as indecomposability. indecomposability. SeeSee chapterchapter 2 for the definition definition of transitivity. SeeSee chapters 55 and 4 forfor thethe definitionsdefinitions ofof decomposability whenwhen kk == 2 and 3.) 6. AssumeAssume thethe hypothesishypothesis andand notation of Exercise 1.2, let X bebe anan object in -'k,-ek, and and p:p: G G + Aut(X) a -'k-representation.&k-representation. A A -ik-equivalence&k-equivalence relation relation on X isis anan equivalenceequivalence relationrelation - on X such that - is is preserved preserved byby thethe operations onon XX ifif kk = 22 oror 33 (i.e. (i.e. ifif yy is is an an n-ary n-ary operation operation onon XX andand 12 Preliminary resultsresults

xi ^--- yiyi thenthen y(xl,y(x1,...... , ,xn) x,) --- y(yl,... y(y1,. , .y,,)). . , y,)). Define Define Grr Gn to to preserve preserve ^-- ifif xx ^-- y y implies implies xgTr xgn -- yglrygn forfor eacheach gg cE G.G. Prove Prove thatthat (a)(a) andand (b)(b) are are equivalent:equivalent: (a)(a) GzrGn preserves preserves no no nontrivial nontrivial gk-equivalencek-equivalence relation onon X.X. (b)(b) is is an an irreducibleirreducible -'k-representation.gk-representation. (See(See chapterschapters 5 andand 44 forfor thethe definition definition ofof anan irreducibleirreducible -'k-repre-gk-repre- sentationsentation whenwhen kk == 2 2and and 3. 3. A A wl &l-representation 1-representation isis irreducibleirreducible if if itit is is primitive,primitive, andand primitivityprimitivity isis defineddefined inin chapterchapter 2.) 2.) 7.7. LetLet n, n, or:0: G G -*+ Aut(X) Aut(X) bebe faithful faithful -'-representations. &-representations. Prove Prove rrn isis quasi- quasi- equivalentequivalent to oro ifif andand only ifif GnG7r isis conjugateconjugate toto GoGa inin Aut(X). 8.8. LetLet GG bebe aa groupgroup and 9 = =(Gi (Gi: : i iE E I) I) a familya family of of subgroups subgroups of of G.G. Prove Prove (1)(1) TheThe geometrygeometry Fr == 1'(G, r(G, O 5F) is connected is connected if ifand and only only if if G G = =(J9,_). (F). (2)(2) ForFor g cE G,G, definedefine gn:gir: rF +- F r by by (Gix)g'r (Gix)gn = = Gixg. Gixg. Prove Prove it n is is a a re- re- presentationpresentation ofof GG asas a a group group of of automorphismsautomorphisms ofof F.r. Permutation representations

Section 5 develops the elementary theorytheory of permutationpermutation representations.representations. TheThe foundation for this theory is the notion of the transitive permutation representation. The transitivetransitive representationsrepresentations play the rolerole ofof thethe indecomposablesindecomposables inin the theory.theory. ItIt will develop thatthat everyevery transitivetransitive permutationpermutation representationrepresentation of aa group G isis equivalentequivalent to a representation by right multiplication on the set of cosets of some subgroup of G. Hence thethe studystudy ofof permutationpermutation representationsrepresentations of G is equivalent to the study of the subgroup structure of G. Section 6 is devoteddevoted to aa proofproof ofof Sylow'sSylow's Theorem.Theorem. TheThe proofproof suppliessupplies aa nice application of the techniques developed in section 5.5. Sylow's Theorem is one of the most important results in finite group theory. It is the firstfirst theorem in the local theory of finite groups. The local theory studies a finite group from the point of view of itsits p-subgroupsp-subgroups and the normalizers of thesethese p-subgroups.p-subgroups.

5 PermutationPermutation representationsrepresentations In this section X is a set, G a group, and n:7r:G G +-* Sym(X)Sym(X) is is a a permutationpermutation rep-representation of G. Recall Sym(X) is the symmetric group on X; that isis Sym(X)Sym(X) is the group of all permutationspermutations of X. Thus Sym(X) is the automorphism group of X in thethe categorycategory of setssets andand functions,functions, and 7rn is a representation in that category. For x cE XX andand aa EE Sym(X) Sym(X) writewrite xa for for thethe imageimage of x underunder a. Notice Notice that, by definition of multiplication in Sym(X):

x(ap) = (xa)p x EE X, a, P p EE Sym(X).Sym(X).

I'll oftenoften suppresssuppress thethe representation 7rn andand writewrite xg xg forfor x(gn),x(g7r), xx E X, g E G.G. One feature of this notation is that: x(gh) = (xg)hx E X,g, h E G. The relation --- onon XX defineddefined by xx --- y y if if and and only only ifif therethere exists g eE GG withwith xg == y y is is an an equivalence equivalence relation relation onon X. X. The The equivalence equivalence classclass ofof xx underunder thisthis relation is xG=fxg:geG} and is called the orbit of xx underunder G. AsAs thethe equivalenceequivalence classesclasses of anan equivalenceequivalence relation partition a set, X is partitioned by thethe orbits ofof G on X. 14 Permutation representations

Let Y be a subset of X. G is said to act on YY if YY is a union of orbits of G. Notice G actsacts on Y precisely when yg E Y for each y E Y, and each gg EE G.G. Further if GG actsacts onon YY thenthen g gly IY isis a permutationpermutation of Y for each g EE G,G, and and thethe restriction map G -f+ Sym(Y) Sym(Y)

gHglyg I+ glr isis a permutation representationrepresentation with with kernelkernel Gy={g(=-G:yg=yGy={g€G:yg=y forall yEY}.~EY).

Hence Gy

G(Y) = {g (g EE G:G: YgYg = Y}, Y),

where Yg == { (yg: yg: y y EE Y}.Y). GYGy and G(Y) are subgroups of G called thethepointwise pointwise stabilizer of Y in G and the global stabilizer ofof YY inin G,G, respectively. respectively. G(Y)G(Y) isis the largest subgroup of G actingacting onon Y.Y. Write GYG~ for the image of G(Y)G(Y) underunder the restriction mapmap onon Y.Y. WeWe havehave seenseen that:that:

(5.1) TheThe restriction restriction map map of of G(Y)G(Y) on on Y Y is aapermutationrepresentation permutation representation of of G(Y) with kernel GyGy and image GY 2 - G(Y)/Gy,G(Y)/Gy, for for each each subset subset Y Y ofof X.X.

For Xx E X write GxG, for G{x}.G{,). Next Next for for S S G c G define

Fix(S) = {x {x EE X:X: xs = x x ffor o r all ss EE S}.S). Fix(S) is thethe setset ofoffuedpoints fixed points of S. Notice Fix(S) = Fix((S)). Also

(5.2) If If HH a9 G G then then G G acts acts on on Fix(H). Fix(H). More More generally generally G G permutes permutes the the orbitsorbits of H ofof cardinalitycardinality c,c, forfor eacheach c.c.

For Y C X,X, I'llI'll sometimessometimes write write CG(Y) CG(Y) and and NG(Y)NG(Y) for Gy and and G(Y),G(Y), respec- respec- tively, andand I'llI'll sometimes write Cx(G) forfor Fix(G).Fix(G). UsuallyUsually thisthis notationnotation willwill be used only when X possesses aa group structure preserved by G.

(5.3) Let P be thethe set of all subsets ofof X.X. ThenThen a: a:G G + - Sym(P) Sym(P) isis aa per- per- mutationmutation representation ofof GG on P wherewhere ga:ga: Y Y + Yg for each g E G andand YcX.Y E X.

7rIT is a a transitivetransitive permutation representation ifif G has just one orbitorbit onon X;X; equivalentlyequivalently forfor eacheach x,x, yy E X there exists gg E G with xgxg = y. y. GG will will alsoalso bebe saidsaid toto bebe transitivetransitive on X.X. Permutation representations 15

Here's oneone wayway to generategenerate transitive representations of G:

(5.4) Let H <5 G.G. Then Then a: GG -+ Sym(G/H) Sym(G/H) is isa transitivea transitive permutation permutation repre- repre- sentation of G on the cosetcoset spacespace G/H,G/H, where

ga: Hx Hx i-+ I+ Hxg.Hxg. a isis thethe representationrepresentation of of GG onon thethe cosetscosets of H by right multiplication. H isis thethe stabilizer of the coset H inin thisthis representation.representation.

We'll soonsoon seesee that that every every transitivetransitive representation ofof G is equivalentequivalent to a representation ofof G by rightright multiplicationmultiplication on the cosets of somesome subgroup.subgroup. But first here is another way to generate permutationpermutation representationsrepresentations ofof G.

(5.5) The map a:a: G -+ Sym(G) Sym(G) is isa apermutation permutation representation representation of of GG on itself, where

ga:xHxgga: x I+ xg x,gEG.x, g E G. a isis thethe representationrepresentation ofof G on itself by conjugation. For For S S c G,G, the the global global stabilizer of S in G isis NG(S),NG(S), while CG(S) is the pointwise stabilizer of S.

Notice that 5.5 is essentially a consequenceconsequence ofof ExerciseExercise 1.3.1.3. Recall Recall NG(S)NG(S) = = (g{g EE G:G: Sg = S}, S), Sg = (sg: {sg: s EE S},S), and CG(S) = = (g{g EE G:G: sg = s s forfor allall s EE S].S}. By 5.3, G is alsoalso representedrepresented on the power set of G, and and evidently,evidently, for S C G,G, thethe setset SGsG = (S9: {Sg: gg EE G} G) ofof conjugatesconjugates ofof SS underunder GG isis thethe orbitorbit of S under G with respect to this representation.

(5.6) Let Y C XX andand g EE G.G. Then Then G(Yg)G(Yg) = G(Y)9 G(Y)g and and GygGya = (GY)g. (Gy)g.

(5.7) Assume 7rn isis aa transitivetransitive representation representation andand let let x x EE XX andand HH = Gx G,. . Then ker(7r) = n Hg = kerH(G) gEG is the largestlargest normal subgroupsubgroup ofof GG containedcontained inin H.H.

(5.8) Assume nit is aa transitive transitive permutation permutation representation, representation, let let x Ex X,E X, H H= = G,, G, and letlet aa be the representation of G on the cosetscosets ofof HH by right multiplication. Define B:8: G/H G/H + - XX Hg i-+I-+ xg. Then PB is anan equivalenceequivalence of the permutation representations a andand 7r.n. 16 Permutation representationsrepresentations

Proof. WeWe mustmust showshow ,B/3 is a well-defined bijectionbijection ofof G/HG/H withwith XX andand that,that, forfor each a, g E G,G, (Ha)O(ga)(Ha)B(ga) == (Ha)gnf. (Ha)gnp. Both Both computations computations are are straight- straight- forward.forward.

(5.9)(5.9) (1) EveryEvery transitivetransitive permutationpermutation representation of G isis equivalentequivalent to aa representationrepresentation ofof GG byby rightright multiplication multiplication on on thethe cosetscosets ofof somesome subgroup. subgroup. (2)(2) IfIf jr':n': GG ->+= Sym(X') Sym(X1) and and rr n are are transitive transitive representations, representations, xx E X, andand x' EE X',X', thenthen n 7t is is equivalent equivalent to to jr' n' if if and and only only if if GxG, isis conjugateconjugate toto G,,,G,t in G.

Proof.Proof. Part Part (1) (1) follows follows from from 5.8.5.8. AssumeAssume thethe hypothesis of (2). IfIf, B: 8: XX +=-> X' isis an equivalence of njr and n' thenthen Gx G, == GxA GXB and,and, byby 5.6,5.6, GxpGXB isis conjugateconjugate to Gx,G,? in G. Conversely if G,IGx isis conjugate to GxG, in G, then by 5.6 there is y EE X withwith GyG, == Gx' G,I andand by by 5.8 5.8 bothboth jr n and and jr' n' are are equivalent equivalent to to thethe representation representation ofof GG onon the the cosetscosets of of Gx'G,. andand hence hence equivalentequivalent to to eacheach other.other.

LetLet (Xi: i EE I) be be thethe orbitsorbits of G onon XX andand ni thethe restrictionrestriction of n toto Xi.Xi. ByBy 5.1,5.1, niq isis a a permutation permutation representation ofof GG on Xi and, as Xi is anan orbitorbit of G,G, nini isis even even a a transitive transitive representation. representation. (ni:(ni: i i c=E I) I) is the family of transitivetransitive constituentsconstituents ofof nn and wewe writewrite n n == Eic]xi,, ni.q. Evidently:Evidently:

(5.10)(5.10) TheThe transitive transitive constituentsconstituents (ni(xi: : ii E I) of of nn are are uniquely uniquely determined determined by by n,n, and and if if n' n' is is a apermutation permutation representation representation of of G G with with transitive transitive constituents constituents (jr:(nj: j jE EJ), J), thenthen n nis isequivalent equivalent to to n' n' if ifand and only only if if there there is is a a bijection bijection aa! ofof II withwith J suchsuch that that ntani6( isis equivalent equivalent to ni for each i E I.

SoSo the studystudy ofof permutation representations isis effectivelyeffectively reduced toto thethe studystudy ofof transitivetransitive representations, representations, and and 5.9 5.9 says says in in turn turn that that thethe transitive transitive permutation permutation representationsrepresentations ofof aa groupgroup areare determineddetermined byby itsits subgroupsubgroup structure.structure. TheThe transitivetransitive representations playplay thethe rolerole ofof thethe indecomposable permuta-permuta- tiontion representations.representations. For For example example see see Exercise Exercise 1.5. 1.5.

(5.11)(5.11) IfIf GG is is transitive transitive on on X X then then X X has has cardinality cardinality IG IG : :Gx G, IJ for each x EE X.X.

Proof.Proof. This This is is a a consequence consequence of of 5.8.5.8.

(5.12)(5.12) LetLet SS CG G. G. Then Then S S has has exactlyexactly IGJG: NG(S)ING(S)I conjugates conjugates inin G.G.

Proof.Proof. We We observed observed earlier earlier that that G G is is transitively transitively representedrepresented onon the the setset SGSG ofof conjugatesconjugates of of SS via via conjugation,conjugation, while, while, by by 5.5, 5.5, NG(S) NG(S) isis the the stabilizer stabilizer of of S S with with respectrespect to to this this representation, representation, so so the the lemma lemma follows follows from from 5.11. 5.11. Permutation representations 17

Let p bebe aa primeprime andand recallrecall thatthat aa p-groupp-group isis aa groupgroup whosewhose orderorder isis aa powerpower of p.

(5.13) If G is aa p-groupp-group thenthen allall orbits of G on X have order a power ofof p.p.

Proof. ThisThis followsfollows fromfrom 5.115.1 1 andand thethe factfact thatthat thethe indexindex of any subgroup of G divides the order of G. (5.14) LetLet GG be a p-group and assume XX isis finite.finite. Then Then 1x1 IXI - - (Fix(G)IIFix(G)I mod p. Proof. As As the the fixed fixed points points of of G G are are its its orbits orbits of of lengthlength 1, 1,5.14 5.14 follows from 5.13.5.13.

Here are a couple applications of 5.14:

(5.15) LetLet G and H be p-groupsp-groups with H # 1 and let a: G -+ Aut(H) be a group homomorphism. ThenThen CH CH(G)(G) # ; 1.1.

Proof. a isis also also a a permutation permutation representationrepresentation ofof G on H.H. By By 5.14, 5.14, CHI (HI-. JFix(G)IIFix(G)I mod mod p. p. ButBut Fix(G)Fix(G) == CH(G) in this representation, while while IHI IHI =- - 00 mod p as H isis aa p-group with H # 1. So CH(G)Cn(G) # 1, as claimed.

(5.16) If G is a p-group with G # 1, then Z(G) # 1.

Proof. ApplyApply 5.15 5.15 to to the the representation representation of of GG onon itselfitself by conjugation and recall Z(G) = CG(G). CG(G).

The following technical lemmalemma will will bebe usedused inin the next section to prove Sylow's Theorem:

(5.17) LetLet XX bebe finitefinite and assume forfor eacheach xx E X that therethere existsexists aa p-subgroupp-subgroup P(x) ofof GG suchsuch that {x}{x) == Fix(P(x)). Then (1) GG isis transitive transitive onon X,X, andand (2) 1/XI XI zz - 1 modp.mod p.

Proof. Let X == Y Y + + Z Z be be a apartition partition of of XX with with GG acting acting onon YY andand Z.Z. LetLet Y # 0 andand pickpick y EE Y.Y. For V = YY or Z and H <_( GG denote denote byby Fixv(H)Fixv(H) the the fixed pointspoints ofof HH on V. ByBy hypothesishypothesis (y) (y} = = Fix(P(y)),Fix(P(y)), so 1 = (Fixy(P(y))IIFixy(P(y))l andand0= 0 = IFix,(P(y))I.Hence, IFix,(P(y))l. Hence, by5.14,IY)by 5.14, IYI r 1modpandIZI1 mod p and I ZI -. 00modp.But mod p. But if ZZ isis nonempty, nonempty, then, then, by by symmetry, symmetry, I Y[YI I r 1 mod p,p, aa contradiction.contradiction. ThusThus 18 Permutation representations

Y = X X and,and, as as IYIIY 1 =-E 1 mod p, (2)(2) holds. Since we could have chosen Y to be an orbit of G on X, (1) (1) holds.holds.

Let Q bebe aa partitionpartition of X.X. QQ isis G-invariant G-invariant if if GG permutes permutes thethe membersmembers of Q. Equivalently regardregard QQ asas aa subsetsubset ofof thethe powerpower set set PP of X and represent G on PP asas inin 5.3;5.3; thenthen QQ isis G-invariantG-invariant if G acts on the subset Q of P withwith respect to this representation.representation. InIn particularparticular notice thatthat ifif Q is G-invariant then there is a natural permutation representation representation of of G G on on Q. Q. QQ isis nontrivial ifif QQ0{{x}:xEX)andQ0{X}. # {Ix):xE XI and Q # 1x1. Let G be transitive on X. G is imprimitive on X if there exists a nontrivialnontrivial G- invariant partition QQ ofof X.X. In In thisthis event event Q Q is is said said to to be be a a system system ofof imprimitivityimprimitivity . for G on X. GG isis primitive onon XX ifif itit isis transitivetransitive andand notnot imprimitive.imprimitive.

(5.18) Let G be transitive on X and y E X. (1) IfIf Q is a system of imprimitivityimprimitivity forfor GG onon XX andand yy E Y E Q, then GG isis transitive onon Q, the stabilizer H ofof YY inin GG isis aa proper subgroupsubgroup of GG properlyproperly containing G,,Gy, Y Y is is anan orbitorbit ofof H onon X,X, IXI 1x1 = IYIIQI, IYllQl, IQIlQl = IGIG: : HI,HI, and IYIIYl == IH:IH:GyI. Gyl. (2) IfIf GyGy < HH << G G then then Q Q = ={Yg: {Yg: gg E E G) G) is is a a system system of of imprimitivityimprimitivity for G on X, where Y == yH andand HH isis the the stabilizer stabilizer of of YY in G.

The proof is left as an exercise. As aa directdirect consequenceconsequence of of 5.185.18 wewe have:have:

(5.19) Let G be transitive on X and xX E X.X. Then G is primitive onon X if and only if GxG, is a maximal subgroup of G.

Let G be finite and transitive onon X, let I1x1 X I> > 1,1, and and let let x x EE X.X. Then there isis a sequence G, Gx = = Ho Ho ( < HI Hl 5< .... . < ( H H,, = =G Gwith with H, Hi maximal maximal in in Hi+i. Hi+l. This This gives rise to a family of primitive permutation representations:representations: the representations of Hi+IHi+l on the cosets of Hi. This family of primitive representationsrepresentations cancan be used to investigate thethe representation nit of G on X.X. From this point of view the primitive representations playplay thethe rolerole ofof irre-irreducible permutation representations. SeeSee alsoalso Exercise 1.6.1.6. I close this section with two useful lemmas. The proofs are left as exercises.

(5.20) Let G be transitive onon X,X, xX E E X, X, and and H H 5< G. Then H isis transitivetransitive on X if and onlyonly if if GG == G,GxH. H.

(5.21) Let G be transitivetransitive on on X, X, x X E E X, X, H H = = G,Gx andand KK 5< H.H. ThenThen NG(K)NG(K) is transitive onon Fix(K) Fix(K) if if and and only only if if lCG KG n fl H H = = KK". H. Sylow's Theorem 19

6 Sylow'sSylow's Theorem In this section G is a finite group.group. IfIf nn isis aa positivepositive integerinteger andand pp aa prime,prime, writewrite n,np for thethe highest powerpower ofof pp dividing n.n. n,np isis thethe p-part p-part of n.n. A Sylow p-subgroup ofof GG is is a a subgroup subgroup of of G G of of order order I GI G I I,. p. Write Sy1p(G)Syl, (G) for the set of Sylow p-subgroups of G. In this sectionsection wewe prove:prove:

Sylow's Theorem. LetLet GG bebe aa finitefinite group and p a prime. Then (1) Sylp(G)Syl,(G) isis nonempty. nonempty. (2) G G acts acts transitively transitively onon Sylp(G)Syl,(G) viavia conjugation.conjugation. (3) ISylp(G)IISyl,(G)I = I IG G :: NG(P)ING(P)I r - 11 modmod p forfor PP EE Sylp(G).Syl,(G). (4) EveryEvery p-subgroupp-subgroup ofof GG is is contained contained inin somesome SylowSylow p-subgroup of G.

Let Pr bebe thethe setset ofof allall p-subgroupsp-subgroups ofof G and Q0 thethe setset ofof allall maximalmaximal p- subgroups of G; that is partiallypartially orderorder rr byby inclusioninclusion and let Q0 bebe thethe maximal members of this partially ordered set.set. ItIt follows fromfrom 5.35.3 andand 5.55.5 thatthat GG is represented as a permutation group via conjugation on the power set of G, and it is evidentevident thatthat G acts on r,F, Q,2, andand Sy1p(G)Sylp(G) with respect to this representation.representation. Let R EE 0.Q. Claim Claim R R is is the the uniqueunique pointpoint of 0Q fixed fixed byby the subgroup R of G. For if R fixes Q E 0Q then, then, byby 5.5,5.5, RR <5 NG NG(Q), (Q), so, by 1.7.2, RQRQ 5< G and,and, by 1.7.3, [RQlIRQI = = IRIIQIIIRIRIIQI/IR n n Ql. Q. Thus Thus RQ RQ E E r, F, so, so, as as R R 5 < RQRQ 2? QQ andand R, Q EE S2,C2, wewe concludeconclude RR == RRQ Q = Q. Q. So So the the claim claim is is established.established. I've shown thatthat forfor eacheach RR EE Q0 therethere is a p-subgroup P(R) ofof GG suchsuch thatthat R is the unique pointpoint of of Q0 fixed byby P(R); namely R = P(R).P(R). SoSo it it followsfollows from 5.17 that: (i) GG isis transitivetransitive on Q,0, andand (ii) IQ1Q1 I-1modE 1 modp. p. Let P EE 0Q and and suppose suppose I I P I=I = I IG G II,. p.Then Then P P E Syl,(G),Sylp(G), so, as G is transitivetransitive on 0Q and and G G acts acts on on Sylp Syl,(G), (G), wewe have have S2 C2 C E Sylp Syl,(G). (G). OnOn the the other other hand hand as as I RI RI I divides JGJI G forI for each each R RE EQ 0 it it is is clear clear Syl,(G) Sylp(G) G c 0,Q, so so SZ Q = Sylp(G).Syl,(G). ThusThus (i) implies parts (1) and (2)(2) ofof Sylow'sSylow's Theorem,Theorem, whilewhile (ii)(ii) andand 5.125.12 implyimply part (3). Evidently eacheach membermember of of r r is contained inin aa membermember ofof Q,0, so, as QSZ = = Syl,(G),Sylp(G), the fourth part of Sylow's Theorem holds.holds. So to complete the proof of Sylow'sSylow's Theorem it remains to show I PI PI == IGII G 1, p for P EE 0.Q. Assume Assume otherwiseotherwise and letlet MM == NA(P). NG(P). By By (i), (i), (ii), (ii), and and 5.12,5.12, IGI G :MI : M GI- 11 mod mod p,p, so so I /MIpM I p= = I IGl,,G I p,and and hence hence pp divides I[MIPI. M/P I. Therefore, Therefore, by Cauchy's Theorem (Exercise(Exercise 2.3), 2.3), there there exists exists a asubgroup subgroup RIP R/P of of MIP M/P of order p.p. But But now now I I R I = I I R/PRIP I I I P P I Iis is a apower power of of p, p, so so P P< < RR EE r,I', contradicting PES2.P E Q. This completes the proof of Sylow's Theorem. 20 Permutation representations

Next a few consequences of Sylow'sSylow's Theorem.Theorem.

(6.1)(6.1) Let P EE Sylp(G).SylP(G). Then PP 5a GG ifif andand onlyonly ifif PP is is the the unique unique Sylow Sylow p-subgroup ofof G.G.

Proof. ThisThis isis becausebecause GG acts acts transitively transitively onon Sylp(G)Sylp(G) via via conjugationconjugation withwith NG(P) thethe stabilizerstabilizer ofof PP inin this this representation.representation.

LemmaLemma 6.16.1 andand thethe numerical restrictions in part (3)(3) ofof Sylow'sSylow's TheoremTheorem cancan bebe usedused toto showshow groupsgroups ofof certaincertain ordersorders have have normalnormal SylowSylow groups.groups. SeeSee forfor exampleexample ExercisesExercises 2.52.5 andand 2.6.2.6.

(6.2)(6.2) (Frattini(Frattini Argument) LetLet HH 5a G and P EE Sylp(H).Syl,(H). ThenThen GG == HNG(P). HNG(P).

Proof. ApplyApply 5.20 5.20 to to thethe representationrepresentation of G on Sylp(H),SyI,(H), usingusing Sylow'sSylow's The-The- oremorem to get H transitivetransitive onon Sylp(H).Syl,(H).

ActuallyActually Lemma 6.2 is aa specialspecial casecase ofof thethe followingfollowing lemma,lemma, which has aa similarsimilar proof,proof, andand whichwhich II alsoalso refer refer toto asas aa FrattiniFrattini Argument: Argument:

(6.3)(6.3) (Frattini(Frattini Argument)Argument) Let Let K K be be a a group, group, H H 5 a K, K, andand XX EC H. Then K = HNKHNK (X) if and only if XKx == X X H. H. Indeed Indeed H H has has I KI K : HNK: HNK (X)(X) I 1 orbits of equal lengthlength on on XK,xK, withwith representativesrepresentatives (X-Y:(XY: y y E c Y),Y), wherewhere Y isis a setset ofof cosetcoset representativesrepresentatives forfor HNK(X)HNK(X) in in K.K.

(6.4)(6.4) Let H a5 GG andand P EE Sylp(G). Syl,(G). Then P n HH EE Sylp(H).Syl,(H).

ExercisesExercises for chapter 22 1.1. ProveProve Lemma Lemma 5.18. 5.18. 2.2. ProveProve Lemmas Lemmas 5.20 5.20 andand 5.21.5.21. 3.3. ProveProve Cauchy's Cauchy's Theorem: Theorem: Let Let G G be be a a finite finite group group and and pp a a prime prime divisor divisor ofof (GI.G 1. ThenThen GG contains an element of of orderorder p.p. (Hint: (Hint: Prove Prove p p divides divides I CG1 CG(x)l (x) forfor somesome xx EE G#. G'. ThenThen proceed proceed by by inductioninduction onon !GI.) 4.4. LetLet G G be be a a finite finite group group and and pp a a prime. prime. Prove: Prove: (1)(1) IfIf G/Z(G)G/Z(G) is is cyclic, cyclic, then then G G is is abelian. abelian. (2)(2) IfIf GJGI A = p2 P2 then then G G = 2 Zp2 Zpz or or Ep2. Ep2. 5.5. (1) (1) LetLet IGI /GI == pem, pem, p p >> m, m, p p prime, prime, (p, (p, m) m) = = 1. 1. Prove Prove G G has has a anormal normal SylowSylow p-subgroup.p-subgroup. (2)(2) LetLet IGI I G I == pq, pq, p andp and q prime.q prime. Prove Prove G G has has a anormal normal Sylow Sylow p-subgroup p-subgroup oror a a normal normal Sylow Sylow q-subgroup. q-subgroup. Sylow's Theorem 21

6.6. LetLet I IG G II = pq2, pq2, where where p p and and q q are are distinct distinct primes. primes. Prove Prove one one of of the the following following holds:holds: (1)(1) qq > > p pand and G Ghas has a anormal normal Sylow Sylow q-group. q-group. (2)(2) pp > >q qand and G G has has a anormal normal Sylow Sylow p-group. p-group. (3)(3) IIG G I I= = 12 and G has a normal SylowSylow 2-group.2-group. 7.7. LetLet G G act act transitively transitively onon aa setset X,X, xx EE X,X, andand PP E E Sylp(G,,). Syl,(G,). ProveProve NG(P)Nc(P) isis transitivetransitive onon Fix(P).Fix(P). 8.8. ProveProve Lemmas Lemmas 6.3 6.3 and and 6.4. 6.4. 9.9. ProveProve that that if if G G has has just just oneone Sylow Sylow p-subgroupp-subgroup forfor eacheach pp E E 7r(G), n(G), thenthen GG isis thethe directdirect productproduct ofof itsits SylowSylow p-subgroups.p-subgroups. 3

Representations of groupsgroups on groups

Chapter 3 investigates representations in the category of groupsgroups and homomor-homomorphisms, with emphasis on the normal and subnormal subgroups of groups. In sectionsection 7 the conceptconcept ofof anan irreducibleirreducible representation representation is is defined,defined, andand thethe Jordan-Holder Theorem is established. As a consequence,consequence, the composition factors of a finite group are seen to be an invariant of the group, and these composition factors areare simple.simple. The question arises as to how much the structure of a groupgroup is controlledcontrolled by its composition factors. Certainly many nonisomorphic groupsgroups cancan havehave the same set of compositioncomposition factors, so control is far from complete. To To investigate this question further we must consider extensions of a group G by aa groupgroup A.A. Section 10 studies split extensions and introduces semidirect products. Section 9 investigates solvable and nilpotent groups. For finite groups this amounts to the study of groups all of whose composition factors are,are, inin the first case, of prime order and, in the second, of order p forfor somesome fixedfixed prime p. Commutators, characteristic subgroups, minimal normal subgroups, central products, and wreath products areare alsoalso studied.studied.

7 NormalNormal series In this section G andand AA areare groups,groups, andand n:n: A A -p+ Aut(G) Aut(G) isis aa representation representation of AA inin the the category category ofof groupsgroups andand homomorphisms.homomorphisms. I'llI'll also also saysay thatthat A A acts as a groupgroup ofof automorphismsautomorphisms on G.G. ObserveObserve thatthat 7rn is alsoalso aa permuta-permutation representation, so we can use thethe terminology,terminology, notation, and results from chapter 2.2. A normal series of length n for G is aa seriesseries 1=Go

(7.1) LetLet (Gi(Gi: : 00 j< i j< n) be an A-invariant normal series and H an A-invariant subgroup of G. Then (1) TheThe restriction restriction 7rn lH:I H: A A + -a Aut(H) Aut(H) isis aa representationrepresentation ofof AA onon H.H. (2) (Gi(Gi flfl H: H: 0 0 < j i i< j n) n) is is an an A-invariant A-invariant normalnormal series for H. (3) IfIf HH < GG thenTrG/H: then TGIH: A-4A + Aut(G/H) Aut(G/H) is isa arepresentation, representation, where where a7rG/H: anc/ff: Hg t+HH(ga) H(ga) for for a aE E A, A, g g E E G.G. Normal series 23

(4)(4) IfIf HH < 9 G G then then (GiH/H: (GiH/H: 0 0 < 5 i i< _< n) n) is is an an A-invariant A-invariant normalnormal seriesseries for G/H.GIH. (5)(5) IfIf XX is is an an A-invariant A-invariant subgroup ofof GG andand HH 9< G,G, thenthen Xx fl HH isis anan A-invariant normal subgroup ofof X and XH/HXHIH isis anan A-invariantA-invariant subgroup of G/HGIH which which isis A-isomorphicA-isomorphic toto X/(XX/(X fln H).H).

A subgroup H H ofof GG is is subnormal in in GG if if there exists a series H == Go Go < 9 G G1 1 5 . . . <-a GG, == G. G. Write Write H H < 9 < 9 G G to to indicate indicate H H is is subnormal subnormal inin G.

(7.2)(7.2) IfIfX X ==Go Go < 9 G1 G1 49 ... . . < 9 G. G, = =G, G, then theneitherx eitherX = = G G or or (XG) (XG) <_< G,_1Gn-1 # G, soso (Xo)(xG)# G.

(7.3)(7.3) LetLet XX andand HH bebe A-invariant A-invariant subgroupssubgroups of G with X <9 <

(2) XXnHSH. fl H < < H. (3) IfIf HH <9 G G then then XH/HXHIH 9< <9 G/H.GIH.

Proof. PartPart (1) (1) follows follows from from 7.2 7.2 and and induction induction on on the the length length n n ofof aa subnormal subnormal series X = Go GO

The family of factorsfactors of a normal series (Gi: 0 (< i <( n) isis thethe familyfamily ofof factor groups (Gi+1/Gi: 0 0 <5 ii

for each 0 <5 ii <_< m, m, and and some some j j(i).(i ). The representation nit is said to be irreducibleirreducible if GG andand 11 areare thethe onlyonly A-A- invariant normalnormal subgroupssubgroups of of G.G. We alsoalso saysay that G isis A-simple.A-simple. An A- composition series for G is a normal series (Gi:(Gi: 00 5< ii (< n) maximal subject to being A-invariant and toto Gi # Gi+1Gi+I for 0 (< i (< n. OfOf particular importance is the case A = 1. 1. G G is is said said toto be simple ifif it is 1-simple. Similarly the the composition series for GG areare itsits 1-composition1-composition series. series.

(7.4) IfIf GG isis finite finite thenthen GG possessespossesses anan A-compositionA-composition series.

(7.5) An A-invariantA-invariant normal series is an A-composition seriesseries ifif and only ifif each of its factorsfactors isis aa nontrivialnontrivial A-simpleA-simple group.group. 24 Representations of groups on groups

Jordan-Holder Theorem.Theorem. Let Let (Gi(Gi: : 0 5< ii 5< n)n) and (Hi:(Hi: 00 5< ii 5< m) m) bebe A- composition series series for for G. G. ThenThen n n == mm andand therethere existsexists a permutation aa ofof {i:0{i: 0 i< i

Proof. Let H = HmW1and k = min{i min{i: : Gi 5 H). I'll show:show:

(a) G/HGIH is is A-isomorphic A-isomorphic to to Gk/Gk_1,Gk/Gk-l, and and (b) (Xi(Xi == Gia Gi, nfl H:H: 0 0 < 5 i i< c n) n) is is an an A-composition A-composition series series for for HH with with Xi+1 Xi+1 /Xi/ Xi A-isomorphic toto Giatl/GiaGi,,+1 / Gi,, for for 0 50 k-1. 1.

Suppose (a) andand (b)(b) hold.hold. By By induction induction on on n, n, n n - - 1 == m - 1, 1, and and therethere isis aa permutation j3,8 ofof {i:{i:0 0 5 < ii c< n - 11 1) with Xip+1/XiflXiS+l/Xia A-isomorphic to Hi+l/Hi.Hi+1/Hi. Hence n == m and the permutation aa of ti:{i: 0 <5 i << n) n} defined defined belowbelow does the trick in the Jordan-Holder Theorem: Theorem: is=i,Ba ifi

So it remainsremains toto establishestablish (a)(a) and and (b). (b). First, First, as as Gi Gi 5 < H H forfor ii c< k, Gi = Gia Gia = Xi, so certainly Xi/X1_1,Xi/XiPl, is A-isomorphic to Gi,,Gia/ /Gia_1. Gia-1. If Gk fln H -$ Gk-1,Gk_1, then,then, asas (Gk(Gk nn H)Gk-l/Gk-lH)Gk_1/Gk_1 is an A-invariant normal subgroup ofof thethe A-simpleA-simple groupgroup Gk/Gk-1,Gk/Gk-1, we havehave GkGk = = (Gk(Gk fl n H)Gk-1H)Gk_1 i< H, contrary to thethe definition ofof k.k. SoSo GkGk fln HH = Gk_1.Gk-1. On the other hand,hand, ifif jj >2 k, k, then Gj -$ H, soso aa similarsimilar argumentargument using 7.3 shows G ==HG1, HGj, andand hencehence G/HGIH =HGj/H = HGj/H is isA-isomorphic A-isomorphic to to Gj Gj/(Gj /(G1 nn H).H). In In particular Gj/(GjGj /(Gj n H) is A-simple,A-simple, so,so, if if G Gj n n H H 5< GGj_1, j-1, thenthen GGj_i j-1 /(Gl(Gj nn H)H) is is a a proper A-invariant normal subgroup ofof thethe A-simpleA-simple group group G Gj j/(G /(Gj fln H), and hence G j-1j_1 = = GjGj fln H <5 H.H. But But thenthen jj == k k by by definition definition ofof k.k. Moreover G/H is is A-isomorphic A-isomorphic to Gk/(Gk n H) = Gk/Gk-1, Gk/Gkel, soSO (a) (a) holds.holds. By thelastparagraph,the last paragraph, GjnHGj nH Gj_1Gjel for jj >> k.k. So, as above,above, Gj G3 = _ (Gjn(Gj n H) Gj-1,Gj_1, and hencehence Gj/GjP1Gj/G1_1 is A-isomorphic to to (Gj (Gj n n H)/(Gj-1H)/(G1_1 nn H), com-completing the proof of (b).

The Jordan-Holder Theorem sayssays that the factors of an A-composition series of G are (up to equivalence and order) independent of the series, and hence are an invariant of thethe representationrepresentation n.n. These factors are the composition factorsfactors of the representation n. IfIf AA = = 1, 1, these these factors factors are are the the composition composition factors of G. Characteristic subgroups subgroups andand commutatorscommutators 25

(7.6)(7.6) Let XX bebe anan A-invariant A-invariant subnormalsubnormal subgroupsubgroup of the finite finite groupgroup G.G. Then (1)(1) TheThe A-composition A-composition factors factors of of XX are are a a subfamily subfamily of of thethe A-composition A-composition factors of G.G. (2)(2) IfIf XX < 9 G G then then the the A-composition A-composition factorsfactors of GG are are thethe unionunion ofof thethe A-composition factors ofof XX andand G/X.G/X.

Proof. ThereThere is is an an A-invariant A-invariant normalnormal seriesseries containingcontaining X by 7.3, and as G is finite this series isis containedcontained inin aa maximalmaximal A-invariantA-invariant series.series. ThusThus therethere isis anan A-composition seriesseries throughthrough X,X, so so that that the the result result is is clear. clear.

8 CharacteristicCharacteristic subgroups subgroups andand commutatorscommutators A subgroup H ofof aa groupgroup GG isis characteristiccharacteristic in in GG if if H H is is Aut(G)-invariant. Aut(G)-invariant. Write H charchar GG toto indicateindicate thatthat HH isis a a characteristiccharacteristic subgroup of G.

(8.1)(8.1) (1)(1) If H charchar KK andand KK charchar G,G, thenthen HH charchar G.G. (2) If H charchar K and K

A group GG isis characteristicallycharacteristically simple simple ifif GG andand 11 are are the the onlyonly characteristiccharacteristic subgroups of G.G. A minimal normal subgroup of G is a minimal member of the set of nonidentity normal subgroups of G,G, partiallypartially orderedordered by inclusion.inclusion.

(8.2)(8.2) IfIf 11 # G is aa characteristicallycharacteristically simple finite group, then G isis thethe directdirect product of isomorphic simple subgroups.

Proof. LetLet HH be be a a minimal minimal normal normal subgroupsubgroup of GG andand MM maximalmaximal subjectsubject to MM <9 GG andand MM thethe direct direct productproduct ofof imagesimages ofof HH underunder Aut(G).Aut(G). NowNow X = (Ha: (Ha: aa E E Aut(G)) Aut(G)) is is characteristic characteristic in in G,G, so so by by hypothesishypothesis X = G. G. Hence, Hence, if M #0 G,G, therethere isis aa Ec Aut(G) withwith Ha Ha M.M. As As Ha Ha is isa aminimal minimal normal normal sub-subgroup of G and Ha fln M <9 G, G, wewe conclude Ha fln M M == 1. 1. ButBut then M < M xx Ha < 9 G, G, contradicting contradicting the the maximalitymaximality ofof M. So G = M. M. ThusThus GG = H H x x K K for for some some K K < 5 G, G, so so every every normal normal subgroupsubgroup of H isis alsoalso normalnormal in G.G. ThusThus HH isis simple simple byby minimalityminimality of H, and and thethe lemmalemma is established.

(8.3)(8.3) MinimalMinimal normalnormal subgroupssubgroups are are characteristicallycharacteristically simple. simple.

Proof. ThisThis follows follows from from 8.1.2.8.1.2. 26 Representations of groupsups on groups

(8.4) (1)(1) TheThe simplesimple abelianabelian groupsgroups areare thethe groups of prime order. (2) If G isis characteristicallycharacteristically simple, finite, and abelian, thenthen G 2= EpllEP" forfor some prime p andand somesome integerinteger n.

For x, y cE G,G, write write [x,[x, y]y] for for the the group group elementelement x-1X-'~-'X~. y-1xy. [x, [x, y]y] is is the the com- com- mutator of x andand y.y. ForFor X,X, Y Y <5 G,G, define define

[X, Y]Y] = ([x, ([x, y]:y]: xx E E X,X, yy EE Y). For z cE Z <5 GG write write [x,[x, y,y, z] z] for for [[x,[[x, y],y], z]z] and and [X,[X, Y,Y, Z] for [[X,[[X, Y],Y], Z].

(8.5) Let GG bebe aa group,group, a, b, b, c c E E G, and and X,X, YY <5 G.G. Then Then (1)(1) [a,[a, b]b]= =1 1 if if andandonly only ifif abab=ba. = ba. (2)(2) [X, [X, Y]Y] == 1 1 if if andand only only ifif xyxy == yx yx forfor all all x x E E X X andand yy c E Y. Y. (3)(3) IfIf a:a: GG - + H His ais group a group homomorphism homomorphism then then [a, [a, b]ab]a == [act, [aa, bet]ba] and and [X, Y]a = [Xa, [Xa, Yet].Ya]. (4)(4) [ab,[ab, c] = [a,[a, c]b[b,clb[b, C]c] and [a, bc]be] == [a, c][a, bIC.b]'. (5)(5) X X < 5 NG(Y) Nc(Y) if if and and only only if if [X,[X, Y]Y] <5 Y. Y. (6)(6) [X, [X, Y]Yl = = [Y, [Y, X]XI :a9 (X,(X, Y).Y).

Proof. II proveprove (6)(6) and and leaveleave thethe otherother partsparts asas exercises.exercises. Notice [a, b]-1b]-' = [b, a], soso [X,[X, Y]Y] = [Y, [Y, X].XI. Further,Further, toto proveprove [X,[X, Y]Y] :a9 (X,(X, Y),Y), itit willwill sufficesuffice to showshow [x, y]ZylZ EE [X, Y] foreachxfor each x E X, Yy E Y, andz and Z EE XUX U Y.Y. AsAs [x,[x, y]-' y]-1 = [y, x], we may assumeassume zz E Y. But, But, byby (4),(4), [x, [x, ylZ y]Z = = [x,[x, z]-'[x,z]-1[x, yz] E [X,[X, Y],Y], so the proof is complete.complete.

(8.6)(8.6) LetLet GG bebe aa group,group, x,x, y y E E G,G, and and assumeassume zz = [x, [x, y]y] centralizes centralizes xx andand y.y. Then (1)(1) [xn[xn, ym]ym] =Znm= znm for for all all n, n, m m E7L. E Z. (2) (yx)' (yx)" = =Zn(n-1)12ynxn ~"(~-')/~y"x" forfor all all 0 0 < 5 n n E E 71. Z.

Proof. WithoutWithout lossloss GG == (x, (x, y),y), so so z z EE Z(G).Z(G). zz == [x, [x, y]y] so so xy xY = = xz. xz. Then, Then, forfor nn EE 1,Z, (xn)y(xn)Y = (xy)n (xY)" = (xZ)n (xz)" = xnZn xnzn as z E Z(G).Z(G). Thus [x",[xn, y]y] = = Zn. zn. Similarly Similarly [x,[x, ym]ym] ==zm, Zm, soSo [xn,[xn, ym]=ym] _ [x, [x, ym]nymIn ==zmn, Zmn, andand (1)(1) holds. Part (2)(2) is is establishedestablished byby inductioninduction on on n.n. Namely Namely (yx)n+1 (YX)"+' == (yx)nyx (yx)"yx == Zn(n-1)12ynxnyx,Zn(n-1)/2 yn x n yx, while while byby (1) xny = yxfZn, yxnzn, soso thatthat thethe resultresult holds.holds.

(8.7) (Three-Subgroup(Three-Subgroup Lemma) Let X, Y,Y, Z be subgroups of a group G withwith [X,Y,Z]=[Y,Z,X]=1.Then[X,Y, Z]= [Y,Z, XI = 1. Then [Z,X,Y]=1.[Z,X, Y]= 1.

Proof. Let Let x x E E X, X, y y cE Y, Y, andand zz cE Z. Z. A A straightforward straightforward calculation calculation shows: shows: =x-1y-1xz-1x-1yxy-1zy z)-1a(y, (*)(*) [x, y-1,y-', zlyzlY =x-'~-~xz-~x-'~x~-'z~=a(x,=a(x, y, y, z)-'~(~,z,z, x), Solvable and nilpotent groups 27 wherea(u,where a (u, v, w)=uwu-'vu.~pplyingthepermutations(x,w) = uw u-1 vu. Applying the permutations (x, y, z)and(x,z) and (x, z, y) to (*) and taking the product of (*) with these two images, we conclude:

[x, Y-'z]y[Y, z-1, x]Z[z, x-1, Y]X =1.

As [X, Y, Z]Z] =_ [Y, Z,Z, XIX] = =1, 1, also [x, y-1,y-', z]z] _= [y, [y, z-1,z-', x]x] == 1, 1, so so byby (**)(**) we get [z, x-',x-1, y]y] == 1. 1. Finally Finally asas [Z,[Z, X] XI is is generated generated byby the commutators [z, x-'1,x-1], z E Z, xX EE X,X, itit followsfollows from from 8.5.1 8.5.1 that that y y centralizes centralizes [Z,[Z, XI.X]. But then,then, byby 8.5.2,8.5.2, [Z, X, Y]Y]=1. = 1.

The commutator group or derived group of a group G is the subgroup GMG(') = [G, G].GI. Extend thethe notationnotation recursivelyrecursively andand define define G(") G(") = [G(n-1), [G("-'), G(n-1)]G("-')I for n > 1. DefineDefine G(O) G(0) = = G G and and Gm G-= = f°_nzl 1 G(OG(').

(8.8) Let G be a group and H 5< G. Then (1) H(n)H(") 5

A group G isis perfectpe$ect ifif GG == G(1). G(').

(8.9) Let XX andand L be subgroups of a groupgroup GG withwith LL perfectperfect andand [X,[X, L,L, L] = 1.1. Then [X, L] = 1. 1.

Proof. [L,[L, X, X, L]L] = = [X, [X, L, L, L] L] and and byby hypothesishypothesis bothboth areare 1.1. So by the Three- Subgroup Lemma,Lemma, [L,[L, L,L, XIX] ==1. 1. Butby But by hypothesis L = [L,[L , L], L], so [L[L, , X]XI = 1. 1.

9 SolvableSolvable and nilpotent groups A group G is solvable ifif it possesses aa normal seriesseries whose factors areare abelian.abelian.

(9.1) AA groupgroup G G is is solvable solvable if if andand onlyonly ifif Gin)G(") = 1 1 for somesome positive integer n.

Proof. IfIf G(n)G(") = 1, 1, then then (G(n-`):(G("-'): 0 5< i <5 n)n) isis aa normalnormal seriesseries withwith abelianabelian fac-factors by 8.8.4.8.8.4. Conversely if (G1:(Gi: 00 5< ii 5< n) is suchsuch a series then, by 8.8.48.8.4 andand induction on i,i, G(`)G(') 5< Gn_i,GnWi, soso G(n)G(") == 1.

(9.2) AA finitefinite groupgroup isis solvablesolvable ifif andand onlyonly ifif allall itsits composition factorsfactors areare of prime order. 28 Representations of groups on groups

Proof. IfIf allall compositioncomposition factors are of prime order then a composition series for GG isis aa normalnormal series series all all of of whosewhose factorsfactors are are abelian. abelian. Conversely Conversely ifif (Gi(Gi: : 0 0

(9.3) (1)(1) Subgroups andand homomorphic imagesimages ofof solvablesolvable groups are are solvable. (2) IfIf HH <9 G G with with HH and and G/HG/H solvable, solvable, then then GG is is solvable. solvable.

(9.4) Solvable minimal normal subgroups of finitefinite groupsgroups areare elementaryelementary abelian p-groups.

Proof. LetLet G G be be finite finite and and MM a a solvable solvable minimalminimal normalnormal subgroupsubgroup of G. By 9.1 and solvability of M,M, M(1) M(') # M. Next, by 8.3, MM isis characteristicallycharacteristically simple. simple. So, as M(')MM char M, wewe concludeconclude M(l)M(') == 1. Thus M is abelianabelian by 8.8.8.8. Then 8.4.2 completes the proof.

Define L1(G)L 1(G)= = G,G, and,and, proceeding proceeding recursively, recursively, define define L,(G) Ln(G) = = [L,-l [L,-1(G), (G), GIG] for 1 < n Ec Z. GG isis saidsaid toto bebe nilpotentnilpotent ifif LnL,(G) (G) = 1 for some 11 5< n E Z.Z. The class of a nilpotentnilpotent groupgroup isis m m -- 1,1, where m = =min{i: min{i : Li(G) Li(G) == 11. 1).

(9.5) (1)(1) Ln(G)L,(G) charchar G G for for each each 1 1

Proof. PartPart (1) (1) follows follows from from 8.5.38.5.3 byby induction on n. Then (1) and 8.5.5 imply (2) while 8.5.1 and 8.5.3 imply (3).

Define Zo(G)Z0(G) ==1 1 and and proceeding proceeding recursively recursively define define ZnZ, (G) to be the preimagepreimage in G of Z(G/Zn-1(G))Z(G/Z,-l (G)) for for 1 1 <, < n n EE Z. Evidently Z,(G)Zn(G) char G.

(9.6) G isis nilpotent nilpotent if and only if GG = Zn(G)Z,(G) for some 0

Proof. II firstfirst showshow thatthat ifif GG is is nilpotent nilpotent of of classclass mm then then L,n+i-iLm+l-,(G) (G) < 5 ZiZi(G) (G) for 0 0 and Lm+2-i(G)L.+2-i(G)

Next let's see that ifif Z,(G)Zn (G) = = G G for for some some 0 0 5 > 0 0 andand LiLi(G) (G) < 5 zn-i+,(G)Zn_i+l (G) then Li+l(G)Li+1(G) = {Li[Li(G), (G), GIG] 5< {Zn-i+l[zn-i+l(G), (G), G]GI

(9.7) 11 # G is nilpotent of class mm ifif andand onlyonly ifif G/Z(G)G/Z(G) isis nilpotentnilpotent of class m-1.m - 1.

Proof.Proof ThisThis is is a a direct direct consequence consequence ofof 9.6.

(9.8) p-groups areare nilpotent.

Proof. LetLet G G be be a a minimal minimal countercounter example. Then certainly G # 1, so,so, byby 5.16, Z(G) # 1. Hence, by minimalityminimality of of G,G, G/Z(G)G/Z(G) isis nilpotent,nilpotent, so, by 9.7, G is nilpotent, contrary to the choice of G.

(9.9) Let GG bebe nilpotentnilpotent of class m. Then subgroups and homomorphichomomorphic imagesimages of G areare nilpotentnilpotent ofof classclass atat mostmost m.m.

(9.10) If G isis nilpotentnilpotent and H isis aa properproper subgroupsubgroup of G, then H isis properproper inin NG(H>.NG (H).

Proof. Assume NG(H)NC(H) = = H < G. Then Z(G) _(< NG(H)NC(H) = = H, so H* < G*G* == G/Z(G). By By 9.79.7 and and inductioninduction onon thethe nilpotencenilpotence class of G,G, H*H* < NG*(H*). But, as Z(G) <5 H,H, NG.(H*) NG*(H*) = NG(H)*, NG(H)*, so H << NC(H), NG(H), aa contradiction. contradiction.

(9.11) A finite group isis nilpotent ifif and only ifif itit is the direct product of its Sylow groups.

Proof. TheThe direct direct product product ofof nilpotentnilpotent groupsgroups is nilpotent, so by 9.8 the directdirect product of p-groups is is nilpotent. nilpotent. Conversely Conversely let let G G be be nilpotent; nilpotent; we we wish wish toto showshow G is the direct product of itsits SylowSylow groups.groups. LetLet PP E Syl,(G).Syl,(G). ByBy ExerciseExercise 2.9 it suffices toto showshow PP a< G.G. IfIf not, M = NG NG(P) (P) < G, so,so, by 9.10, M < NG(M). Nc(M). But, as P < M,M, {P){P} == SylP(M), Syl,(M), soso PP charchar M. M. Hence Hence NG(M)NG(M) < 5 NG(P) = M, a contradiction.

10 SemidirectSemidirect products In this sectionsection AA andand GG areare groupsgroups andand rr:n: AA + Aut(G) is a representation of A as a group of group automorphisms of G. 30 Representations ofofgroups groups on groups

Let H a9 G. G. AA complementcomplement toto H in G isis aa subgroupsubgroup K of G withwith G = HK and H f1n KK ==1. 1. G G is is said said to to be be anan extensionextension ofof aa group X by aa group YY if there exists HH 9a G with H Z- X X andand G/HG/H - ZY. Y. The The extension extension is is saidsaid toto splitsplit if HH has a complement inin G. The following construction cancan bebe used to describe split extensions. Let S be the set product A x GG and and definedefine aa binarybinary operation on S by

(a,g)(b,h)=(ab,gb"h)(a, g)(b, h) = (ab, gbrh) a,a,bEA,g,hEG b E A, g, h E G where gbkgb" denotes the image of G underunder the automorphismautomorphism b7r bn of G. We call S the semidirectproductsemidirect product of GG byby AA withwith respect toto 7r.n. Denote S by S(A, G, G, 7r). n).

(10.1) (1) SS = = S(A, S(A, G, G, rr) n) is is a a group. group. (2) TheThe mapsmaps u~:A +-+ S andand o-G:u~: GG +-+ SS are injectiveinjective groupgroup homomor-homomorphisms, wherewhere orA: o;l: aa Hi-+ (a,(a, 1) and aG:u~: g i-+H (1, g). (3) GaGGuG a 9 S S and and AorA A- is a complement toto GuGGoo in in S.S. (4) (1,g)fa'1?=(1,ga")forgEG,aEA.(1, g)(atl)= (1, gar) for g E G, a E A.

Observe that if nit isis thethe trivialtrivial homomorphismhomomorphism then thethe semidirectsemidirect productproduct isis just the direct product of A and G.

(10.2) Let H be a group, G 9< H, and B a complementto to GGin in H. Let a: B -++ Aut(G) bebe the conjugation map (i.e., ba: g g -++ gb gb for forb b E E B,B, gg EE G;G; see see Exercise 1.3). DefineDefine B:,B: S(B, S(B, G, G, a)a) +-+ H byby (b,(b, g),B g)B = bg. bg. ThenThen ,B B is an isomorphism.

We see from 10.1 and 10.210.2 that the semidirect productsproducts ofof G by A areare precisely the split extensions of G by A. Moreover the representation defining thethe semidirect product is a conjugation map. Under the hypotheses ofof 10.2, I'llI'll say that H isis aa semidirectsemidirect product of G by B. Formally this means the mapmap ,BB ofof 10.2 isis anan isomorphism.isomorphism.

(10.3) Let SiSi = S(A1, S(Ai, GIGi,, ,rixi),), i ==1, 1,2, 2, be be semidirectsemidirect products. products. ThenThen therethere ex-exists anan isomorphismisomorphism 0: @: SlS1+ -+ S2S2 withwith A1orA,0=A2orA2 Aiu~,@= Az~A, and G1orG,0=G2o'o2Gluc1@= Gz~G, if and only if nl7r1 is is quasiequivalent quasiequivalent to to n2 n2 in in thethe categorycategory ofof groupsgroups andand homo-homomorphisms.

It isis notnot difficultdifficult toto seesee that that semidirectsemidirect products products Si S1= = S(A, S(A, G, G, 7r1) nl) andand S2S2 = S(G, A, n2)ir2) cancan be isomorphic without Jr1nl being quasiequivalent toto n2.72. To investigate just when SlS1 andand S2S2 are isomorphic we need to know more about how Aut(SS)Aut(Si) acts acts on on its its normal normal subgroupssubgroups isomorphicisomorphic toto G,G, and how thethe stabilizer in Aut(Si) of such a subgroup actsacts on itsits complements.complements. This latterlatter questionquestion isis considered in chapter 6. Semidirect products 31

It's alsoalso easy toto cook up nonsplit extensions, and it isis of interestinterest toto generategenerate conditions which insure that an extension splits. The following is perhaps the most important such condition:

(10.4) (Gaschutz'(Gaschiitz' Theorem)Theorem) Let Let p p bebe aa prime, V an abelianabelian normalnormal p-subgroupp-subgroup of a finite group G, and P EE Syl,(G).Sylp(G). Then G splits over V if and only ifif P splits over V.V.

Proof. Notice V 5< P. HenceHence if H isis aa complement complement to VV inin G G thenthen byby thethe Modular PropertyProperty of of Groups, Groups, 1.14, 1.14, P P = = P P nfl GG == PP fl VH=VH = V(P V(P fl H) andand P fln H H isis aa complement complement to V in P. Conversely suppose Q is aa complementcomplement toto VV in in P. P. LetLet C?G == G/ VV and observe P Z Q r= Q.Q. LetLet XX bebe aa setset ofof cosetcoset representatives for V in G. Then the map x H x f isis a a bijection bijection of of XX withwith GC? andand I denote the inverse ofof thisthis mapmap by by a a H H xa.xu. Then

(i) xaxb ==xaby(a, xaby(a, b)b) forfor a, b b E G, and for some y(a,y (a, b) b) E E V.V. Next usingusing associativityassociativity inin GG andand G G we we have have xabcy(a, xabcy (a, bc)y(b, bc)y (b, c) c) = =xaxbCY xaxb, y (b, c)C) ==XU(X~XC) xa (xbxc) = (xaxb)xc (~a~b)~c ==xab~(a, xab y (a, ~)XCb)xc =x~~x~Y(~,= xabxcY (a, b)XCb)x` =xabcy(ab,= xabc y (ab, c)y c)y (a, b)xb)q, from from which which we we conclude:conclude:

(ii) y(ab, c)y(a, b)x1= b)XC F y(a, bc)y(b,,c)bc)y(b, c) for all a, b, b, cc EE G. Now choose X = Q QY, Y, where Y isis a set ofof cosetcoset representativesrepresentatives forfor PP in G. Then, for g E Q and a E G, xgaxsa == gx,,gxa, so:

(iii) xgxs=g = g and y(g,y(g,a)=l a) = 1 forforallg~Q all g E Q and UEG.a E G.

Now (ii) and (iii) imply:

(iv) y(gb,c)=y(b,c)y(gb, c) = y(b, c) forallb,c~Gfor all b, c E G and and ~EQ.g E Q. Next for Cc EE GG definedefine p(c),B(c) = = n,,yI1yEpy(y, y(7, c).c). ByBy (iv),(iv), p(c),B(c) is is independent independent ofof thethe choice of the set Y? of coset representatives. But if b E G then Yb?b is another set of coset representativesrepresentatives forfor Q in G, so:

(v) p(c)NO= = flfl y(yb, y(gb, c)c) for all b,b, cc EE G. ~€9yFk As V is abelian we conclude from (ii) that

(flv(bc)) (flv(b))_fj Y(y, bc)(flYbc)) yEY yEY yEY yEY 32 Representations of groups on groups and then appealing to (v) we obtain

(vi) j(c)j(b)j?(~)j?(b)~~ ', ==,(bc)y(b, j?(bc)y(b, c)~c)' for all b, c E G,G, where m = jGIG :: P1.PI. As PP EE Sylp(G),Syl,(G), (m,(m, p)p) = = 1. 1. Thus Thus m m is is invertible invertible mod mod I VI V 1.I. HenceHence we can define a(c)a(c) == j?(~)-~-',,6(c)-'"-' , forfor cc EE G. G. Then Then taking taking the the -m-' -m-1 powerpower ofof (vi)(vi) wewe obtain:

(vii) a(c)a(b)a(~)a(b)~~ , =a(bc)y(b, c)-1c)-' for allall b, cc EE G.

Finally define y,y,, = = x,a(a)xaa(a) for aii EE G andand setset HH = = {y,,: {y,: a E G).G}. HH will bebe shown to be a complement to V in G. This will complete the proof. It suffices to showshow YbYcyby, = = yb, ybc for for all all b, b, c c E E G. G. But But yby, ybyC =xba(b)xca(c) =xba(b)x0a(c) =_

xbx,a(b)XCa(c) = xbcy(b,xbxà(b)Xà(c)=xbcY(b, c)a(b)xca(c) = Ybca(b~)-lc)a(b)xù(c)=ybca(bc)-ly(b,y (b, c)a(b)xca(c). c)a(b)X°a(c). Then, Then, as VV isis abelian,abelian, (vii) (vii) implies implies ybYC yby, = Ybc, yb,, as desired.

The Schur-Zassenhaus TheoremTheorem inin sectionsection 18 is another another useful result on splitting.

11 Central productsproducts and wreath products (11.1) LetLet {G1:{Gi: 11 _(< ii _(< n)n} be a set of subgroups of G. Then thethe followingfollowing areare equivalent:

(1) GG=(Gi:l(i(n)and[Gi,Gj]=lfori#j. = (Gi: 1

If eithereither of thethe equivalentequivalent conditionsconditions of of 11.111.1 holds, holds, then then G G isis saidsaid toto be a central productproduct of the subgroups Gi, 1 (i< i < n. Notice that the kernel of the homomorphism nit of 11.111.1 is contained in the centre of G1 x . . . x G,.G.

(11.2) Let (Gi(Gi: : 11 _(< ii _(< n) be a family of groups such that Z(G1)Z(G 1)S - Z(G1)Z(Gi) andand Aut(Z(Gi))=AutAut(G,)(Z(Gi))Aut(Z(Gi)) = AU~~"~(~,)(Z(G~))for for 1 1 (

Proof. Adopt the notation of 11.111.1 andand identify Gi withwith Di.Di. By By hypothesishypothesis there are isomorphisms ai: Z(D1) + Z(Di), 11 5

Next assume G is a central product of the G,Gi with Z(GZ(G1) 1) = = Z(Gi)Z(G,) for each i,i, and and letlet 7r:n: D + G bebe thethe surjectivesurjective homomorphism homomorphism suppliedsupplied byby 11.1.11.1. LetLet Pi:PI: Z(D1)Z(Dl) + Z(Di)Z(Dl) be be thethe isomorphismisomorphism whichwhich isis thethe compositioncomposition of of irIz(o,):nlziD,): Z(D1) +- Z(G1) Z(G1) and and (7r(z(o;))-1: (n[Z(D,))-l: Z(G,) + Z(Di).Z(D,). Observe Observe

ker(7r)ker(n) == (z(z-1pi):(z(z-'~i): zz E Z(D1), 1 5< i <5 n) = =A A is a complement toto Z(Di) in Z forfor eacheach i,i, and and ofof coursecourse GG =2 D/A.DIA. To To com-complete the proof I exhibit y E Aut(D)Aut(D) with DiyDi y = = DiDi and Ey ==A. A. NoticeNotice yy induces an isomorphism ofof DIEDIE with D/ADIA mapping mapping GiGi to to Gi,Gi, demonstrating demonstrating uniqueness. LetLet 8i =(a1)`1,Bi:= (ai)-lBi: Z(Di)Z(D,) + Z(D,),Z(Di), so SO that that 8i ai E E Aut(Z(Di)).Aut(Z(Di)). By By hypothesishypothesis there isis yiyi EE Aut(Di) with with yi yi IIz(D,) z(oi) == ai.8i. Define y: D + DD byby (x1,...,xn)H(x1,x2Y2,...,xnYn) and observe y EE Aut(D) with (z(z-1ai))y(~(z-la~))~ == z(z-1,Bi), z(z-'pi), so that Ey = =A. A. ThusThus the proof isis complete.complete.

Under the hypotheses of 11.2, we say G isis thethe centralcentral productproduct ofof the the groupsgroups G,Gi withwith identifiedidentijied centers,centers, and and write write G G = = G G1 1 ** G2 G2 * * .. . . * Gn.G,. Let L be aa groupgroup andand 7r:n: G + Sym(X) aa permutation representation ofof GG onon

X = {1, (1, ... . ,, n}.n). Form the directdirect productproduct DD of n copies of L. G acts as a group of automorphisms of D viavia thethe representationrepresentation aa defineddefined by by ga: (xi,(XI, ...... , , xn) x,) -H (xig-'n, (xlg-I,, ...... , , xng-I,).xng-in). The wreath product ofof LL byby GG (with(with respect respect to to 7r) n) is defined to be the semidirect product S(G, D, a).a). The wreath productproduct is is denoteddenoted by by Lwr Lwr G G or or Lwr,G Lwr,G or LwrrLwr,G. G.

(11.3) Let WW = Lwr, Lwr,G G bebe thethe wreathwreath productproduct ofof LL byby GG with with respectrespect toto Jr.n. Then (1) WW isis aa semidirectsemidirect product ofof D by G wherewhere D = L1 x . . . x LnL, is a direct product of n copies of L. (2) GG permutes permutes A A = = {Li {Li: : 11 < ii <5 nn) } via conjugation and thethe permutationpermutation representation ofof G on A isis equivalentequivalent to r.n. That That is is (L1)g (Li)g = = Lign Lign for for eacheach gEGandI

Exercises for chapter 33 1. LetLet GG and and A A be be finite finite groups groups with with (I (IG G 1, I, ((A A1) 1)= = 1, 1, assume assume A A is is representedrepresented on GG asas a a group group of of automorphisms,automorphisms, andand (Gi(Gi: : 0 (< i

AA centralizescentralizes G. ProduceProduce a countercounter example when (1(I A A 1, G(GI) j) #01. 1. (Hint: (Hint: ReduceReduce toto thethe casecase wherewhere AA isis aa p-group p-group and and use use 5.14.) 5.14.) 2.2. LetLet G G be be a a finite finite group, group, p p aa prime, prime, and and XX aa p-subgroup p-subgroup of of G.G. Prove Prove (1)(1) Either XX EE Sylp(G) oror XX isis properlyproperly containedcontained inin aa SylowSylow p-subgroupp-subgroup ofof NG(X).NG(X). (2)(2) IfIf GG isis aa p-groupp-group andand XX isis aa maximalmaximal subgroupsubgroup of G, thenthen X 4a GG andand JG:X1(G:XI=p. =p. 3.3. Prove Prove lemmas lemmas 10.1 10.1 and and 10.3. 10.3. Exhibit Exhibit a a nonsplit nonsplit extension. extension. 4.4. Let Let p p and and q q be be primes primes with with p p > > q. q. Prove Prove every every group group of of orderorder pqpq is is a a split split extensionextension of of 7LPZp byby Z,.7Lq. Up Up to to isomorphism, isomorphism, how how manymany groupsgroups of order pq exist?exist? (Hint:(Hint: UseUse 10.3 10.3 and and Exercise Exercise 1.7. 1.7. Prove Prove Aut(Zp)Aut(Z,) isis cycliccyclic ofof orderorder pp -- 1. 1. You You may may use use the the fact fact that that the the multiplicative multiplicative group group of of a a finite finite field field isis cyclic.)cyclic.) 5.5. LetLet G G be be a a central central product product of of n n copies copies G1 Gi, , 1 5< i <5 n,n, ofof aa perfect group L and letlet aa! be be an an automorphism automorphism of of G G of of order order n npermuting permuting {G1 {Gi :: 1 1 < 5 ii <5 n} n) transi-transitively.tively. Prove CG(a!)CG (a)= =KZ KZ where where K K = = ~~(a!)(') CG (aP)Z - L/ Uu forfor somesome Uu <5 Z(L) Z(L) andand Z=CZ(G)(a).Z = CZ(G)(~). FurtherFurther NA,t(G)(GI) NAut(~)(G1) nn C(K)

Linear representations

Chapter 4 developsdevelops thethe elementaryelementary theorytheory ofof linearlinear representations. representations. LinearLinear representations are are discusseddiscussed fromfrom thethe pointpoint of view of modules over the group ring. Irreducibility and indecomposability areare defined,defined, andand we find that the Jordan-Holder TheoremTheorem holdsholds forfor finitefinite dimensionaldimensional linearlinear representations.representations. Maschke's Theorem is established in section section 12.12. Maschke'sMaschke's TheoremTheorem sayssays that, if G isis aa finitefinite groupgroup andand FF aa field field whosewhose characteristiccharacteristic does not divide the orderorder of G,G, then then thethe indecomposableindecomposable representationsrepresentations of of G G overover FF areare irreducible. Section 1313 explores the connection between finite dimensional linear representations and matrices. There is also a discussion of the special linear group, the general linear group, and the corresponding projective groups. In particular we find that the special linear group is generatedgenerated by itsits transvectionstransvections and is almost always perfect. Section 1414 contains a discussiondiscussion of the dualdual representationrepresentation which will be needed in section 17.17.

12 ModulesModules over the group ring Section 12 studies linear representations overover aa fieldfield FF using the groupgroup ringring ofof G over F. This This requires requires an an elementaryelementary knowledge knowledge ofof modulesmodules over rings. One reference for this material is chapter 3 of Lang [La]. Throughout section 12, V will be a vector spacespace overover F.F. The group of automorphisms of V in the category ofof vectorvector spacesspaces andand F-linearF-linear transformations is thethe generalgeneral linear group GL(V). Assume n:n: G +-* GL(V) isis a representation of G in thisthis category.category. Such representations will be called FG-representations and V will be called the representation module forfor n.n. Representation modules for FG-representations will will be termedtermed FG-modules.FG-modules. Let R = F[G]F [GI bebe thethe vectorvector spacespace over F withwith basisbasis GG andand definedefine multiplication on R toto bebe thethe linearlinear extensionextension of the multiplication ofof G. Hence a typical member of R is of the formform xgEGagg, wherewhere ag a, c-E FF and at most aa finite number of the coefficients a,ag are nonzero. Multiplication becomes

(ag)(bhh)= agbhgh. a h a,hEG 336 6 Linear representations

As is well known (and easy to check), this multiplication makes R into a ring with identity and, asas thethe multiplicationmultiplication on.on R commutes with scalar multiplication from F, RR isis eveneven an F-algebra. R R = F[G] F[G] is is the the group group ringring oror groupgroup algebra of G over F. Observe that V becomes a (right) R-module under the scalar multiplication:

v(Eagg)= Eag(v(gn))vEV, Eagg c R R-module Conversely if U is an R-module then we have a representationrepresentation a: G -+ GL(U) defined byby u(ga)u(ga) = ug, ug, wherewhere ugug is the module product ofof uu Ec U by g Ec R. Further ifif 7r,:ni: G -+ GL(Vi),GL(V1), ii == 1,1,2, 2, are FG-representations then then ,B: B: V1 -+ - V2 V2 is an equivalence of the representations precisely precisely when when ,BB isis an isomorphismisomorphism of the corresponding R-modules.R-modules. Indeed Indeed y: y: V1 Vl -+ V2Vz is is anan FG-homomorphismFG-homomorphism if and only if y is an R-module homomorphism of the corresponding R-modules. Here y:y: V1Vl -+ V2 is is defined defined to to be bean an FG-homomorphism FG-homomorphismif if yy isis anan F-linearF-linear mapmap commuting with the actions of G inin thethe sensesense that that v(gnl)yv(gn1)y = v v y y (gn2) for each v E Vl,V1, and and g g E c G.G. InIn the terminology of section 4, the FG-homomorphisms are the G-morphisms. The upshot of these observations is that the study of FG-representations is equivalent toto thethe studystudy ofof modulesmodules forfor thethe groupgroup ringring FIG]F[G] = R. II willwill taketake both points of view and appeal to various standard theorems on modules over rings. Lang [La][La] is a reference forfor suchsuch results.results. Observe alsoalso thatthat V is an abelian group underunder additionaddition andand n n is a representation of G on V in the category of groups and homomorphisms. Indeed n induces a representation

n': F#F' x GG -+ Aut(V) in thatthat categorycategory defineddefined byby (a,(a, g)n': vv i-->H av(gn),av(g7r), forfor aa E F',F#, g cE G..G. HereHere F'F# is the multiplicativemultiplicative groupgroup of F. Two FG-representations n andand or a are equivalent if and only ifif n'jr' andand a' are are equivalent, equivalent, so so we we cancan useuse thethe resultsresults ofof chapterchapter 33 to study FG-representations. In the case where F isis aa fieldfield ofof primeprime orderorder wewe can say even more.

(12.1) Let F bebe thethe fieldfield ofof integersintegers modulo p forfor somesome prime p andand assumeassume V is of finite dimension. Then (1) VV isis anan elementaryelementary abelian p-group and nn isis aa representationrepresentation ofof G in the category of groups and homomorphisms. (2) IfIf UU is is an an elementaryelementary abelian abelian p-group p-group writtenwritten additively, additively, thenthen UU isis a vector space F U U over over F, F,where where scalar scalar multiplication multiplication is is defined defined by by ((p) ((p) + + n)u n)u == nu,nu, nEl,uEU.~E~,uEU. (3) GL(FGL(FU) U) is equal to thethe groupgroup Aut(U)Aut(U) ofof groupgroup automorphismsautomorphisms of U.U. Indeed if W is an elementary abelian p-group then the group homomorphisms from UU intointo W W are are precisely precisely the the F-linear F-linear transformations transformations from from F U F intoU into F W F W.W. Modules over the group ring 37

(4) TheThe vector vector spacespace FFV V defineddefined using the construction in part (2) is precisely the vector space V.

As a consequence of 12.1, if F isis a field of prime order, thethe FG-representationsFG-representations are the same as the representations of G on elementary abelian p-groups. A vector subspace U of VV isis anan FG-submoduleFG-submodule of VV ifif UU isis G-invariant. G-invariant. U is anan FG-submodule if and only if U is anan R-submoduleR-submodule of the R-module V. From 7.1 therethere areare groupgroup representationsrepresentations of of F#F# xx G on UU andand V/U.V/U. TheseThese representations areare alsoalso FG-representations andand theythey correspondcorrespond toto thethe R- R- modules U andand V/V/ U. V is irreducible or simple if 0 and V are the only FG-submodules. A composition series for VV isis aa seriesseries O=Va

(12.2) (Jordan-Holder TheoremTheorem forfor FG-modules)FG-modules) Let V bebe aa finitefinite dimensional FG-module. Then VV possessespossesses aa compositioncomposition seriesseries andand thethe composi-composition factors are independent (up to order andand equivalence)equivalence) of the choicechoice of composition series.

TheTherestrictions restrictions nini = = 7r n I Iv,/v,_, v,/v,-,, , 0 < ii <5 n, n, of of 7r n to thethecomposition composition seriesseries (Vi: (Vi: 0 <5 i <5 n) n) of of a afinite finite dimensional dimensional FG-representation FG-representation 7rn areare calledcalled thethe irreducibleirreducible constituents of jr.n. They They areare defineddefined only up to orderorder andand equivalenceequivalence but, sub-subject toto thisthis constraint,constraint, theythey areare wellwell defineddefined and unique by the Jordan-Holder Theorem. V is decomposabledecomposable if there exist proper FG-submodules U and W ofof VV with V = U U ®@ W.W. Otherwise Otherwise V V isis indecomposable. indecomposable. I'll I'll write write 7r n = = 7r1 nl + n2 if V = Vl Vl ®@ V2V2 with VlVl and and V2V2 FG-submodules FG-submodules of of VV and and 7r nlv, I v; is equivalent to ni. ObserveObserve that if a = al a1 + + a2 a2 is is an an FG-representation FG-representation withwith ai equivalentequivalent to ni forfor ii =1= 1and and 2, 2, then then 7r n isis equivalent toto a.a. As in section 4, an FG-module V V isis said said to to bebe thethe extensionextension ofof aa module XX byby a module Y ifif therethere exists exists a a submodule submodule U U of of V V with with U U Z = XX andand V/UV/ U Z= Y.Y. A complement to U in V is an FG-submoduleFG-submodule W with V = U U ®@ W. The extension is said to split if U possesses a complement in V. As in chapter 3, we wish to investigate when extensions split.split. 38 Linear representations

An R-module VV is is cyclic cyclic if if V V = = x xR R = = f {xr:xr: rre E R]R } for for some some xx E V. Equiva- lently V = (xG) is is generatedgenerated asas aa vectorvector space by the images of x under G. The element xx is said to be a generator for the cyclic module V. Notice that irreducible modules are cyclic.

(12.3) (1) If V ==xR x R is is cyclic cyclic then then the the map map r ri-+ H xrxr is is surjective surjective R-module R-module homomorphism from R onto VV with kernel A(x) = f (r r E R: xrxr = 01. 0). (2) Homomorphic imagesimages ofof cycliccyclic modulesmodules areare cyclic,cyclic, soso thethe cycliccyclic R- modules are precisely the homomorphic imagesimages of R. (3) V is irreducible if and only if A(x) is a maximal right ideal of R.

Given R-modules UU and V, HomR(U,HomR(U, V)V) denotes thethe set of all R-moduleR-module homomorphisms of of UU into V.V. HomR(U,HomR(U, V) is anan abelianabelian groupgroup underunder thethe following definition of addition:addition:

u(a +,8)+ /?) == ua ua + + u,8 u/? u E U,U, a,a B/? EE HomR(U, V). If R is commutative, HomR(U, V) is even an R-module when scalar multiplication is defined by u(ar) = (ur)au E U, r E R, a E HomR(U, V). Finally HomRHomR(V, (V, V)V) = = EndR(V)EndR (V) is is a a ring, ring, where where multiplicationmultiplication isis defineddefined toto be composition. That is

u (a,8) _ (ua),8u E V, a, ,8 E EndR (V ). In the language of of sectionsection 4, 4, HomR Hom~ (U, V)V) == MorG Morc (U, V)-V).

(12.4) (Schur's Lemma) LetLet U and VV bebe R-modulesR-modules and a EE HomR(U,HomR(U, V). V). Then (1) IfIf UU isis simplesimple either aa= = 00 oror aa isis an an injection.injection. (2) IfIf VV isis simplesimple either a = 0 0 oror a isis aa surjection.surjection. (3) IfIf UU andand VV are are simplesimple thenthen eithereither a == 0 0 oror aa is is an an isomorphism. isomorphism. (4) IfIf VV isis simplesimple then then EndR(V)EndR(V) isis aa division ring.ring.

The module V isis a a semisimplesemisimple R-module ifif V is thethe directdirect sumsum ofof simplesimple submodules. The The soclesocle ofof V is the submodulesubmodule Soc(V) generated byby all the simple submodules of V.V.

(12.5) Assume Q is a setset of simplesimple submodules ofof V and A C_ Q such thatthat V = (0)(Q) and and (A)(A) = ®AEA eAEA A.A. ThenThen there there exists exists r rG c SZ0 withwith AA CE Fr suchsuch thatthat V == ®bEra,, B.B. Modules over the group ring 39

Proof. Let SS bebe thethe set of r CE Q C2 withwith AA C_E Fr andand (F)(r) = ®BErB. @,,,B. PartiallyPartially order S by inclusion. Check that ifif C is a chain in S then UrECU,,, Fr is anan upper bound for C in S.S. Hence by Zorn's Lemma there isis aa maximalmaximal membermember r r of S. Finally proveprove VV == (r).(F).

(12.6) The followingfollowing are equivalent: (1) VV isis semisimple.semisimple. (2) Vv ==Soc(V). Soc(V). (3) VV splits splits over over everyevery submodulesubmodule of V.V.

Proof. TheThe equivalence equivalence of of (1)(1) andand (2)(2) follows follows from from 12.5.12.5. Assume (3)(3) holds but VV # Soc(V). By (3) there is a complementcomplement U to Soc(V)Soc(V) in V. Let xX EE u#,U#, II a maximalmaximal right idealideal of R containingcontaining AA(x) (x) =_ (r (r E R:R: xrx r == 01, and W the image of I inin xx R under the homomorphism of 12.3.1.12.3.1. By (3) there is a complement Z to W W inin V.V. ByBy the the Modular Modular PropertyProperty ofof Groups,Groups, 1.14, 1.14, ZnxR=MZ fl x R = M isacomplementtois a complement to WinxR.ThenM W in x R. Then M ZxR/W,soMissimple x R/ W, so M is simple by 12.3.3. HenceHence MM 5< Soc(V), so 0 +$ M M <5 Soc(V) Soc(V) fl n U U = = 0, 0, a a contradiction. contradiction. Thus (3) implies (2).(2). Finally assume VV isis semisimplesernisimple and UU isis aa submodule of VV withwith no complement in V. Now Soc(U) = ®AEoA @,,,A forfor some some set set of of simplesimple submodules,submodules, so so byby 12.4 therethere isis aa set set r r ofof simplesimple submodules submodules of of V V with with A A E C r F and and V V = = ®BEr B. Then W == (F(r -- A) A) is is a acomplement complement to to Soc(U)Soc(U) inin V.V. Hence U #+ Soc(U). Soc(U). ByBy the Modular Property ofof Groups,Groups, 1.14,1.14, UU ='SOC(U)= Soc(U) CB® (U fl W). Thus U flfl WW has no simple submodules.submodules. Choose Xx EE (U(U nfl w)#W )O so so that that x xhas has nonzero nonzero projection projection on on a a minimal minimal numbernumber n of members ofof r,r, and and let A E rF suchsuch thatthat xa xa +# 0, 0, wherewhere u:a: xRx R -++ A A isis thethe projection of x RR ontoonto A.A. ForFor 0 0 +0 yy EEX x R,R, thethe setset supp(y)supp(y) ofof membersmembers of of rr upon which y projects nontrivially is a subset of supp(x). Thus by minimality of n, supp(x) = supp(y). supp(y). ThereforeTherefore a: x x R +-+ A is an injection, and hence, by 12.4.2, a is an isomorphism. ButBut thenthenx x R Z A is simple, whereas it has already been observed that UU fln WW hashas nono simplesimple submodules. submodules.

(12.7) Submodules andand homomorphic imagesimages of semisimple modules are semisimple.

(12.8) Assume G is finite, letlet U be an FG-submoduleFG-submodule ofof V,V, andand if if char(F) char(F) == p >> 0 0 assumeassume there is an FP-complement W to U in V for some P EE Syl,(G).Sylp(G). Then V splits over U.

Proof. LetLet WW bebe aa vectorvector subspace ofof V with V == U @® W, andand ifif char(F)char(F) == p >> 0 0 choose choose W W to to be be P-invariantP-invariant forfor somesome PP EE Sylp(G).Syl,(G). If char (F)(F) = 0 let 40 Linear representations

P = 1. 1. LetLet XX bebe aa setset of of cosetcoset representatives representatives for P inin GG and and let let 7r: n: V V -+ UU be the projection of V on U withwith respectrespect to the decomposition V == U ®@ W.W. Let n == jGI G : PIP) and definedefine 8:9: VV -+ -+ VV byby 8 9 = = (Ex,,(r_xex x-'nx)/n,x-17rx)/n, where the sum takes placeplace in in EndF(V). EndF(V). As As (p, (p, n) n)=1, = 1, l/n1ln existsexists in F. AlsoAlso x, x-',x-1, andand n are in EndF(V), so 80 isis aa well-definedwell-defined membermember of of End~(v). EndF(V).As As W W is is P-P- invariant, hn h7r = = nh 7rh for for all all h hE EP, P, so so if ifx xH H h,hx is a map from XX intointo P then 6' 6'0 = (EXEX(Ex,, x-'(h,)-'nh,x)/n.x-1(hx)-17rhxx)/n. ThatThat is is 9 6' isis independentindependent of the choice of coset representatives XX ofof PP in G. Claim 06 E EndR(V).EndR (V). As As the the multiplication multiplication inin RR is aa linearlinear extensionextension of that in G and 80 E EndF(V),EndF(V), it suffices toto showshow g6 g0 = = BgOg for for all all g g E E G. G. But But go gO = = (Ex,, (rxex (~~-')-'nx~-')~/n(xg-1)-17rxg-1)g/n == Og Bg asas Xg-1 xg-' is is a a set set of of cosetcoset representativesrepresentatives for P inin GG and 80 is independent of the choice of X. It remains to observe that, that, as as 7rn isis thethe identityidentity on U, as W = ker(7r), ker(n), and as U is G-invariant, we also have 08 the identity on U and U = V VB. O. HenceHence B202 = = 6.0. There-There- fore V = VO VB @® ker(6).ker(0). AsAs 80 E EndR(G),EndR(G), ker(6)ker(0) isis an FG-submodule. Hence, as U = V0,V8, ker(0)ker(8) is aa complementcomplement to U inin V.V. That is VV splitssplits overover U.U.

(12.9) (Maschke's Theorem)Theorem) AssumeAssume GG isis a finite group and char(F) doesdoes not divide the order of G.G. ThenThen everyevery FG-moduleFG-module isis semisimplesemisimple andand everyevery FG-extension splits.splits.

Proof. ThisThis isis aa directdirect consequenceconsequence of 12.612.6 and 12.8.12.8.

Using 12.9 and notation and terminology introduced earlier in this section we have:

(12.10) Assume G is aa finitefinite group and char(F) does not divide the order of G. Let 7r:n: GG -+--> GL(V) GL(V) bebe aa finitefinite dimensionaldimensional FG-representation.FG-representation. Then (1) 7rn = yi_1 E;=, 7ri ni is is the the sum sum of of itsits irreducibleirreducible constituents (xi:(7ri: 1 1 ( < i i 5< r). (2) If If aa = = Ei_1 xi=, aiai is isanother another finite finite dimensional dimensional FG-representation FG-representation withwith irreducible constituentsconstituents (ai: (a1: 1 1 ( < ii <( s)s) then then 7rn isis equivalent toto aa ifif and only if r = s s andand there is a permutation aa of {1,{1,2, 2, ...... ,, r} r) with niit equivalentequivalent toto aivai, for each i. So, in this special case, the study of FG-representations is essentially reduced to the study of irreducible FG-representations. Let VV bebe aa semisimplesemisimple R-module R-module and and SS aa simple simple R-module. R-module. The The homoge- homoge- neous component of of V V determined determined by by S Sis is(U: (U: U U ( < V,V, U U Z= S). V V isis homoge-homogeneous if it is generated byby isomorphic simple simple submodules.submodules.

(12.11) Let V be a semisimple R-module. Then Modules over the group ring 414 1

(1)(1) IfIf VV is is homogeneous homogeneous thenthen everyevery pairpair ofof simplesimple submodulessubmodules ofof V is isomorphic. (2)(2) VV is is the the direct direct sum sum of of itsits homogeneous homogeneous components.components.

Proof.Proof. As As V V is is semisimple, semisimple, V V = = ®A,12 eAGQ AA for for some some set set S2 S2 of simple submodules of V.V. Let TT bebe aa simplesimple submodule submodule of of VV andand supp(T)supp(T) the the set set of of submodulessubmodules in in SZS2 upon upon which which TT projects nontrivially. If A E supp(T)supp(T) then the projection mapmap a: TT -->-+ AA is an isomorphism byby Schur's Lemma. But if VV isis homogeneoushomogeneous then, by 12.5, 12.5, we may choose A Z= SS forfor somesome simplesimple R-moduleR-module S andand allall A E S2.0. HenceHence (1) (1) holds.holds. Similarly if SS isis aa simplesimple R-module, R-module, HH the the homogeneous homogeneous component component ofof V determined by S, and K thethe submodulesubmodule of VV generatedgenerated by thethe remainingremaining homogeneous components, then,then, asas V isis semisimple, VV == H + K.K. FurtherFurther ifif HHnK#O,wemaychooseT f1 K 0, we may choose T 5< HHnKby f1 K by 12.7.Butnow,by(l),SZ12.7. But now, by (1), S = T T Z= QQ for some simplesimple R-moduleR-module QQ determiningdetermining a a homogeneoushomogeneous componentcomponent distinct distinct from that ofof S,S, aa contradiction.contradiction.

(12.12)(12.12) LetLet HH <9 G G and and U U a a simple simple FH-submodule FH-submodule of of V.V. ThenThen (1) UgUg isis a a simple simple FH-submodule FH-submodule of of VV for for eacheach gg EE G.G. (2) IfIf gg E E CG(H)CG(H) thenthen UU isis FH-isomorphicFH-isomorphic toto Ug.Ug. (3)(3) IfIf XX and and Y Y are are isomorphic isomorphic FH-submodules FH-submodules of of VV thenthen Xg Xg andand YgYg areare FH-isomorphic submodules for each g Ec GG'!

(12.13)(12.13) (Clifford's(Clifford's Theorem)Theorem) Let Let V V be be a afinite finite dimensional dimensional irreducible irreducible FG- FG- module and H <9 G.G. ThenThen (1)(1) VV is is a a semisimple semisimple FH-module. FH-module. (2)(2) GG acts acts transitively transitively on on thethe FH-homogeneousFH-homogeneous components components of of V. V. (3)(3) LetLet U U be be an an FH-homogeneous FH-homogeneous component component ofof V. V. ThenThen NG(U)NG(U) is is irre-irreducible on U and HCG(H) < 5 NG(U).NG(U).

Proof. By By 12.12.1, 12.12.1, G G acts acts on on the the socle socle of of V,V, regarded regarded asas anan FH-module.FH-module. ThusThus (1) holds by 12.612.6 andand thethe irreducibleirreducible action of G. ByBy 12.12.3,12.12.3, GG permutespermutes thethe homogeneous FH-components of of V.V. Then,Then, byby 12.11.212.11.2 and and thethe irreducibleirreducible action of G, GG isis transitivetransitive onon thosethose homogeneoushomogeneous components.components. ByBy 12.12.2,12.12.2, HCG(H) actsacts onon eacheach homogeneoushomogeneous component.component. Then 12.11.212.11.2 and the irreducible action of G completes the proof of (3).

Observe that n7r cancan bebe extendedextended toto aa representationrepresentation ofof RR on V (that is to anan F-algebra homomorphism of of R R into into EndF(V)) EndF(V)) via via n: 7r: C > agg agg H H C> ag(g7r). ag(gn). Indeed for r EE R and v E V,V, v(r7r)v(rn) = yr vr is is just just thethe imageimage of of vv under the module 42 Linear representations product of v byby r in the R-module V.V. Further

ker(n)ker(ir) == (r EE R: yrvr =O= 0 for all vv EE V1.V). V is said to be a faithfulfaithful R-moduleR-module ifif nit is an injectioninjection on R. As G generates R as an F-algebra, R,rRn is is the the subalgebra subalgebra of of EndF(V)EndF(V) generated generated byby G7r.Gn. We We callcall RzrRn the enveloping algebraalgebra of the representation n.

(12.14) EndF(V) = CEfldF(v)(R7r) = CEndF(v)(GJr).

(12.15) If G is finite andand nit is irreducible thenthen Z(Gn)Z(Gir) isis cyclic.

Proof. Let E =EndF(V).= EndF (V). AsAs 7rn isis irreducible,irreducible, D ==EndF(V) EndR(V) isis aa divisiondivision ring by Schur's Lemma. Z = Z(G7r)Z(Gn) 5< CE(Gn)CE(G7r) = = D D by by 12.14.12.14. Also Also D D 5< CE(Gn)CE(GTr) ( < CE(Z),CE(Z), soso Z <5 Z(D).Z(D). Thus Thus the the sub-division-ringsub-division-ring K of D genera-generated by Z isis aa field.field. NowNow ZZ isis a a finite finite subgroup subgroup ofof thethe multiplicativemultiplicative group of the field K, and hence K isis cyclic.cyclic.

I conclude this section by recording two results whose proofs can be found in section 3 of chapter 1717 of Lang [La].[La].

(12.16) LetLet 7r:n: G +-* GL(V) GL(V) be be anan irreducible finite dimensionaldimensional FG-representa-FG-representation. Then RnRir is isomorphic as an F-algebra toto thethe ringring of all m by m matrices over the division ring EndFG(V)EndFG (V)= = D, where m = dimD dimD(V). (V). Further Further F isis in the centre of D.

(12.17) (Burnside) Assume F isis algebraicallyalgebraically closedclosed andand 7r:n: G +-* GL(V) GL(V) isis an irreducible finitefinite dimensionaldimensional FG-representation.FG-representation. ThenThen EndF(V) EndF(V) == RnRir and F=EndFG(V).F = EndFG(V).

13 TheThe generalgeneral linearlinear groupgroup andand specialspecial linear groupgroup In this section F isis aa field,field, nn isis aa positive positive integer,integer, andand VV isis anan n-dimensionaln-dimensional vector space overover F. Recall the group of vector space automorphisms of V is the general linearlinear groupgroup GL(V). As the isomorphismisomorphism type of V depends only on n andand F, the the samesame isis true for GL(V), so we can also write GLn(F)GL"(F) for GL(V).

(13.1) LetLet FnxnF""" denote thethe F-algebra ofof allall nn by n matricesmatrices over F, letlet X = (xi(xl ...... ,x") xn) be an ordered basis forfor V, and forfor gg EE EndF(V) EndF(V) let let Mx(g) Mx(g) =_ (gij)(gig) bebe the the matrix matrix defined defined by by xig xi g= = Cj >j giggijxj, xj, gijgij EE F.F. Then (1) The map Mx: gg H Mx(g) Mx(g) is is an an F-algebra F-algebra isomorphism isomorphism ofof EndF(V)EndF(V) with FnXn.F"'HenceHence the the map map restricts restricts to toa groupa group isomorphism isomorphism of of GL(V) GL(V) with with thethe group ofof allall nonsingularnonsingular nn byby nn matricesmatrices overover F.F. TheThe general linearlinear groupgroup andand specialspecial linearlinear groupgroup 4343

(2)(2) LetLet YY = (yl,(yl, ...... , , yn)be be a a secondsecond ordered ordered basis basis ofof V,V, letlet hh bebe thethe uniqueunique elementelement of GL(V)GL(V) withwith xihxih = = yi, yi , 1 1

BecauseBecause ofof 13.1,13.1, we we cancan thinkthink ofof subgroupssubgroups of of GL(V)GL(V) asas groupsgroups ofof matricesmatrices ifif wewe choose.choose. II taketake thisthis pointpoint ofof viewview whenwhen itit isis profitable.profitable. SimilarlySimilarly anan FG-FG- representationrepresentation itn onon VV can can be be thoughtthought of of asas aa homomorphismhomomorphism fromfrom GG intointo thethe groupgroup of of allall nn byby nn nonsingularnonsingular matricesmatrices overover G,G, by by composingcomposing it n withwith thethe iso-iso- morphismmorphism Mx. itn isis equivalentequivalent to ir':n': GG -*+ GL(V) GL(V) ifif andand onlyonly if 7r'=n' = irh* nh* for for somesome h h EE GL(V), GL(V), and and by by 13.1.2 13.1.2 this this happens happens precisely precisely when whennfMx Jr'Mx = Tr= nMxB*Mx B * forfor somesome nonsingularnonsingular matrixmatrix B. B. This This gives gives a a notion notion of of equivalence equivalence for for `matrix 'matrix representations'.representations'. Namely two homomorphisms a a and a' ofof GG intointo thethe groupgroup ofof allall nn byby nn nonsingularnonsingular matrices matrices overover FF are are equivalent equivalent if if therethere exists exists aa non- non- singularsingular matrixmatrix BB withwith a'a' = = a a B*. B*. Let'sLet's seesee next next whatwhat thethe notions notions ofof reducibilityreducibility andand decomposabilitydecomposability corre- corre- spondspond toto fromfrom thethe point point of of viewview ofof matrices.matrices.

(13.2)(13.2) LetLet 7r: n: G G +GL(V) GL(V) bebe an an FG-representation,FG-representation, UU anan FG-submoduleFG-submodule ofof V,Vv,~=V/U, = V/ U, andandX=(xi: X = (xi: 11 5< i <5 n)n)a a basis basis forfor VV with Y =(xi:= (xi: 11 5< ii <5 m) aa basisbasis forfor U.U. Then,Then, for for g g E E G, G,

[My(g7r l u) 0 Mx(gn) = L A(g) Mx(g7r(g7r v) forfor somesome n - m m by by m m matrix matrix A(g). A(g).

OfOf coursecourse there there is is aa suitable suitable converse converse to to 13.2. 13.2.

(13.3) LetLet ir:n: G G -*+ GL(V) GL(V) be be an an FG-representation,FG-representation, U U and and WW FG-submodulesFG-submodules ofof VV withwith V=UV = U ® @I W,W, and and X X = = (xi (xi: : 1 1 < 5 i < n) aa basisbasis for V V suchsuch that Y=Y = (xi: 1 5< ii <5 m)m) andand ZZ == (xi: (xi: m < ii < 5 n) n) are are basis basis for for U U and and W, W, respectively. respectively. ThenThen forfor gg EE G,G,

My(gnly) 0 Mx(gn) [ fi Mz(gn)w)

AgainAgain therethere isis aa suitable suitable converse converse to to 13.3.13.3. RecallRecall the the notionnotion ofof geometry geometry defined defined in in sectionsection 3. 3. We We associateassociate a a geometrygeometry PG(V)PG(V) toto V,V, calledcalled the the projective projective geometrygeometry ofof' V. V. TheThe objectsobjects ofof PG(V)PG(V) are are the proper nonzero subspacessubspaces of V,V, withwith incidenceincidence defined defined byby inclusion.inclusion. IfIf UU 44 Linear representations

is a subspace ofof V,V, thethe projective projective dimension dimension of of U U is is Pdim(U) Pdim(U) = = dimF(U) dimF(U) - - 1.1. The type function for PG(V) is the projective dimension function

Pdim: PG(V) + I= {0,{O,1, 1,. ...,. . , n - 1). 1).

PG(V) is said to be of dimensiondimension nn -- 1. 1. The subspaces of projective dimension 0, 1,1, and n - 2 2 are are referred referred to to as as points, points, lines,lines, and and hyperplanes,hyperplanes, respectively.respectively. Forg EEGL(V)define GL(V) definegP: gP:PG(V)- PG(V) + PG(V)bygP: PG(V) by gP: UU Ht-, Ug, Ug, for for U U EPG(V).E PG(V). Evidently P: GL(V)GL(V) + Aut(PG(V)) is a representation ofof GL(V) in the category of geometries. (See the discussion in section 3.) Denote the image of

GL(V) under P byby PGL(V).PGL(V). PGL(V)PGL(V) isis the the projective projective generalgeneral linear linear group.group. , The notation PGL,(F)PGLn(F) isis alsoalso usedused forfor PGL(V).PGL(V). A scalar transformation of VVisa is a member member gg ofof EndF(V)EndF(V) suchsuch thatthat vgvg = av av for all v in V and some a in F independentindependent of v. A scalar matrix is a matrix of the form al,aI, a a EE F, where where II is is the the identity identity matrix.matrix.

(13.4) (1) Z(EndF(V))Z(EndF(V)) isis the the set set ofof scalarscalar transformations. transformations. The image of Z(EndF(V)) under Mx isis thethe setset of scalar matrices. (2) Z(GL(V))Z(GL(V)) is is the the set set of of nonzerononzero scalarscalar transformations.transformations. (3) Z(GL(V))=ker(P).Z(GL(V)) = ker(P).

By 13.4, the projective general linear group PGL(V) is isomorphic to to the group of all nn byby n n nonsingular nonsingular matricesmatrices modulomodulo thethe subgroupsubgroup ofof scalarscalar matrices.matrices. OftenOften it will be convenient to regard these groups as the same. Given any FG-representation n:n: G + GL(V), n cancan bebe composedcomposed with P to obtain aa homomorphismhomomorphismnP: nP: G + PGL(V). Observe that nP is is aa represen-representation of G on the projective geometry PG(V). For y E EndF(V)EndF (V) definedefine thethe determinant determinant ofof yy toto be be det(x) det(x) == det(Mx(y)).det(Mx (y)). That is the determinant of y isis thethe determinantdeterminant ofof itsits associatedassociated matrix.matrix. SimilarlySimilarly define thethe tracetrace of y to bebe Tr(y)Tr(y) = Tr(Mx Tr(Mx(y)). (y)). SoSo thethe tracetrace ofof y is the trace of its associated matrix. If A is a matrix and BB isis aa nonsingularnonsingular matrixmatrix thenthen det(~~)det(AB) == det(A)det(A) and Tr(AB)T~(A~) == Tr(A), so det(y) and Tr(y)Tr(y) are independentindependent ofof the choicechoice of basis X by 13.1.2.13.1.2. Define thethe specialspecial linear group SL(V) to be the set of elements of GL(V) of determinant 1. TheThe determinant mapmap isis a homomorphism ofof GL(V) onto the multiplicative groupgroup ofof F withwith SL(V)SL(V) thethe kernelkernel ofof thisthis homomorphism,homomorphism, soso SL(V) is a normal subgroupsubgroup ofof GL(V)GL(V) and and GL(V)/SL(V) GL(V)/SL(V) Z= F#.F'. AlsoAlso writewrite SL,(F)SLn(F) for SL(V). The image of SL(V) under P isis denoteddenoted by PSL(V)PSL(V) or PSLnPSL,(F). (F). The group PSL(V) is the projectiveprojective special linear group. SometimesSometimes PSLn(F)PSL,(F) is denoted by L,(F).Ln(F). The general linear groupgroup and specialspecial linearlinear group group 45

ProveProve thethe nextnext lemmalemma forfor GL(V)GL(V) andand thenthen useuse 5.205.20 to to showshow thethe resultresult holds holds forfor SL(V).SL(V). See See section section 15 15 for for the the definition definition of of 2-transitivity.2-transitivity.

(13.5) SL(V)SL(V) is is 2-transitive 2-transitive onon thethe pointspoints ofof PGL(V).PGL(V).

ForFor v in V and a inin EndF(V), [v, a] = va va -- v vis is the the commutator commutator of of vv with with a.a. ThisThis correspondscorresponds with the notion of commutatorcommutator in section 8. Indeed we can form the semidirectsemidirect product of V by GL(V) with respect to the natural representation, and in this group thethe twotwo notionsnotions agree.agree. Similarly,Similarly, for for G G 5< GL(V),

[V, G]GI = ([v, ([v, g]:gl: v EE V,V, gg EE G) and, for g E G, [V, glg] =_ [V, [V, (g)].(g)l. AA transvectiontransvection is an elementelement t of GL(V)GL(V) such that [V, t] is aa pointpoint ofof PG(V), Cv(t) is is aa hyperplane ofof PG(V),PG(V), andand [V,[V, t] t] I

1 00

MX(t)= 0 1 0

1 0 1 so evidently t isis ofof determinant 11 andand 13.1.213.1.2 impliesimplies GL(V)GL(V) isis transitivetransitive onon itsits transvections. WriteWrite diag(al,diag(al, ...., . . ,an) a,) forfor the the diagonal diagonal matrixmatrix whosewhose (i,(i, i)-thi)-th entry isis ai.ai. IfIf n n >> 22let let A = {diag(1, a, 1, ... , 1): a E F#} and ififn=2 n = 2 let let

A = (diag(a,{diag(a, a): a EE F#).F') .

Then A <5 CGL(v)(t) CGL(~)(t) and either det: A + F#F' isis aa surjectionsurjection oror n = 2 2 andand somesome element of F isis notnot aa squaresquare in F.F. In In the the first first casecase GL(V)GL(V) = ASL(V), ASL(V), so, as GL(V) is transitivetransitive on its transvections, so isis SL(V)SL(V) by 5.20. FurtherFurther ifif nn > 22 and s isis thethe transvectiontransvection with Cv(s)Cv(s) == Cv(t) andand [xn,[x,, s]s] =x2,=x2, then st is alsoalso a transvection. So,So, asas SL(V)SL(V) isis transitivetransitive on on its its transvections, transvections, t =t =s-1(st) s-'(st) is a commutator of of SL(V).SL(V). On On the the otherother hand, hand, ifif nn == 2, 2, then then for forb b E E F#F' letlet t t(b) (b) be the transvection with x2t(b)x2t(b) == x2x2 + +bxl bxl andand gg = = diag(a,diag(a, a-I).a-1). Then t(b)gt(b)9 = t(a2b). (a2_1)-l' Further ifif IFF >I >3 3then then a a canbe can be chosen chosen with with a2a2# 1. Thus,Thus, settingsetting bb == (a2- I)-', we have [t(b), g] = t, t, and and againagain tt isis a a commutator commutator of of SL(V).SL(V). 46 Linear representations

We have shown:

(13.6) (1)(1) Transvections Transvections are are of of determinantdeterminant 1.1. (2) TheThe transvectionstransvections form a conjugacy class of GL(V). (3) EitherEither thethe transvectionstransvections form a conjugacy class ofof SL(V) or n = 2 and F containscontains nonsquares. (4) IfIf IFIF II >> 3 3 or or n n >> 2, 2, then then each each transvection transvection is is in in the the commutatorcommutator groupgroup of SL(V).

(13.7) SL(V)SYV) is is generatedgenerated by by itsits transvections.transvections.

Proof.P r o o fLet . Let i-2 0be be the the set set of of n-tuples n-tuples o w = = (XI, (x1,...... ,, x,-l,x,,-,, (xn))(x,)) suchsuch that that (x (xl l , ...... , , x,)xn) is a basis for V. Let TT bebe thethe subgroupsubgroup of GG == SL(V) SL(V) generatedgenerated by the transvections ofof G.G. I'llI'll show T isis transitivetransitive onon a.0. Then, Then, byby 5.20,5.20, GG = TG,,,. TG,. But GU,G, = = 1,1, soso thethe lemmalemma holds. It remains toto showshow TT isis transitivetransitive on on a. 0. Pick aa = (yi,(yl, ...... , ,Yn-1, y,-1, (Yn))(y,)) E i-20 such thatthat aa 04 wT,oT, yiyi =xi forfor i i< i m, m, and, and, subject subject to to these these constraints, constraints, with m maximal. LetLet U=(xi:iU = (xi: i Yy=y,+l, = Y,n+i, andand W=W = (U,x,(U, x, y).y). Then dim(W/ U) = k = 1 1 or 2. Suppose kk == 22 andand letlet HH be a hyperplanehyperplane of of VV containing containing U U and and x x - - y but not x.x. LetLet tt bebe the the transvection transvection with with axis axis H H and and [y, [y, t] t] = = x x - - y.y. ThenThen ytyt == x and xixit t =xi= xi for i <( m, m, soso atat E E wT oT byby maximalitymaximality ofof m.m. Then a EE wT,oT, contrarycontrary to to the choice of a. Sok=l.Supposem=n-l.Ask=l,ax-ySo k = 1. Suppose m = n - 1. As k = 1, ax - y EE UUforsomea for some a E F#.F#.As As rnm == nn -- 1 1 andand a ## w, o, ax -- y y# #0. 0.So So there there is is a atransvection transvection tt withwith axisaxis UU and [y,[y, t]t] =ax = ax -- y.y. Now at = w, o, contradicting a a 4 0 oT.off. SoSo mm << n - 1, 1, and and hence there is z E V - W. W. An An argument argument in in the the lastlast paragraphparagraph showsshows there are transvections s s and and t t withwith U U (< Ca(t)Cv(t) flrl Cv(s),Cv(s), ys = z, and zt =x. ButBut nownow xist =xi forfor i i< 5 m m and and yst yst = = x, x, contradicting contradicting the the choicechoice of a.

(13.8) If n >2 2 2 thenthen SL,(F)SL,(F) is is perfect perfect unless unless n n = = 2 2 and and IF IF I I == 2 2 oror 3.3.

ProojProof. LetLet G G = = SLn(F). SL,(F). By By 13.7 13.7 it it suffices suffices to to showshow transvections are containedcontained in GM,G('), andand thisthis follows from 13.6.4.13.6.4.

14 TheThe dual representation In sectionsection 14, 14, VV continuescontinues to bebe an an n-dimensionaln-dimensional vector space over F andand 7r:n: GG +-+ GL(V) is an FG-representation of a group G. Let (K:(Vi: -oo -oo << i i< < oo) oo) be be a asequence sequence of of FG-modulesFG-modules andand ...-+ V_1 VoC"') V1-+ ... The dualdual representation 47 a sequence of FG-homomorphisms. TheThe latterlatter sequencesequence isis saidsaid toto bebe exactexact ifif ker(ai+1)ker(ai+l) == KaiViai for each i. A shortshort exactexact sequencesequence isis anan exactexact sequencesequence of the formform 00 + -+ U U 3 - VV 5 W +-+ 0. The maps 0 -++ U U andand W W --> -t O0 areare forcedforced to be trivial. Observe that the hypothesis that the sequence is exactexact isis equivalentequivalent to requiring thatthat aa be an injection, /?$ a surjection, andand UaUa = ker(p).ker(/?). Hence Hence W E- V/ V/ Ua Ua and and the the sequence sequence is essentially 0 0 + -+ Ua Ua +-+ V +-+ V/ Ua +-+ 0 with Ua -++ V V inclusioninclusion and V +-+ V/Ua the the natural natural map. map. TheThe sequence sequence isis saidsaid to split if V splitssplits overover Ua. As As isis well well known, known, thethe sequencesequence splitssplits ifif andand onlyonly ifif there is y EE HomFG(W,HomFG(W, V) with y/?yp =1,= 1, and and this this condition condition isis equivalentequivalent in turn to the existence of S6 E HomFG(V, U) with a6aS = 1. 1. Let VV* * = = HomF(V,HomF (V, F) F) andand recallrecall fromfrom sectionsection 13 thatthat V* isis aa vectorvector spacespace over F. WeWe callcall V* V* the the dual dual space space of of V. V. It It is is well well known known that that n n= =dimF dimF(V*). (V * ). If UU isis an an F-space F-space and and a EHomF(U,a €HomF(U, V) V) define define a* a* EHomF(V*,€HomF(V*, U*) by XU*xa* =ax, x x EE V*.V*.

(14.1) LetLet U,U, V,V, andand WW be finite dimensionaldimensional F-spaces,F-spaces, a EE HomF(U, V),V), and ,B/? E E HomF(V, HomF(V, W). Then (1) The mapmap y i-+H y*y* is is an an F-space F-space isomorphism isomorphism ofof HomF(U,HomF(U, V) with HomF(V*, U*). (2) (afi)*(a/?)* = /?*a*. (3) IfIf U4U 3 V V - 5W Wis exactis exact then then so so is is W* W* V*V*-* % U*.U*.

If 7r:n: GG +-+ GL(V) is anan FG-representationFG-representation then, then, fromfrom 14.1,14.1, 7r*: n *: G + GL(V*) is also an FG-representation, where 7r*:n*: gg HH (g-17r)*.(g-ln)*. TheThe representation representation 7r* n* is called the dual ofof 7rn and thethe representationrepresentation modulemodule V*V* of 7r*n* isis called the dual of the representation module VV ofof 7r.n. Given aa basisbasis X X = = (xi:(xi :1 1 _( < i i5 < n)n) forfor V,V, thethe dual dual basis basis 2 X = = (ai (zi :: 1 1 5 < ii 5< n) of X isis defineddefined by by xiii: : xjxj i-+ H SidJij. .Notice Notice xi Tiaixiaifi is is thethe uniqueunique membermember ofof V*V* mapping xi to ai for each i.i .

(14.2) LetLet 7r:n: G +-+ GL(V) be an FG-representation and X a basis for V. Then MX(g7r*)M2 (gn *) = = (M~(MX(g7r)-1)T(gn)-l)T, where BTB~ denotes the transpose of aa matrix B.B.

By 14.2, if 7rn is viewed as a matrix representation, then then 7r*n* is justjust the composition of n7r withwith thethe transpose-inversetranspose-inverse mapmap onon GL,(F).GLn(F). As thethe transpose-inverse map is of order 2, we conclude

(14.3) (7r(n*)* *)* isis equivalent equivalent to to n rr for for each each finite finite dimensional dimensional FG-representationFG-representation 7r. n.

Thereere is a more concrete way to see this. 48 Linear representationsrepresentations

(14.4) Let U and VV be finitefinite dimensional F-spaces. Then (1) For For each each v v EE V V there there exists exists a a uniqueunique element vBv0 E (V*)* withwith xu0xv9 == vx for all x E V*.V*. (2) TheThe map map 0:0: vv H vB v0 isis anan F-isomorphismF-isomorphism ofof VV with with (V*)*.(V*)*. (3) ForFor eacheach aa E E HomF(U,HomF(U, V),V), aBa0 = B(a*)*. @(a*)*. (4) 00 defines defines an an equivalence equivalence of of 7r n andand (7r*)*. (n *)*.

Proof. To proveprove (1) (1) let let X X= (xi= (xi: : 1 5 1 i

Notice 14.4 gives a constructiveconstructive proof of 14.3. It will also bebe usefuluseful inin thethe proofproof of the next lemma.

(14.5) Let G be aa groupgroup andand U,U, V,V, andand WW finite finite dimensional dimensional FG-modules.FG-modules. Then (1) HomFG(U*,HOmFG(U*, V*)V*) = (a*: {a*: a EE HomFG(V, HomFc(V, U)}.U)}. B (2) 0 + U -% 4 V +- W W + 0 is an exact sequence of FG-modules ifif andand only B* if 00 + WW* * -+ V V* * % UU* * -+ -+ 0 0 is.is. TheThe first sequence splitssplits ifif and only if thethe second splits.splits. (3) VV is is irreducible, irreducible, indecomposable, indecomposable, semisimple, semisimple, and and homogeneous if and only if V*V* has the respective property. property.

Proof. PartPart (1)(1) followsfollows fromfrom 14.1.114.1.1 andand 14.1.2. 14.1.2. TheThe firstfirst partpart ofof (2)(2) followsfollows from 14.1.3 and 14.4. The second part follows fromfrom thethe remarkremark about splitting at the beginning of this section and 14.1.2.14.1.2. Part (3) follows from (2), since the properties in (3) can be described in terms of exactexact sequences and thethe splitting of such sequences.

(14.6) (1)(1) Let Let a: a: V V --* + U U be be an an FG-homomorphismFG-homomorphism of of finitefinite dimensional modules. Then [G, U] _(< VVa a if andand onlyonly ifif ker(a*)ker(a*) < ( CU. CU*(G), (G), whilewhile [G,[G, V*]V*] _(< U*a* if and onlyonly if if ker(a) ker(a) _(< Cv(G).CV(G). (2) IfIf UU isis aa finitefinite dimensional FG-module thenthen U == [U, GIG] if and only if Cu*(G)CU.(G) = 0, while UU* * == [U*,[U*, G]GI if and only if CU(G) = = 0. The dual representation 49

Proof.Proof. We We have have the the exact exact sequence sequence V-0-'> U - U/Va --> 0 so,so, byby 14.1.3,14.1.3,

0--> (U/Va)*--> isis alsoalso exact.exact. Let 7rn be thethe representationrepresentation of of GG onon U/U/ Va.Va. Then [G, U] <5 VaVa ifif andand onlyonly ifif GnGn = = 1. 1. This This is is equivalent equivalent to to Gn*Gn* == 1 1 which which in in turn turn holds holds ifif andand onlyonly if GG centralizescentralizes (U/Va)*.(U/ Va)*. As As (U/Va)*(U/ Va)* is is FG-isomorphic FG-isomorphic to to ker(a*) ker(a*) by by thethe exactnessexactness ofof thethe secondsecond series series above, above, thethe firstfirst partpart ofof (1)(1) holds,holds, while while the the secondsecond follows follows from from the the first first and and 14.4.3.14.4.3. LetLet UU # [U,[U, G]GI = V V andanda: a: V -+--> U Utheinclusion. theinclusion. By By(1),ker(a*) U/CU(G) be be thethe naturalnatural map.map. ByBy (1),(I), [G,[G, U*]U*] < 5 (U/CU(G))*,8*. (U/CU(G))*B*. AsAs ,B B isis notnot anan injection, injection, ,B* B* is not aa surjectionsurjection by by 14.5.2.14.5.2. SoSo U*U* # [G,[G, U*]. Applying 14.4.314.4.3 we see that ifif 00 # CU.(G)CU*(G) thenthen UU # [G,[G, U], U], completingcompleting the the proofproof ofof (2).(2).

TheThe charactercharacter ofof anan FG-representation FG-representation 7rn isis thethe mapmap X:X: GG -+ FF defined defined byby X~(g) (g) =Tr(gn).=Tr(g7r). RememberRemember Tr(gn) Tr(g r) isis thethe tracetrace of the matrix Mx(gn) andand isis independentindependent of of thethe choicechoice of of thethe basisbasis X X for for'ihe 'the representation representation module module of of 7r.n.

(14.7)(14.7) LetLet 7rn bebe an an FG-representation,FG-representation, n*7r* thethe dualdual ofof n,r, and and Xx andand X*x * thethe characterscharacters of of 7rn andand 7r*,n*, respectively. Then X*(g)x*(g) = X(g-1) x(~-') forfor each each g g E E G. G.

Proof.Proof. ByBy 14.2, 14.2, MX(g7r*) M2(gn*) = Mx(g-17r~~(~-'n)~, )T, soso thethe lemma follows as Tr(A) =_ Tr(AT)T~(A~) for each n by nn matrixmatrix A.A.

SinceSince characterscharacters have have nownow beenbeen introducedintroduced I I shouldshould probablyprobably recordrecord two two more more propertiesproperties ofof characterscharacters which which areare immediateimmediate from from 13.113.1 and and thethe factfact thatthat con-con- jugatejugate matricesmatrices have have thethe samesame trace. trace.

(14.8)(14.8) (1)(1) EquivalentEquivalent FG-representations FG-representations have have the the same same character. character. (2) IfIf X x is is the the character character of of anan FG-representation FG-representation then then X (gh)(gh) == X x (g) (g) forfor eacheach g,g, hh EE G.G.

Remarks. TheThe stuff stuff in in sections sections 12 12 and and 13 13 is is pretty pretty basic basic but but sectionsection 14 14 is is moremore specialized. Section Section 14 14 is is included here to prepare the way forfor thel-cohomologythe l-cohomology inin sectionsection 17.17. ThatThat sectionsection isis alsoalso specialized.specialized. BothBoth cancan bebe safelysafely skippedskipped oror 50 Linear representations postponed by the casual reader. If so, lemma 17.10 must be assumed in proving the Schur-Zassenhaus Theorem inin section 18. ButBut that'sthat's no problem. The reader who is not familiar with the theory of modules overover ringsrings mightmight want to bone up on modules before beginning section 12.12.

Exercises forfor chapterchapter 4 1. LetLet GG bebe aa finite finite subgroup subgroup ofof GL(V),GL(V), wherewhere VV isis aa finite finite dimensional dimensional vector space over aa fieldfield FF with (char (F), IGI)\GI) = 1. 1. ProveProve (1) VV = [G, [G, V]Vl ®@ Cv(G).Cv(G). (2) If GG isis abelianabelian then V == (CV(D):(Cv(D):D D EE A), where A is the set of subgroups DD ofof GG with G/DG/D cyclic.cyclic. (3) IfIf G =Z Ep",E,., nn >> 0, 0, and and VV = = [V, [V, G],GI, then V == ®HEr@,,, Cv(H),CV(H), wherewhere r isis the the setset ofof subgroupssubgroups of G ofof indexindex p. 2. LetLet VV be be aa finite finite dimensional dimensional vector space over a field F, g EE EndF(V),EndF(V), and U a g-invariant subspace of V.V. Prove (1) gg centralizescentralizes V/UV/ U ifif andand onlyonly ifif [V,[V, g] 5< U. (2) The The mapmap vv i-+H [v,[v, g] g] is is a a surjective surjective linearlinear transformationtransformation of V onto [V, g] with kernel CV(g).Cv(g). (3) dimF(V)dim~(V) = dimF(CV(g))dim~(Cv(g)>+ +dimF([V, dim~([V, gl).g]) 3. Let Let G G bebe aa finite finite group, group, FF a afield field of of prime prime characteristic characteristic p,p, and and .7r n anan irreducible FG-representation. Prove Prove Op(G7r)O,(Gn) = 1. 1. 4. LetLet FF be be a a field, field, rr andand qq bebe primes,primes, XX aa groupgroup of order r acting irreducibly on a noncyclic elementaryelementary abelianabelian q-groupq-group Q,Q, and V == [V, Q] a faithful irreducible FXQ-module. ThenThen dimF(V)dimF(V) == rkrk where k == dimF(Cv(X))dimF(CV(X)) = dimF(Cv(H))dimF(CV(H)) forfor some hyperplanehyperplane HH ofof Q. 5. LetLet a a: : GG -->+ SymSym (I)(I) be be a apermutation permutation representation representation of of aa finite finite group group G G onon afinitea finite setset I, I,and and letlet FF bebe aa field field and and V V an an F-space F-space with with basis basis X X = = (xi(xi :: i iE E I). I). The FG-representation n7r inducedinduced byby aa is the representation on V with g7r:gn: xi xi HH xigaxiga for each g E G and i E I. V V is called thethepemzutation permutation modulemodule of a. Let Let Xx bebe thethe charactercharacter of of 7r.n. Prove (1) X(g)~(g) isis the the number number of of fixedfixed pointspoints ofof gotga on I forfor each each g g E E G.G. (2) (>gEG(CgsG X(g))/IGIx(g))/lG I isis the the number number ofof fixedfixed points of G on I. (g)2)/ (3) IfIf GG isis transitive onon I I thenthen (C,,,(EgEG Xx (~)')/IGI GI Iis is the the permutation permutation rankrank of G on I. (See (See sectionsection 1515 forfor the definition of permutation rank.)rank.) 6. AssumeAssume thethe hypothesishypothesis of of the previous exercise with G transitive onon I.I. Let zZ == G-+iEICis[xi, Xi, Z Z= = (z), (z), and

U={(xi: =O,a1 EF . IEI iEI U is the core of the permutation module V.V. Prove (1) ZZ == CV Cv(G) (G) and and UU == [V, [V, G].GI. The dual representation 551 1

(2) If If WW isis anan FG-module,FG-module, i E I, H = Gi Gi isis thethe stabilizerstabilizer in G ofof i,i, w w E Cw(H),Cw (H), and and WW == (wG), (w G) , then then there there is is a a surjective surjective homomorphismhomomorphism of V onto W. (3) AssumeAssume pp == char(F) char(F) is is a aprime prime divisor divisor of of 111. I I I. Then V does not split over U, V does notnot split split over over 2, Z,and and if ifOP(G) O' (G) = = G G then then H H '(G, 1(G, U/Z) U/Z) ## 0. (See section 17 for a discussion of thethe 1-cohomology groupgroup H';H1; in particular use 17.11.)17.1 1.) 7. LetLet F F be be a a field, field, U U a a 2-dimensional 2-dimensional vector vector spacespace overover FF with with basisbasis {x,{x, y}, y), G = GL(U), andand VV = F[x, F [x, y] y] the the polynomial polynomial ring ring inin x and y over F.F. Prove (1) Irn isis anan FG-representationFG-representation on VV where Irn is defined by f (x, y)g7ry)gn == ff(xg, (xg, yg)yg)for for f EE VV and andg g EE G.G. (2) GG acts acts on on the the (n (n + + 1)-dimensional 1)-dimensional subspacesubspace Vn Vn of homogeneous polynomials of degreedegree n.n. LetLet nnnn be be the the restriction restriction of of 7r n toto Vn. Vn. (3) IfIf char(F)char(F) = = p p > > 0, 0, prove prove 7rn n,, is is not irreducible forfor pp 5< n $ -1- 1 mod mod p, but nn7rn is is irreducible irreducible for for 0 0 5 < nn << p. (4) ker(7rn)ker(nn) isis thethe groupgroup ofof scalar scalar transformations transformations a a I Iof of U U with with a a EE FF and an=1. (Hint: InIn (3)(3) letlet TT be the group of transvectionstransvections in G with center (x) and for i <5 n n letlet Mi Mi bebe thethe subspace ofof MM == Vn generated generated by by yjxn-1, y1xn-j, 0 05 < j j 5< i. Prove [ylxn-l[yixn-1 +Mi-2, +Mi_2, T]TI = Mi_1/Mi_2 Mi-l/Mi-z for all 1 i< ii <5 n. n. Conclude Conclude Mo Mo is contained in any nonzero FG-submodule of M andand then, as Mo isis conjugate conjugate to (yn)(yn) underunder G,G, concludeconclude MiMI isis containedcontqined inin anyany suchsuch submodule submodule for all i.) 8. LetLetVbeavectorspaceoverafieldFandO=Vo~Vl~~~ V be a vector space over a field F and 0 = Vo < V1 5< VnVn = = VVa a sequence of subspaces. Let Let G G bebe aa subgroup of of GL(V)GL(V) centralizing centralizing Vi+1 Vi+' / ViVl for eacheach i,i, 00 5< i < n.n. ProveProve (1) GG isis nilpotentnilpotent of class at mostmost nn -- 1.1. (2) If If 00 # U U is is a a G-invariantG-invariant subspace ofof VV then then CU Cu(G) (G) # # 0 andand [U,[U, GIG] < U. 9. LetLet VV bebe anan n-dimensionaln-dimensional vector spacespace over over a a field field F F with with n n 2> 2, G = GL(V),X=(xi:15iGL(V), X = (xi: 1 2 2 then then NG NG(Y) (Y) = NG(H). 52 Linear representations

(4) IfIf FF is is finite finite of of characteristiccharacteristic p thenthen UU EE Sylp(G).Syl,(G). (5) BB == NG(U). NG(U). Indeed Indeed VV Vi is the unique object of type i fixed by U. (6) TheThe residue residue Fsrs of of aa flag flag S S of of corankcorank 11 is is isomorphicisomorphic toto thethe projective line over F, and (G5)rs(Gs)~s2 PGL2(F). (See (See sectionsection 33 forfor the definitiondefinition of residue.) 10. Let FF bebe aa field field andand FI- == F F U U {oo} {co} the projectiveprojective lineline over over F. F. Let Let G G == GL2(F)GL(F) be the group of invertible 22 byby 22 matricesmatrices overover F,F, and for

a1,1 a1,2 A = (aij) = (=- GL2(F) a2,1 a2,2 define @(A):O(A):I- F +-+ FI- by a1,1z + a2,1 O(A): z H a1,2z + a2,2 where byby conventionconvention alto a/oo = 0 0 forfor a EE F#F# andand ooo(A)co@(A) == ai, al,l/al,2. i /a1,2. PickPick aa basis B = {x1, {xl, x2} for aa 2-dimensional2-dimensional vectorvector spacespace VV overover F,F, andand identifyidentify GL(V) with G via the isomorphism MB: : GL(V) GL(V) +-+ G. Let Q!2 be the points of the projective geometry of V. Prove (1) ForFor AA EE G,G, O(A) @(A) isis aa permutation permutation ofof 1,.r. (2) G*G* == {O(A): {@(A): A EE G}G} isis aa subgroupsubgroup of Sym(F),Sym(r), and @:0: G +-+ G* is a surjective groupgroup homomorphismhomomorphism with with kernel kernel Z(G), Z(G), soso @0 induces an isomorphism 6:: Gc -->. + G*, where G = PGL(V).PGL(V). (3) Define Define a: Q!2 +-+ Fr byby a: Fx1 Fxl i-±H cooo andand a: a: F(hxlF(a.x1 +x2) +x2) H H h? for ha. EE F.F. Then aa is a bijectionbijection suchsuch thatthat (wg)a (tog)a == (oa)$(g) for each Coo E E !20 and g EE G.c. Hence Hence a a is is an an equivalence equivalence of of thethe permutationpermutation representationsrepresentations of OC on Q!2 and G* on r.F. 5

Permutation groups

This chapter derives a number of properties of the alternating and symmetric groupsoups AnA, and S,S ofof finitefinite degree degree n. n. ForFor exampleexample the the conjugacyconjugacy ofof elementselements in in An and S,S,, is determined, and it is shown that A,An is simple if n >2 5.5. Section Section 15 also contains a brief discussion of multiplymultiply transitivetransitive permutation groups.groups. Section 16 studies rank 3 permutation groups.groups.

15 TheThe symmetric and alternating groups Let X be aa setset andand SS thethe symmetricsymmetric group on X. AA permutation group on XX is a subgroup of S. Let G be a permutation group onon X. In this section X is assumed to be of finite order n. Thus S is of order n!, so S and GG areare finite.finite. Suppose gg E S and letlet HH = (g). (g). ThenThen g g isis of finite orderorder m m and and H H == {gl:{g': 00 5< i << m m}. }. Further H has a finitefinite numbernumber of of orbits orbits (x, (xi H: H: 1 1 5 < ii 5< k),k), and the orbit xix, HH is of finite orderorder I,.li. LetLet H,Hl = Hx;H,, bebe thethe stabilizerstabilizer in H ofof xi.x,. ByBy 5.11,5.1 1, I,li = I IH:H,I,so,asH H : Hi 1, so, as H = = (g)iscyclic,H,(g) is cyclic, Hi = = (gli)and{g-':Os(gli) and {gJ : 0

,xig11,-1). -1). glx,~gIXix= = (xl,(XZ, xzg,xig, xlg2,xig2, ... -.9 Xzg

This notation indicates that g:g: x,xig' g-' i-+H x,xi gj+lgJ+1 for for 0 0 5 < j j << 1,li - 1 1 and and g: xix, glg-lgli -1H H xi.x, . The last fact holds as gbgl fixes fixes xi. Further, as the orbits of H partition X, we can describe the actionaction of g onon XX withwith thethe followingfollowing notation:notation:

xkgit-1). (x1, xlg, ... , xlg1 -1) (X2, X29, .. -X29 12-1) ... (xk, xkg, .. ,

This is the cyclecycle notation for the permutation g.g. It describes g, and the description is unique up to a choice ofof representative xixi forfor thethe ithith orbit and the ordering ofof thethe orbits.orbits. For For example,example, if if XX is is the the set set of of integers integers { {1,2, 1 , 2, ...... , ,n}, n}, the the representative xixi couldcould be be chosenchosen toto bebe thethe minimalminimal membermember of of thethe iith th orbit,orbit, and the orbits ordered so thatthat xlx1 < x2 < . . . < xk.xk. If so, g can be written uniquely in cycle notation, and conversely eacheach partitionpartition of X and each ordering of the partition and the members of the partition, subject to these constraints, defines some member of S in thethe cyclecycle notation. By convention the terms (xi) corresponding toto orbitsorbits ofof HH of length 11 are omitted. Thus for example ifif n = 55 wewe would would writewrite gg == (1, (1,2)(3,4)(5) 2)(3, 4)(5) asas 54 Permutation groups g = (1, (1,2)(3,4). 2)(3, 4). Notice g isis still still uniquely uniquely describeddescribed in thisthis modified modified cycle notation. Subject to thisthis convention,convention, gi g1 = = (xi,(xi, xig,xt g, . ..., . . ,x!xig''-') gt;-l) isis a member of S. The elements gl, ...... , ,gk gk areare called the cycles of g. Also g is said to be a cycle if H has at most one orbit of length greater than 1. Notice the two uses of the term 'cycle'`cycle' are compatible. Given a subset A of X let Mov(A) be the set ofof pointspoints of X moved by A. HereHere x in X is moved by A if ax # x for some aa E A. Notice Mov(A)Mov(A) == Mov((A)) and X is the disjoint union of Mov(A) and Fix(A). Cycles cc andand d inin SS areare saidsaid to be disjoint if Mov(c) fl Mov(d) isis empty.empty.

(15.1) Let A, B gC S withwith Mov(A)Mov(A) nfl Mov(B)Mov(B) empty.empty. Then Then ab ab = = baba for all a~Aandb~aEAandbEB.B.

(15.2) Let glgi, , ...,. . . ,gr gr bebe the nontrivial cyclescycles of of g g EE s'.S. ThenThen (1) gjgigigj =gjgi= gigs forifor i ## j. (2) g = gigl .. . . grg, is the product in S of its nontrivial cycles.cycles. (3) If g = cl cl ... . .cs . cs with with {cl, {cl, .... ., .cs) , c,) a aset set of of nontrivial nontrivial disjoint disjoint cycles cycles thenthen {cl,...,C5)Iclt ..-,cs)= _ Igl,(gl,...,g,). -..,gr). (4) The order of g is the least common multiple of the lengths of its cycles.

By 15.2, each member of S# can be written uniquely asas the product ofof nontrivial disjoint cycles, and these cycles commute, so the order of thethe productproduct is immaterial. For g cE S S define define Cycg Cyc, to to be be the the function function from from 7L+ Z+ intointo 1L Z such that Cycg(i)Cyc,(i) is the numbernumber of cyclescycles ofof gg of lengthlength i.i. Permutations g and h are said to have the same cycle structurestructure if Cyc,Cycg = Cych.Cyc,.

(15.3) Let g, h E S with g =(al,...,aa)(bl,...,bp)... Then (1) ghg' = (alh,...(alh, . . ,. ,aah)(blh,a,h)(blh, ...... , , bah). bah)...... (2) ss EE SS isis conjugateconjugate toto g inin SS ifif andand only ififs s and and gg havehave thethe samesame cyclecycle structure.

A transposition is an element of S moving exactly twotwo pointspoints ofof X.X. That is a transposition is a cycle of lengthlength 2.2.

(15.4) S isis generated by itsits transpositions.transpositions. The symmetric andand alternatingalternating groups 55

Proof. By 15.2 itit suffices suffices toto showshow eacheach cyclecycle isis aa product product ofof transpositions.transpositions. But (1,2,(1, 2, ...., . . ,m) m) == (1, (1,2)(1,3). 2)(1, 3) ... . .(l, (1, m). m).

A permutation is said to bebe anan even even permutation if it cancan bebe writtenwritten asas the the product of an even number of transpositions, and to be an odd permutationpermutation if it can be written as the product of an odd number of transpositions. DenoteDenote byby Alt(X) the setset ofof allall eveneven permutations of X.

(15.5) (1) Alt(X)Alt(X) isis aa normalnormal subgroupsubgroup ofof Sym(X) ofof indexindex 2. (2) AA permutation permutation is is eveneven ifif andand onlyonly ifif itit hashas anan eveneven numbernumber of cycles of even length. A permutation is odd if and only if it has an odd number of cycles of even length.

Proof. Without lossloss X = (1,{I, 2, ...... , , n). nj. ConsiderConsider the polynomial ringring R == 7L[xl,Z[xl, ...... , , xn]x,] in nn variablesvariables xixi overover the the ring ring 1L Z of integers.integers. ForFor Ss EE SS definedefine sa:Rsa: R+ R byR by f(x1,...,xn)sa f(x1,. . . ,x,)sa = f(x1s,...,xn5).Check= f(xls,. . .,x,,). Checkthat that a:S-+ a:S+ Aut(R)Aut(R) is a representation of S inin thethe category of rings and ring homomorphisms. Consider thethe polynomialpolynomial P(xl, P(x1, ...... , x,)xn) == P EE RR defineddefined by P = P(x1,P(xl,...... , x,)x'n) = = nH (X(xj j -xi).-xi). l'i

The group Alt(X) is the alternating groupgroup onon X.X. EvidentlyEvidently the isomorphismisomorphism type of Sym(X) andand Alt(X) dependsdepends onlyonly onon thethe cardinalitycardinality of X,X, soso wewe maymay write SnS, and AnA, for Sym(X)Sym(X) andand Alt(X),Alt(X), respectively,respectively, when IX(XI I == n.n. The groups SnS, and AnA, are the symmetric andand alternating groups ofof degreedegree n.n. If m is aa positivepositive integer, denote byby XmX1 thethe setset product of m copiescopies of X.X. If GG isis aa permutationpermutation groupgroup on X, thenthen GG isis also also aa permutation permutation groupgroup on X'Xm via via g: g :(x1, (xl , ...... , ,x,n) x,) H (x1g,(xlg, ...... , ,xng) xmg) for for gg cE G G and and xixi cE X.X. Assume Assume GG is transitive onon X. Then the orbits of G on X2x2 are are calledcalled the orbitals of G.G. One orbital is thethe diagonaldiagonal orbitalorbital {(x, x):x): xx Ec X}.X). The permutationpermutation rank of a transitive permutation group G is defined to be the number of orbitals of G. 56 Permutation groups

(15.6) Let G be a transitivetransitive permutationpermutation group group on on X, X, x X E E G, G,and and (xi (xi : 1: 15 < i i5 < r)r) representatives for for the the action action of of H H = = G,Gx onon X.X. ThenThen {(x,{(x, xi): xi): 1 1 _( < ii 5< r} is a set of representatives forfor thethe orbitalsorbitals ofof GG andand (x,(x, y) E (x, xi)G ifif andand onlyonly ifif y EE xi H. In particular r is the permutation rank of G.

The regularpermutationregular permutation representationrepresentation of of aa group H isis the representation of H on itself byby right right multiplication. multiplication. ApermutationA permutation representation representation n:n :H H +-)- Sym(Y) is semiregularsemiregular if and only if the identityidentity elementelement isis thethe onlyonly elementelement ofof H fixing points of Y. EquivalentlyEquivalently H,Hy = 11 for for allall yy inin Y.Y.

(15.8) A permutation representation nn ofof finitefinite degreedegree isis semiregularsemiregular if and , only if the transitive constituentsconstituents ofof nn are regular.

A regular normal subgroup of G is a normal subgroup of G which is regular on X.

(15.9) Let G be transitivetransitive onon X, X, x x EE X,X, andand H H I_(< G. Then H isis regularregular on X if and only if G,Gx isis a complement toto H in G.

(15.10) Let H bebe thethe splitsplit extensionextension of a normal subgroup K by a subgroup A of H andand let jrn be the representation ofof HH on thethe cosetscosets of of A. A. Then Then K K E = KnKit and KitKn is is aa regularregular normal subgroup ofof Hn.Hn.

(15.11) Let H bebe aa regularregular normalnormal subgroup of G and Xx E X. Then the map a: H + X defined byby a:a: h h H xh xh is is an an equivalence equivalence of of thethe representationrepresentation of G,Gx onon HH via conjugation with the representation of G,Gx on X.

Given apositivea positive integerinteger m, G is said toto bebe m-transitive m-transitive onon XX ifif GG actsacts transitivelytransitively on the subset of XmX' consistingconsisting of of the the m-tuples m-tuples all all of of whosewhose entriesentries are are distinct.distinct. Notice G is 2-transitive if and only if it is transitive of permutationpermutation rank 2. Also m-transitivity implies k-transitivity forfor kk 5< m.m.

(15.12) (1)(1) Let m >2 22 andand xx EE X.X. Then Then GG is is m-transitive m-transitive on X ifif andand only if G is transitive andand G,Gx isis (m(m -- 1)-transitive 1)-transitive on X - {x}. {x}. (2) Sym(X)Sym(X) isis n-transitiven-transitive on X. (3) Alt(X)Alt(X) is (n - 2)-transitive 2)-transitive on on X.X. (4) IfIf GG isis (n(n - 2)-transitive 2)-transitive onon XX thenthen GG == Sym(X) Sym(X) or or Alt(X). Alt(X).

Proof. TheThe firstfirst threethree statementsstatements areare straightforward.straightforward. Prove the fourth by induction on n using (1) and the following observation: IfIf G is transitive on X and GxG, = Sym(X Sym(X), )x or Alt(X),Alt(X )x then then G G == Sym(X) or Alt(X), respectively. The observation followsfollows from from 5.20 5.20 plus plus the the fact fact that, that, if if G, GX 5 < Alt(X) Alt(X) and and n n 2> 4, The symmetric andand alternatingalternating groups 57 then G = (Gy: (G,: Yy EE X)X) <5 Alt(X), Alt(X), since since Alt(X) Alt(X) a 1 Sym(X) Sym(X) and and G G is is 2-transitive 2-transitive on X.

(15.13) Let G bebe m-transitivem-transitive on X and H aa regularregular normalnormal subgroupsubgroup of G. Then (1) IfIf mm == 2 2 then then n n is is a apower power ofof somesome primeprime p andand HH isis an an elementaryelementary abelian p-group. (2) IfIf mm == 3 3 then then eithereither nn isis aa power of 22 oror nn == 3 and G = Sym(X). Sym(X). (3) IfIfm m >24thenm 4 then m =4=nandG= 4 = n and G ==Sym(X). Sym(X).

Proof.Pro05 WeWe may take mm 2> 2. Let xx E X and K = Gx G,. . By 15.12, KK isis (m(m - - 1)-1)- transitive onon XX -- {x}, {x}, and then, byby 15.11, KK acts (m(m -- 1)-transitively 1)-transitively on H#H' via conjugation. In particular K isis transitive on H#.H'. Let p be a prime divisor of n. As H isis regular onon X,X, nn == I HIH 1. I. So by Cauchy's Theorem therethere isis hh EE HH of order p. Thus,Thus, as KK isis transitivetransitive on H#,H', everyevery element ofof H'H# is of orderorder p. WeWe concludeconclude from Cauchy's Theorem thatthat HH is a p-group. So n = JHJ IHI isis aa powerpower of p. ByBy 9.8,9.8, HH isis solvablesolvable andand as KK is transitive onon H',H#, H isis aa minimalminimal normal subgroup of G. So, by 9.4, H isis elementaryntary abelian. Thiss completes thethe proof proof of of (I), (1), so so we we may may take take m m 2 > 3. 3. Let Let y y = = xh.xh. By 15.2, KYK, is is transitive transitive on on X X -- {x, {x, y}y } and so, by 15.11,15.11, KYK, == CK(h)CK(h) isis transitivetransitive on H -- {1, (1, h} h} via via conjugation.conjugation. But CK(h)CK(h) CK((h)), so either (h) == 11, (1, h}h} or n = 3. 3. InIn thethe firstfirst case p = 2 2 and and inin thethe secondsecond G == Sym(X) Sym(X) by by 15.12.4.15.12.4. This completes thethe proofproof of of (2), (2), so so we we may may take take m m 2 > 4. 4. Let Let g g E E H H -- (h).(h). By 15.1115.11 and 15.12, CK(g)CK(g) f? nC,y(h) CK(h) = = J isis transitivetransitive onon HH - {1, (1, g,g, h)h) viavia conjugation. ButBut JJ centralizes gh,gh, so n = 4.4. HenceHence G == Sym(X) Sym(X) by by 15.12.4.15.12.4.

Recall the definitiondefinition of aa primitive representation from from sectionsection 5. 5.

(15.14) 2-transitive representations areare primitive.

(15.15) If GG isis primitiveprimitive onon XX andand 11 # H a GG thenthen HH isis transitive transitive on X and G = Gx G,H H forforeachx each x E X.

Pro05Proof. LetLet Xx E X. By 5.19, M == GX G, is a maximalmaximal subgroup ofof G, while as H a1 G, G, MHMH isis aa subgroupsubgroup of G containing MM by 1.7. ThusThus MHMH == M or G. In the latter casecase HH is transitive onon X X byby 5.20. 5.20. In In the the former former H H 5 < M,M, soso HH 5< ker(r)ker(n) by by 5.7,5.7, wherewhere rn is is the the representation representation of of GG onon X.X. But, as G <5 S,S, jr n is is the the identity map on G andand inin particular isis faithful.faithful. This is impossible asas HH # 1.

(15.16) The alternating groupgroup A, A isis simplesimple if nn 2> 5.5. 558 8 Permutation groups

Proof. Let n >2 5,5, GG == Alt(X), Alt(X), andand 11 # HH a _a G. G. We We mustmust showshow H = G. G. By By 15.12, G is (n - 2)-transitive 2)-transitive on on X,X, so,so, byby 15.1415.14 and 15.15, H is transitive on XandG=X and G = KKH,whereK H, where K ==G,andx Gx and X EcX.If1 X. If 1 == HHnKthen,by15.9,His fl K then, by 15.9, H is regular on X. But this contradicts 15.13 and thethe factfact thatthat GG isis (n(n -- 2)-transitive with n >>_ 5.5. So 110 # H H fl n K. K. But But K K = = Alt(Y), Alt(Y), where where YY = X X -- (x), (x), so, so, by by induction induction onon n, either KK is simplesimple or or n n = = 5.5. InIn thethe former former case, case, as as 1 1# ¢H H fl flK K _a 4 K,K, KK = HnKH fl K 4< H,H,soG so G == KH KH = =H. H.Thuswemaytaken Thus we may take n == 5. 5.HereatleastHflK Here at least H fl K isistransitiveony transitive on Y by 15.12, 15.14,15.14,and and 15.15.15.15.So4 So 4 == I lYlY (= = (H lHnK fI K : (H:(HnK),I fI K)y I for y EE YY byby 5.11.5.11. SimilarlySimilarly 55 == (X (XI ( = = ( (HH :: H H fln K KI, (, soso 2020 dividesdivides the order of H.ButH. But (S(IS1 == 5! = 120 120while while IS:IS: G(=GI = 2, 2,so so IGI IGI == 60. 60. Thus, Thus, as as IHI [HI dividesdivides . IGI,(G(, I1H( H I == 20 or 60. In the latter casecase HH == G, G, so so we we may may assume assume the the former. former. By Exercise 2.6,2.6, HH has a unique SylowSylow 5-group 5-group P. P. HenceHence PP char H _a4 G,G, so, by 8.1, P 4a G. G. This This is is impossible impossible as as we we have have just just shownshown that that 44 dividesdivides thethe order of any nontrivial normal subgroup of G.G.

(15.17) (Jordan)(Jordan) Let GG bebe aa primitiveprimitive permutation group on aa finitefinite set X and suppose Y is aa nonemptynonempty subset of X such thatthat IX( X - - YY I (> > 1 andand GyGy isis transitivetransitive ononX X -- Y. Y. ThenThen (1) GG isis 2-transitiveZ-transitive onon X, andand (2) ifif GyGy is is primitiveprimitive on X - Y Y thenthen GxG, is primitiveprimitive onon XX - - (x}(x) for xx cE X.X.

Proof. LetLet Fr == X X - -Y Y and and induct induct on on (Y(. I Y I. IfIf (YI1 Y I == 1 1 the the resultresult isis trivial.trivial. SoSo assume IIY Y I (> > 11 andand let let x x andand yy bebe distinctdistinct pointspoints ofof Y. ByBy ExerciseExercise 5.5, there is ggEGwith E G with xxEYgandyVYg.LetH=(Gy,Gyg)andS2=FUFg.Then E Yg and y $ Yg. Let H = (Gy,GYg) and C2 = r U rg. Then H <_( Gx G, andandH H acts actsonC2.Suppose on 0. Suppose I Irlr I >> (Y IYI.Thenrnrgisnonemptyso 1. Then r fl rg is nonempty so HH is transitive on on C2. 0. AsAs HH 5< G,G,, r rU U(y} (y} is is contained contained inin anan orbit of G,.G. SinceSince this holds forfor eacheach yy E Y - (x}, (x}, G G is is 2-transitive Ztransitive onon X byby 15.12.1.15.12.1. Further if Gy is primitive onon rF and Q is a G,-invariantGx-invariant partition partition of of X' X' = = X - {x}, (x), then then ffor o rZ Z E E Q Q either either IZ (Z f? flr[ r (4 < 1 1 or or r r _Gc Z. As 1171Irl 2 > (Y((Y (it it follows follows that that IZ((Z (= = 11 or IX'(.IX'I. Hence G,Gx is primitive on XI.X'. So assume IrlIr( <_( (Y(lYI andand let a, y y bebe distinctdistinct pointspoints of F.r. By By ExerciseExercise 5.55.5 thereishthereis h EE G Gwithy with y E c rh rhbut but a a $V rh. rh.LetK Let K == (Gy,(Gy,Gyh),F' GYh),r' = = r rUrh, U rh, and Y' == X - F'. r'. Then Then KK <_( Gy,, Gy/, andand asas y cE rr fl f? Fh,rh, K K is is transitive transitive on F'r' and Irl( F > ( >Ir' IF'-r(.As - rl. As Irl(rI 4< (Y(IYI and andy y E c r rflFh,wehave n rh, we have Y' # 0.IfGy0. If Gy is is primitive on rF then, as (I rF 1 (> > I (r' r' -- F r 1,1, K is primitive onon r'F' byby anan argumentargument in the lastlast paragraph. So,So, replacing replacing Y Y by by Y' Y' and and applying applying induction induction on on I Y IY 1, I, thethe result holds.

Jordan's TheoremTheorem isis a useful tool for investigating finite alternatingalternating groups and symmetric groups. See for exampleexample Exercises 5.6,5.6,5.7, 5.7, andand 16.2.16.2. Rank 3 permutation groups 59

1616 RankRank 33 permutationpermutation groups groups In this section GG isis aa transitivetransitive permutation groupgroup on aa finitefinite set X of order n. Recall thethe definitiondefinition of an orbital in the preceding section. Given an orbital A of G, the paired orbital AP of A is

AP = {(y, x): (x, y) E A).

Evidently APAp is an orbital ofof GG with (AP)P(Ap)p = A.A. TheThe orbitalorbital AA isis saidsaid toto bebe self pairedpaired if A = AP. AP.

(16.1)(16.1) (1)(1) A A nondiagonal nondiagonal orbitalorbital A A ofof GG is is selfself pairedpaired ifif andand only if (x, y) isis aa cycle in some g E G, for (x, y) EE A.A. (2)(2) G G possesses possesses a a nondiagonal nondiagonal self self paired paired orbital orbital if if andand onlyonly ifif GG isis ofof eveneven order. (3)(3) If If GG is is of of even even order order and and (permutation) (permutation) rank rank 3 3 then then both both nondiagonal nondiagonal orbitals of GG areare selfself paired.paired.

Recall the definitiondefinition ofof aa graphgraph fromfrom sectionsection 3. 3.

(16.2)(16.2) LetLet AA bebe aa self self pairedpaired orbitalorbital ofof G.G. ThenThen AA isis aa symmetricsymmetric relation relation on X, so g-9 = (X, (X, A)A) isis a a graph graph and and GG is is a a group group of of automorphismsautomorphisms of of -9g transitive on the edgesedges ofof -9.g. 61,<

In the remainder of this section assume GG isis ofof eveneven order and of permutation rank 3. Hence G has two nondiagonal orbitals A A andand F.r. ByBy 16.1.316.1.3 bothboth AA andand Fr are self paired. For xx cE X,X, Gx G, hashas twotwo orbits A(x) andand F(x)r(x) onon XX -- {x}, {x), wherewhere A(x) = {y(y cE X:X: (x, (x, y) cE A}A) and r(x) == {y {y c E X: X: (x, (x, y)y) cE r); r); thisthis holds holds byby 15.6. ByBy 16.2, g' = = (X, (X, A)A) is isa grapha graph and and G G is is a agroup group of of automorphisms automorphisms of of -9g transitivetransitive on the edges of -9.g. Notice A(x) isis thethe setset of vertices adjacent to x in thisthis gaph.graph. Defined Define x1 = = {x)UA(x), {x}UA(x),kk = = lA(x)l,lIA(x)l,l == IF(x)I,,1Ir(x)l,h == IA(x)nA(y)IIA(x)nA(y)l forfor y EE A(x),A(x), andand µp == I A(x)nlA(x)n A(z)IA(z)l for for Z Z E E F(x). r(x). AsAs Gx G, isis transitive transitive on on A(x)A(x) and F(x),Ar(x),h andand µp are are well well defined. defined. AsAs GG isis transitive transitive onon XX thesethese definitionsdefinitions are independent of the choice of x.

(16.3) (1) n=1+k+l. (2) µl = k(k - A - 1).

Proof. PartPart (1) (1) is is trivial. trivial. To To proveprove (2)(2) countcount IQIIC2I inin twotwo differentdifferent ways,ways, where

0S2 == {(a, {(a, b)b) :: b, x x Ec A(a),A(a), b cE F(x)}r(x)) for fixed xx Ec X.X. 60 Permutation groups

(16.4) IfIf kk <5 1 1 then then thethe followingfollowing areare equivalent:equivalent:

(1) GG isis imprimitive. imprimitive. (2) h=k-1.a,=k-1. (3) ap == 0. 0. (4) x1xL = y1 y' forfor y cE A(x),A(x), {(x1)g:{(x')~: g Ec G} = QC2 is aa systemsystem of ofimprimitivity imprimitivity for G, andand G isis 2-transitive2-transitive on Q.C2.

Proof.Proof: IfIf SS isis aa systemsystem ofof imprimitivityimprimitivity for G and x cE 90 EE S,S, then, then, byby 5.18,5.18, GxG, actsacts onon 8.9. So,So, asas G,Gx is is transitivetransitive on on A(x) A(x) and and r(x), F(x), either either 0 9= = x'x1 oror 09 = (x) {x) UU F(x).r(x). By By 5.18,5.18, 101101 dividesdivides n, n, so so as ask k 5< 1,1,e 9 == x1.x'. NowNOW (4)(4) holds. If1f . x1x' == y1 y' thenthen (2) (2) and and (3) (3) hold. hold. Also Also (2) (2) is is equivalent equivalent to to (3).(3). Finally Finally ifif (2)(2) holds then x'x1 = y1, y', soso Q C2 is is a a system system of of imprimitivityimprimitivity forfor GG andand hencehence (1)(1) holds.

(16.5) If pµ 0# 0 0 or or k k then then GG isis primitive.primitive.

Proof.Proof: ByBy 16.416.4 wewe maymay take take k k> > 1. 1. Let Let ,2 µ= = 1 l - - (k(k --,X h -- 1).1). If ,2µ = 0 0 then,then, by 16.3.2, kk == µ,p, contrary contrary to to hypothesis.hypothesis. Hence,Hence, by 16.416.4 and symmetry between A and F,r, G G isis primitive.primitive.

(16.6) If G is primitiveprimitive then then ' isis connected.connected.

Proof. By By 5.19, 5.19, Gx G, isis maximal maximal in in G, G, so so G G = = _ (Gx,(G,, Gy)G,) for x # y. Now Exercise 5.2 completes the proof.

(16.7) Assume G is primitive. Then either

(1) k=landp=h+l k=land µ=,k+l=k/2, =k/2,or or (2) d == (X (A - µ)2 p)2 + + 4(k 4(k - - µ) p) is is a asquare square and and settingsetting D == 2k 2k ++ (A (A - µ)(k p)(k + + 1) 1) we we have:have: (a) d1/2d112 divides D butbut 2d1j22d1I2 does not if n is even, while (b) 2d1/22d1f2 divides D ifif nn isis odd.odd.

Proof:Proof. LetLet AA bebe thethe incidence matrix for s?.-9. ThatThat isis AA = (axy)(a,,) is the n by n matrix whose rowsrows andand columnscolumns areare indexedindexed by by X, X, andand withwith a,, axy = = 1 ifif (x, y) is an edge of -9' while a,,axy == 00 otherwise.otherwise. Let B bebe thethe incidenceincidence matrixmatrix forfor (X, r),F), J thenthe n by by n n matrix matrix all all of of whosewhose entries are 1, andand II thethen n byby n identityidentity matrix. Observe:

(i) AA isis symmetric.symmetric. (ii) I + A ++ BB == J. Rank 3 permutation groups 61

(iii) AJ = kJ,kJ, soso (A -- kI)J == 0. 0. (iv) A2A2 = kIkI + +AA+pB. ,kA + AB.

The first two statements are immediate from the definitions. AsAs lA(x)l(A(x)e == k, each row of A has k entriesentries equal to 1. 1. ThusThus (iii)(iii) holds. By (i)(i) thethe (x,(x, y)-thy)-th entry of A2 is the inner product ofof thethe xthxth and yth rows of A. But this inner product justjust counts IA(x)lA(x) fln A(y)e, A(y)l, so (iv)(iv) holds. Next (ii), (iii), and (iv) imply:

(v)(A-kI)(A2-(,l-A)A-(k-ll)I)=0,(v) (A - ~I)(A'- (A - p)A - (k - p)I) = 0, so the minimal polynomial ofof AA dividesdivides p(x) _ (x -k)(x2 -()-A)x -(k-bl)). The roots ofof p(x) areare k, s,s, andand t,t, where where ss == ((A ((A - A) p) ++ d1/2)/2 d1I2)/2 and t = ((),((A - A) p) - - d1/2)/2. d1I2)/2. LetLet me me be be the the multiplicity multiplicity of of thethe eigenvalue eigenvalue e.e. Claim mkMk = = 1.1. Indeed Indeed for for c c= = (el, (cl, . ...., . , c,), cA = kckc ifif andand only if ci = clcl for all i. ToTo prove this we can taketake IIcll c1 >>_ Icilcil for each i.i. As cAcA = kc,kc, kc1kcl = Ziciail,Eiciail, so,so, asas exactlyexactly k of the ail areare 11 andand thethe restrest areare 0, c1cl = cici forfor eacheach i cE A(1).A(1). ButBut nownow asas -9 isis connectedconnected it follows thatthat clcl == ci forfor allall ii EE X,X, completing the proof of the claim.claim. As mkMk == 11 itit followsfollows that:that: (vi)m,.+mt =n - 1 =k+1.

Also, as A is of trace 0: (vii) kk ++ m,sm,s ++ mtt = 0.0. Now (vi) and (vii) imply: (viii) m,m,. = = ((k ((k + + l)t l)t + + k)/(t k)/(t -- s).s).

Of coursecourse t t -- s == -d1/2, -dl/', t t= = ((A ((A - -A) p) - d1/2)/2,- d1I2)/2, and and ms m, is is an an integer, integer. Thus

(ix) (Dd-'I2(Dd-1/2 - (k (k ++ l))/2 1))/2 is is an an integer, integer, where where D ==2k+(A-A)(k+1). 2k + (A - p)(k + I). In particular eithereither dd isis aa square square or or D D == 0. IfIf DD = 0 0 thenthen Ap == A h + 1 1 andand 1 = k,k, so 16.7.1 holds by 16.3.2. If d is a square then 16.7.216.7.2 holds by (ix).(ix).

Remarks. WielandtWielandt [Wi[Wi 2]21 isis aa goodgood placeplace toto find find moremore informationinformation aboutabout permutation groups. The material in section 16 comes from Higman [Hi]. Section 16 is somewhat technical and can be safely omitted by the novice. On the other hand the results in section 15 are reasonably basic. 62 Permutation groups

Exercises for chapter 5

1. (1) Prove A5 has no no faithful faithful permutationpermutation representationrepresentation of degreedegree lessless than 5. (2) Prove Prove that,that, upup toto equivalence,equivalence, A5A5 has unique transitive representations of degree 5 and 6. Prove both are 2-transitive. (3) Prove Prove Aut(A5) = S5.S5. (4) ProveProve therethere areare exactlyexactly twotwo conjugacyconjugacy classes ofof subgroupssubgroups ofof S6S6 isomorphic to A5.A5. Prove thethe samesame for for A6- Ag. 2. LetLet GG bebe aa transitivetransitive permutationpermutation groupgroup onon X,X, A a self paired orbital ofof G, (x,(x, y)y) EE A,A, #-9 = = (X, A)E) the graph on X determineddetermined byby A, and H = (G,,(G, Gy). G,). ProveProve (1) xHXH U U yH yH is is the the connected connected componentcomponent of of -99 containingcontaining x,x, andand (2) -9 isis connected connected preciselyprecisely whenwhen oneone of of thethe followingfollowing holds:holds: (i) G G = H, H, or or (ii) IIG G ::HI H 4= = 22 andgand ' is is bipartite bipartite with withpartition.{xH, partition. {xH, yH}yH) (i.e. (i.e. {xH, {xH, yH}yH] is aa partitionpartition of of X X and and A A 2 c_ (xH (xH x x y yH) H) UU (y (yH H xx xxH)). H)). 3. LetLet SS andand AA bebe thethe symmetricsymmetric and and alternatingalternating groups of degree n, respectively, andand let let a a E A. ProveProve aAaA # 0 as preciselyprecisely when (*) holds (*) Cyc,(2m)Cyca(2m) == 0 and Cyc,(2mCyca(2m -- 1) 1) < 5 1 1 for for each each positive positive integerinteger m. inwhichcaseasin which case as = = aA aAU UbAforsomeb bAforsomebE E Awith(a) Awith(a) == (b)andlaAI(b) andlaAI _= (bAl. IbAI. 4. Let Let A A be be thethe alternatingalternating group on a set X of finite orderorder nn > 33 andand let VV be the set of subsets of X of order 2. Prove (1) A A isis aa rankrank 33 permutation group on V.V. (2) AA, isis a amaximal maximal subgroup subgroup of of AA forfor Vv E V. (3) IfIf GG isis aa primitive rank 3 groupgroup onon a a set set Q 2 of orderorder 1010 then then G G %- A5A5 or S5 andand thethe representationrepresentation ofof GG onon S20 isis equivalentequivalent to its representation on V.V. 5. LetLet GG bebe aa primitive primitive permutationpermutation groupgroup on a finite set X and let xx andand yy be distinct points of X. Let Y be a nonempty proper subset of X and define S(x) = {Yg: {Yg: gg E G, x EE Yg}Yg} andand T(x)T(x) = nzEs(x)n,,,,,, Z.2. Prove (1) T(x)T(x) == {x), {XI, andand (2) therethere exists gg E G with x E Yg but yy 60 Yg. 6. LetLet GG bebe aa primitive primitive permutationpermutation groupgroup on a set X of finite order n. Prove (1) If Y EC X such thatthat GyG y is is primitive primitive on on X X- Y-Y and and IYI (Y = =m m withwith 11 5< m <5 nn -- 2, 2, then then G G is is (m (m ++ 1)-transitive 1)-transitive on X. (2) IfIf GG contains contains a a transposition transposition or or aa cyclecycle of length 3 then G contains the alternating group on X. (3) If a,a, b b EE G with with IIMov(a) Mov(a) n fl Mov(b)J Mov(b)l = =1 1then then [a, [a, b] b] is is aa cycle ofof length 3. (4) LetLet YY C2 XX bebe ofof minimalminimal order order subjectsubject to GyGy == 1. 1. Prove Prove eithereither GG contains the alternating group on X or IIY Y I (<5 n/2.n/2. RankRank 33 permutation permutation groupsgroups 6363

(5)(5) If If G G does does not not contain contain the the alternating alternating group group on on X, X, prove prove j Sym(X)iSym(X) : :G GI I >3 [(n[(n ++ 1)/2]! 1>/21! 7.7. Let Let X X be be a aset set of of finite finite order order n n <5 5, 5, A A the the alternating alternating group group on on X,X, and and GG a a properproper subgroup subgroup of of A. A. Prove Prove one one of of the the following following holds: holds: (1)(1) JA:IA :G) GI >> n. n. (2)(2) (A:GI1A:GI =nand=nand G=Ax G =A, forsomefor somex xEX. E X. (3)(3) (A:GI1A:Gl =n=6andGSA5. =n=6andG-A5. 8.8. ProveProve that that either either (1)(1) An A, hashas a aunique unique conjugacy conjugacy class class of of subgroups subgroups isomorphic isomorphic to to A,-1, oror (2)(2) nn = = 6 6and and An A, hashas exactly exactly two two such such classes. classes. 6

Extensions ofof groups and modulesmodules

Chapter 6 considers various questions about extensions of groups and modules, most particularlyparticularly the conjugacy of complements to some fixed normal subgroup in a splitsplit group extension. Suppose G is representedrepresented on an abelianabelian group or F-space VV andand formform thethe semidirectsemidirect productproduct GV.GV. Section 1717 shows there is a . bijection between thethe set of conjugacy classes of complements toto V in GVGV and the 1-cohomology groupgroup H1(G,H1(G, V).V). IfIf V is an F-space so is H1(G,H'(G, V).V). Moreover ifif C"(G)Cv(G) = 0 0 there there is is a a largest largest member member ofof the class of FG-modules U such that CU(G)Cu(G) = 0 0 and and U U is is the the extension extension ofof VV byby aa modulemodule centralized by G. Indeed it turns out that if UU isis thethe largestlargest membermember of thisthis classclass thenthen U/VEU/V = H1(G, V). Further thethe dual of the statement isis also true: thatthat is if V = [V, [V, G] GI thenthen therethere isis aa largest FG-module U* such thatthat U*U* == [U*,[U*, G] GI andand U* is the extension ofof an FG-module Z by VV withwith GG centralizingcentralizing Z.Z. In In thisthis case ZZ E= H1(G,H1(G, V*). V*). These results together with Maschke's Theorem are then used to prove the Schur-Zassenhaus Theorem, which givesgives reasonablyreasonably completecomplete informationinformation about extensions ofof aa finite groupgroup B by a finite group A when the orders of A and B areare relativelyrelatively prime.prime. TheThe Schur-ZassenhausSchur-Zassenhaus TheoremTheorem isis thenthen used used to prove Phillip Hall's extended extended Sylow Theorem for solvable groups. Hall's Theorem supplies a good illustration of how restrictions on the compositioncomposition factors of a finitefinite groupgroup cancan bebe usedused toto derive derive strong strong information information aboutabout thethe group.group. I have chosen to discuss 1-cohomology1-cohomology from a group theoretical point of view. Homological algebra is kept to a minimum. Still the arguments in section 17 have a different flavor than most in this book.

17 1-cohomology1-cohomology In this section p is a prime,prime, F isis aa field of characteristiccharacteristic p, VV isis an abelian group written additively, G is a finite group, and ir:n: G -->+ Aut(V) isis aa representation of G on V.V. Form the semidirect productproduct S(G,S(G, V, n)7r) of of VV byby GG with respect toto nit and identify G and VV with subgroups of of S(G,S(G, V, V, 7r) n) via via thethe injectionsinjections of of 10.1.10.1. Then Then S(G, V,V, 7r)n) = GVGV with with VV 4 9 GV GV and and G G is is a a complement complement to to VV to to GV. GV. A cocycle fromfrom G G intointo VV isis a function y:y: GG +- V V satisfying satisfying the the cocycle cocycle condition (gh)y == (gy)p` (gylh + byhy g, h E G. 1-cohomology 65

Notice the cocycle condition forces each cocycle toto mapmap thethe identityidentity ofof GG to the identity of V.V. Let F(G,r(G, V) V) denote denote the the setset ofof cocycles from GG toto V andand makemake F(G, r(G, V)V) into a group (again written additively)additively) via:

g(Y+S)=gy+gSg(y + 6) = gy + g6 y, 6S EE UG, I'(G,V),g V), g EEG. G. Let A = Aut(G Aut(G V) V) bebe the group ofof automorphismsautomorphisms ofof GV,GV, and and let let U(G, U(G, V) V) = _ CA(GV/CA(GV/V) v) nn CA(V). CA(V). If V is a vector space overover FF and n7r isis anan FG-representationFG-representation thenthen F(G,r(G, V)V) is also a vector space over F via:via:

day)g(ay) = a(gy) Yy EE r(G,UG, V), V), g EE G, a E F.

(17.1) For y EE r(G,F(G, V) V) define define S(y)S(y) = = fggy: {ggy g : gE EG}. G}. Then Then the the map map S: S: y y r-* I-+ S(y) isis aa bijectionbijection of r(G,F(G, V) V) with with the the set set of of complementscomplements to to VV in in GV.GV.

Proof. TheThe cocyclecocycle condition condition sayssays thatthat S(y)S(y) is is a a subgroup subgroup of of GV.GV. EvidentlyEvidently S(y) isis a a complement complement to V inin GVGV andand SS is is injective. injective. ConverselyConversely ifif C is a complement then,then, forfor gg E G,G, gVgV nn Cc containscontains a uniqueunique element gv andand y: g r-*I-+ v is a cocycle withwith S(y) S(y) == C.

(17.2) For y EE I'F(G, (G, V) define y y0: 4: GGV V -+--i GVG V byby YO: gv H g((gY) + v). Then thethe mapmap 4: 0: y y I-+H yO y4 is is aa groupgroup isomorphism of of F(G, r(G, V) with U(G, V).V ForForu u E U(G, V)V)andg and g E G,G,u&':g uO-1: g I-+r-* g-'gU.g-lg"

Proof. TheThe cocyclecocycle condition impliesimplies y4yO isis aa homomorphism.homomorphism. (-y)4 (-y)O is an inverse forfor yyO, 4, soso yyO 4 E Aut(GAut(GV). V). By definition yyO 4 centralizescentralizes VV andand GGV/ V/ V, so y4yO E U(G,U(G, V). V). ForFor Uu EE UU definedefine ul/r:u+: gg I-+r-* g-'gU.g-lg". AnAn easyeasy checkcheck showsshows u+ui/r isis a cocyclecocycle andand +i1r = = &', 0-1, so 40 is an isomorphism.

(17.3) U(G,U(G, V) V) acts acts regularly regularly onon thethe setset ofof complementscomplements toto VV inin GGV. V. Indeed G"GU = S(uo-1) S(u4-') forfor eacheach u EE U(G,U(G, V).V).

Proof. AsAs UU = = U(G, U(G, V) V) acts acts on on V V it it permutes permutes the the complements complements to VV in GV. By definitiondkfinition ofof thethe mapsmaps 4 0 andand S,S, GUG" = S(uO-1)S(u4-') forfor uu EE U.U. Hence,Hence, asas SS andand 40 areare bijections, the actionaction of UU isis regular.regular.

Lemmas 17.1 and 17.317.3 givegive descriptionsdescriptions ofof the complementscomplements toto V inin GV in termsterms of r(G,F(G, V) V) and and U(G, U(G, V), V), while while 17.2 17.2 and and 17.3 17.3 give give thethe corres-correspondence between the twotwo descriptions.descriptions. TheThe nextnext few few lemmaslemmas describedescribe thethe 66 Extensions of groups and modules representations of G on r(G,F(G, V)V) andand U(G,U(G, V),V), andand showshow these representations are equivalent.

(17.4)(17.4) For gg E G andand yy E r(G,F(G, V) define yg: yg: G G + - V V by by h(yg)h(y8) == [(hg-' [(hg-i)y]g, )y]g, for h EE G.G. ThenThen Frit isis aa representationrepresentation of of G G on on r(G, F(G, V),V), wherewhere git: gfr: y y HH yg.yg. If trn isis anan FG-representation so is ft.5.

(17.5) For vv EE V definedefine va: va: G G + -+ V byby g(va)g(va) == [g,[g, v] = v v -- vg. vg. Then Then the the mapmap a: v v H vava isis aa G-homomorphismG-homomorphism of of VV into F(G,r(G, V) V) withwith kernelkernel Cv(G).CV(G). IfIf 7rn is an FG-representation then a isis anan FG-homomorphism.FG-homomorphism.

Proof. LemmaLemma 8.5.4 8.5.4 says says vu va is is a a cocycle. cocycle. The The rest rest isis straightforward. straightforward.

IfIf VV isis anan FG-moduleFG-module then so is F(G,r(G, V) V) andand 0# inducesinduces anan F-spaceF-space structurestructure onon U(G,U(G, V)V) whichwhich makesmakes 0# intointo an an F-space F-space isomorphism. isomorphism. ThatThat is,is, forfor uu EE U(G, V)V) andand a EE F,F, auau = = (a(uo-1))O. (a(u4-I))#. EquivalentlyEquivalently if gug" = gvgv thenthen g°"gaU = g(av).g(av). ThisThis isis thethe F-spaceF-space structure structure on on U(G,U(G, V) V) implicitimplicit in in thethe remainderremainder ofof the section.section.

(17.6) Let c:c: GG V V + - Aut(G Aut(G V)V) and d: Aut(GAut(G V) V) n f1 N(V) N(V) + - Aut(U(G,Aut(U(G, V))V)) be the conjugation maps. Then a V- F(G, V)

U(G, V) is aa commutativecommutative diagram, the maps c, a, and and 0# areare G-homomorphisms, G-homomorphisms, andand if VV isis anan FG-moduleFG-module thethe mapsmaps areare FG-homomorphisms. HereHere #,0, a,a, and frit are defineddefined in 17.2,17.2, 17.5,17.5, andand 17.4, 17.4, respectively,respectively, and Jr,n, it,Fr, andand cdcd are thethe representations of G on V, F(G,r(G, V), V), andand U(G,U(G, V), V), respectively. respectively.

Proof. LetLet v,v, ww EE V,V, Uu E U(G, V),V), and g, h E G. Then vc: gw g°w = g(v - vg + w) = (gw)(va)O, so c = ao a# and and the the diagram diagram commutes. commutes. By 17.5, a isis aa G-homomorphism,G-homomorphism, and and eveneven an FG-homomorphism if V is an FG-module. By 17.2,17.2,# 0 isis aa homomorphism.homomorphism. Next ugcd:agcd:hv by H H (h~)~-"g(hv)g-'"g= = hg-'ugv.hg-'"gv. HenceHence (hv)(yg#) (hv)(ygo) = = h(hy8h(hyg ++ v) _= h(((hg-i)y)gh(((ha ')y)g + + v)v) == (hg(hg-l(hg-')y)gv '(ha ')y)gv = ((hg((hg-l)(y#))gv ')(yo))gv = (h~)((~#)g~~),(hv)((yO)gcd) soSO ygoyg# = (yO)gcd,(y#)gcd, and therefore #0 isis aa G-homomorphism.G-homomorphism. 1-cohomology 67

Suppose V isis anan FG-module. FG-module. ForFor a aE E F, F, (au)g`d (au)gcd = = ((a(uO-1))c)g`d ((a(u@-'))@)gcd = ((a(u@-'))g)@((a(uO-1))9)0 = = (a((ug`d)0-1))O (a((ugcd)@-I))@ == a(ugcd), a(ugcd), so so cd cd is is anan FG-representation.FG-representation. Finally, asas cc = aoa@ andand a a and and 0 @ are are G-homomorphisms G-homomorphisms (or (or FG-homomor- FG-homomorphisms), so is c.c. Therefore thethe proof isis complete.complete.

By 17.6,17.6, F(G,r(G, V)/V)/ VaVa =S U(G, U(G, V)/ V)/ Vc. Vc. The The first jirst cohomology cohomology groupgroup ofof thethe representation n7r isis H'(G,H 1(G, V) V) 2 = 1'(G,r(G, V)/V)/Va V a =S U U(G, (G, V)/V)/Vc. V c. This is anan additive groupgroup and, if V is anan FG-module, H1H'(G, (G, V) is even a vector space over F. The next lemma says that H1H1(G, (G, V) is in one to one correspondence with the setset of conjugacyconjugacy classesclasses of of complementscomplements to to VV in in GV. GV.

(17.7) HH1(G, 1 (G, V) acts regularly on thethe set of conjugacyconjugacy classesclasses of complementscomplements to V in GV via (Gwu)v (Vc)u: (G")'(Gw)v HH (G"')' U,u, wW EE U(G,U(G, V).

In particular the number of conjugacy classesclasses ofof complementscomplements toto VV inin GVGV is IH1(G,V)IIH1(G, V)I.

Proof. ThisThis isis aa consequenceconsequence ofof 17.317.3 and the fact thatthat U(G, V) acts on V.

ii (17.8) [Gcd,[Gcd, U(G, U(G, V)] V)] < 5 Vc. Vc.

Proof. For g EE G,G, Uu E U(G, V), [gcd, u]u] =_ [gc, [gc, u] u] EE U(G,U(G, V) V) f1n GVc, as U(G, V)V) andand GVcGVc areare normalnormal in Aut(GV). ButBut U(G,U(G, V)V) nfl GVcGVc = Vc.Vc.

(17.9) AssumeAssume eithereither Cv(G) Cv(G) = = 0 or G == OP(G). Op(G). Then Then CU(G,v)(Gcd)Cu(~,v)(Gcd)= = 0 and ifif CVCv(G)(G) == 0 then c:c: VV +- U(G, U(G, V) V) is is an an injection. injection.

Proof. IfIf Uu E E CU(G,v)(Gcd) CU(G,v)(G~d) thenthen 1 == [u, Gcd] = [u,[u, Gc], so [u, GIG] < ker(c) = Z(GV)Z(GV) == CZ(G)(V) Cz(~)(V) Xx Cv(G). I'll showshow [u, GIG] <( G, G, so so that, that, byby 17.3, uu = 1, 1, and and hencehence CU(G,v)(Gcd)CU(~,v)(Gcd) == 0. If Cv(G)CV(G) = 0 0 thenthen [u,[u, G] GI <5 CZ(G)(V) CZ(G)(V)5 < G. G. IfIf G = OP(G)OP(G) then GG = Op(GZ(GV))OP(GZ(GV)) char GZ(GV), soso [u,[u, G] GI <5 G. G. So So the the claim claim is is established.established. As CV(G)Cv(G) is the kernel of c,c, cc isis anan injectioninjection if if Cv(G) Cv(G) == 0.

(17.10) If G isis aa finitefinite p'-group and VV anan FG-module, thenthen H1(G,H '(G, V) = 0.0. Hence G isis transitive on the complements to VV in GVGV in this case. 68 Extensions of groups and modules

Proof. ByBy Maschke'sMaschke's Theorem,Theorem, thethe FG-module U == U(G, U(G, V) V) splits splits over over V Vc. c. Let W be an FG-complementFG-complement toto VcV c in in U. U. By By 17.8, 17.8, [Gcd, [Gcd, W]W] 5< W nfl Vcv c = 0.0. So, byby 17.9, WW == 0. Thus H'(G,H 1(G, V)V) G- W W = = 0, 0, and and 17.7 17.7 completes completes the the proof.proof.

Lemma 17.1017.10 will be used to proveprove thethe Schur-ZassenhausSchur-Zassenhaus Theorem in the next section.

(17.11) Let V bebe anan FG-module, FG-module, B: 0: VV +- W W an an injective injective FG-homomorphism, FG-homomorphism, and assume [G, W] <5 V18VB andand Cw(G)Cw(G) = 0. 0. Then Then therethere existsexists anan injective FG- homomorphism y:y: WW +- U(G, U(G, V) V) making making the the following following diagram diagram commute: commute:

W--> U(G, V) Y

In particular dimF(W/dimF(W/ VB)V,B) 5 < dimF(~'dimF(H1(G,(G, V)).

Proof. LetLet r'n' bebe the the representation representation of of GG on on W W and and consider consider the the semidirect semidirect products H == S(F# S(F# x x G, G, W, W, 7r') n') and and S S = = S(F# S(F' x xG, G, V, V, 7r). IT). As As ,B B is is anan injec-injective FG-homomorphismFG-homomorphism it induces by 10.310.3 an injective group homomorphism B:,J: S + H which is the identity onon F'F# xx G. As [G, W] <5 V,8,VB, (GV),B(GV)B 4 H, so the conjugationconjugation mapmap ee ofof HH onon (G(GV)B V),B composed composed with with (,!-I)*: (0-1)*: Aut((G Aut((G V)B) V),B) + - Aut(GV) = AA mapsmaps HH intointo A. A. AsAs Cw(G)Cw(G) = = 0, 0, the the restriction restriction y y of of e(,B-1)* e(Bw')* to WW isis anan injectioninjection of WW intointo U = U(G, U(G, V).V). AsAs yy isis the the composition composition ofof (F'(F# x G)-homomorphisms,G)-homomorphisms, y isis anan (F#(F' xx G)-homomorphism,G)-homomorphism, and hence isis aa G-homomorphism preserving the multiplication by F. InIn particular if the multiplication u Hi-+ ua,ua, aa eE F#,F', Uu EE UU ofof F#F' onon UU isis thatthat ofof thethe F-space structure on U defined earlier, then y is an FG-homomorphism. ForFor thisthis wewe need to show gUa gU' == gaugaU for gg Ee G.G. ButBut ifif gUgU == gv then gaUgau = = gav, whilewhile = ga-'uaga-'ua -= guagun = (gv)a(guy = gavgav asas [a,[a, g] g] == 1 1 and and Vava = av. It remainsremains to showshow ,ByBy = c. c. KeepingKeeping inin mindmind thatthat ,B:B: S + H isis anan injec-injective homomorphism trivial trivial on on F# F# x x G,G, we seesee thatthat forfor v, ww E V and g E G we have (gw)vB)' (gw)°Py = = (((gw)~)VB)~-' (((gw)P)' )P-1 == (gv(gvBw~)~-l wP)fi-1 == (g[g, v~lw~)B-'vP]wP)f-1 = g[g, v]w = g"wgVw == (gw)v (gw)' == (gw)"`. (g~)~'. SoSO ,ByBy == c c as as desired.desired.

Lemma 17.1117.1 1 sayssays thatthat if if CV(G)Cv(G) = 00 thenthen U(G,U(G, V) V) isis thethe largest extension W of V such thatthat [W,[W, GI G] 5< V and Cw(G) = 0.0. Recall the definition of the dual V* of a finite dimensional FG-module V given in section 14, and define U*(G, V)V) toto bebe (U(G,(U(G, V*))*. V*))*. 1-cohomology 69

(17.12) Let V be aa finitefinite dimensional dimensional FG-module, FG-module, B: 0: W W + - V V a a surjective surjective FG- homomorphism, and assume ker(,B)ker(B) 5 < Cw(G)CW(G) andand WW == [W,[W, G].GI. ThenThen therethere exists aa surjectivesurjective FG-homomorphism FG-homomorphism y: y: U*(G, U*(G, V) V) + - W W making making thethe following diagram commute:commute:

Here e* isis thethe dualdual ofof the the conjugation conjugation map map e: e: V* V* + - U(G,U(G, V*) V*) and and e*e* is a surjective FG-homomorphism with H'(G, V*)V*) = E ker(e*) ker(e*) < 5 CU.(G,v)(G). Cv*(c,v)(G). Thus dimF(ker(,B))dimF(ker(B)) 5 < dimF(H1(~,dimF(H'(G, V*)).V*)).

Proof. AsAs WW = = [W, [W, G], GI, CW.(G) Cw*(G) == 0 0by by 14.6. 14.6. Similarly, Similarly, as as ker(,B) ker(B) <5 CwCw(G), (G), [G, WW*] *] 5< VV*B* *,B* by by 14.6. 14.6. Let Let e: e:V* V* + - U(G,U(G, V*) V*) be be thethe conjugationconjugation map. map. ByBy 17.11 therethere isis an injective FG-homomorphismFG-homomorphism S:S: W* W* + -+ U(G,U(G, V*) such that the diagram commutes:

W* -U(G, V*) 6

Then applying 14.1.2 and 14.4.3 we concludeconclude thethe followingfollowing diagramdiagram commutes,commutes,

where y == S*. S*. As As S6 is injective, y isis surjective,surjective, so e* = y,8yB isis surjectivesurjective since j3fi is surjective byby hypothesis.hypothesis. H'(G, H1 (G, V*) V*) = = U(G, U(G, V*)/V*e,V*)/ V*e, so so H'(G, H'(G, V*) E= H'H'(G, (G, V*)*V*)* =E ker(e*) ker(e*) by 14.5. As [U(G, V*), V*), G]GI <5 V*e,V*e, G G centralizes centralizes ker(e*) ker(e*) by 14.6.14.6.

Lemma 17.12 says thatthat ifif V = [V,[V, G] GI thenthen U*(G,U*(G, V) V) is is the the largest largest extension extension W of aa modulemodule ZZ byby VV withwith ZZ 5< CW(G)Cw(G) and W == [W,[W, G].GI. 70 Extensions of groups and modules

18 CoprimeCoprime action (18.1) (Schur-Zassenhaus(Schur-Zassenhaus Theorem) Theorem) Let Let G G bebe aa finite group,group, H 9< G, andand assume

(i)(9 (IHI,(IHl, IG/HI)IGIHI) == 1, 1, and and (ii) either either H oror G/HG/H is is solvable. solvable. Then (1)(1) G splits over H, and and (2)(2) G is transitive on the complements toto HH in G.

Proof. LetLet G G be be a a minimal minimal counterexample. counterexample. Assume Assume firstfirst that HH is is solvable solvable and let M be a minimal normal subgroup of G contained in H. By 9.4, M is an elementary abelian p-groupp-group forfor somesome prime prime p. p. LetLet G G == G/M. By By minimalityminimality of G, there is a complement X to H inin GG andand GG isis transitivetransitive onon the complements to H. AlsoAlso ifif Y is a complementcomplement toto HH inin G then YF is a complementcomplement toto HH in G, so it suffices toto showshow XX splitssplits over MM and X is transitive on itsits complementscomplements to M. Hence, by minimalityminimality of of G, G, G G = = X and HH = M. M. Now Now Gaschutz'Gaschiitz' Theorem,Theorem, 10.4, says that G splits over H. Let Y be a complement toto H in G. By 12.112.1 the conjugation map map c:c: YY -+ Aut(M) isis a FY-representation, where F isis thethe fieldfield ofof orderorder p.p. Hence Hence by by 17.10,17.10, G G is is transitive transitive onon thethe complementscomplements totoH=MinG. H = M in G. Assume next thatthat GIG/HH isis solvable. Let G* = G/HGIH and and K* K *a a minimal minimal normal normal subgroupsubgroup of G*. By 9.4, K* is an elementary abelian p-group for some prime p. LetLet PP E E Sylp(K) Sylp(K) and and observeobserve that P isis aa complementcomplement toto H in K.K. ByBy aa Frattini Argument, 6.2, 6.2, G G == KNG(P), soso asas KK == HP, also G = HNG(P).HNG(P). IfIfYisacomplementtoHinGthenK Y is a complement to H in G then K = = KnG= K fl G = KnHYK fl HY = H(KH(KnY) fl Y) by the Modular Property of Groups, 1.14. ThenThen RR == KK fln YY isis aa complementcomplement toto HH inin K,K, so so R R E E Sylp(K). Sylp(K). Hence, Hence, by by Sylow'sSylow's Theorem, Theorem, there there isis k EE KK withwith Rk=P.AsK

Remarks. TheThe OddOdd OrderOrder TheoremTheorem ofof FeitFeit andand ThompsonThompson [FT][FT] sayssays thatthat groups ofof oddodd orderorder are are solvable. solvable. Notice Notice that that if if(I A(IAl, 1, IIGJ) G 1) = = 1 then eithereither IAIA I or I IGIG I isis odd,odd, soso AA oror GG isis solvable. solvable. ThusThus thethe OddOdd OrderOrder TheoremTheorem sayssays hypothesis (ii) of the Schur-Zassenhaus Theorem can be removed. Coprime action 771 1

Let n be be aa setset of primes. TheThe n-partn part ofof aa positive integerinteger n n is is n, n, = r[pen n,,, pep,pep, where nppeprlppep = n n is is thethe primeprime factorizationfactorization of of n. GivenGiven a finite group G, n(G)n (G) denotes the set of prime factors of II GG I. 1.G G is is a a n-group n -group if if n(G) r (G) is is aa subset of 7r.n. A Hall 7rn-subgroup -subgroup of G isis aa subgroupsubgroup of of orderorder I /GI,.G Jn .n' n' denotes the set of primes not in nn. . The following lemma gives a useful characterization of Hall n-subgroups.

(18.2) H is aa HallHall n-subgroupn-subgroup of the finite groupgroup GG ifif and onlyonly ifif HH is a a n-subgroup ofof GG andand IG:(G: HI, HI, = 1.1.

(18.3) Let GG bebe aa finitefinite groupgroup andand HH aa 7r n-subgroup -subgroup ofof G. Then (1) IfIf a: G -+ Ga is is aa homomorphismhomomorphism then Ha is is aa rr-subgroupn-subgroup of Ga. If H isis a Hall n-subgroup of G and a isis surjective then Ha is is aa HallHall n-subgroupn-subgroup of Ga. (2) If H isis aa HallHall n-subgroup ofof GG andand HH (< K (< G then H isis a a HallHall 7r-subgroupn-subgroup of K.

(18.4) IfIf pp andand qq areare distinct primes and H and K are HallHall p'-p'- andand 9'-subgroupsq'-subgroups of a finite group G, respectively,respectively, then (1) G=HK,andG =HK,and (2) HH fln K K is is a a Hall Hall (p,(p, q}'-subgroup 9)'-subgroup ofof G,G, aa Hall p'-subgroup of K, and a Hall 9'-subgroupq'-subgroup of H.

Proof. ThisThis follows follows from from 1.7.3.1.7.3.

(18.5) (Phillip(Phillip Hall's Theorem) Let Let G G be be a a finite finite solvable solvable groupgroup and n a a setset of primes. Then (1) GG possesses possesses a a HallHall n-subgroup.n-subgroup. (2) GG acts acts transitively transitively onon itsits HallHall n-subgroupsn-subgroups viavia conjugation.conjugation. (3) AnyAny n-subgroupn-subgroup of of GG is is contained contained in in some some Hall Hall 7r-subgroup n-subgroup ofof G.G.

Proof. LetLet G G be be a a minimal minimal counterexample counterexample and and MM aa minimal normal subgroup of G. By 9.4,9.4, MM isis aa p-groupp-group forfor somesome prime prime p. p. Let Let G* G* = = GIM.G/M. By By minimalityminimality of G, G*G* satisfiessatisfies the theorem. In particular G*G* possesses aa HallHall n-subgroupn-subgroup H*. AlsoAlso ifif XX isis aa n-subgroupn-subgroup of of GG then then so so is is X*, X*, so so X* X* is is contained contained inin somesome conjugate HH*g *g ofof HH*, *, andand hencehence X X 5< HgHg. . Suppose pP E n. Then Then HH is is a a Hall Hall n-subgroupn-subgroup of of G.G. Further Further XX isis containedcontained in the Hall n-subgroup Hg of G. IndeedIndeed if X isis aa Hall n-subgroup of G, thenthen 1x1IXJ =_ IH91lHgl soso XX = HgHg isis aa conjugateconjugate of H. Thus we may assume p 6 n. By By thethe Schur-Zassenhaus Schur-Zassenhaus Theorem,Theorem, 18.1,18.1, there is a complementcomplement KK to M in H, andand H isis transitivetransitive onon suchsuch complements. 72 Extensions of groupsgroups and modules

But as (H*I H*J I = IIGlnt G I,' these complements are preciselyprecisely thethe HallHall 7r-subgroups n-subgroups of GG containedcontained in H.H. Therefore Therefore GG possesses possesses HallHall 7r-subgroups n-subgroups andand HH isis transitive onon thethe HallHall n-subgroups7r-subgroups of of G G contained contained in in H. H. So,So, asas Xg 5< H,H, X X is conjugate toto K ifif XX isis aa Hall Hall 7r-subgroup n-subgroup of G. This shows G isis transi-transitive on its Hall 7r-subgroups.n-subgroups. FinallyFinally ifif H # G then, by minimality ofof G, X is contained in a HallHall 7r-subgroupn-subgroup ofof Hg, whichwhich isis also also aa Hall Hall 7r-subgroup n-subgroup of G. SoassumeH = = G.ThenGG.ThenG ==KMsoXM=XMf1G KMsoXM=XMnG =XMf1KM==XMnKM= (XMn(XMf K)MK)M by by thethe Modular Modular Property Property ofof Groups,Groups, 1.14.1.14. Now X and XMn K areare Hall 7r-subgroupsn-subgroups ofof XM,XM, so,so, by by (2),(2), there there is is yy EE XMXM withwith XYXY = == XM XMl n K K 5< K.K. Thus the proof is complete.

A converse of Phillip Hall's Theorem also holds, as we will see soon. In particular the hypothesis of solvabilitysolvability is necessary to insureinsure thethe validityvalidity of thethe theorem. Hall's Theorem isis aa goodgood exampleexample of how restrictions on the composition factors of a finite groupgroup cancan leadlead toto significant restrictionsrestrictions ofof thethe globalglobal structure of the group. Now to a proofproof of aa converseconverse toto PhillipPhillip Hall's Theorem.Theorem. The proof will appeal toto Bumside'sBurnside's paqb-~heorempagb-Theorem that finite groups G with 17r(G)IJn(G)I = = 2 are solvable. TheThe proofproof ofof Bumside'sBurnside's Theorem will bebe postponedpostponed untiluntil thethe chapterchapter on character theory. theory.

(18.6) LetLet G G be be aafinite finite group possessingpossessing aHalla Hall pl-subgroupp'-subgroup for each p EE 7r(G).x(G). Then G isis solvable.solvable.

Proof. LetLet GG bebe aa minimalminimal counterexample.counterexample. ByBy 9.8,9.8, GG is not a p-group. By Burnside's pogopaqa Theorem, Theorem, 35.13, 35.13, I7r(G)) In(G)J # 2. Thus j7r(G)IIn(G)I > 2.2. Let p EE 7rn(G) (G) andand H H aa HallHall pl-subgroupp'-subgroup ofof G.G. ByBy 18.4,18.4, HH nfl K is a Hall ql-q'- subgroup ofof H for each prime q distinct from p and each Hall ql-subgroupq'-subgroup of K ofof G.G. Therefore Therefore HH satisfies satisfies the the hypothesishypothesis of the lemma, and hence H isis solvable by minimality of G. Let M be aa minimalminimal normalnormal subgroupsubgroup ofof H.H. By 9.4,9.4, M is an r-group for somesome primeprimer. r. As 17r(G)I[n(G)I > > 22 therethere is is q q EE n(G)7r(G) - - (p, r}. r }. Let K be a Hall qf-subgroupq'-subgroup of G. As q 54# r, K contains a Sylow r-subgroupr-subgroup of G. Hence, by Sylow's Theorem, M is containedcontained in some conjugate of K, which wemaywe may take to be K. As q # p, G = HK HKby by 18.4. So,So, asas M M 5 a H,H, XX = (M~)(MG) = (MHK) (M~~) == (MK)(MK) <5 K. K. Hence, Hence, as as subgroups subgroups of of solvablesolvable groupsgroups are solvable, X is solvable. OfOf coursecourse XX

(18.7) (Coprime Action) LetLet A A and and H H be be finite finite groups groups with with (IA (I A 1, 1, I1HI) HI) == 1.1. Assume A is represented as a group of automorphisms ofof HH and either A or H isis solvable.solvable. Let p bebe aa prime.prime. Then (1) There exists an A-invariant SylowSylow p-subgroupp-subgroup ofof H.H. (2) CH(A) is transitive on the A-invariant SylowSylow p-subgroupsp-subgroups of G. (3) Every A-invariant p-subgroupp-subgroup ofof H isis contained contained inin anan A-invariantA-invariant Sylow p-subgroup of H. (4) Let K bebean an A-invariant normalnormal subgroup subgroup of of H H and and H* H* = = HIK.H/K. Then CH*(A) == NH*(A) = CH(A)*.CH(A)*.

Proof. FormForm the the semidirect semidirect product product GG ofof HH byby AA with with respectrespect toto the representation of A on H, and identify A and H with subgroups of G via the injections of 10.1. Then H

Again the hypothesis that A or G is solvable can bebe removedremoved fromfrom thethe statementstatement of 18.7, modulo the Odd OrderOrder Theorem.Theorem.

Remarks. TheThe material material in in section section 18 18 is is basic,basic, whilewhile thatthat inin section 1717 is more specialized. Thus the reader may wish to skip or postpone section 17. If so, 17.10 must be assumed in provingproving thethe Schur-Zassenhaus Theorem. 74 Extensions of groups and modules

Exercises for chapter 66 1. Let AA bebe aa solvable solvable group group acting acting on on G G = = XY XY withwith YY a 9 G, G, X X and and YA- YA- invariant, and (IAA,(I A(, IGI) 1GI) = = 1.1. Then Then CG(A)CG(A) == CX(A)Cy(A).Cx(A)Cy(A). 2. ProveProve 18.7 18.7 with with pp replacedreplaced byby aa setset ofof primes n,jr, under the assumption thatthat H isis solvable.solvable. 3. LetLet GG be the alternating groupgroup onon a a set set I I of of finite finite order order n n> > 2, 2, let let F F == GF(2), let V be thethe permutation module ofof thethe representationrepresentation on on I,I, and define Z and the core U of VV asas inin Exercise 4.6. Prove (1) 0,0, Z,Z, U,U, andand VV areare thethe onlyonly FG-submodules ofof V.V. InIn particularparticular U U == (U + Z)/ZZ)/Z is is an an irreducible irreducible FG-module. FG-module. (2) IfIf n isis odd proveprove H'(G,H1(G, U)U) == 0. (3) IfIf nn isis even prove V is an indecomposableindecomposable FG-module,FG-module, H'(G,H 1 (G, U) U) 2 = F, and V = U(G,U(G, U).D). (Hint: InIn (2)(2) andand (3)(3) letlet H be the stabilizer ofof a point xx of I andand proceedproceed by induction onon n. n. If If n n is is odd odd prove prove H H centralizes centralizes w wE cU(G, U(G, 0) U) - - U,0, and appeal to Exercise 4.6.4.6. If n is even prove H centralizescentralizes aa complementcomplement WW to U inin U(G,U(G, U)0) andand W W is is a a hyperplane hyperplane of of CU(G,CLI(G,~)(Hy) U)(Hy)for for x x# 0 y y EE I. Use this to concludeconclude dim(dim(W) W) 5< 1.)1.) 4. (Alperin-Gorenstein)(Alperin-Gorenstein) LetLet FF be be a a field field ofof characteristiccharacteristic p, GG aa finitefinite group, V an FG-module,FG-module, and A a G-invariant collection of pf-subgroupsp'-subgroups such such that:that: (1) VV=[V,X]foreach = [V, XI for each XXEA,and E A, and (2) thethe graphgraph on A obtained by joiningjoining XX toto Y ifif [X,[X, Y]Y] == 11 is connected. Prove H'(G,H1(G, V) == 0. 5. LetLet AA be an abelian r-group acting onon anan rf-groupr'-group G.G. ThenThen GG == (CG(B):(CG(B): B <( A, A, A/BAIBcyclic). cyclic). 7

Spaces with formsforms

Chapter 7 considers pairs (V, f) where where VV is is a a finite finite dimensionaldimensional vector vector spacespace over a field F andand f isis a a nontrivial nontrivial sesquilinear, sesquilinear, bilinear, bilinear, or quadratic form on V. We'll be primarily interested inin thethe situationsituation wherewhere Aut(V, Aut(V, f) f) is large; in that event f satisfiessatisfies one one of of severalseveral symmetry symmetry conditions conditions (cf. (cf. Exercises Exercises 7.9,7.9, 7.10,7.10,9.1, 9.1, and 9.9). Under suitable restrictions on F, such such pairspairs areare determineddetermined up to isomorphism inin sections 19,19, 20,20, andand 21.21. ForFor exampleexample if if FF is finite the isomorphism types are listed explicitly in section 21. It turns out that such spaces satisfy the Witt property: that is, if X and Y are subobjects ofof (V,(V, f)f) andand a:a: XX ->-+ YY isis anan isomorphism,isomorphism, then aa extendsextends to an automorphism of (V, f ). As a resultresult thethe representationrepresentation ofof Aut(V, Aut(V, f)f) onon (V, ff) ) is particularlyparticularly useful inin investigatinginvestigating Aut(V,Aut(V, ff). ). The groups Aut(V, ff), ), certain certain normal normal subgroupssubgroups of of thesethese groups,groups, andand theirtheir images under thethe projectiveprojective map of section 1313 are called the classical getups.groups. Section 22 derives various properties of the classical groups. For example for suitable fields they are essentially generated by their transvections or reflections, and are essentially perfect. ItIt will develop muchmuch laterlater inin sectionsection 4141 thatthat if G is a perfect finitefinite classical group then the projective group PGPG is simple. Conversely the Classification TheoremTheorem for finite simple groups says that, byby some measure, most of the finitefinite simple groupsgroups areare classicalclassical groups.groups. This chapter is one of the longest and most complicated in the book. More- over the material covered here is in somesome sensesense specializedspecialized andand tangentialtangential toto much of thethe other other materialmaterial in this book.book. Still,Still, asas I'veI've indicated, indicated, thethe classi-classical groups and their representations on the associated spaces (V, Q) areare veryvery important, soso thethe efforteffort seemsseems warranted.warranted.

19 Bilinear,Bilinear, sesquilinear, andand quadraticquadratic forms In thisthis sectionsection VV isis anan n-dimensionaln-dimensional vector spacespace overover aa fieldfield FF andand 08 isis an automorphism ofof F. A sesquilinear formform on V with respect to 89 is aa mapmap f: VV x V->FsuchV -+ F such that, that, forforallx,y,zeV all x, y, z E Vand and all all aa EE F:

.ff(x (x +Y,z)+ y, z) = .f f(x, (x, z) Z) + + .f f(y,z) (y, z) f(ax,.f (ax, y)=af(x,y) = of (x, y) .ff (x9(x, Y y ++ z)z) = .ff (x,(x, y)Y) + ff (x,(x, z)z) .ff (x,(x, ay) = ae f.f (x, (x, y).y) The form f isis saidsaid toto be bilinear if 80 = 1.1. Usually I'llI'll write (x,(x, y)y) forfor ff (x, y). 76 Spaces with formsforms

I'll alwaysalways assumeassume 08 isis ofof orderorder atat mostmost 2.2. ThereThere isis littlelittle lossloss ofof generalitygenerality in this assumption since we are interested in forms with big symmetry groups. Exercises 7.10 and 9.9 make this commentcomment moremore precise. f isis symmetric symmetric ifif ff isis bilinearbilinear and ff (x,(x, y)y) = = ff (y,(y, x) x) for for all all x, x, y y in in V. V. f f is skew symmetricsymmetric if iff f isis bilinearbilinear and and f f (x, (x, y) y) = _ - -ff (y,(y, x) forfor all x, y inin V.V. Finally f isis hermitian hermitian symmetricsymmetric if 08 is an involution andand ff (x, y) == f (y,(y, x)8x)' for all x, y in V.V. I'llI'll alwaysalways assumeassume that f hashas one one of of these these three three symmetrysymmetry conditions. One consequence of this assumption isis that

(*)(* For allall x, x, y y in in V, V, f(x, f(x,y)=0 y)=Oif if andandonly only if iff(y,x)=O. f(y,x)=0. On the other hand if (*) holds thenthen ExerciseExercise 7.107.10 showsshows that that ff (essentially)(essentially) . satisfies oneone of the three symmetry conditions. FurtherFurther ifif our form has a big group of automorphisms then Exercises 7.9 and 9.1 show we may as well take f toto satisfy satisfy one one ofof thethe conditions.conditions. IfIff f (x,(x, y)y) == 0 0 I'll I'll writewrite x x 1I y y and and say say that that x x andand yy areare orthogonal. For For X C5 VV define X'={vEV:x±v forallxEX} and observe that XX' 1-is is aa subspacesubspace of V and that XX' 1-= = (x)'.(X )1. IndeedIndeed

(19.1) For X x EE V, V, x' x1-= = ker(a), ker(a), where where a a E E HomF(V, HomF(V, F) F) is is defineddefined by by yaya = (y, x). Hence dim(xL)dim(x1-) >_ > nn - 1 1 with with equality equality preciselyprecisely when x 0$ V1-.v'.

VVL -L is is called called the the radical radical of of V. V. Write Write Rad(V) Rad(V) forfor v'.V J-.We We say say f f isis nondegenerate nondegenerate if Rad(V) = 0. 0. The form f willwill bebe saidsaid toto bebe orthogonalorthogonal ifif ff is is nondegenerate nondegenerate and and sym-symmetric, and if in addition, whenwhen char(F)char(F) == 2, ff (x, x) = 0 for all x inin V.V. TheThe form f isis said said to to bebe symplecticsymplectic iffif f isis nondegenerate nondegenerate and and skewskew symmetric, and in addition whenwhen char(F) char(F) == 2,2, ff (x,(x, x)x) == 00 forfor allall x x inin V. FinallyFinally ff isis saidsaid to be unitary ififf f isis nondegenerate nondegenerate and and hermitianhermitian symmetric.symmetric. A few words to motivate these definitions. I've already indicated why the symmetry assumptions areare appropriate.appropriate. ForFor anyany spacespace V, V == Rad(V)Rad(V) @® U for some subspace UU suchsuch thatthat thethe restrictionrestriction of of ff toto UU isis nondegenerate. nondegenerate. Thus there is little loss in assumingassuming f toto be be nondegenerate. nondegenerate. Besides, Besides, fromfrom Exercise 9.1, if (V, f) admits admits an an irreducible irreducible groupgroup ofof automorphismsautomorphisms then (essentially) f isis forcedforced toto be nondegenerate. Observe thatthat ifif char(F)char(F) == 2 then symmetry and skew symmetrysymmetry are thethe same.same. Also,Also, ifif char(F)char(F) # 2 and ff is skew symmetric, notice f (x,(x, x) x) = = 0 0 for for all all x x EE V.V. ThisThis motivatesmotivates thethe requirement thatthat f (x,(x, x) == 0 0 forfor allall x E V when char(F) = 2 2 andand f isis orthogonal orthogonal or or symplectic. symplectic. Indeed Exercise 7.9.3 shows that this assumption leads to little loss of generality and is alwaysalways satisfiedsatisfied ifif (V,(V, f)f) admits admits anan irreducible groupgroup of automorphisms. Bilinear, sesquilinear, and quadratic forms 77

(19.2) Let f bebe nondegenerate nondegenerate and U <5 V.V. ThenThen dim(U1) dim(^') == codim(U). codim(U).

Proof. The The proof proof is is by by inductioninduction on on m = dim(U). dim(U). The The lemma lemma is is trivial trivial if if mm = 0, 0, so take mm > 0.0. Then as f isis nondegenerate nondegenerate there exists xx EE VV -- U1.u'. ByBy 19.1,19.1, W == U nfl x'x1 is a hyperplane ofof U.U. ByBy induction induction on on m, m, dim(wL) dim(W1) = = n n - - mm ++ 1. Let uU E E U U - - W; thenthen U'U1= = W1 W' fln u1.u'. AsAs xx E W'W1 - u1, u', U1U' isis a ahyperplane hyperplane of W'W1 by 19.1. SoSo dim(^')dim(U1)=dim(W1) = dim(wL) - 1 1 = = n n - -m, m, completing completing the the proof. proof.

A vector x E VV isis isotropicisotropic if if ff (x,(x, x)x) == 0.0. I'veI've alreadyalready observedobserved that, that, if if f f is skew symmetric and char(F) isis notnot 2,2, thenthen everyevery vector is isotropic. Recall that this is part of the defining hypothesis of a symplectic or orthogonal form when char(F) = 2, A subspace U of V is totally isotropic ifif thethe restriction restriction of of f f to U is trivial, or equivalently if if U U 5< U1.u'. UU isis nondegenerate ifif thethe restrictionrestriction of of f f toto U is nondegenerate, oror equivalently equivalently U u n fl U' U1= = Rad(U) = 0.

(19.3) Let ff bebe nondegenerate nondegenerate and and UU a a subspace subspace of of V.V. ThenThen (1) UU isis nondegenerate nondegenerate if and only ifif V == U @® u'.U1. (2) (U1)1(u')' = U. U. (3) IfIf UU isis totally totally isotropicisotropic thenthen each complement to U in U'U1 isis nondegenerate. (4) IfIf UU isis totallytotally isotropic thenthen dim(U)dim(U) < n/2.

Proof. PartsParts (1),(I), (2),(2), andand (4)(4) areare easyeasy consequencesconsequences of 19.2,19.2, while (2) im-implies (3).(3).

Assume ff isis symmetric.symmetric. A quadratic formform onon VV associated to to f f is aa mapmap Q: V +-* F suchsuch that,that, for all x, y E V and aa EE F,F, Q(ax) = a2Q(x) a2e(x) andand Q(x + Y) = Q(x) + Q(Y) + f (x, Y). Observe thatthat if char(F) # 2 this definition forcesforces Q(x)Q(x) = = ff (x, x)/2,x)/2, so the quadratic formform isis uniquely determineddetermined by by f,f, andand hencehence addsadds nono newnew infor-information. On the otherother hand,hand, ifif char(F)char(F) = = 2 2 therethere are are many many quadratic quadratic formsforms associated toto f.f. ObserveObserve also that f isis uniquely uniquely determineddetermined by Q, since f(x, Y) = Q(x +Y) - Q(x) - Q(y) A symplectic spacespace (V,(V, f)f) isis aa pairpair consistingconsisting of a vector space V and a symplectic form form f f on V. A unitary spacespace isis aa pair pair (V, (V, f) f) with ff aa unitaryunitary form. An orthogonal space is a pair (V, Q) where QQ isis aa quadraticquadratic form on V with associated bilinear formform f.f. 78 Spaces withwith forms

In the remainder of this section assume (V,(V, f)f) isis aa symplecticsymplectic oror unitaryunitary space, or Q Q isis aa quadraticquadratic formform onon VV withwith associatedassociated orthogonal formform ff and (V, Q) isis anan orthogonalorthogonal space.space. The typetype of VV isis symplectic,symplectic, unitary,unitary, or orthogonal, respectively. A vector v in V is singular if vv isis isotropicisotropic and also Q(v) = 0 when VV is an orthogonal space.space. AA subspacesubspace U U ofof V is totally singular if U isis totallytotally isotropic and also each vector of U isis singular.singular. TheThe WittWitt index of V isis thethe maximum dimension of a totally singular subspace of V.V. NoticeNotice 19.319.3 says says thethe Witt index ofof V is atat mostmost n/2.n/2. An isometry ofof spacesspaces (V,(V, f)f) andand (U,(U, g)g) isis aa nonsingular nonsingular linearlinear transfor-transformation a:a: V +-* U such that g(xa, ya) = = f f (x, (x, y) y) forfor allall x, x, y y E E V.V. AA similar-similarity is a a nonsingularnonsingular linearlinear transformationtransformation a: a: V V + -+ U U suchsuch thatthat g(xa,g(xa, ya)ya) = h(a).l(a) ff (x,(x, y) for all x, y E V and some A(a)h(a) EE F'F# independentindependent of of x x andand y.y. If (V, Q) and (U,(U, P)P) are orthogonalorthogonal I'll I'll also also require require P(xa) P(xa) = = Q(x) Q(x) or or P(xa) P(xa) = h(a)Q(x),X(a) Q(x), inin thethe respective respective case. case. Forms Forms f f and gg (or(or P and Q) on VV areare said to be equivalentequivalent if if (V, (V, f) f) and (V, g)g) (or(or (V,(V, Q)Q) andand (U,(U, P))P)) are isometric.isometric. f f and g (or P andand Q)Q) are are similar similar if if thethe corresponding corresponding spaces are similar. O(V,O(V, f)f) (or O(V, Q)) denotes the group ofof isometricsisometries of of the the space, space, while while A(V, O(V, f) f) (or A(V, Q)) denotes thethe groupgroup of of similarities. similarities. Evidently Evidently O(V, O(V, f) f) 9 a A(V, L(V, ff). ). LetLetX=(xi:l X = (xi : 1 p < ii p< n)n)beabasisofV.Define be a basis of V. Define J= J = J(X,J(X, f)f)tobethenbyn to be the n byn matrix J == (Jib) (Jij) withwith Ji1Jij = f (xi,(xi, xj).xi). ObserveObserve that J uniquelyuniquely determinesdetermines the form f.f. Suppose YY = = (yi: (yi 1: 1p j aijxj, aiJaij E F, and A = (aid). (aij). Set AeAB = = (at)(ar) and and letlet ATAT be the transpose of A. Observe AeTAOT=ATe = ATO and and J(Y, J(Y, f) f) =AJA~'.= AJATB. Further

(19.4) AA formform gg on V is similar similar to f ifif and and only only if if there there exists exists aa basis basis Y == (yi:(yi :1 1 5 < i i 5< n) ofof VV withwith J(Y,J(Y, g)g) = = AJ(X, h J(X, f)f) for for some some X h EE F#.F'. Equiva- lence holds precisely whenwhen hA can can be be chosen chosen to to be be 1. 1. If If Q Q andand PP are quadratic forms associated toto ff andand g,g, respectively,respectively, then Q isis similarsimilar toto PP precisely whenwhen Y cancan be be chosen chosen so so that that J(Y, J(Y, g) g)=).J(X, = hJ(X, f) f) and P(yi)P(yi)=).Q(xi) = hQ(xi) forfor eacheach i, with Xh == 1 in case of equivalence.

Proof. LetLeta:(V, a: (V, f)f)+(V,g)beasimilarityandletyi=xiaandY=(yi: -* (V, g) be a similarity and let yi = xi a and Y = (yi : 1 p< i

(19.5) AA formform gg on VV is similar to f ifif andand only only if if J(X,J(X, g) g) _ =,kA hA J J(X,(X, f f)AT )ATe O for some nonsingular matrix AA and some Ah E F',F#, with hX = = 1 inin casecase ofof equivalence. equivalence. Bilineal;Bilinear, sesquilineal;sesquilinear, andand quadratic formsforms 79

Proof. ThisThis follows follows from from 19.419.4 and and thethe remarkremark immediately preceding it.

One can seesee fromfrom thethe precedingpreceding discussiondiscussion that that equivalenceequivalence ofof formsforms corre- corre- sponds to equivalenceequivalence ofof thethe associatedassociated defining defining matricesmatrices ofof thethe forms.forms. Given aa E GL(V)GL(V) definedefine Mx(a)MX(a) toto bebe thethe nn by n matrixmatrix (aij) defined defined byby xia = CEj ajjxj.aijxj.

(19.6) LetLet aa E GL(V).GL(V). Then Then a a E E A(V, A(V, f) f) if if andand onlyonly ifif (xis,(xia, xja)xja) ==.l(a) h(a) (x1,(xi, xi)xj) forfor allall ii and j, andand somesome .l(a)h(a) EE F#, with a EE O(V, f) precisely precisely whenwhen A(a)h(a) ==1. 1. If If (V,(V, Q)Q) is is orthogonalorthogonal then a E A(V, Q) if and onlyonly ifif aa E A(V,A(V, f)f) and Q(x;a)Q(xia) = = A(a)Q(x;) h(a)Q(xi) for for each each i,i, with with A(a)h(a) =1= 1 for for equivalence. equivalence.

(19.7) LetLet a E GL(V),GL(V), AA = = Mx(a), Mx (a), and and J J = = J(X,J(X, f).f ). Then aa E O(V,O(V, f) f) if and only if JJ = = AJATB. AJA~O. Ua E A(V,A(V, f)f) ifif andand onlyonly ifif AJh J = = AJATB A JA~' forfor somesome Ah EE F#. P.

(19.8) If If VV isis not not a a symplectic symplectic space space then then VV contains contains a a nonsingular nonsingular vector. vector.

Proof. AssumeAssumeotherwise.Letx otherwise. Letx E V# and and YE E V V - - xL.~hen x1. Then1 1 = = ((x, ((x, y)-lx, y)-lx, y), so without lossloss (x,(x, y)y) = = 1.1. NowNow (ax (ax + + by,by, axax ++ by) =abs= abe + baebas asas xx andand yy are singular andand (x, (x, y) y) = = 1.1. If If char(F) char(F) # # 2 tdketake aa == bb ==1 1 to get ax + by nonsingular. IfIf char(F)char(F) = 2 and V is unitary taketake a a == 1 andand bb ## be.be. Finally if V is orthogonal and char(F) = 2 then as x and y are singular, Q(x) = Q(y)Q(y) = 0, 0, soso Q(x + y) y) _= (x, (x, y) y) = = 1, 1, and and hence hence xx ++ y y is is nonsingular. nonsingular.

(19.9) AssumeAssume V V is is not symplectic, andifand if VV is is orthogonal orthogonal assume assume char(F) char(F) # # 2. Then there exists aa basisbasis XX == (xi(x1: : 1 1 5 < ii < n)n) ofof VV suchsuch that the members of X are nonsingular andand distinctdistinct membersmembers are are orthogonal.orthogonal.

Proof. By 19.8there19.8 thereisanonisotropicvectorxl is anonisotropicvector x1EV.By E V. By 19.3, 19.3,V=(xl)®(x1)1 V = (xl)@(xl)' with (x1)1(xl)l nondegenerate. By induction on n there is aa correspondingcorresponding basis (xi:l(xi: 1 <

A basis like the oneone ofof 19.919.9 will will be be termed termed an an orthogonal orthogonal basis.basis. AnAn orthonormalorthonormal basis for V isis aa basisbasis X X such such that that J(X, J(X, f) f) == I.

(19.10) If VV isis unitaryunitary thenthen (x,(x, x) isis inin thethe fixedfixed fieldfield Fix(O)Fix(8) of 80 for each XEV.xEV. 80 Spaces with formsforms

(19.11) AssumeAssume V V is is not symplectic, and if V isis orthogonal orthogonal assumeassume char(F)char(F) #0 2. Assume further thatthat thethe fixedfixed fieldfield Fix(@) Fix(9)of of @ 0 satisfies satisfies Fix(@) Fix(O)= _ {aas: {aae:a a EE F].F). Then (1) V V possesses possesses anan orthonormalorthonormal basis.basis. (2) AllAll formsforms onon VV ofof eacheach ofof thethe prescribedprescribed types types areare equivalent.equivalent.

Proof. NoticeNotice (1) (1) and and 19.4 19.4 imply imply (2).(2). ToTo prove (1),(I), choose an orthogonal basisbasis X as inin 19.9. ThenThen by by hypothesis hypothesis and and Lemma Lemma 19.10, 19.10, (xi, (xi, xi) xi) =a:('+') = ail+e> for some a,ai EE F#. Now replacing xi by aixi, wewe obtainobtain ourour orthonormalorthonormal basis.basis.

V is a hyperbolic planeplane if n = 2 and VV possesses a basis X = (x1, (XI, x2) such that xl andand x2 are singular and (xl(xl, , x2) x2) = 1. 1. Such a basis will be termed a hyperbolic pair.

(19.12) LetLet x x EE V#V be singular andand y y E c V -- x1.x'. ThenThen (x,(x, y) = U U isis aa hyperbolic plane and x is contained in a hyperbolic pair of U.

Proof. LetLet bb = (x, (x, y)-B.y)-s. ThenThen (x,(x, by)by) = 1, 1, so so withoutwithout loss (x, y) = 1.1. Observe U is nondegenerate, so if y is singular we are done. Thus we may assume each member of U - (x) (x) is is nonsingular, nonsingular, so so in in particular particular V V is is not not symplectic.symplectic. Thus,Thus, unless V isis orthogonal orthogonal and and char(F) char(F) = = 2,0 2, 0 # 0 (ax(ax ++y,y, axax + y) = a a +a"+as + +(y, (y, Y).y). However if if char(F) char(F) #0 2 we may take aa == -(y, y)/2,y)/2, and and use use 19.10 19.10 to to obtainobtain a contradiction. Thus char(F)char(F) = 2. Suppose V isis unitary.unitary. Let Let d d E E F F -- Fix(O).Fix(@). ThenThen ee = = dd ++ dsdB #0 0.0. Let c == (y, y)/e andand a = cd. cd. By 19.10,19.10, c Ec Fix(O),Fix(@), so so aa ++ asae == ce == (y, y), and hence axax + y y isis singular.singular. This leaves thethe casecase VV orthogonal.orthogonal. Then Then choosing choosing a a= = Q(y),Q(y), axax + y is singular, completing the proof.

Here's an immediate corollary to 19.1219.12 and 19.4:

(19.13) Let dim(V) = 2. 2. IfIf V#V# possesses a singular vector, then V is a hyperbolic plane. In particular, up to equivalence, there is a unique nondegenerate form on V of each type possessing aa nontrivialnontrivial singularsingular vector.vector.

(19.14) Let U be a totally singular subspacesubspace of of V,V, R R = = (ri: (ri 1: 15 < i i 5< m) a basis for U, and W a complement toto U in U1.u'. ThenThen therethere exists S = (si: 1 5< i <5 m)m) : VV suchsuch thatthat ri,ri, si si isis a a hyperbolic hyperbolic pairpair forfor thethe hyperbolichyperbolic planeplane Ui == (ri,(ri, si)si) andand W'W1 is thethe orthogonal orthogonal directdirect sumsum ofof thethe planesplanes (Ui:(Ui: 1 1 5< i <5 m).m). Witt'sWitt 's Lemma 81

Proof. ByB~ 19.319.3 we maymay take take w W = = 0. 0. Thus Thus u U = = u'. U1. ~et LetUo Uo = = (r,: (r,: 1 1 < < ii 5< m). By 19.2, U is a hyperplane ofof (~0)~.(Uo)'. ThenThen there exists a complement U1Ul = (rl,(r1, sl)Si) to to Uo inin U: Uo and by 19.3, U1 isis nondegenerate.nondegenerate. By 19.12 we may assume rlrl, ,s1 sl isis aa hyperbolichyperbolic pair pair forfor U1. U1. ByBy 19.3,19.3, V V = = U1 U1 ® @ Ul u:. L. Finally by induction on m we may choose (si:(s1: 1 1 < < ii <_( m)m) inin (U1)1(~1)' to satisfysatisfy the lemma.

Define V to be hyperbolic if V is the orthogonalorthogonal direct sum of hyperbolic planes. A hyperbolic basisbasis forfor aa hyperbolichyperbolic space space V V is is a abasis basis X X = = (xi:(x1: 1 1 5 < ii 5< m)m) such that V V isis thethe orthogonalorthogonal sum sum ofof thethe hyperbolic hyperbolic planes planes (x2i_1, (x2i-1, x20x2i) withwith hyperbolic pairpair x2i-1,x2i_1, xi. xt. We We say say VV isis dejinitedefinite ifif V possesses no nontrivial singular vectors. As a consequence of 19.1419.14 and 19.319.3 wewe have:

(19.15) LetLet UU bebe aa maximal hyperbolic subspace of V. ThenThen V = U U ® @ U1U' andand U1U' isis definite.definite. Moreover every totally singular subspace of VV of dimensiondimension mm is contained in a hyperbolic subspace ofof dimensiondimension 2m2m andand WittWitt indexindex m.m.

(19.16) All symplectic spaces are hyperbolic. hyperbolic. In particular particular allall symplecticsy mplectic spaces are of even dimension and, up to equivalence, each space of even dimension admits a unique symplectic form.

Proof. ThisThis isis immediate immediate fromfrom 19.15,19.15, 19.4,19.4,'and 11,and the fact that all vectors in a symplectic space are singular.

If char(F) == 2 2 and and (V,(V, Q)Q) isis orthogonalorthogonal thenthen (V,(V, f)f) is symplectic, wherewhere f f is the bilinear form determineddetermined by Q.Q. HenceHence byby 19.16:19.16:

(19.17) If VV is orthogonal and char(F)char(F) = 2, 2, then then VV is is of of eveneven dimension.dimension.

20 Witt'sWitt's LemmaLemma This section is devoted to a proof of Witt's Lemma. I feel Witt's LemmaLemma is probably the most important result in the theory of spaces with forms. Here it is:

Witt's Lemma.Lemma. LetLet V V be be an an orthogonal, orthogonal, symplectic, symplectic, oror unitaryunitary space.space. Let U and W be subspaces of V and suppose a: U + W is an isometry. Then a extends to an isometry of V.

Before proving Witt's Lemma let me interject anan aside.aside. Define an object X inin a category i'6? to to possess possess the the Witt Wittproperty property if, whenever Y and Z are subobjects 82 Spaces with formsforms of X and a: Y Y +-+ Z isis anan isomorphism,isomorphism, then a extendsextends toto anan automorphismautomorphism of X. Witt's LemmaLemma sayssays thatthat orthogonalorthogonal spaces, spaces, symplecticsymplectic spaces, spaces, andand unitaryunitary spaces have the Witt propertyproperty in thethe category ofof spacesspaces with formsforms and isome-isome- tries. All objects in the categorycategory of setssets andand functionsfunctions havehave thethe WittWitt property.property. But in most categories few objects have the Witt property; those that do areare very well behaved indeed.indeed. IfIf X is an object with the Witt property and G isis its group of automorphisms, thenthen thethe representation ofof G on X isis usuallyusually anan excellent tool for studying G.G. Now to the proofproof ofof Witt'sWitt's Lemma.Lemma. Continue thethe hypothesishypothesis andand notationnotation ofof the previous section. The proof involves a numbernumber ofof steps.steps. AssumeAssume thethe lemmalemma is false and letlet V be a counterexample with n minimal.

(20.1) Let H <5 U U and and suppose suppose a a/HI H extendsextends to an isometry /3P of V.V. ThenThen y == a$-1: a/3-': U U -+ + WEB-' W/3-' isis anan isometryisometry with yJHYIH = = 1,1, andand a extendsextends to an isometry of V ifif and only ifif y does.

(20.2) AssumeAssume 0 0: # H H < 5 U U with with H H nondegenerate. nondegenerate. ThenThen (1) If H'H1 G (~a)'(Ha)1 thenthen a a extends extends to to an an isometryisometry of V.V. (2) IfIf HaHa == H H then then a aextends extends to to an an isometry isometry of V.

Proof. Notice (1)(1) impliesimplies (2). (2). As As H H isis nondegenerate,nondegenerate, so so is isHa, Ha, and and V V == H @I®H1 H' == Ha Ha ®(Ha)1. @I(H~)'. LetLet $: /3: H1 H' -++ (Ha)1 (~a)' beanbe anisometry. isometry. By By minimality minimality of n, (ajunH±)$-1(alunH~)/3-' extends to to an an isometry isometry y yof of H'. H1. Then Then yB: y$: HI H1 +-+ (~a)'(Ha)1 is anan isometryisometry extendingextending a alunHl I unH± andand al~a I H+ + y/3 y,B is is an an isometry isometry of of VV extending a.

(20.3) If H isis aa totallytotally singularsingular subspacesubspace of Rad(U) and K aa complementcomplement to H inin Rad(U), then there existexist subspacessubspaces U' U' and and W' W' of of V V with with K K == Rad(U1)Rad(U') and U <5 U'U' suchsuch that a extends toto anan isometry isometry a: a: U' U' -+ W'.W'. If If U U = = H' H1 then U'U'=V. = v.

Proof. Let (ri:(r1:1 15 < ii <5 m)m) be be aa basisbasis for H, XX aa complementcomplement to to H H inin H' H1 containing K,K, andand X'X' aa complement to to Ha Ha inin (~a)'(Ha)1 containingcontaining (X fl U)a. By 19.14 therethere is is (si: (s;:1 1 5 < ii 5< m) and (sj:(s': 1 <5 ii <5 m) m) such such that that XX' 1 and (X')1(x')' are the orthogonal sum of hyperbolic planesplanes (ri, si)s;) and (r;(ria, a, s'),s,'), respectively.respectively. Extend a to

U'= (U,si:1(U,si:1 i

(20.4) VV isis notnot symplectic.symplectic.

Proof. ByBy 20.320.3 we may assume UU isis nondegenerate.nondegenerate. As As U U g = W, dim(U)dim(U) == dim(W), so dim(U1)dim(^') == dim(W1-). dim(wL). Hence,Hence, by 19.16,19.16, U1-U' E= W1-.w'. Then 20.2 contradicts the choice of V as a counterexample.

(20.5) If therethere existsexists a a totally totally singular singular subspace subspace 0 # 0: H H == Ha of of Rad(U)Rad(U) then a extendsextends toto V.V.

Proof. Let L = H1H' andand LE == L/H. LIH. Then Then f f(or (or Q) Q) induces induces a a form form f ofof typetype f (or(or Q)Q) on on L defineddefined by f(f,f (x, y)ji) = f f (x, (x, y) y) and and thethe inducedinduced map ti:a: 0U + Ww is anan isometry,isometry, so, by minimality ofof n,n, tiit extends toto anan isometryisometry fiP ofof E.L. Let X be a basis of L with Xx flfl HH andand xX fl n U U bases bases for for H H and and U, U, respectively,respectively, and letlet /3 P Ee GL(L) bebe aa map map with with /3lu 8ju = = a a andand 3 xj8 = = ffi xp for for x x E e XX -- U.U. ByBy construction (x, Y) = (x, Y) = (0, 0) = (xp, YP) for x, y eE X, X, so so ,B /3 isis an an isometry isometry of of L. L. Now, Now, by by 20.3, 20.3, ,B,/3, andand hencehence also a, extends to an isometry of V.V.

(20.6) Assume H is a hyperplanehyperplane of of U U with with al~ aIH= = 1. AssumeAssume also also that that H H = = 00 if V is unitary or char(F) # 2. 2. ThenThen a a extends extends to to anan isometryisometry of V.V.

Proof. Let u eE UU -- H H and and set set K K = = U U + +W. W. Assume Assume a a does does not not extend.extend. Suppose UU == W.W. ThenThen Rad(U)Rad(U) #O 0 0by by 20.2,20.2, andand asas aa actsacts onon Rad(U),Rad(U), Rad(U) is not totally singular byby 20.5.20.5. ThusThus char(F)char(F) = 2 2 andand VV isis orthogonal.orthogonal. AsAsadoesnotextend,alu a does not extend, a I u# #1,sou 1, so u# 0 ua.Nowua ua. Now ua == au+h,forsomeaau +h, for some a E F# andandh h E H.H.AsaIH As a IH == 1,l,aactsonX= a acts on X = (u,(u,h).By20.2,Rad(X)#O.Henceas h). By 20.2, Rad(X) 0. Hence as each member of V# isis isotropicisotropic (because(because VV isis orthogonalorthogonal and and char(F) char(F) = 2),2), XX is totally isotropic.isotropic. HenceHence asas Q(u)Q(u) == Q(ua), zz = u u ++ ua ua is is singular. singular. Therefore Therefore either X isis totallytotally singularsingular or (z)(z) is is thethe uniqueunique singularsingular pointpoint in X,X, andand hencehence is a-invariant. By 20.2,20.2, HH containscontains nono nondegeneratenondegenerate subspaces, so f IIH H = =0 0 by 19.12.19.12. Thus h E Rad(U),Rad(U), so,so, byby 20.5,20.5, hh is nonsingular. So So (z)(z) = _ (za)(za) is singular and z 4 H. HenceHence we may assume z = u. Again by 20.5 there is h' EE H -z'.-z1. NowNow a a acts acts on on X'= X' = (h', (h', z) z) and, and, as as X' X' is is nondegenerate, nondegenerate, 20.2 20.2 suppliessupplies a contradiction. So UU #: W. W. Let Let cc == 1 1 if if (u,(u, ua) ua) = = 0 0and and c c = = (u, (u, ua)B/(u, ua)'/(u, ua)ua) otherwise. otherwise. Observe wewe cancan extend extend a a to to an an isometry isometry a' a' of of K K with with (ua)al (ua)a' = cu, cu, by by 19.6.19.6. So, by thethe firstfirst argument argument in the previous previous paragraph, char(F)char(F) == 22 and V is orthogonal. By By definition definition of ofa', a', a' a'fixes fixes z z= = u u + + ua. ua. NowNow H' H'= = (H, z) is a 84 Spaces withwith forms hyperplane of K with a'(H'a' (H' = 1,1, so,SO, by the previous paragraph, a',a', and hence alsoalso a, extendsextends toto anan isometryisometry of V.V.

(20.7)(20.7) VV isis orthogonalorthogonal and char(F) = 2. 2.

Proof.Proof. Assume Assume not. not. By By 20.3, 20.3, we we may may take take U U toto bebe nondegenerate.nondegenerate. By By 19.819.8 and 20.4 there is a nonsingular pointpoint LL inin U. By 20.6 applied toto LL in the role ofof U,U, aa I 1 Lextends extends to to an an isometry isometry ofof V. Then by 20.1 we may take aa I 1 L= = 1.1. But nownow 20.220.2 suppliessupplies aa contradiction.contradiction.

WeWe are now in a positionposition to completecomplete the proofproof ofof Witt'sWitt's Lemma.Lemma. ChooseChoose U of minimalminimal dimension so that an isometry a: U U -*-+ WW doesdoes notnot extendextend toto V.V. Let Let H bebe aa hyperplanehyperplane of U.U. ByBy minimalityminimality of U,U, ajHal~ extendsextends toto anan isometryisometry of V,V, soso byby 20.1 20.1 we we maymay taketake aa I 1 HH = 1.1. Now Now 20.620.6 suppliessupplies a a contradiction contradiction and and completescompletes the the proof.proof.

II closeclose thisthis sectionsection with somesome corollaries to Witt's Lemma.Lemma.

(20.8)(20.8) (1)(1) The The isometry isometry group group of of V V is is transitive transitive on on the the maximal maximal totally totally singular singular subspacessubspaces of V,V, andand onon thethe maximalmaximal hyperbolichyperbolic subspaces of V.V. (2)(2) V V is is the the orthogonal orthogonal direct direct sumsum ofof aa hyperbolichyperbolic spacespace HH andand a a definite definite space.space. Moreover H isis aa maximal maximal hyperbolic hyperbolic space space and and this this decomposition decomposition is is unique up to an isometry of V.V. (3)(3) TheThe dimension dimension of of a a maximal maximal hyperbolic hyperbolic subspace subspace of of VV isis twicetwice thethe WittWitt indexindex ofof V.V.

Proof.Proof. TheseThese remarks remarks are are a a consequence consequence of of Witt'sWitt's LemmaLemma andand 19.15.19.15.

(20.9)(20.9) (1) IfIf KK isis a a quadratic quadratic GaloisGalois extension ofof FF and NFN;: : KK -+-* F isis thethe normnorm ofof KK overover F,F, then then (K,(K, NF) N;) isis a a2-dimensional Zdimensional definite definite orthogonal orthogonal space space overover F.F. (2)(2) Every Every 2-dimensionalz-dimensional definitedefinite orthogonal space over F isis similarsimilar toto a a spacespace (K,(K, NF) N;) forfor some some quadratic quadratic GaloisGalois extensionextension K ofof F.F.

Proof.Proof. IfIf KK isis aa quadraticquadratic Galois extension extension of of F F then then Gal(K/F) Gal(K/F) = = (u)(a) isis of orderorder 2 and N;(a)NF (a) == aa'aaa for for a ac K.E K. It isIt straightforwardis straightforward to to prove prove (K, (K, NF N;) ) isis aa definitedefinite orthogonalorthogonal space. space. NextNext aa proofproof ofof (2).(2). LetLet (V,(V, Q) Q) be be aa definite definite orthogonalorthogonal space and {x,(x, y) aa basisbasis forfor V.V. IfIf char(F)char(F) 0 # 2 2then then by by 19.9 19.9 we we can can choose choose (x, (x, y) y) = = 0, 0, while while if if char(F)char(F) = = 2 2choose choose (x, (x, y) y) = = 1. 1.Replacing Replacing Q Q by by a ascalar scalar multiple multiple if if necessary,necessary, wewe can assume Q(x)Q(x) = = 1.1. Let Let Q(y) Q(y) = = b b and and P(t) p(t) = = t2 t2+t(x,+t(x, y)+b,y) +b, so so that that PP isis Spaces over finite fields 85 a quadratic polynomial overover F.F. As V is definite, P is irreducible. Let K be the splitting field forfor PP over F and c cE KK a a root root of of P.P. Then Then the the map map x x i-+ H 1, 1, yy i-+H c induces an isometry of (V, Q) with (K,(K, NFN:). ).

(20.10) Assume FF is is algebraically algebraically closed. closed. ThenThen (1) IfIf char(F)char(F) 0 # 2 2then, then, up up to to equivalence, equivalence, V V admitsadmits a a unique nondegenerate quadratic form. Moreover V has an an orthonormal orthonormal basis with respectrespect to thatthat form. (2) IfIf char(F)char(F) = = 22 then then V V admits admits a a nondegenerate nondegenerate quadratic quadratic form form ifif and only if n isis even.even. The formform isis determineddetermined up to equivalence and V is a hyperbolic space with respectrespect toto thisthis form.form.

Proof. PartPart (1) (1) follows follows from from 19.11.19.1 1. ToTo proveprove part (2)(2) itit sufficessuffices by 19.1719.17 and 20.8 to take VV anan orthogonalorthogonal space space of of dimensiondimension 2 2 and and prove prove V V isis not not definite. definite. But as F is is algebraically algebraically closed closed itit possessespossesses nono quadraticquadratic extensions, so V is not definite by 20.9.20.9.

(20.11) If VV isis anan orthogonalorthogonal spacespace of dimension at least 2 then V has a nondegenerate 2-dimensionalZdimensional subspace.subspace.

Proof. If char(F) 0# 22 this this is is a a consequence consequence ofof 19.9. If If char(F)char(F) == 22 thenthen by 19.1619.16 the underlyingunderlying symplecticsymplectic space isi~ hyperbolic and hence possesses a hyperbolichyperbolic plane, whichwhich is a a nondegeneratenondegenerate subspacesubspace of thethe orthogonal orthogonal space V.V.

21 SpacesSpaces overover finitefinite fields In this section the hypothesis and notation of section 1919 continue; in particularparticular V is an orthogonal, symplectic, oror unitaryunitary spacespace overover F.F. In addition assume F is a finite field of characteristic p.

(21.1) Assume n == 2. 2. Then Then up up to to equivalence equivalence there there isis a unique nondegenerate definite quadraticquadratic form form Q Q onon V.V. FurtherFurther therethere is is aa basisbasis XX = {x,(x, y} y) ofof VV such that:that:

(1) IfIf p isis odd then (x,(x, y)y) == 0, Q(x) = 1,1, andand -Q(y) isis a a generator generator of F'.F#. (2) If p = 2 2 thenthen (x, y)y) == 1, Q(x) == 1, Q(y) == b, and P(t)P(t) = t2t2 + t + bb isis an irreducible polynomial over F.F.

Proof. ByBy 20.9 20.9 andand itsits proof,proof, QQ is is at at least least similar similar to to suchsuch aa form.form. ItIt isis thenthen anan easy exercise to prove formsforms similarsimilar to to QQ are eveneven equivalentequivalent to to Q.Q. AsAs FF is finite, it has a unique quadratic extension, so Q isis uniqueunique by 20.9.2. 86 Spaces with formsforms

Denote byby D = D+ D+ and and QQ = = D_ D- the the (isometry (isometry type type of of the)the) hyperbolic hyperbolic plane and the 2-dimensional definite orthogonal space over F, respectively. respectively. WriteWrite Dmek Qkfor for the the orthogonal orthogonal directdirect sumsum ofof mm copiescopies ofof DD with k copies of Q.

(21.2) LetLet FF be be a a finite finite field.field. ThenThen (1) DmDm isis aa hyperbolichyperbolic space space of of WittWitt indexindex m. (2) Dii-1Dm-' QQ isis ofof Witt index mm -- 1. 1. (3) D2mDZm isis isometricisometric to to Q2m.eZm. (4) Every Every 2m-dimensional2m-dimensional orthogonal orthogonal space over F isis isometricisometric to to exactlyexactly one of DmDm or Dm-1Dm-l Q.

Proof. ByBy constructionconstruction Dm Dm is hyperbolic and, by 19.3.4,19.3.4, Dm is of Witt index m. By constructionconstruction Q Q is is of of Witt Witt index index 0. 0.Let Let V V2 =Dm-' Dm-'Q, Q, Dm-' Dm-1 2 = U U 5< V, and QQ =2 U1=U' = W. W. LetLet X X be be a amaximal maximal totally totally singularsingular subspacesubspace of U. As dim(X) =m =m - 1,X1= 1, X' =X x® @W. W. ForFor W w E EW#, W', Q(w)~(w) # 0,0, so,so, for for xx EE X Q(x + w) == Q(w) Q(w) # #0. Thus0. Thus X is X a is maximal a maximal totally totally singular singular subspace subspace of x',X1, so X is also aa maximal totally singular subspace of V. Hence, by 20.8, gm-1Dm-1 QQ isis notnot isometric to Dm, as they have different Witt indices. Let V V bebe aa 2m-dimensional2m-dimensional orthogonal spacespace overover F.F. By 20.120.11 1 V has aa nondegenerate planeplane U.U. ByBy 21.121.1 andand induction induction on on m, m, U U 2 = DD or Q and U1U' 2 gm-1Dm-1 oror Dm-'Q.Dm-2 Q. So So V V 2= Dm,Dm, Dm-1Q,Dm-'Q, or 0"-'Q2.Dm-2 Q2.Thus Thus to complete the proof of 21.2 it remains to show Q2Q2 2 0'.D2. This will follow from 20.8 if we can show Q2 hashas a2-dimensional a 2-dimensional totally totally singular singular subspace. subspace. So So take take U U2 = U' U1 2= Q.Q. Let {x,(x, y}y} andand (u,{u, v} v} be be bases bases for for U U andand u',U1, respectively.respectively. If If char(F) char(F) == 22 then byby 21.121.1 we we may may choose choose (x, (x, y) y)= =(u, (u, v) v)= = 1, 1, Q(x) Q(x) = = Q(u) = 1,1, andand Q(y) = Q(v). Q(v). Then Then (x(x + + u, u, y y + + v) v) is is a atotally totally singular singular plane plane ofof V.V. SoSo taketake char(F) toto bebe odd.odd. Then by 21.121.1 we may take x, y, u,u, andand vv toto bebe orthogonalorthogonal with Q(u)Q(u) = _ - Q(x) Q(x) and Q(v) =_ - Q(y). Then again (x + u, y + v) is a totally singular plane.

If F isis finitefinite and n == 2m 2m is is even, even, then then 21.2 21.2 sayssays that,that, upup toto equivalence,equivalence, there are exactly two quadratic forms on V, andand that thethe correspondingcorresponding orthogonal spaces have Witt indexindex mm and m - 1, 1, respectively. respectively. DefineDefine thethe signsign ofof thesethese spaces toto bebe ++1 1 and --1, 1, respectively, respectively, andand write sgn(Q) or sgn(V) for the sign of the space. Thus the isometry type of anan even dimensional orthogonal spacespace over a finite field isis determineddetermined by its sign. If V is an orthogonal spacespace ofof oddodd dimension overover F,F, then,then, by 19.17,19.17, char(F)char(F) is is odd. odd. Let's.Let's look look atat suchsuch spacesspaces next.

(21.3) Let VV be anan orthogonalorthogonal spacespace ofof oddodd dimensiondimension over a finite field F. Then V possesses a hyperplane which is hyperbolic. 86 Spaces with forms Spaces over finite fields 87

Denote by D = D+ and Q = D_ the (isometry type of the) hyperbolic plane Proof. The proof is by induction on n. The remark is trivial if n = 1, so take and the 2-dimensional definite orthogonal space over F, respectively. Write n > 3. By 19.9, V possesses a nondegenerate subspace U of codimension 2. Dm Qk for the orthogonal direct sum of m copies of D with k copies of Q. By induction on n, U possesses a hyperbolic hyperplane K. If n > 3 then, by induction on n, Kl possesses a hyperbolic plane W. Then K ® W is a (21.2) Let F be a finite field. Then hyperbolic hyperplane of V. (1) Dm is a hyperbolic space of Witt index m. Son = 3. Choose a basis X = (x;:1 < i < 3) for V as in 19.9. We may (2) Dni-1 Q is of Witt index m -1. assume V is definite. Thus (x1, x2) is definite, and hence possesses a vector (3) Den` is isometric to Q2rn. y such that Q(y) has the same quadratic character as -Q(x3). Thus there is (4) Every 2m-dimensional orthogonal space over F is isometric to exactly a E F with a2Q(y) _ -Q(x3). Now ay + x3 is singular and 19.12 completes one of D' or D'-' Q. the proof.

Proof. By construction Dm is hyperbolic and, by 19.3.4, Dm is of Witt index If V is an odd dimensional orthogonal space over a finite field then, by 21.3, m. By construction Q is of Witt index 0. Let V = Dm- IQ, Dm-1 - U < V possesses a hyperbolic hyperplane H, and, by 20.8, H is determined up to V, and Q = U1= W. Let X be a maximal totally singular subspace of U. conjugacy under the isometry group of V. Let x be a generator of Hl and As dim(X) =m - 1, Xl = X ® W. For W E W#, Q(w) 0 0, so, for x E X define the sign of V (or Q) to be +1 if Q(x) is a quadratic residue in F, and Q(x + w) = Q(w) 0. Thus X is a maximal totally singular subspace -1 if Q(x) is not a quadratic residue. Then evidently when n is odd there are of X', so X is also a maximal totally singular subspace of V. Hence, by 20.8, orthogonal spaces of sign s = +1 and -1, and, by the uniqueness of H (up to Dm-1 Q is not isometric to Dm, as they have different Witt indices. conjugacy), these spaces are not isometric. On the other hand if c is a generator Let V be a 2m-dimensional orthogonal space over F. By 20.11 V has a for the multiplicative group F# of F, then c Q is similar to Q under the scalar nondegenerate plane U. By 21.1 and induction on m, U - D or Q and Ul - transformation cl. Moreover (H, Q) is similar to (H, cQ), so (H, cQ) is also Drn-1 or Dm_2Q. So V - Dm, Dm-1 Q, or Drn-2 Q2. Thus to complete the hyperbolic. Hence, as (cQ)(x) = cQ(x) has different quadratic character from proof of 21.2 it remains to show Q2 - D2. This will follow from 20.8 if we can Q(x), sgn(cQ) : sgn(Q). Thus we have shown: show Q2 has a 2-dimensional totally singular subspace. So take U = Ul = Q. Let (x, y} and (u, v} be bases for U and Ul, respectively. If char(F)= 2 then by 21.1 we may choose (x, y) = (u, v) = 1, Q(x) = Q(u) = 1, and (21.4) Let F be a field of odd order, n an odd integer, and c a generator of the Q(y) = Q(v). Then (x + u, y + v) is a totally singular plane of V. So take multiplicative group F# of F. Then char(F) to be odd. Then by 21.1 we may take x, y, u, and v to be orthogonal (1) Up to equivalence there are exactly two nondegenerate quadratic forms with Q(u) _ - Q(x) and Q(v) _Q(y). Then again (x + u, y + v) is a totally Q and c Q on an n-dimensional vector space V over F. singular plane. (2) sgn(Q) = +1 and sgn(cQ) _ -1. (3) Q and cQ are similar via the scalar transformation cl, so O(V, Q) _ If F is finite and n = 2m is even, then 21.2 says that, up to equivalence, there O(V, cQ). are exactly two quadratic forms on V, and that the corresponding orthogonal spaces have Witt index m and m - 1, respectively. Define the sign of these (21.5) Let F be a finite field of square order. Then up to equivalence V admits spaces to be +1 and -1, respectively, and write sgn(Q) or sgn(V) for the sign a unique unitary form f. Further (V, f) possesses an orthonormal basis. of the space. Thus the isometry type of an even dimensional orthogonal space over a finite field is determined by its sign. If V is an orthogonal space of odd Proof. As F is of square order it possesses a unique automorphism 0 of order dimension over F, then, by 19.17, char(F) is odd. Let's look at such spaces 2. Moreover Fix(0) = f aae: a E F}, so 19.11 completes the proof. next.

(21.3) Let V be an orthogonal space of odd dimension over a finite field F. The final lemma of this section summarizes some of the previous lemmas in Then V possesses a hyperplane which is hyperbolic. this chapter, and provides a complete description of forms over finite fields. 8888 Spaces with formsforms

(21.6) Let V be an n-dimensional space overover aa finitefinite fieldfield FF of order q and characteristic p. Then (1) VV admits admits aa symplecticsymplectic form f ifif andand onlyonly if nn isis even,even, in which casecase ff is unique up to equivalence andand (V,(V, f)f) isis hyperbolic.hyperbolic. (2) VV admits admits aa unitary form f ifif andand onlyonly ifif qq isis aa square,square, in which case ff is unique up toto equivalence andand (V,(V, f)f) has has aa orthonormalorthonormal basis. (3) IfIf nn isis eveneven thenthen VV admitsadmits exactlyexactly two equivalence classes of nondegenerate quadratic forms. Two forms are equivalent precisely when they have the same sign. IfIf P isis suchsuch aa formform then (V, P) isis isometricisometric to 0""Dn/2 or 0tnJ2)-lD(n/2)-1 Q of sign ++1 1 and -1,- 1, respectively. respectively. (4) IfIf nn isis oddodd then V admits a nondegenerate quadratic formform preciselyprecisely whenwhen p isis odd,odd, inin whichwhich casecase therethere areare two equivalenceequivalence classes of forms. All forms are similar.

22 TheThe classicalclassical groups In section 22, continue to assume the hypothesis and notation of section 19.19. Section 22 considersconsiders thethe isometry isometry groups groups O(V, O(V, f) f) and O(V,O(V, Q), certain normalnormal subgroups of these groups, and the images of such groups under the projective map P ofof sectionsection 13.13. Notice that one can also regard the general linear group as the isometryisometry group group O(V, O(V, f), f), where ff isis thethe trivial form f (u, v) = 0 0 for for allall u, v cE V.V. TheThe groupsgroups GG andand PG,PG, as as G G ranges ranges over over certain certain normalnormal subgroups of O(V, f),f), areare calledcalled thethe classicalclassical groupsgroups (where(where f isis trivial, trivial, orthogonal, orthogonal, symplectic, or unitary). We'll be particularly concerned with classical groups over finite fields. Observe that if two spacesspaces areare isometricisometric then theirtheir isometryisometry groups are isomorphic. This is a special case of anan observationobservation made inin sectionsection 2.2. AsAs aa matter of factfact thethe isometryisometry groups are isomorphic if the spaces are only similar, which is relevant because of 21.6.4. The upshot of these observations is that in discussing the classical groups we need only concern ourselves with forms up to similarity. Recall, from 19.16 thatthat if n is eveneven there is, up toto equivalence,equivalence, a unique symplectic formform f onon V.V. WriteWrite Sp(V)Sp(V) for for the the isometryisometry groupgroup O(V, ff). ). Sp(V)Sp(V) is the symplectic group group on on V.V. AsAs VV isis determineddetermined by by n n and and F,F, I'll also write Spn(F) for Sp(V). Spn(F)Sp,(F) isis thethe n-dimensionaln-dimensional symplectic groupgroup overover F.F. If f isis unitary unitary thenthen O(V,O(V, f)f) is is called called aa unitaryunitary group.group. Similarly ifif Q is a nondegenerate quadratic form then O(V, Q)Q) is an orthogonal group.group. InIn generalgeneral there are a number of similaritysimilarity classes of forms on V and hence more than one unitary group or orthogonalorthogonal group on V. Lemma 21.6 givesgives preciseprecise informationinformation when F isis finite;finite; we will consider that case in a moment. In any event I'll write GU(V) or O(V) for aa unitaryunitary oror orthogonalorthogonal groupgroup onon V,V, respectively,respectively, eveneven The classical groups 89 thoughthough therethere maymay be moremore than one suchsuch group.group. GU(V) isis (the)(the) generalgeneral uni-uni- tarytary group.group. WriteWrite SU(V)SU(V) and and SO(V)SO(V) for for SL(V) SL(V) fl n GU(V) GU(V) andand SL(V)SL(V) fl n O(V), O(V), respectivelyrespectively (recall(recall thethe specialspecial linear group SL(V) isis defineddefined and discussed in sectionsection 13).13). SU(V) andand SO(V)SO(V) areare thethe specialspecial unitaryunitary groupgroup andand specialspecial or-or- thogonalthogonal group,group, respectively.respectively. Write Q(V)Q(V) forfor thethe commutatorcommutator groupgroup ofof O(V).O(V). SupposeSuppose for the moment that F == GF(q) GF(q) is is the the finite finite field field of of order order q. q. Then Then writewrite SpnSp, (q) for Sp,(F).Spn (F). Also,Also, fromfrom 21.6, there is a unitaryunitary formform on V precisely whenwhen q == r2 r2 is is a asquare, square, in in which which case case the the form form is is unique, unique, andand II write write GUn(r) GU,(r) andand SU,(r) forfor GU(V) GU(V) andand SU(V).SU(V). NoticeNotice r # (F1,IF/, rather rather rr == IF11/2 IF^''^ inin thethe unitaryunitary case.case. IfIf nn isis oddodd therethere isis anan orthogonalorthogonal form on VV onlyonly whenwhen qq isis odd,odd, inin whichwhich casecase allall suchsuch formsforms areare similarsimilar andand II writewrite On(q),O,(q), SOn(q),SO,(q), andand Qn(q)Q,(q) forfor O(V),O(V), SO(V),SO(V), andand Q(V).Q(V). Finally Finally if if nn is is even even then then up up to to equivalence equivalence there there areare just two nondegenerate quadraticquadratic forms forms QeQ, onon V,V, distinguisheddistinguished by by thethe sign sign sgn(QE)sgn(Q,) = sE == +1 +1 or or -1 -1 of ofthe the form. form. Write Write 01(q), Oi(q), SO' SOi(q), (q), and and Q Qi(q) '(q) for for the the correspondingcorresponding groups. groups. For each group G we can restrict thethe representationrepresentation P: P: GL(V) GL(V) +-* PGL(V) of GL(V) onon the projective space PG(V)PG(V) toto GG and obtainobtain thethe imageimage PGPG of G which is aa groupgroup ofof automorphismsautomorphisms of of thethe projectiveprojective spacespace PG(V). ThusThus forfor example we obtainobtain thethe groupsgroups PSpn(q),PSp,(q), PGUn(r),PGU,(r), POn(q),POi(q), PS2n(q), PQi(q), etc. It will develop much later that the groups PSp,(q),PSpn(q), PSU,(r),PSUn(r), and PQi(q)PQ (q) are are simple simple unless n andand qq areare small. small. In In thisthis section section we we prove prove these these groupsgroups are are (usually) (usually) perfect (i.e.(i.e. eacheach groupgroup isis itsits ownown commutatorcommutator group).group). This factfact togethertogether withwith Exercise 7.8 isis usedused inin 43.1143.1 1 to to establishestablish the thg simplicitysimplicity of of thethe groups. groups. Recall from section 1313 that subspaces of V of dimension 1,2,1, 2, andand nn -- 1 1 are are called points, lines, andand hyperplanes, respectively, respectively, and in general subspaces of V are objects of thethe projectiveprojective space space PG(V). PG(V). If If V V has has a a form form f for or Q,Q, then from sectionsection 1919 wewe havehave aa notionnotion ofof totallytotally singularsingular andand nondegeneratenondegenerate subspace,subspace, and hence totally singular and nondegenerate points, lines, and hyperplanes.

(22.1)(22.1) If g E O(V, f)f) then then Cv(g) = [V, [V, g]1. glL.

Proof. Let U = Cv(g). Cv(g). For For u EE UU andand vv EE V,V, (u, v) = (ug,(ug, vg)vg) == (u, (u, vg). vg). Thus as

v + u1u' == {x (x E E VV: : (x, U)u) = (v,(v, u)]u)} we have vg E v + u1.u'. HenceHence

[a,[v, g] g] E En u1u' == U1. u'. UEUuEU Therefore [V,[V, g] g] 5< U1.uL. But, But, by by ExerciseExercise 4.2.34.2.3 andand 19.2,19.2, dim([V, g]) g]) =_ dim(U1),dim(uL), so so thethe proofproof isis complete.complete. 90 Spaces with formsforms

Recall the definition of a transvection inin sectionsection 13. I'llI'll prove the next two lemmas together.together.

(22.2) O(V, f) (or (or O(V, O(V, Q)) Q)) contains contains a a transvection transvection ifif andand onlyonly ifif eacheach ofof thethe following holds:

(1) V V possesses possesses isotropicisotropic points. points. (2) IfIf (V,(V, Q)Q) isis orthogonalorthogonal thenthen char(F)char(F) = 2.2.

(22.3) Let G = O(V, O(V, f) f) (or (or O(V, O(V, Q)) Q)) and and assume assume t tis is a a transvection transvection inin G.G. Then (1) U = [V, [V, t]t] isis anan isotropic pointpoint andand Cv(t) Cv(t) == U1.UL. (2) Let AUAu be the set of transvections with with center center U U = = (u) (u) and and let let R R == RURu = (AU)(Au) be the root group oft.of t. ThenThen R#R# == Au, forfor eacheach r EE RR and and Yy EE VV we havehave yryr = y y ++ ar(y, a,(y, u)uu)u for for some some ar a, EE F, F, and and one one of of the the following following holds:

(i) (V, (V, f) is is symplectic symplectic andand thethe map r i--H ara, is is an an isomorphism isomorphism ofof RR withwith the additive groupgroup ofof F.F. (ii) (V,(V, f) f) is is unitary unitary and and r r i-- H are are is is an an isomorphism isomorphism of of R R with with Fix(9), Fix(@), wherewhere e E F# withwith ee ee = = -e. -e. (iii) (V, (V, Q)Q) is orthogonal, char(F)char(F) == 2, R =E Z2, Z2, U U is is nonsingular,nonsingular, andand a,at = Q(u)-'Q(u)-l. (3) If (V, f) is is symplectic symplectic or or unitary unitary thenthen each singular point is thethe centercenter ofof a transvection andand G isis transitivetransitive onon thethe rootroot groupsgroups of of transvections. transvections. If If (V, (V, Q)Q) is orthogonal each nonsingular pointpoint isis thethe centercenter ofof aa uniqueunique transvection.transvection. (4) Assume either either (V, (V, f) f) isis symplecticsymplectic and and H H = = G or (V, f) is is unitaryunitary and H == SU(V). SU(V). Then Then one one of of thethe following following holds:holds: (i) R 5< H(').W) (ii) nn == 2 2 and and IFix(9)!(Fix(0)I

Now to the proof of 22.222.2 andand 22.3.22.3. FirstFirst 22.3.122.3.1 followsfollows fromfrom 22.122.1 andand thethe definition of a transvection. InIn particularparticular ifif G possesses a transvection then V possesses isotropic points,points, soso wewe may may assume assume U U == (u) is an isotropic point of V. By 19.12 therethere is is an an isotropic isotropic vector vector x xE E V V - - U1U' withwith x, uu aa hyperbolichyperbolic pair in thethe hyperbolichyperbolic hyperplane hyperplane W W = = (u, x).x). LetLet XX = (u,{u, x} x} CE Y Y bebe aa basisbasis for V with Y -- X X aa basis for W1.WL. Let t == to t, bebe thethe transvectiontransvection in in GL(V)GL(V) withwith Cv(t) == U1 U' andand xtxt = = x x+ + au, au, where where a a is is some some fixed fixed member ofof F#.P. Then,Then, by 19.6, tt E G if and only ifif (vlt, vet)v2t) == (vl, (vl, v2) v2) forfor allall v1,vl, v2 E X, and if (V, Q) is orthogonal alsoalso Q(vt)Q(vt) = Q(v)Q(v) for for allall vv EE X.X. By By constructionconstruction ifif sufficessuffices toto The classical groups 91

check thesethese equalitiesequalities when when vl vl == xx == V2, v2, and,and, if if (V,(V, Q) Q) is is orthogonal, orthogonal, forfor v = x. x. The The check check reduces reduces toto aa verificationverification that: (')(*) a + sae&ae == 0, 0, and and ifif (V,(V, Q) is orthogonal also also a a= = - -Q(u)-1 Q(u)-' # 0, where &E = = -1 ifif (V,(V, f)f) is is symplectic, symplectic, and and sE == +1 +1 otherwise. otherwise.

If (V, f) is is symplectic symplectic then(*)then (*) holds for eacheach a a E F',F#, soso A"Au = {ta: {t,: a EE F#}.F'). Also the map a i--H tot, isis an an isomorphismisomorphism of RR withwith the additive group ofof F. If (V,(V, f) is is unitary unitary thenthen (*)(") hashas aa solutionsolution ifif andand only if there exists ee E F#F# with ee = -e, -e, inin which which case case a a is is a asolution solution to to (*) (*) preciselyprecisely when a = be be withwith b E Fix(@).Fix(B). Observe there is c E FF with c # ceco and,and, setting setting e e= = c c- - ce, co, ee ee = = -e. -e. Finally if (V, Q) is orthogonal thenthen aa isis a solution toto (*)(*) if if andand only only if if a a == Q(u) # 0 and char(F) = 2.2. ObserveObserve in eacheach case case t,: ta: y yH H y + a(y, u)u for each y EE X,X, andand hence also for each y E V.V. So 22.2 and the first two parts of 22.3 are established. The transitivity statement in 22.3.3 followsfollows from Witt's Lemma,Lemma, soso itit remainsremains toto establishestablish 22.3.4. 22.3.4. Assume the hypothesis of of 22.3.4.22.3.4. LetLet LL bebe thethe group generated byby thethe transvec-transvections with centerscenters in in W.W. W'W1 <5 Cv(L),Cv(L), so so L L is is faithful faithful onon W W and and hencehence L <5 O(W, O(W, f). f). Now, Now, by by Exercise Exercise 7.17.1 andand 13.7,13.7, L -E SL2(Fix(0)). SL2(Fix(@)). Then,Then, by by 13.6.4, eithereither RR 5< LML(') <5 HM,H('), or or IFix(0)I /Fix(@)/ <5 3, 3, and and we we maymay assumeassume the latter with n > 2.2. If (V, f) is is unitary unitary letlet vv bebe aa nonsingularnonsingular vectorvector inin W'W1 andand ZZ = (W,(W, v), v), while if (V, f) is is symplectic symplectic let let ZZ bebe a a nondegenerate nondegenerate subspacesubspace containing W of dimension 44 oror 6,6, forfor IIFIFI == 3 or 2, respectively.resp$ctively. Let Let K K == CH(Z1),c~(z'), soSO that K = Sp(Z) Sp(Z) or or SU(Z). SU(Z). If If RR < 5 KM K(') then then the the proof proof is is complete, complete, so so without without lossloss V = Z. Z. But But now now ExerciseExercise 7.1 7.1 completescompletes the the proof.proof.

(22.4) AssumeAssume either either (V, (V, f) f) is is symplectic symplectic and and G G= = O(V, O(V, f) f) == Sp(V),Sp(V), or (V, f)f) is is unitaryunitary of dimension atat least least 2, 2, Fix(@) Fix(B)= = {aae:{aae:a aE E F),F}, andand GG == SL(V) nfl O(V, f)f) = = SU(V). SU(V). Then Then eithereither

(1) GG is is generated generated byby thethe transvectionstransvections inin O(V,O(V, ff), ), oror (2) (V,(V, f)f) is is unitary, unitary, II FIF I == 4, and nn = 3.3.

Proof. IfIf (V,(V, f)f) is is symplectic symplectic let let Fr == V# V' andand 0 S-2 the the set set of of hyperbolic hyperbolic bases bases of V. If (V, f) is is unitaryunitary letlet F = {v E V: (v, v) = 1}.

Let T be the group generated byby thethe transvectionstransvections in in O(V, O(V, f f). ). I'll show:show:

(i) TT is is transitive transitive on on Fr unlessunless 22.4.222.4.2 holds,holds, andand (ii) TT isis transitivetransitive on S-20 ifif (V,(V, f)f) is is symplectic. symplectic. 92 Spaces with formsforms

Observe that the stabilizer in G of w E 7 is trivial, soso (ii)(ii) implies implies T T == G. Observe also that byby ExerciseExercise 7.17.1 thethe lemmalemma holdsholds ifif nn == 2,2, so we may assume n>2.n > 2. Suppose (i)(i) holds.holds. IfIf (V,(V, f)f) is unitary and xx E Fr thenthen GXG, = SU(xL),su(xL), and, and, as n >> 2,2, inductioninduction on n implies G,GX i < TT unless nn == 44 andand IFSI FI = 4,4, wherewhere Exercise 7.37.3 sayssays thethe samesame thing. thing. Hence, Hence, by by (i) (i) and and 5.20, 5.20, G G = = T.T. SoSo taketake (V, f)f) symplecticsymplectic and and let let X X = = (x,:(x1: 1 15 < i i 5< n)n) andand YY == (y,:(yi: 11 5 4.4. Let u, vv be a hyperbolic basis forforyL.~emaytakex yL.We may take= x=yy+u.~henn'= +u. Then xL=(u,y-v)and (u,y -v)andifz=au+y-v, if z=au +y-v, then (y, z)z) nfl yLyL == (au(au -- v). v). It It suffices suffices to to choosechoose z so thatthat zz andand au au -- v v areare nonsingular. Equivalently,Equivalently, if ifT T == TTL~(~) ,,(e) is the trace of FF overover Fix(B),Fix(@), then T(a) # 0 or 1. Hence asas IF1IFS >> 4 andand T: F -*+ Fix(B) Fix(@) isis surjective, surjective, we can choose z as desired.

A field FF is perjfectperfect if if char(F) char(F) == 0 or char(F)char(F) = pp >> 0 0 and and the the p-power p-power mapmap is a surjection from from F F ontoonto FF (i.e. F == F"). FP). For For example example finite finite fieldsfields and and algebraically closed fields are perfect.

(22.5) Let F bebe aa perfect perfect fieldfield ofof characteristiccharacteristic 2 and (V,(V, Q)Q) orthogonalorthogonal of dimension atat leastleast 4.4. LetLet (u) be a nonsingular point of V, t the transvection with center (u),(u), andand GG == O(V,O(V, Q).Q). ThenThen (1) G(u)G(,, = CG(t)CG(t) is is representedrepresented asas Sp(uL/(u))Sp(u-/(u)) on uL/(u), with with (t)(t) the the kernel of this representation. (2) CG(t)CG(t) isis transitivetransitive on the nonsingular points inin uL distinct fromfrom (u). (3) IfIf IIFI FI > 22 thenthen uL = [u [uL, L, CG(t)].Ca(t)]. (4) EitherEither G G isis generated by, and isis transitivetransitive on, on, its its transvections, transvections, or or n n = = 4, IF1IFS == 2, andand sgn(Q)sgn(Q) == +l.+1. The classical groups 93

Proof. LetLet UU == (u) (u) and and H H = = Gu. GU. AsAs t ist isthe the unique unique transvection transvection with with centercenter U, H == CG(t) Cc(t) and and t tis is the the kernel kernel ofof thetherepresentation representation 7tTr of of H H on on u'/ UL/ UU = M.M. Observe that,that, if if ff isis thethe bilinear form onon V defined by by Q,Q, thenthen (M,(M, 7)f) isis aa symplectic space, space, where where f(Z, 1(x, 5,) y) = = ff (x,(x, y).y). Let Let. Z EE M#M# andand W = (x, (x, u). u). As As F isis perfect we can choose u with Q(u) = 1, 1, and and WW contains contains a a unique singular point (w).(w ) . For a EE F#,F#, let let ta t, bebe thethe transvection with with center (aw(aw + + u). u) . ThenThen ta7r tan is a transvectiontransvection onon M M with with center center w W and, and, for for y yE E M, M, ?(tan) y(ta7r) = = 5, y + +a2(y, a2(ji, w)w#)a by 22.3.2. Thus,Thus, as as F F == F2, RnR7t == (tan: (tan: a a E EF#) F#) is isthe the full full root root group group of of tan in in s~(M)Sp(M) by 22.3.2. So, byby 22.4,22.4, H7tH7r == Sp(M).Sp(M). ThereforeTherefore (1) (1) is established. Also [y, t,]ta] = a(y,a(y, w)(aw w)(aw ++ u) u) by by 22.3.2, 22.3.2, so, so, if if IFII FI >> 2, 2, then, then, choosingchoosing y V6 wL,wL, we havehave uu E [y, R], so, as M = [M, [M, H7r], Hn], (3) (3) holds. holds. ToTo prove (2) observe thatthat H7rH7t is transitive on M# by (1),(I), so it suffices to show NH(W) is transitive on the set Fr ofof pointspoints ofof WW distinctdistinct from (u) and and (w).(w ). LetLet w,ii, , v 6 bebe aa hyperbolichyperbolic pair in M. By (I),(1), forfor a a E F# therethere isis g,ga EE HH withwith wg,wga = = awaw and vg,vga == a-1v.a-'v. Hence (ga:(g,: a E F#)F#) isis transitivetransitive on F.r . It remains toto prove (4). Let TT bebe thethe groupgroup generatedgenerated byby thethe transvectionstransvections in G. We've seen thatthat HH <( T, T, so so to to proveprove TT == G G it it suffices suffices by by 5.205.20 toto showshow T is transitivetransitive on the set AA ofof nonsingularnonsingular points of V.V. ThisThis willwiU alsoalso showshow G is transitive on its transvections. Let Z be aa secondsecond nonsingular point. We must show ZZ EE UTU T. T. This This follows follows from from (2) (2) if if (Z (Z + + u)~U)L containscontains a nonsingular point distinct fromfrom UU andand Z,Z, soso assumeassume otherwise. otherwise. Then Then U U + + Z Z == (U + Z)L z)' and U and Z are the only nonsingularnonsingular point?point§ in in U U + + Z.Z. ThisThis forcesforces n n == 4 and IF)JFI = 2. 2. LetLet A = Rad(U Rad(U ++ Z). Z). If If BB is is aa nonsingularnonsingular pointpoint in in A' AL - - (U + Z) then U + B B andand ZZ + + B B are are nondegenerate, nondegenerate, so so U, U, Z Z E E BT BT and and hence hence Z ZE EUT U T. T. Thus no such pointpoint exists,exists, which which forces forces sgn(Q) sgn(Q) = = +l.+1.

If char(F) ;# 2 2and and (V, (V, Q) Q) is isorthogonal, orthogonal, then then a areflection rejection on on V V is is an an element element r inin O(V,O(V, Q)Q) suchsuch thatthat [V,[V, r]r] isis aa pointpoint ofof V.V. [V,[V, r]r] is is calledcalled thethe centercenter ofof r.r.

(22.6) Let char(F) # 2 and (V, Q) orthogonal.orthogonal. Then (1) IfIf rr isis a a reflection reflection onon VV thenthen r ris is an an involution, involution, [V, [V, r] r] is is nonsingular, nonsingular, and Cv(r)Cv(r) = [V, r]-L.r]'. (2) If If UU = = (u) (u) is is a anonsingular nonsingular point point of of V V then then there there exists exists a a unique unique reflec- reflec- tion r,r,, onon VV withwith center center U. U. Indeed Indeed xr, xra = = x x - - (x, u)u/Q(u) forfor eacheach x E V.

Proof. Let r bebe aa reflectionreflection onon V. ThenThen [V,[V, r] r] = (v)(v) is is aa point. point. ByBy 22.1,22.1, Cv(r)Cv (r) = vL, v', soso vL v' isis aa hyperplane hyperplane byby 19.2.19.2. By By 22.2,22.2, rr isis not not aa transvection,transvection, soso v V6 v'vL andand hencehence v v is is nonsingular. nonsingular. Next Next vr yr = = av av for for some some 1 10# aa E F#F# and and Q(v) = Q(vr)Q(vr) == Q(av) Q(av) = = a2Q(v), a2e(v), so so a a = = -1. -1. Hence Hence r ris is an an involution. involution. 94 Spaces with formsforms

Conversely letlet U = (u) (u) be be a a nonsingular nonsingular point.point. ByBy (1)(1) therethere isis atat mostmost oneone reflection withwith center U, whilewhile aa straightforwardstraightforward calculationcalculation showsshows thethe mapmap listed in (2) is such a reflection.

The proof of the following lemma comes essentially from 1.5.1 onon pagepage 19 ofof Chevalley [Ch 2].21.

(22.7) Let (V,(V, Q) Q) bebe anan orthogonalorthogonal space.space. Then either

(1) O(V,O(V, Q) Q) isis generated generated by by itsits transvectionstransvections oror reflections,reflections, or (2) IF1 FI =I = 2, 2, n n = = 4, 4, and and sgn(Q) sgn(Q) = = ++1. 1.

Proof. If n == 2 2or or IFS 1 FI == 2 the2 the result result follows follows from from Exercise Exercise 7.27.2 oror 22.5.4,22.5.4, respectively. If If n n == 1,1, thenthen char(F) ## 2 2and and O(V, O(V, Q) Q) is is generated generated byby thethe reflection -I.-I. ThusThus we may taketaken n > 22 andand IF)IF I >> 2.2. Let T bebe thethe groupgroup generatedgenerated by all transvections oror reflectionsreflections in in G G == O(V, Q) and suppose hh E G - T. T. Pick Pick h h so so that that dim(Cv(h))dim(Cv(h)) isis maximalmaximal in the coset hhT. T. Suppose yY E E VV with with z z= = [y,[y,h] h]nonsingular. nonsingular. yh yh = = y y +z +z and and Q(yh) Q(yh) == Q(y),Q(y), so Q(z)Q(z)+(y,+ (y, z)=O. InInparticulary particular y $ 0 z' z1and andy+z y +z = = yy-(y, - (y, z)z/Q(z)z)z/Q(z) == yrz, yr,, where r,rZ isis the transvection or or reflectionreflection with with center center (2). (z). Thus Thus yh yh == yrz,yr,, h]L so y EE Cv(hrz).Cv(hr,). By 22.1, Cv(h)Cv(h) = [V,[V,h]' CE z1z' == Cv(rz), Cv(r,), soso Cv(h)Cv(h) <5 Cv(hrz).Cv(hr,). Hence dim(Cvdim(Cv(hrr)) (hr,)) > dim((Cv(h),dim((Cv(h), y)) > > dim(Cv(h)), dim(Cv (h)), contrarycontrary toto the choice of h. Therefore [V, h] is totally singular. I claim next that T isis transitivetransitive onon thethe maximal totally singular subspaces of V. Assume not and pickpick two suchsuch spacesspaces M and N such that M 0$ NTNT and, and, subject subject to to thisthis constraint,constraint, with dim(M fln N) maximal. ThenThen MM ## NN soso MM << M M + + N, N, and and hence hence by by maximality maximality of of MM therethere is a nonsingular vectorvector xx == m + n n EE M M + + N. N. As As x x is is nonsingular, nonsingular, (m, (m, n)n) 54 # 0.0. Also Q(x)Q(x) = = (m, n), n), soso mr, mrx = = -n -n EE N, while MM fln NN 5< x1xL == Cv(rx). Cv(r,). ThusThus M n N << (M (M n n N, N, n) n) < 5 Mrx Mr, n nN, N, and and then, then, by by maximalitymaximality of M n N, MrxMr, EE NR. But now M EE NR, contrary to the choice of M and N. So the claim isis established.established. Next there is a maximalmaximal totallytotally singular subspace M with [V,[V,h] h] 5 < M.M. ThenThen M' M1( < [V,[V,hlL h]1 == Cv(h). LetLet H H == Cc(M1)CG(ML) n Cc(V/M1),CG(v/ML), soSO that that hh E E H. H. As As T T is is transitive transitive on on maximal maximal totally totally singularsingular subspaces, itit followsfollows thatthat GG == NG(M)TNG(M)T and and eacheach membermember of GG hashas aa T-cosetT-coset representative in in HgH9 forfor somesome g g EE G.G. ThenThen asas H H I!< NG(M),NG(M), HT HT

V then h == h1h2 hlh2 with with h1hl EE O(U, O(U, Q) Q) and and h2 h2 E E O(U1-, O(uL, Q).Q). By By inductioninduction onon n, h,hi EE T,T, so so h h E E T. T. Hence Hence h h acts acts on on no no such such subspace. subspace. In In particular particular C1, Cv(h) (h) isis totallytotally isotropic,isotropic, so, asas [V,[V, h] is is totallytotally singular,singular, 22.122.1 andand 19.3.219.3.2 implyimply Cv(h)Cv(h) = = [V, [V, h]h] = = M, M, and and V Vis ishyperbolic. hyperbolic. Further, Further, for for V v E E V, V, [v,[v, h]h] E E v1-, vL, sincesince (v,(v, [v,[v, h])h]) isis notnot nondegenerate.nondegenerate. By 19.1419.14 there is a totally singularsingular subspace subspace N N of of V V with with VV == MM ®@ N.N. Let xlx1 E M#.M'. As M == [V, [V, h]h] _ = [N, [N, h],h], ¢:4: n n r+ I-+ [n,[n, h]h] is is a avector vector space space isomorphismisomorphism ofof N and M. Hence there isis y2 E N with xlx1 = [y2, [y2, h]. By thethe lastlast remarkremark ofof thethe previousprevious paragraph, (X1,(xl, y2) y2) = = 0. As ¢@ isis anan isomorphism,isomorphism, (N(N -- x1 x:)@ )¢ ¢ gy2 y$, L, soSO there isis yly1 E N -- xi x: with with ¢(y1) @(yl) = = X2 x2 0 $y2 yj!-. L. LetLet XX == (x1, (xl, x2,x2, Y1, yl, Y2).y2). Then Then VlV1 = = (X)(X) isis nondegeneratenondegenerate andand h-invariant,h-invariant, so Vv = V1.vl. For a EE F#F# define define ah EE GL(V)GL(V) byby v(ah)v(ah) = = v v+ + a[v, a[v, h]h] for for each each v v E E V. V. Notice Q(v(ah)) = Q(v)Q(v) as as [v,[v, h] E E v1 v' fln M. M. Therefore Therefore ah EE H.H. Indeed Indeed settingsetting XaX, = (axi, (axl, x2,xp, a-1y1,a-lyl, y2) wewe havehave Mx(h) MX(h) = = Mx,(ah)MXa(ah) and and J(X, J(X, f) f) =_ J(Xa,J(X,, f). So So the the elementelement g EE GL(V)GL(V) with Xg = Xa X, isis inin GG by by 19.619.6 andand hg = ah. ah. Now, Now, asas IFSI FI > 2, we cancan choosechoose a a with with a a - - 1 # 0.0. Then Then [h, g] =_ h-'hgh-lhg == ((-1)h)(ah) ((-l)h)(ah) = (a= -(a 1)h - l)hE hG, E hG, contrarycontrary to toan an earlier earlier remark. remark. TheThe proof is complete.

Let (V, Q) Q) bebe anan orthogonal space. We next constructconstruct thethe ClifSordClifford algebra algebra CC = C(Q) ofof (V,(V, Q). Q). TheThe treatment treatment here here will will belabbreviated. be4abbreviated. ForFor aa moremore completecomplete discussion seesee chapterchapter 2 2 in in ChevalleyChevalley [Ch[Ch 1].11. CC is is thethe tensortensor algebraalgebra (cf. (cf. LangLang [La], chapter 16, sectionsection 5)5) of of V,V,modulo modulo the the relations relations x x @ ® x x- - Q(x)lQ(x)1 = 0, 0, forfor x EE V.V. ForFor our our purposes purposes it it will will suffice suffice to to knowknow thethe following:following:

(22.8) LetLet (V,(V, Q) Q) bebe anan orthogonalorthogonal spacespace with ordered basis X. Choose X to be orthogonal if char(F) ## 2 2and and choose choose X X to to be be a a hyperbolic hyperbolic basis basis of of thethe un-un- derlying symplectic spacespace (V,(V, f) f) if char(F) = = 2. 2. Let Let CC = = C(Q) C(Q) bebe the the Clifford Clifford algebra of (V, Q). Q). Then Then CC is is an an F-algebraF-algebra with with the the following following properties: properties:

(1) ThereThere is anan injective F-linearF-linear mapmap p:p: VV +-- C C such such that that CC is is generatedgenerated asas an F-algebra by Vp.Vp. Write ex for xp if x EE X.X. (2) For S == {x1, {xl, ...... ,, x,"} x,} EC X with xlx1 < . -. < x, write es == ex, ex, ... . . exm. . ex,,,. Then (es: S S EC X) is a basis forfor CC overover F.F. In particularparticular e, eH = = 11 = = 1, and dimF(C) = 2". 2". (3) ForFor u, v E V,V, (up)2 == Q(u)Q(u) 1 and upvpupvp ++ vpup = (u,(u, v)v) . 1. (4) LetLet ClCi bebe thethe subspacesubspace of C spannedspanned by the vectors es, S CE X,X, ISI IS1 = imod 2, i == 0, 0,l. 1. Then Then {C0, {Co, C1C1} } isis aa gradinggrading of C. That is C == Co Co ® @ C1 C1 and C;CjCiCj C Ci+j,Ci+j, forfor i, i, j j EE {0, {0, 11, 11, wherewhere i ++ j j is is read read mod mod 2. 2. 96 Spaces with formsforms

(5) If uu is is a a nonsingular nonsingular vector of V thenthen up isis aa unitunit inin CC with with inverseinverse Q(u)-lup,Q(u)-'up, and,and, forfor vv cE V, V, (vp)uP == -(vru)p, -(vru)p, where where ru r, is is the the reflection reflection or transvection inin O(V, Q) with center (u). (6) The The CliffordClifford group G of C isis thethe subgroupsubgroup ofof unitsunits inin CC whichwhich permutepermute Vp via conjugation. The representation 7rn of G on VV (subject(subject to to thethe iden-iden- tification ofof V with Vp via p) isis aa representationrepresentation of G onon (V,(V, Q)Q) withwith GnGir =O(V,=O(V, Q) if n isis eveneven and and GirGn = SO(V, Q)Q) ifif nn is odd. IfIf u is a a nonsingular vectorvector in in V V then then upn up7r = = -r,. -ru. (7) ker(,-r)ker(n) is thethe setset ofof units units in in Z(C) Z(C) andand Z(C)Z(C) = F F1 1 oror FF 1 1 + Fex for for n n eveneven or odd, respectively. IfIf n n is is odd, odd, no no unit unit in in C C induces induces -I -I on C by conjugation. (8) There is an involutory algebraalgebra antiisomorphismantiisomorphism t tof of C C such such that that est est = ex,,e,,, ...... ex, for each S = {x1, {XI, ...,. . . xm, x,} } CG X.X.

Proof. I'llI'll sketch sketch a aproof. proof. If char(F)char(F) # 2 the multilinearmultilinear algebracanalgebra can be avoidedavoided as in chapter 5, section 4 of ArtinArtin [Ar].[Ar]. A fullfull treatmenttreatment can be found in Chevalley [Ch 1].11. By definition, C C == T/K wherewhere TT == ®°°o $Eo Ti(V) z(V) is is the the tensor tensor algebra algebra and and K == (x (x ®x @ x - -Q(u)1: Q(u)l: X x E E V). V). In In particular particular To(V) To(V) == F1 F 1and and there there is is aa naturalnatural isomorphism po: po: V V+ -+ Tl(V). T1(V). Then Then p: p:V V+ -+ C C is is the the map map v vH H vpo ++ K induced byby po, and (1) will follow follow from (2),(2), once thatthat part part is is established.established. es = ex,ex, ...... ex,,,ex,n = = xlx1 @ ®. . .-®xm8 x, ++ K, K, so so Ci Ci is is the the image image ofof ®j_iej-, Tj(V)Tj(V) in C. HenceHence (4)(4) followsfollows from (2)(2) andand thethe definitiondefinition of multiplication inin T. The universal property of TT impliesimplies there there is is an an involutory involutory antiisomorphism antiisomorphism to to of T with (xl(xl 8® . . . . 8(9 xm)toxm)to= = x,xm 8 ® . -.® @3 x1. xl. As As to to preserves preserves KK itit induces induces t onon C.C. ThusThus (8)(8) holds.holds. PartPart (3)(3) isis a a direct direct consequence consequence ofof thethe definitiondefinition ofof C, since xx 8® xx - Q(u)1 Q(u)l E E K. K. An An easy easy induction induction argument argument using using (3)(3) showsshows eseTese~ is a linear combination ofof thethe elementselements e~, eR,R R C X,X, forfor eacheach S,S, TT C X,X, so (eR: R R C X) spansspans C.C. UsingUsing thethe universaluniversal property of the tensor algebra, Chevalley shows onon pagepage 39 of [Ch 1]11 that there is a homomorphic image of C of dimension 2", completing thethe proof ofof (2),(2), andand hencehence alsoalso ofof (1)(1) andand (4).(4). I omit this demonstration. Part (5) is a straightforward consequence consequence of of (3). (3). If If char(F) char(F) #0 22 thenthen (3)(3) shows e: e' == (-1)mes (-l),es forfor x x E E X X and and S S C G X, X, where where m m == BSI IS] ifif x $ SandS and mM = ISIS1 I --1 1 if if xx EE S.S. Since Since Xp Xp generates generates C C as as an an F-algebra,F-algebra, (7)(7) followsfollows in this case. If char(F) == 2 2 then then choose choose X X so so that that each each of of itsits membersmembers isis nonsingular. Then (3)(3) showsshows [e,, [ex, es] es] = = 00 ifif S C_E xL, x1, whilewhile [ex, [ex, es] es] = = Q(x)-'es+,Q(x)-les+y if S contains thethe uniqueunique y y in in X X - - x1,xL, where where SS + y y isis thethe symmetricsymmetric difference ofof S with {y}.It It follows follows that that [ex, [ex, C] C] = = (es:(es: S S Gc x1)xL) is is ofof dimensiondimension 2a-1.2'+'. So, as 2"-12n-1 = dim(C)/2dim(C)/2 and and exex is is anan involution, [ex, C]C] = Cc(ex).Cc(ex). ThusThus (7)(7) holds in this case too. Let G and n7r bebe asas inin (6).(6). For For g g EE GG and vV EE V,V, Q(vgn)lQ(vgir)1 == ((vp)g)2((vp)s)' == ((vp)')g((vp)2)g = = Q(v)l,Q(v)1, soso Q(vgn)Q(vgir) = Q(v). Q(v). Hence Hence GirGn <5 O(V, O(V, Q). Q). Let Let Go Go bebe thethe The classical groups 97 subgroupsubgroup of of GG generated generated by by thethe elementselements up up as as u u varies varies over over the the nonsingular nonsingular vectorsvectors of V. By (5),(5), upnup,-r = = -r,,-ru, so,so, by 22.7, GnG,-r = = Gon = O(V, O(V, Q)Q) ifif char(F)char(F) == 2. 2. Further Further if if char(F)char(F) # 22 andand n isis eveneven then -I-1 isis a a product product ofof elementselements -ru,-r,, sosor, ru cE Go7r Gon and, and, again by 22.7,22.7, Gn G7r = = GonGoir = O(V,O(V, Q).Q). FinallyFinally ifif nn isis odd then det(-r,)det(-ru) == +1, +1, so so Goir Gon < 5 SO(V, SO(V, Q), Q), and and then,then, asas O(V, Q) = (-I)(-I) x SO(V,x SO(V, Q), Q), 22.7 22.7 says says G07r Gon = = SO(V, SO(V, Q). Q). Then Then to to complete complete the the proof proof ofof (6)(6) itit remainsremains only to observe that,that, byby (7),(7), -I-1 $ G7r,Gn, so GonG0ir = G7r.Gn.

(22.9)(22.9) LetLet (V,(V, Q)Q) bebe anan orthogonalorthogonal space, C = C(Q) C(Q) its its Clifford Clifford algebra,algebra, GG thethe CliffordClifford groupgroup ofof (V,(V, Q),Q), and and G+G+ = = G G fl n Co Co the the special special Clifford Clifford group. group. Let Let 7rn be the representation of of 22.8.6.22.8.6. ThenThen (1)(1) G+Gf isis a a subgroup subgroup of of GG of of indexindex 2. 2. (2)(2) IfIf char(F)char(F) # 22 thenthen G+G+n 7r = SO(V,SO(V, Q).Q). (3)(3) IfIf char(F)char(F) = = 2 2then then G+ G+n 7r is is of of index index 22 inin O(V,O(V, Q).Q). (4)(4) G+nGf n contains contains no no transvections transvections or or reflections. reflections.

Proof.Proof. Part Part (1) (1) is is a aconsequence consequence of of 22.8.4. 22.8.4. If If nn is is even even then,then, by 22.8.7, ker(7r)ker(n) <5 G+G+ and,and, byby 22.8.6,22.8.6, G,-rGn = O(V, O(V, Q), Q), so so G+ G+n 7r isis ofof indexindex 22 in O(V,O(V, Q).Q). AlsoAlso -1-I E EG+n Gf nby by the the proof proof of of 22.8, 22.8, so so by by 22.8.6 22.8.6 each each transvection transvection or or reflection reflection rur, isis notnot inin G+,-r. G+n. Thus Thus thethe lemmalemma holds holds inin thisthis case case as as reflections reflections areare not not inin SO(V,SO(V, Q).Q). IfIf nn isis oddodd thenthen GirGn == SO(V, SO(V, Q) Q) by by 22.8.6, 22.8.6, while, while, by by 22.8.7,22.8.7, G == G+ker(,-r). G+ker(n). So So again again thethe lemmalemma holds.holds. 9

(22.10)(22.10) LetLet v1, vl, .... . ,. ,vm urn be be nonsingularnonsingular vectorsvectors inin thethe orthogonalorthogonal space space (V,(V, Q)Q) suchsuch that r,,,r,, ...... r,,, r,,,, == 1. 1. Then Then the the product product Q(vl)Q(vl) ... . . Q(vm). Q(vrn) isis aa square square in F.

Proof.Proof. Let Let c c = = vl vl p p... . vm . . vmp. p. AsAs r,,, r,, .... .r,,,, . r, = =1, 1,m mis iseven, even, because because by by 22.9 22.9 there there isis aa subgroupsubgroup of of O(V,O(V, Q) Q) of of index index 2 2 containing containing no no transvection transvection or or reflection. reflection.

HenceHence c7r cn = (-1)m (- l)"r,, r,,, ...... r,,, r,,,=1= 1by by 22.8.6. 22.8.6. So So C c E E ker(7r ker(n). ). Indeed, as as mm isis even, c EE G+G+ fln ker(7r) ker(n) = F#F# . 1 by 22.8.7. So cc = a . 11 for some aa cE F#.F#. LetLet tt bebe thethe antiisomorphismantiisomorphism of 22.8.8. ItIt followsfollows that that c(ct) c(ct) == c2 = a2 a2 . 1. On the other hand c(ct)c(ct) == vi vlp p ... . .vm . v,pvrnp pvm p ...... v1 vlp p asas t isis anan antiisomorphism.antiisomorphism. FurtherFurther (vi (vip)' p)2 = Q Q(vi)l, (vi) 1, soso c(ct)c(ct) _= (Q(v1)... (Q(v1) . Q. . (vm))Q(vrn))l, 1, completingcompleting the the proof. proof.

LetLet F2F2 = = {a2: {a2: aa E EF#} F#} be be the the subgroup subgroup of of squares squares in in F#, F#, and and consider consider the the factor group F#/F2. ForFor example example ifif FF is is a a finitefinite field of oddodd orderorder thenthen F#/F2F#/F2 isis ofof orderorder 2,2, whilewhile ifif FF is is perfect perfect ofof characteristiccharacteristic 22 thenthen F#F# == F2. F2. Define Define a mapmap 9:8: O(V, Q) +-+ F#/F2F'#/F2 asas follows. follows. For For gg E E O(V,O(V, Q),Q), gg == rx, r,, .. . . r,,,,, r,,, for suitablesuitable transvections or reflections r,,rX, withwith centercenter (xi).(xi). (Except in the ex-exceptional case of 22.7, which I'llI'll ignore.) Define B(g)9(g) == Q(xl) ...... Q(xm)F2. Q(x,)F'. Observe first that Q(axi)~(axi) =a=a2Q(xi) 2 Q(xi) E Q(x~)F',Q(xi )F2, SOso the definition of 89 isis in-independent of the choice of generator xi of (xi).(xi). Also Also ifif gg = = ry, r,, .. .. . r,, ryk then 98 Spaces withwith forms

1 = r,,, r,, .... . r,,,,,. r,,,,r, rYk ...... ry,,so,by22.10,ry, , so, by 22. 10,Q(xl). Q (xi) .... . Q(xm>F2Q (xm) F2= = Q(y1).Q (yi) ... . .Q e(yk)F2. (Yk)F2 Thus 0 is independentindependent ofof thethe choicechoice ofof transvectionstransvections andand reflectionsreflections too.too. 00 isis called the spinor norm of O(V, Q). From the preceding discussion it is evident that.

(22.11) TheThe spinorspinor norm 0 is a group homomorphismhomomorphism ofof O(V, O(V, Q)Q) intointo F#/F2.F#/F2.

(22.12) If If QQ isis not not definitedefinite thenthen thethe spinorspinor norm maps G+n surjectively surjectively onto onto F#/F2.

Proof. VV containscontains aa hyperbolichyperbolic plane U andand for each a EE F# there exist u, v cE UU withwith Q(v)Q(v) = 1 1 and and Q(v)Q(v) == a. a. Now Now O(r,ru) = a.

(22.13) Let (V, Q) be a hyperbolic orthogonalorthogonal space, space, let let r r == F(V)r(V) be be thethe setset of maximal totallytotally singular singular subspaces subspaces of of V, V, and and define define a relationa relation -- -on on r r by A --- B B if ifdim(A/(A dim(A/(A fl n B)) B)) is is even. even. Then Then - --is isan an equivalence equivalence relation relation withwith exactly two equivalence classes.

Proof. GivenGiven aa triple A, B, C ofof members of r definedefine 6(A,S(A, B,B, C)C) == dim(A/(A nfl B))B)) ++ dim(A/(A nfl C))C)) + + dim(B/(Bdim(B/(B nfl C)).c)). Observe that the lemma is equivalent to the assertion that S(A,6(A, B, C) isis eveneven for each triple A, B, C from F.r. Assume the lemma is false and pick aa counterexamplecounterexample V with nn minimal.minimal. As V isis hyperbolic,hyperbolic, nn = 2m2m is even. If m = 1 1 then then IF)r I = 2,2, soso the result holds. Thus n > 1.1. LetLet A,A, B,B, CC EE Fr withwith S(A,6(A, B, C)C) odd.odd. Let let^ D ==An~nCandsuppose~ A fl B fl C and suppose D #0.~etU0. Let U == D1~'andu and U == U/ U/D.Aswe D. As we have seen several timestimes already,already, QQ induces a quadratic formform Q on U.0. Further 0U is hyperbolic withwith A,A, B,B, CC E I'(U),r(U), so, so, byby minimalityminimality of V,V, 8(A,6(A, B, C) isis even. As 6(A,S(A, B,B, C)C) = S(A, 6(A, B,B, C)C) wewe havehave aa contradictioncontradiction to the choice of A, B,B, C.HenceC. Hence D D=0. = 0. Suppose next next E E == AA fln BB # 0 and letlet CoCo == (C fln E1)E') + E.E. By By 19.2,19.2, Co E r.F. 0 # E == A A fl n B B fl n Co, Co, so, so, by by a aprevious previous case, case, S(A, 6(A, B,B, Co) Co) is is even.even. But XX nfl CoCo = = (X nfl C)C) ++ EE and XX nfl C C n fl E E == 0 for X = AA andand B,B, so so S(A,6(A, B, C) - S(A,6(A, B, B, Co) Co) mod mod 2,2, againagain a a contradiction.contradiction. Thus 6(A,S(A, B,B, C)C) == 3m.3m. Hence mm isis odd.odd. ThereforeTherefore if if T, T, S S E E r F withwith TT --ti A -- S then A nfl T # 0, so, byby thethe last last case, case, T T -- - S.S. Hence --- is is an an equivalenceequivalence relation. FinallyFinally letlet RR E Fr withwith AA fln RR aa hyperplane hyperplane ofof A. A. ThenThen 00 # A nfl R, so, by the last case, 8(A,6(A, B, R) isis even,even, andand hencehence BB -- R. This showsshows -- - hashas two classes and completes the proof. The classical groups 99

(22.14) Let (V, Q) be a hyperbolic orthogonalorthogonal space,space, let let G G == O(V,O(V, Q),Q), andand let H bebe thethe subgroupsubgroup of of GG preservingpreserving thethe equivalence relation of 22.13. Then (1) IG:IG: HI = 2. 2. (2) HH is is the the image image of of the the special special Clifford Clifford group group under under thethe mapmap ofof 22.8.6.22.8.6. (3) ReflectionsReflections and transvectionstransvections are are in in G G - - H.

Proof. ByBy Witt's Witt's Lemma,Lemma, G G is is transitivetransitive on the setset rr of maximal totally singularsingular subspaces of V, so IG: HI is the number of equivalence classesclasses of of r.r. That is IG:I G: HI H I= = 22 byby 22.13. It'sIt's easy to check thatthat (3)(3) holds. Then, by (1),(I), (3),(3), andand 22.7, H isis thethe subgroupsubgroup of of GG consistingconsisting of of the elements which are the product of an even number of transvections or reflections, while by 22.9 this subgroup is the image of the special Clifford group under the map of 22.8.6.

I close this section with a brief discussion of some geometries associated to the classical groups. A few properties of these geometries are derived in Exercise 7.8, while chapter 1414 investigates these geometries in great detail. Assume the Witt index mm ofof (V,(V, f)f) oror (V,(V, Q) Q) is is positive. positive. InIn thisthis casecase therethere are some interesting geometries associated to the space and preserved by itsits isometry group. The reader may wish toto referrefer toto thethe discussiondiscussion inin sectionsection 33 onon geometries. The polarpolar geometrygeometry r r ofof (V, f) f) oror (V,(V, Q) is the geometry over over I I = (0,10, 1,1, ...... , , m m -- 1 1)} whose objects of type i arearq the totally singular subspaces of V of projective dimension i, withwith incidenceincidence defineddefined byby inclusion.inclusion. EvidentlyEvidently O(V, f) is is represented represented as as a a group group of of automorphisms of F.r. Indeed the similarity group A(V,O(V, f) is is also also soso represented. represented. If (V, Q)Q) isis aa hyperbolichyperbolic orthogonal orthogonal space space there isis anotheranother geometry associated to (V, Q)Q) which which isis inin many ways nicer than the polar geometry. Assume the dimension ofof VV isis atat least 6,6, so that the Witt index m of (V, Q)Q) is is atat leastleast 3.3. The orijlammeoriflamme geometry geometry r rof of (V, (V, Q) Q) is is the the geometry geometry over over I I= = (0,10, 1, 1, ...... ,, m m - - l}1) whose objects of typetype ii < m m -- 2 2 are are the the totally totally singular singular subspaces subspaces of of projective dimension i, and whosewhose objectsobjects of of types types m m - - 1 and mm -- 2 2 areare thethe two equivalence classes of maximal totally singular subspaces of (V, Q)Q) defineddefined byby thethe equivalence relation of 22.13. Incidence is inclusion, except between objects U and W of type mm -- 1 1 andand mm -- 2, 2, which which are are incident incident if if UU flfl W W isis aa hyper- hyper- plane of U and W.W. In this case the subgroup of A(V,O(V, Q) of index 2 preserving the equivalence relation of 22.13 is represented as a group of automorphisms of F.r.

Remarks. TheThe standard standard reference reference forfor muchmuch ofof thethe materialmaterial inin thisthis chapterchapter isis DieudonnCDieudonne [Di].[Di]. In particular this is aa goodgood placeplace toto findfind outout whowho firstfirst proved what in the subject. 100 Spaces withwith forms

We will encounterencounter groups generated by reflectionsreflections again in sectionssections 29 29 and 30.30. Observe that, by 13.8,13.8, 22.4.4, 22.4, and and ExerciseExercise 7.6, almostalmost allall thethe finitefinite classical groups SL,(q),SLn(q), Sp,(q),Spn(q), Qi(q),Q '(q), and SU,(q)SUn(q) are perfect. This fact is used to prove in 43.12 that the projective groups PSL,(q),PSL,,(q), PSp,(q),PSpn(q), PQ:(q),PQ (q), and PSU,(q)PSUn(q) are simple unless n and q areare small.small. Since by some measure most finite simple groups are classical, the study of the classical groups is certainly important. MoreoverMoreover alongalong withwith LieLie theorytheory (cf. chapter 14 and section 47) the representations of the classical groups on theirtheir associated spacesspaces is is thethe bestbest tooltool forfor studyingstudying thethe classical groups.groups. On On thethe other hand the study of the classical groupsgroups isis aa specialspecial topic andand thethe materialmaterial inin thisthis chapter isis technical. Thus the casual reader may wish to skip, or at least postpone, thisthis chapter.chapter.

Exercises for chapter 77 1. LetLet VV be be anan n-dimensionaln-dimensional vector vector spacespace overover F.F. Prove: Prove: (1) IfIf n == 2 2 then then SL(V) = Sp(V).Sp(V). (2) If 98 isis anan automorphismautomorphism of of F F of of order order 2, 2, n n= = 2, and (V,(V, f)f) isis aa hyperbolic unitary space,space, then then SL(V) SL(V) n n O(V,O(V, f) f) =E SL2(Fix(9)). SLz(Fix(8)). (3) LetLet IFI FI I = q2q2 < < oo oo andand (V,(V, f) f) a a3-dimensional 3-dimensional unitary unitary spacespace over F. Then there exists a basis X of V such that

0 0

J(X, f) _ 0 1 0

1 0 '01

Moreover if P consists of those g E SU(V) with

1 a c Mx(g) = 0 1 b

0 0 1

thenthen PI PI 1 = q3 q3 and and [P,[P, P]PI is is a aroot root group group of of SU(V).SU(V). (4)(4) LetLet IF!IF] == q q< oo< ooand and (V, (V, f) f)a 2m-dimensional a 2m-dimensional symplectic symplectic space space over F with m > 1.1. ThenThen there existsexists aa basis basis X X = = (xi:(xi: 11 F:< i <5 2m)2m) such that

J(X, f) _-IM 0 Im0/ where InI, isis thethe mm byby mm identity identity matrix.matrix. MoreoverMoreover if PP consistsconsists ofof The classical groups 101

those g cE Sp(V)Sp(V) with

1 0 0 Mx(g) = a12._2 0 b c I where a andand cc areare columncolumn andand rowrow vectors,vectors, respectively,respectively, thenthen II P II == q2'-1.qZm-'. If q is odd, [P, P]PI isis aa root group of Sp(V). IfIf qq = 2 2 anand d mm == 33 then P containscontains aa transvectiontransvection and and P P= = [P,[P, HIH] where HH = Sp(U)Sp(U) and U == (XI, (xr, x2m)1~2,)'. 2. LetLet (V,(V, Q)Q) bebe aa 2-dimensional2-dimensional orthogonal space over F.F. Prove: (1) O(V,O(V, Q) Q) is is the the semidirectsemidirect product of aa subgroupsubgroup HH byby (r)(r) E= Z2, Z2, where r invertsinverts H. (2) EachEach elementelement of O(V, Q) - H H is is a atransvection transvection or or reflection. reflection. InIn particular O(V, Q)Q) isis generatedgenerated byby suchsuch elements.elements. (3) If If O(V,O(V, Q)Q) is is hyperbolic hyperbolic thenthen HH is is isomorphic isomorphic to to the the multiplicative multiplicative group F# of F. (4) IfIf (V,(V, Q) Q) is is definite definite then then therethere existsexists a quadratic Galois extension K of F suchsuch that (V,(V, Q)Q) is is similar similar to to (K, (K, N;) NF) and and H H E - {a{a EE K:K: aae aae = = 11,I), where (0)(0) == Gal(K/F).Gal(K/F). 3. LetLet (V,(V, f)f) be be a a4-dimensional 4-dimensional unitary unitary spacespace overover a field F ofof orderorder 4, XX an orthonormal basis basis for for V, V, A A = _ {(x):{(x):xx cE X},XI, and G == SU(V). SU(V). Prove Prove N~(A)~NG(A)° -E S4, S4, Go GA = EE27, EZ7, andand NG(A) NG(A) s is generated generated by by transvections.transvections. Let D EE A, TT the subgroup of G generated bybi the transvections in G, and FI? the set ofof conjugatesconjugates of of A A under under NG(D). NG(D). Prove Prove N~(D)~ NG(D)r2 - A4 andand )Gr I Gr1 = 54.54. Prove Nc(D)NG(D) 5< T. 4. LetLet q q be be a a prime prime power. power. ProveProve (1) Z(GU,,(q))Z(GU,(q)) andand GU,(q)/SU,(q) are cyclic of orderorder qq + 1.1. (2) Z(SU,(q))Z(SU,(q)) andand PGU,(q)/ PGU,(q)/ U,,(q)U,(q) areare cyclic cyclic of order (q + 1,1, n). 5. Assume Assume thethe hypothesishypothesis and notation of ExerciseExercise 4.74.7 withwith char(F) char(F) # 2. Let W = V3, V3, a == 7r3, n3, and define Q: W -->+ F by

e(ax2Q(ax2 ++ bxy + cy2) == b2 -- 4ac. 4ac. Prove (1) QQ is is a a nondegenerate nondegenerate quadratic quadratic form form on on WW with with bilinear bilinear form form (ax (ax2 2 + bxy + cy2,rx2 rx2 + + sxysxy + + ty2)ty2) == 2bs2bs - - 4(at ++ cr). (2) For each g g E E G, G, ga ga is isa similaritya similarity of (W,of (W, Q) Q)with with h(ga) A(ga) = det(g)2. = det(g)2. (3) (Ga)S(Ga)S is thethe group group A(W, A(W, Q) Q) of of all all similarities similarities of (W,of (W, Q), Q), where where S is S is the group of scalar maps on W.W. (4) UpUp toto similarity,similarity, (W,(W, Q)Q) isis thethe unique unique 3-dimensional3-dimensional nondefinitenondefinite or-orthogonal space over F. 102 Spaces with formsforms

(5) IfIf FF is is finite finite or or algebraically algebraically closed closed every every 3-dimensional 3-dimensional orthogonal orthogonal space over FF is similar to (W, Q). (6) (rrh:(rrh: rr EE R,R, h h EE Get) Ga) = = O(W, A(W, Q)(l) Q)") =S L2(F), L2(F), where where RR isis the the setset of reflections in O(W, Q). 6. LetLet (V, (V, Q) Q) bebe an an n-dimensionaln-dimensional orthogonal orthogonal space over aa fieldfield F F withwith nn 2> 3. (1) AssumeAssume (V,(V, Q)Q) isis not definite andand ifif I IFFI I5 < 33 andand n n 5< 44 assumeassume nn == 4 and sgn(Q) _= -1. - 1.Prove Prove the the following following subgroups subgroups are are equal.equal. (i)(9 n(V,Q(V, Q).Q). (ii) TheThe kernelkernel inin G+G+n 7r of the spinor norm, where G+ is thethe specialspecial Clifford group. (iii) (rrg:(rrg: r r reflection reflection or or transvection,transvection, g EE O(V,O(V, Q)),Q)). (2) If char(F) 0# 2 2 prove prove Q(V,Q( V, Q)Q) isis perfectperfect unless I IFIF I = 33 andand eithereither nn=3orn = 3 orn =4andsgn(Q) = 4andsgn(Q) _+1.IfFisfinite = +l.If F is finiteproveO(V, prove O(V, Q)/Q(V,Q)/ Q(V, Q)Q) S E4, andand -I-I EE Q(V, Q(V, Q) Q) if if and and only only ifif n is eveneven and and sgn(Q) sgn(Q) r - IFI"/z1 F In/' mod 4. (3) IfIf char(F) == 2 2 and and F F is is perfect, perfect, prove prove either either Q(V,Q(V, Q)Q) isis perfectperfect and IO(V, Q): Q(V,SZ(V, Q)i Q)j = = 2 2 or or IF1 IFI = = 2, 2, n n = = 4,4, andand sgn(Q) sgn(Q) == +l.+1. (Hint: To proveprove Q(V,Q(V, Q) perfect useuse (1) and showshow rrgrrg is contained in a perfect subgroup of O(V, Q) for each reflection or transvectiontransvection r and each g EE G.G. TowardToward that end use Exercise 7.5 in (2) and 22.5 in (3).) 7. LetLet G G bebe a a permutation permutation group onon a set I ofof finitefinite order order n andand VV thethe permutation modulemodule forfor GG overover FF with G-invariant basisbasis X X == (x1:(xi: i i EE I).I). Define aa bilinearbilinear formform ff onon VV byby ff (x1, (xi, xi)xj) == Sjj Sij (the (the Kronecker Kronecker delta)delta) for i,i, jj eE I.I. Let Let z z= =F_ZEI CiEIxi, U thethe corecore ofof thethe permutationpermutation modulemodule V, andand iiV = V/(z).V/(z). If If char(F)char(F) = = 2 2define define a a quadratic quadratic form Q on U by Q(1Q(F_aixi) a;x1) = _ 1 E a:a? + Ei

group of the spacespace or (V, (V, Q)Q) aa hyperbolichyperbolic orthogonalorthogonal space with m >2 3, 1'r the oriflamme geometry ofof (V, Q), and G thethe subgroupsubgroup ofof O(V,O(V, Q)Q) preserving the the equivalence relation ofof 22.13. Let Z be a maximalmaximal hyperbolichyperbolic subspace of of V,V,X X = = (xi: 11 5< ii <5 2m) 2m) a ahyperbolic hyperbolic basisbasis for Z, Vi == (x2j_1:(xzj-l: 1 1 5 < j j 5< i), and Y == {(x):x E X}. LetLet T T == {K:{Vi: 1 1 5 < i i5 < m) if r is the polar geometry and T == {Vi, {Vi, Vh-l: 1 5< i <5 m, ii ## m - 1)1) if r isis the oriflammeoriflamme geometry, geometry, where where V, VhW1 _1 = = (Vm_2, (V,-2, X2m_3,X~rn-3, x2m). ~2,). Prove (1) IfIf mm == 1 1then then G G is is 2-transitive 2-transitive onon thethe points ofof r,I', while ifif mm > 11 then G isis rankrank 3 on these points. (2) GG isis flagflag transitivetransitive on r.r. (3) TT isis aa flagflag of r ofof typetype I. (4) BB == GT GT is is the the semidirect semidirect productproduct of U with HH = Gy, Gy, where where UU isis the subgroupsubgroup of G centralizing V1,Vl , Vi+1Vi+1/ Vi/ Vi,, 1 51 (or(or O(Z1,O(Z'? Q)).Q)). (6) LetLet ii EE I.I. Then Then eithereither UU fixesfixes a unique objectobject of of type type i iin in r r oror rr isis a polar geometry and V is a hyperbolic orthogonal space. (7) BB == NG(U). NG(U). (8) AssumeAssume F isis finitefinite ofof characteristiccharacteristi9 pp and r isis oriflamme oriflamme ifif VV isis hyperbolic orthogonal. Then UU EE Sy1p(G).Syl,(G). (9) NGNG(y)' (Y) Y is is Z2wrS, 7L2wrSm or or of ofindex index 2 2in in that that group, group, for for r r a polarpolar space oror oriflamme geometry, respectively. (10) LetLet SS bebe aa flagflag ofof corankcorank 11 in T. ThenThen either the residue rsI's of SS isis isomorphic to to the the projective projective line line over over F Fand and (Gs)~. (Gs)r,S - PGL2(F)PGLz(F) or L2(F), or r isis a a polar polar geometry,geometry, S is of type {O,{0, ...... , m m -- 2},21, I'srs isis isomorphic to the set of singular pointspoints ofof W == (v,-~)'/(V,,-,)-L/ V,-l,V,,-,, and either (Gs)1'S (GS)~~S PO(W, f) (or (or PO(W, PO(W, Q)) Q)) or or V V is is hyperbolichyperbolic orthogonal and I1 rsI's 1I == 2. 9. LetLet VV be be aa finite dimensional vector spacespace overover a a field field F F and and f f a nontrivial sesquilinear form on V.V. Then (1) IfIf char(F)char(F) # # 2 2 and and f f is is bilinear bilinear then then f f = = g g+ + h hwhere where g g and and h h are are symsym- metric and skew symmetric formsforms onon V,V, respectively,respectively, and and O(V, O(V, f) f) 5< OW,O(V, g) n O(V,OW, h). (2) IfIf char(F)char(F) == 2 2and and f fis isbilinear bilinear there there exists exists a anontrivial nontrivial symmetricsymmetric bilinear formform gg onon VV withwith O(V, O(V, f) f)

(4) AssumeAssume ff is is sesquilinear sesquilinear with with respect respect to to the the involutioninvolution 08 and f isis skew skew hermitian; that that is is f f(x, (x, y) y) = = --ff (y,(y, x)°x)' for all x,x, yy E V. Prove f f is similar to a hermitian form. (5) IfIff f isis sesquilinear sesquilinear with with respect respect toto anan involutioninvolution 0,8, then there exists aa nontrivial hermitianhermitian symmetricsymmetric form form g g on on V V with with O(V, O(V, f) f) 5< O(V, g). 10. Let F bebe aa fieldfield andand f a asesquilinear sesquilinear form form onon VV withwith respectrespect toto thethe auto-automorphism 8 0 of of F,F, suchsuch thatthat for for all all x, x, y y Ee V, ff (x, y)y) = 00 ifif andand onlyonly if f (y,(y, x) == 0. 0. Prove Prove thatthat eithereither (1) ff (x, (x, x) x) == 0 0 for for all all x x EE V,V, 08 == 1, 1, and and ff is is skew skew symmetric, symmetric, or (2) therethere existsexists x EE VV withwith ff (x, (x, x) x) 0 # 0 0and and one one of of the the following following holds:holds: (a) 08 == 1 1 and and ff is is symmetric. symmetric. (b) 191181 == 2 and f isis similar similar to to a a hermitian hermitian symmetricsymmetric form. form. (c) X0118 1 > 2 and Rad(V) is of codimension 11 inin V.V. p-groups

Chapter 8 investigates p-groups from from two two points points ofof view:view: first through a study of p-groups whichwhich areare extremal extremal with with respect toto one of several parameters (usually connected with p-rank) andand secondsecond through a study of the automorphismautomorphism group of the p-group. Recall that if p isis aa primeprime thenthen the p-rankp-rank ofof aa finite finite groupgroup isis thethe maximummaximum dimension of an elementary abelian p-subgroup, regarded as a vector space over GF(p). Section 23 determines p-groupsp-groups ofof p-rank 1, p-groups in which each normal abelian subgroup is cyclic, and, forfor pp odd, p-groups in which each normal abelian subgroup is of p-rank at most 2. Perhaps most important, the p-groups ofof symplecticsymplectic type are determined (a p-group isis ofof symplecticsymplectic type if each of its characteristic abelian subgroups is cyclic). The Frattini subgroup is introduced to study p-groups and their automor-automorphisms. Most attention is focusedfocused onon p'-groups p'-groups of automorphismsautomorphisms ofof p-groups;p-groups; a variety of results on the action of p'-groups onon p-groupsp-groups appearappear inin sectionsection 24. One very useful result is the ThompsonThompson A,x B Lemma. Also ofof importance is the concept of a critical subgroup.subgroup.

23 ExtremalExtremal p-groups In this section p is a prime and G is a p-group. The Frattini subgroup of a group H isis defineddefined to be the intersection of all maximal subgroups of H. (D(H)@(H) denotes the FrattiniFrattini subgroupsubgroup ofof H.H.

(23.1) (1)(1) '1(H)@(H) charchar H. (2) If X C HH withwith HH == (X, (X, (D(H)), @(H)), then H = (X).(X). (3) If H/@(H)H/(D(H) is is cyclic, then H isis cyclic.cyclic.

(23.2) IfIf GG isis aa p-group p-group then then (D @(G) (G) is is the smallest normalnormal subgroupsubgroup HH of GG such that GIHG/H is is elementaryelementary abelian.

Proof. IfIf MM isis aa maximal maximal subgroupsubgroup of G then, by Exercise 3.2,3.2, MM L]4 G and lGI G : :MI MI = = p, p, so,so, byby 8.8,8.8, G(')G(1 5< M. HenceHence G(1)G(') 5< @(G),t (G), so, so, by by 8.8, 8.8, GI G/@(G) (D (G) is abelian. Also, as GIMG/M =Z Zp, Zp, gP gp E E M M for for each each g g e E G. G. so so 9° gp E E (D @(G). (G). Hence G/(D(G)G/@(G) is elementary abelian. abelian. 106 p-groups

Conversely let let HH a G with G/H = = G* G* elementary elementary abelian. abelian. ThenThen G*G* =_ GTxG *x x ..... x x G*G; withwith GiGfSZ,, = l,, so so setting setting H, Hi=(Gj: = (G1: jj # i),IG:HiI=pi),I G : H, I = p and H = n n Hl. Hi. Thus Thus H, Hi is is maximal maximal in G so H = nn H1 Hi >2 4)(G).@(G).

Observe that, as a consequence ofof 23.2,23.2, aa p-groupp-group G is elementary abelian if and onlyonly ifif @(G)c(G) = 1. 1. Recall that if n is a positivepositive integerinteger then Qn(G) (G) isis the the subgroup subgroup ofof GG generatedgenerated by all elements of order at most pn.

(23.3) Let G = (x) (x) be be cycliccyclic ofof order q = pn pn > > 1 1 and and letlet AA = Aut(G), ThenThen (1) The map a Hi-+ m(a)m(a) isis anan isomorphism ofof AA withwith thethe groupgroup U(q) of units of the integers modulo q, where m(a) isis defineddefined byby xaxu == xm(a) xm(a) for a E A. In particular AA isis abelianabelian of of order order @(q) O(q)= = pn-'(p pn-1(p - - 1).1). (2) The subgroup of A of order pp - 1 1 is is cyclic cyclic andand faithfulfaithful onon Stl(G).nl(G). (3) If p is is oddodd thenthen aa SylowSylow p-groupp-group ofof AA is is cyclic cyclic andand generatedgenerated by by thethe element bb withwith m(b)m(b) = = pp + 1.1. In particularparticular the subgroup ofof AA ofof orderorder pp is generated by the element bob0 withwith m(bo)m(bo) == pn-'pn-1 + 1.1. (4) IfIfq=2thenA=1whileifq=4thenA=(c)~Z2,wherem(c)=-1. q = 2 then A = l while if q = 4 then A = (c) = LL2, where m (c) _ -1. (5) If p == 2 2 andand qq >> 4 4 thenthen AA = (b) (b) xx (c)(c) where where b b isis of of orderorder 2n-2 2n-2 withwith m(b) = 5, and c is of order 22 withwith m(c)m(c) == --1. 1. The The involution involution b0bo in (b)(b) satisfies satisfies m(bo) == 2"-'2n-1 ++ 1 and m(cbo)m(cb0) = = 2"-'2n-1 -- 1.1.

ProoJ:Proof. II leaveleave partpart (1)(1) asas anan exerciseexercise andand observeobserve also that a: a i-+H m(a) mod p isis aa surjectivesurjective homomorphism of A ontoonto U(p) with kernel CA(n1(G)).CA(QI(G)). SO,So, as IU(p)IIU(p) 1 ,I p= = pp - 1= 1 = IU(q)Ip,, IU (q) 1 ,I, thethe subgroupsubgroup of A ofof orderorder p p -- 1 1 is is isomor-isomorphic to U(p) andand faithful on cZa1 (G),1(G), while while ker(a) ker(a) = _ (a{a E E A: A: m(a) m (a) - -1 1 mod p}p) EE Syl,(A).Sylp(A). NextNext U(p) isis thethe multiplicativemultiplicative group of the field of order p, andand hence cyclic, soso (2)(2) holds.holds. ThusThus wewe may may take take q q > > p. p. EvidentlyEvidently if if m(c)m(c) = = -1-1 then c is of order 2. So,So, asas JIAl= A I =22 ifif qq =4,=4, (4)(4) holds.holds. ThusThus wewe cancan assumeassume n >> 1, and n >> 22 if p ==2. 2. ChooseChoose b as as inin (3)(3) or or (5).(5). ThenThen b b EE ker(a), so , bpn-'b2"bP"-' = 1. 1. Thus if p is is oddodd itit remainsremains toto showshow by"-2bpn-' = = bo bo and and if ifp p=2= 2 showshow b2"-3= b0. bo. Observe: k2 p2-+1(p (kpm + 1)Pl)P = (1 + kpm+1 kpm+l + k2p2m+1(p-_ 1)/2)1)/2) mod mod pm+2 pm+2 1 + kpm+1 if m> 1 1 oror pp isis odd. odd. HenceHence asas m(b)=lm(b) =1 + +s s withwith ss=p = p ifif pp is is oddodd andand s ==4 4 ifif pp =2,= 2, it it follows follows thatthat m(bpi-z)m(b~"-')= =1 1 ++ pn-1=m(bo)pn-' = m(bo) ifif pp isis odd, odd, whilewhile (b2"-3m(b2"-3)) = 2"-' = 2n-1 + + 1 == m(bo) ifif pp = 2. So the proof is complete. Extremal p-groups 107

Next the definition of four extremal classes of p-groups.p-groups. The modularp-groupmodular p-group Modp"Mod,. of order pnp' is is the the split split extension extension of of aa cyclic group X = (x) (x) of of orderorder pii-1pn-' =xPi-2+1. by a subgroupsubgroup Y = (y) ofof orderorder p withwith xyxY = XP"-'+I. ModpnMod,. is defined only when n 2> 3, where, by 23.3 and 10.3,10.3, Mode-Mod,. is wellwell defineddefined and determined up to isomorphism. Similar commentscomments holdhold forfor thethe otherother classes.classes. IfIf pp == 2 and n >2 22 the dihedral group D2DT isis the the splitsplit extension of X byby YY withxywithxy == x-'x-1 and x2^-2-1 ifif n >2 44 thethe semidihedralsemidihedral groupgroup SD2SD2?, isis the the split split extension extension withwith xyxy = x~~-'-'. The fourth classclass is is aa class of nonsplit extensions. Let GG bebe thethe splitsplit extensionextension of X = (x) ofof orderorder 21-12"-' 2> 44 by Y == (y) of order 4 with xYxy =x-'.=x-1. NoticeNotice (x2"2, (x2"-', y2)Y2) = Z(G). DefineDefine thethe quaternionquaternion group group Q2nQ2. of order 2" toto bebe thethe group G/(x2ii-2y2).G/(X~~-'y2). The modular,modular, dihedral,dihedral, semidihedral,semidihedral, andand quaternionquaternion groupsgroups areare discusseddiscussed in Exercises 8.28.2 and 8.3.8.3. Observe Observe Mod8Mods = D8.

(23.4) LetLet G G be be a a nonabelian nonabelian group group of of orderorder p'° pn withwith a a cycliccyclic subgroupsubgroup of of indexindex p. ThenThen GG -S Modp,,, Mod,. ,Den, D2", SD2-, SD2., or Q2'.122".

Proof.ProoJ: Notice Notice that, that, as as G G is is nonabelian, nonabelian, n n >2 3 3 by by ExerciseExercise 2.4. Let X = (x) (x) bebe ofof index ppin in G. G. By By ExerciseExercise 3.2, 3.2, X X 4 G.G. As As X X is is abelian abelian butbut GG isis not, X = CG(X) CG(X) by Exercise 2.4. So yy E GG -- X X actsacts nontrivially nontrivially on X. AsAs yPyp E X, y induces an automorphism of X of order p. ByBy 23.3,23.3, Aut(X)Aut(X) hashas aa uniqueunique subgroupsubgroup ofof 4 order p unless pp == 2 2 and and n n > 2 4, 4, where where Aut(X) Aut(Xj has has three three involutions. involutions. In In thethe firstfirst case by 23.3, xYxy = xz for some z of order p in X. InIn thethe remaining case p = 2 2 and XYxy ==x-1z', X-'zE, where where s E= = 1 1or or 0 0and and z zis is the the involution involution inin X.X. Now if the extension splits we may choose y of order p andand byby definitiondefinition GModP,,,G S Mod,. D2,,,, D2", or or SD2,,. SD2.. SoSo assume assume the the extension extension does does notnot split.split. Observe Observe CX(y)Cx(y) = (x")(xp) ifxy if xY =xz,=xz, while while CX(y) Cx(y) = = (z) (z) otherwise. otherwise. Also Also yPyp EE Cx(y). As G does not split over X,X, (y, Cx(y)) doesdoes not split over Cx(y),Cx(y), so, as (y, Cx(y))CX(Y)) isis abelian, it is cyclic. ThusThus Cx(y)Cx(y) = (yP). (yp). HenceHence we we may take ypyP = xP if xYxy = xz and y2y2 = z z otherwise.otherwise. Suppose xyxY =xz.= xz. Then z = [x, [x, y]y] centralizescentralizes x and y, so, by 8.6, (YX-')~(yx-1)P = yy~~-~z~(~-1)12 px-pZp(p-1)/2 = Zp(p-1)/2 Z~(~-1)12, while zp(p-1)/2zP(P-1)/2= =1 1unless unless p == 2. 2. So, So, as as GG doesdoes notnot x2n_22"-2 split,split, p = 2. 2. Here Here z = x and if n 2> 4 then, setting ii == 2n-32i-3 -- 1, 1, (yx` (YX')~ )2 = 1. 1. IfIf n = 3 3 thenthen xyXY = x-1, X-', whichwhich we we handle handle below. below. SoSO p == 2, 2, xy XY ==x-'zE, x-1zE, and y2 == z.z. IfIfs E == 0, 0, then then byby definitiondefinition G S= Q2',Q2., soso taketake sE = 1. 1. Then,Then, asas zz cE Z(G),Z(G), (yx)2 (yx)2 = = y2xyx y2~y~ == zx-1zx ZX-'ZX == 1, 1, so SO the the extension extension doesdoes indeedindeed split.split.

(23.5) LetLet GG bebe aa nonabeliannonabelian p-groupp-group containingcontaining a a cyclic cyclic normal normal subgroup subgroup U U ofof order p"pn withwith CG(U)CG(U) = U. U. ThenThen eithereither 108 p-groups

(1) G Z Dz.+l,Den+I, Q2""+',Q2,,+1,or or SDZn+l,SD2n+I, oror (2) M M = = Cc CG(ul (Z51(U))(u)) =S Mod Modpgt+1 p»+1and and Ep2Ep2 2 - S21(M)Q1(M) char char G.

Proof. LetLet G* G* = G/U. G/U. AsAs UU = = CG(U), CG(U), G*G* = = AutG(U) AutG(U) <( Aut(U). Aut(U). As As GG is nonabelian, G* 0# 1 1andand n n > > 2. 2. If If G* G* is is ofof orderorder pp thenthen thethe lemma holds by 23.4 and Exercise 8.2, so assume IG*IIG*l > p. ThenThen byby 23.323.3 therethere exists exists y* y* EE G*G* ofof orderorder p withwith uYUY == upn-'+',up"-'+', wherewhere UU = (u). (u). Let Let M M = = (y, (y, U).U). ByBy 23.4, 23.4, M M - SModpn+1 Modp.+l and, by Exercise 8.2, E = S21(M) Q1(M)S = Ep2.Epz. It It remainsremains to to showshow EE char G. By 23.3, G* is abelian and either G*G* isis cyclic,cyclic, oror pp == 2 2 and and there there exists exists g* g* E E G* G* with~~=~-'.InthefirstcaseQ~(G*)=M*,soE=Q~(M)=Q~(G)charG.with ug = u-1. In the first case Q, (G*) = M*, so E = Q1(M) = Q1(G) char G. In the second Z51(U)U'(U)= = (u2)(u2)= = ([u,([u, g])g]) and as G* is abelian, G(1)G(')( < U.U. HenceHence G(1)=ZS1(U)G(')=u'(u)or U, and and inin either either case case Z51(U) ul(U) char G('),G(1),so Z31U1(U) (U) charchar G. Therefore EE = = S21(CG Q (CG (Q(S2'' (u))) (U))) charchar G.

A critical subgroup ofof G isis a a characteristic characteristic subgroup HH of GG suchsuch that that 1)(H)@(H) (< Z(H)Z(H) 2> [G,[G,HI H] and CG(H)CG(H) == Z(H).Z(H). ObserveObserve that in particularparticular a critical subgroup is of class at most 2.2.

(23.6) Each p-group possessespossesses aa critical subgroup.subgroup.

Proof. LetLet S S bebe thethe set of characteristiccharacteristic subgroupssubgroups HH ofof GG withwith @(H)(D(H)i < Z(H) Z(H) 2> [G, HI. H]. Let H bebe aa maximalmaximal member ofof S; I claim H isis aa critical critical sub-subgroup of G. AssumeAssume not and let K = CG(H),Cc(H), Z Z == Z(H),Z(H), and define X by X/Z=fil(Z(G/Z))X/Z = S21(Z(G/Z))fl flK/Z. K/Z. Then Then K K-& H andand ZZ= = H flfl K,K, so, so, asas KK a G,G, X 0# Z Z by by 5.15. 5.15. But But notice notice XH XH EE S,S, contradictingcontradicting the maximality of H.

A p-group G is specialspecial if t(G)@(G) == Z(G) Z(G) = = G(1). ~('1. AA special special p-group p-group is is saidsaid toto be extraspecial if its center is cyclic.

(23.7) The center of a special p-group isis elementaryelementary abelian.abelian.

Proof. LetLet GG bebe specialspecial and g, h 6e G. Then gpgP Ee @(G)1(G) = Z(G), Z(G), so, so, byby 8.6.1,8.6.1, 1=1 = [gP, [gp, h] h] _ = [g, [g, h]p. h]P. Hence Hence G(') G(') isis elementary, elementary, so, as Z(G) = GM, G('), thethe lemma holds.

(23.8) LetLet E bebe anan extraspecialextraspecial subgroup ofof G with [G,[G, E]El <( Z(E). Then Then G = ECG(E).

Proof. LetLet Z = (z) (z) = Z(E). Z(E). AsAs E/ZE/Z( < AutG(E)Autc(E) 5 < C == CA"~(E)(E/Z),it suffices to show show E/Z= C.C. Let Let aa E E C C and and (x; (xiZ: Z: 115 < ii in)< n) a basisbasis forfor E/Z.E/Z. ThenThen Extremal p-groups 109

[xi, a]a] = z'izmi for for some some 00 <(mi mi < p,p, and, and, asas E = (xi: (xi: 11 5< i <( n) by 23.1, aa is determined by the integers (m(mi: i :1 I< ( i

A p-group isis saidsaid toto bebe ofof symplecticsymplectic type if it has no noncyclic characteristic abelian subgroups.

(23.9) If G is of symplectic typetype thenthen GG == E * R where

(1) Either Either EE is is extraspecial extraspecial or E == 1, 1, and and (2) EitherEither RR isis cyclic,cyclic, oror RR is is dihedral, dihedral, semidihedral, semidihedral, oror quaternion,quaternion, and of order at least 16.16.

Proof. By 23.6,23.6, G possessespossesses a critical subgroupsubgroup H. H. LetLet UU == Z(H). By hypothesis U is cyclic. Let Z be the subgroup ofof U of order p and G*G* = G/Z. For h, k E H,H,hneU,so[h,k]p=[hp,k]=1,by8.6.Thus hP E U,SO [h, k]P = [hp, k] = 1, by 8.6. Thus H(1) H(') 5 2 p2 and 23.5 applies to R. If IUIUI I= = p2p2 thenthen 7J1(u)Z51(U)= = Z Z ( < Z(G). Z(G). So,So, byby 23.5,23.5, RR isis dihedral,dihedral, quatemionquaternion or modular of order p3, and in particular R is extraspecial. But thenthen G == E*RE* R isis also extraspecial, so so thethe lemmalemma holds. holds. Thus Thus we we may may take take I IUIU I > p2 and assumeassume R satisfiessatisfies 23.5.2.23.5.2. LetLet M M = = CR(Z51(U)). CR(ul(u)). ThenThen NN = = CG CG(U'(U)) (Z51(U)) char char G andand N ==EM. EM. ByBy ExerciseExercise 8.2, 8.2, Q2(N)Q2(N) == EQ2(M) EQ2(M) =7Lp2 Zp x x(E (E * 7Lp2) * Zp2) SO SO Z(Q2(N)) is noncyclic, aa contradiction.contradiction.

(23.10) LetLet EE be anan extraspecialextraspecial p-group, p-group, Z Z = = Z(E), Z(E),and and ,!? E = = EIZ. E/Z. (1) Regard Z as the fieldfield ofof integers integers modulo modulo p p andand ,!? t asas a vector space over Z. Define f:f : ,!? E x k,!? -++ Z Z byby ff (x,(2, y)y) = = [x, [x, y].y]. ThenThen ff isis a a symplectic symplectic form on E, soso (E,(,!?, f) f) is is a a symplectic symplectic space space overover Z. (2) m(E) = 2n 2n isis even.even. (3) If pp=2 = 2define define Q:Q: E,!? -+ Z Zby by Q(x)=x Q(2)= 2. x2. Then Then Q Q is is a aquadratic quadratic formform on t,!? associated associated to to f,f, soso (E, (,!?, Q) Q) isis an an orthogonal orthogonal space space overover Z. 110 p-groups

(4) Let Z (< U <( E. Then Then UU isis extraspecialextraspecial or abelian if and only if ~U is nondegenerate or totally isotropic, respectively. IfIf pp == 2 then U is elementary abelian if and only if U0 isis totallytotally singular.singular.

ProoJProof. AsAs ZZ == c(E), (P(E), k Eis iselementary elementary abelian, abelian, so so by by 12.1 12.1 we we cancan regard Ek asas a vector space over Z in a natural way. Notice that under this convention the group operations on Ek andand ZZ areare written additively. By 8.5.4, [xy, z] = [x, z]Y[y,zIY [y, zlz] =

[x, z][y,z][y, z],z], with with thethe latterlatter equalityequality holding holding as as E E is is of of class class 2. 2. This This says says f f is , linear in its firstfirst variable and a similarsimilar argumentargument givesgives linearitylinearity inin thethe secondsecond variable. AsAs ZZ == Z(E),Z(E), f isis nondegenerate. nondegenerate. [x,[x, y] =_ [y, [y, x]-1,XI-', or,or, inin additiveadditive notation, ff (K,(z, y)y") = - f f (y, (y", z).2). Thus Thus (1) (1) holds. holds. Notice (1) and 19.16 implyimply (2).(2). LetLet pp =2.=2. ByBy 8.6,8.6, (xy)2(x~)~ =x2y2[x,=X~~~[X, y],y], or, or, in additive notation, notation, Q(K Q(. ++ y")y) == Q(z)Q(K) + Q(y)Q(y") ++ f f (i,(2, y).7). Thus Thus (3) (3) holds.holds. TheThe proof of (4) is straightforward.

(23.11) AssumeAssume pp isis odd odd and and GG is is of of class class at at most most 2. 2. Then Then S21(G) Q1(G) is of exponent p.

ProoJProof. Let x andand yy bebe elementselements of G of order p. Then [x, y] = z E Z(G).Z(G). By 8.6.1, zPzp = [x",[xp, y]y] = = 1, 1, so, so, by by 8.6.2,8.6.2, (xy)p(xy)P =xpypZp(p-1)12 = xpypzp(p-')I2 = 1. 1.

(23.12) LetLet pp bebe odd and E an extraspecial p-group. Then 01(E)Q1(E) isis of of exponent p and index at mostmost pp inin E.E. If Q101(E)(E) ## E Ethen then S21(E) Q1 (E) == X X x xEo EO wherewhere X X = Z 7Lp Zp and Eo is of order pp or extraspecial, and E = El El * * Eo Eo with with ElEl = Z Modp3. Modp3.

Proof.ProoJ LetLet YY=Q1(E). = 01(E). ByBy 23.11,23.11, Y Y isis of of exponent exponent p. p. Suppose Suppose I EIE:YI=p. : Y) = p. Then, inin the notation of of 23.10,23.10, f f is a a hyperplanehyperplane of E, andand hencehence of odd dimension, so, as all symplectic spaces areare of even dimension, Y P is degenerate. Let RI? be a point inin ~ad(Y).Rad(Y).As As fiLRI is a hyperplane ofof E,E, I?'Rl == Y. P. Hence, Hence, by 23.10.4, Y = = CE(R).CE(R). AsAs YY isis of of exponent exponent p, p, RR == XX x ZZ forfor somesome XX ofof orderorder p. LetLet EoEo bebe aa complement to Rfi inin Y.E. By 19.319.3 and 23.10.4, Eo is extraspecial or Eo == Z; of course Y == X x E0.Eo. Let Let ElEl = = CE(EO). CE(Eo). By 19.319.3 and 23.10, El is extraspecial. AsAs YY # # E, El El > > 01(E1) Ql(E1) so,so, by by 23.4, 23.4, E1= El Z Modp3. Modp3. It remains to show IIE E :: Y YI I ( < p. p. Let Let u, u, v v E E E E and and UU == (u,(u, v). v). It It suffices suffices toto show IU: Q1(U)I01(U)I ( < p.p. If U isis abelianabelian thisthis holdsholds becausebecause Z = (D (P(E). (E).If If UU isis nonabelian appeal to 23.4.

By 23.10 an extraspecial p-group p-group is is of of order order p1+2n for somesome positive integerinteger p. IfIf pp is is odd, odd, denote denote byby p1+2n an extraspecial p-group of exponent pp and order p 1+2n 21+2n.21 +2denotesdenotes any extraspecial 2-group2-group ofof orderorder 21+2n; 21+2"; by 1.131.13 Extremal p-groups 11111 1 there areare no suchsuch groups of exponent 2. Write D8Di QrQ8 for a central product of n copies ofof D8Ds with m copies of Q8,Q8, and all centers identified.

(23.13) LetLet p p bebe an an oddodd primeprime andand n aa positive integer. Then up toto isomorphismisomorphism there is a unique extraspecial p-group EE of of orderorder p1+2n and exponentexponent p.p. E isis the central product of n copies of pl+z

Proof. ByBy 23.10 23.10 and and 19.16,19.16, E E is is a a central central product product of of nn extraspecialextraspecial subgroupssubgroups Ei,E1, 1 (< ii (< n, of order p3 and center Z = Z(E). Z(E). NowNow ExerciseExercise 8.7 8.7 completescompletes the proof.

(23.14) Let nn bebe aa positivepositive integer.integer. Then up to isomorphism D8D: andand D8-1D,"-' Q8 are the unique extraspecial groups of order 22n+122"+'. DiD8 has 2-rank n + 1 1 whilewhile D:-'D8-1 Q8 hashas 2-rank n, so the groups are not isomorphic.

Proof. By 23.1023.10 andand 21.2,21.2, EE isis a a central central product product ofof nn extraspecial extraspecial groupsgroups Ei,E1, 11 (

(23.15) Let A be a maximal abelian normal subgroupsubgroup ofof GG and Z == S2(A).Q1(A). Then (1) A=CA(A).A = CG(A). (2) (cG(CG(A/Z)(A/z) n C(Z)(1)C(Z)(') 5

ProoJProof. Let CC = CcCG (A). AA 5< CC L! < G,G, soso ifif CC ## AA there isis D/AD/A ofof orderorder pp in Z(G/A) n C/A.C/A. Then Then DD < i? GG and and D D is is abelian abelian by by Exercise Exercise 2.4, 2.4, contradict- contradict- ing thethe maximalitymaximality ofof A.A. AA straightforward straightforward calculationcalculation showsshows (CG(A/Z)(CG(A/z> n C(Z))(')C(Z))(' <5 C(A),C(A), soso (1)(1) impliesimplies (2). LetLet p be odd, x ofof orderorder pp inin CG(Z)CG(Z) and X = (x, (x, A).A). Let Let YY == (x, (x, CACA((x, ((X,Z)/Z)). Z)/Z)). Then YY is of class at most 2, so, by 23.11,23.1 1, W = 01(Y)Q1(Y) is of exponentexponent p. ThusThus WW = (x, (x, Z).Z). But But WW charchar YY soso Nx(Y)Nx (Y) (< Nx(W)Nx (W) = = Y, so so Y == X and (3) holds.

(23.16) Let p bebe anan oddodd primeprime andand ZZ a a maximal maximal elementary elementary abelianabelian normalnormal subgroup of G. Then Z = 01(CG Q1(CG(Z)). (Z)). 112 p-groups

Proof. LetLet XX == S21(CG(Z)). Q1(CG(Z)).I'll I'll show XX is of exponent p.p. Hence if X # Z then therethere is is D/ZD/Z of order pp in Z(G/Z) nn X/Z X/Z and, and, by by ExerciseExercise 2.4, D isis elementary abelian, contradicting thethe maximality ofof Z. Let A be aa maximalmaximal abelianabelian normal subgroup of G containingcontaining Z. ThenThen Z = S21(A) Q1(A) by by maximalitymaximality of of Z.Z. By 23.15.3, [X, A] A] 5< Z, soso byby 23.15.2,23.15.2, x(')XM 5 < A. Choose U < X of minimal order subject toto U == Q1(U)Q, (U) and U not of exponent p.p. Then therethere existexist x x andand yy inin UU ofof orderorder pp with xyxy not of order p. By minimality ofof U,U, UU == (x, y). By 7.2,7.2, V == (xU)(xu) #0 U,U, soso V V isis of of expo-exponent p. Hence [x, y] E VV isis ofof orderorder at at most most p, p, so, so, as as x(') X" <( A, A, [x,[x, y]y] EE Z. As X I.< C(Z), UU is is ofof exponentexponent pp byby 23.11,23.1 1, contrary contrary to to thethe choicechoice ofof U.U.

(23.17) LetLet pp bebe anan oddodd primeprime andand assumeassume GG containscontains nono normalnormal abelian abelian subgroup of rankrank 3. ThenThen G is of p-rank atat most 2.

Proof. By Exercise 8.48.4 we may assumeassume Ep2Epz Z= ZZ 9a G. LetLet HH = = CG(Z) CG(Z) and Ep3 S= AA 5 < G.G. ThenThen ]A:IA: A A n n HI HI 5 3. Thus m(H) >2 3. 3. However However byby hypothesis ZZ isis aa maximalmaximal elementaryelementary abelian abelian normalnormal subgroup of G, so Z = c21(H) Q1(H) by 23.16.

24 CoprimeCoprime actionaction on p-groups In this section p is a prime, G isis aa p-group,p-group, andand AA isis aa p'-group ofof automor-automorphisms of G, unless unless the conditions are explicitly relaxed as in the Thompson A x BB Lemma.Lemma.

(24.1) AA isis faithfulfaithful on on GI G/@(G). (D (G).

Proof. SupposeSuppose b b c E A A centralizes centralizes GI G/@(G). (D (G). We wishwish toto showshow bb == 1. IfIf notnot therethere is a prime q andand aa nontrivial power of b which is a q-element and centralizes G/(D(G),G/@(G), so without lossloss bb isis aa q-element. q-element. Let Let BB = = (b) and and gg Ec G. Then B acts on the coset X =g@(G).=g(D(G). ByBy 5.14,5.14, mm = JXJmod IXlmod q, where rnm is the number of fixedfixedpointsofBonX,and,as/XI=~@(G)~isapowerofp,~X~ points of B on X, and, as X j = J(D(G) l is a power of p, X +i 0 Omod0 mod q,soq, so B centralizes somesome x EE X. Hence BB centralizescentralizes a a setset YY ofof cosetcoset representativesrepresentatives for @(G)c(G) inin G,G, so, so, by by 23.1.2,23.1.2, GG = (Y) (Y) <5 CG(B). CG(B). Hence B = 1, 1, completing thethe proof.

(24.2) (Thompson(Thompson A xx BB Lemma). Lemma). Let Let AB AB bebe aa finitefinite groupgroup representedrepresented as a group of automorphisms of of a a p-group p-group G,G, withwith [A,[A, B] B]= = 11= = [A, [A, CG(B)],CG(B)], B a p-group, and A = OP(A). Op(A). Then [A, GIG] = 1.1.

Proof. FormForm the the semidirect semidirect product product H H of of G G by by AB AB andand identify identify AB AB andand GG with with subgroups ofof H.H. We maymay assumeassume [A,[A,GI G] # 0 11 so,so, asas AA = = OP(A),OP(A),[X, [X,GI G] # 0 11 Coprime action on p-groups 113

for somesome p'-subgrouppl-subgroup X ofof A,A, andand replacingreplacing A by X we may assume A isis a p'-group.pf-group. G 9< H = GBA withwith AA 5 < NH(B), NH(B), so so GBGB is is aa normal p-subgroup ofof H, and, replacing G byby GB,GB, wewe maymay assumeassume BB < ( G. G. Then Then B B < ( Q Q = = CG CG(A), (A), soso CG(Q) <( CG(B). Also CG(B) (< Q by hypothesis, so Cc(Q)CG(Q) (< Q. As [A, GIG] 0# 1,1, Q Q 0 #G, G, so, so, by by Exercise Exercise 2.2, 2.2, Q Q is is properly properly contained contained inin NG(Q). So,SO, by definition of Q, [A, NG(Q)] 0# 1,1, and and hencehence wewe maymay assumeassume QQgG.

(24.3)(24.3) IfIf GG is is abelian abelian then then A A is is faithful faithful on on c21(G). Ql(G).

Proof. WithoutWithout loss, A centralizes Q1(G).S21(G). Let Let X X be be of of orderorder pp in G andand G* = G/X. G/X. ByBy ExerciseExercise 3.1,3.1, AA isis faithfulfaithful on G*,G*, so,so, by induction on the order of G,G, AA isis faithfulfaithful onon Q,Q1(G*), (G*), and hence without loss G*G* = Q, Q1(G*). (G*). Now, by 12.1 and Exercise 4.1.1, we may take CG*CG.(A) (A) = =1. 1. Thus, by 18.7.4,18.7.4, X == CG Cc(A), (A), so XX = S21(G). Q1(G). Hence,Hence, asas GG is abelian, 1.11 implies G isis cyclic.cyclic. NowNow 23.323.3 supplies aa contradiction.contradiction.

! (24.4)(24.4) G = [G, [G, A]CG(A).A]CG(A).

Proof.Proof. LetLet G* G* =G/4(G).= G/Q(G). By By 23.2, 23.2, G* G* is isan an elementary elementary abelian abelian p-group, p-group, so, so, by ExerciseExercise 4.1,4.1, G*G* = = [G*, [G*, A]A] Xx CG.(A).CG*(A). By 8.5.3,8.5.3, [G*,[G*, A]A] = [G, [G, A]*A]* and,and, by 18.7.4,18.7.4, CG.(A) CG*(A) = CG(A)*. CG(A)*. Hence GG == ([G, ([G, A], A], CG(A), CG(A), (D Q(G)), (G)), so, by 23.1, G = ([G, ([G, A],A], CG(A)). Finally, byby 8.5.6, [G, A]Al 9

(24.5)(24.5) [G,[G, A] == [G, A, A].Al

Proof.Proof. LetLet H H = = [G, [G, A]. A]. By By 8.5.6, 8.5.6, H H < 9G G and and [H, [H, Al A] :< 9 H. H. Thus Thus CG CG(A) (A) acts acts on [H,[H, A], A],so[H, so [H, A]A] <9 HCG HCG(A)= (A) = G.G.Next Next H = = [H, [H, A]CH A]CH(A)so[G, (A) so [G, A]A] <( [H, [H, A] A] by 8.5. But of course [H, A] (< [G, A] as H <( G. G.

(24.6)(24.6) IfIf GG is is abelian abelian then then G G = = [G, [G, A] A] X x CG Cc(A). (A). 114 p-groups

Proof. LetLet GG be be a aminimal minimal counterexample counterexample and and X X = = [S21(G), [Q1(G), A]. A]. ByBy 24.3,24.3, X #0 1 and, by 12.112.1 andand ExerciseExercise 4.1.1, CxCx(A) (A) == 1.1. ByBy minimalityminimality of G, C([G,Allx)(A)C([G,AI/x)(A)= =1, 1, soso CIG,Al(A)C[G,AI(A)= =1. 1. NowNow 24.424.4 completes the proof.

(24.7) If G = [G, [G, A] A] andand AA centralizescentralizes every every characteristiccharacteristic abelian subgroup of G, thenthen GG isis specialspecial and and Z(G)Z(G) = = CG CG(A). (A).

Proof. AsAs AA centralizescentralizes eacheach characteristiccharacteristic abelian subgroup of G, soso doesdoes G == [G, A]A] by Exercise 3.6.3.6. ThusThus Z == Z(G) isis the the uniqueunique maximal maximal characteristic abelian subgroup. [Z2(G),[Z2(G), G, G]GI = 1,1, so,so, byby thethe Three-Subgroup Three-Subgroup Lemma, Z2(G) centralizes GM.~('1. Hence Z2(G)Z2(G) f1fl G(')GM is abelian, and therefore contained inin Z,Z, so G(')Gf' < ( Z. Z. By By 24.6,24.6, G/G1>GIG(') = (Z/Gf1)(z/G(')) xx [GIG('),[GIG('), A]A] so, as as GG == [G, [G, A], Z = GO).G('). Finally supposesuppose GG hashas exponent pnp" >> p.p. Let 1, g, h E G. By 8.6, [gp"-'[gp"-', , hpn-I]V_'] _= [gp°, tgp", hp"-Z] hpn-'1 == 1, SOso 73'W-'(G) 1 (G) is abelian and hence U""(G)?5n-1(G) 5 < Z. Z. But But then then G/Zis G/Z is of of exponent exponent p. p. So So Z Z = = @(G). I(G).

(24.8) IfIf pp isis odd odd then then A A is is faithful faithful on on S21(G). Q1(G).

Proof. ChooseChoose GG to to be be a aminimal minimal counterexample counterexample and and let let a a E E A# A' centralizecentralize S21(G).Q1(G). By By 24.524.5 andand minimality ofof G, G == [G, a].a]. By 24.3,24.3, aa centralizescentralizes each characteristic abelian abelian subgroupsubgroup of of G, G, so,so, byby 24.7, G isis special special withwith Z = Z(G) = CG(a). ByBy 23.7, Z == Q1(G).S21(G). Let Let g g E E G G - - Z,Z, zz == gpgp and v == [g, gVa].g-°]. ThenThenz, z, vv EE ZZ == S21(G), Ql(G), soso vpvp ==1. 1. Notice that, as ZZ == CG(a),CG (a), (g-')P (g-Q)p == gg_a zz-1 vp(p-1)/2 z-', Z-1,and and gg-a = h h 04 ZZ byby 18.7.4.18.7.4. Now, Now, by 8.6, hPhp = ~z-'vp(p-')/~ = 1, 1, contradicting Z Z = = SZ Q 1(G).'(G).

(24.9) Let H bebe aa criticalcritical subgroup of G. Then (1) A is faithful onon H.H. (2) If p isis odd then AA isis faithfulfaithful onon S21(H),Ql(H), and therethere exists a critical subgroup H ofof GG such such that that SZ1(H) Q1(H) contains each each elementelement of of orderorder pin p in CG(SZ CG(Q1 1(H)).(H)).

Proof. ByBy definitiondefinition of H, CG(H)CG(H) <5 H,H, so so by by thethe ThompsonThompson A x BB LemmaLemma (applied to `A''A' = CA(H) CA(H) andand `B''B' == H), H), CA CA(H) (H) == 1. 1. Thus Thus (1)(1) holds.holds. PartPart (1)(1) and 24.8 imply the first statement in (2). To prove thethe second, choosechoose H with L = S21(H) Ql(H) maximal. It suffices to show Y == Q~(CG(L))01(CG (L)) I< L. Assume not and let VV bebe aa maximal maximal elementaryelementary abelianabelian normal subgroup ofof Y. By 23.16, V = S21(Cy(V)),Q1(CY(V)), SO,so, as YY -$ L, VV $ L. Thus vV f1fl Z2(Y) Zz(Y) $ L, so S21(Z2(Y))Q1(Z2(Y)) == K $ L. By 23.123.11, 1, K is of exponentexponent p, so X $ L, whereX/Z(L) = = Z(G/Z(L)) Z(G/Z(L))l fl (K/Z(L)). NowNow definedefine S asas inin thethe proofproof ofof 23.6.23.6. ThenThen XLXL E SS so,so, by thethe proofproof of 23.6, XL is contained in aa criticalcritical subgroup subgroup C C of of G. G. But But L L<

Remarks.Remarks. TheThe discussion discussion of of p-groupsp-groups in in this this chapter chapter is is essentially essentially the the same same asas Gorenstein's Gorenstein's treatment treatment ofof p-groupsp-groups [Gor [Gor 4], 41, which which was was influenced influenced in in turn turn byby lecturelecture notes notes of of PhillipPhillip Hall. Hall. P.P. Hall originally classified classified the p-groupsp-groups of of symplecticsymplectic type. type. The The notion notion of of a a `critical'critical subgroup'subgroup' isis duedue toto J.J. ThompsonThompson asas isis of course the Thompson A x BB Lemma.Lemma. AlmostAlmost all all of of the the material material in in this this chapterchapter is is basic basic and and belongsbelongs in in the the repertoirerepertoire ofof anyany finitefinite group group theorist.theorist. ForFor thethe simple simple group group theorist theorist it it represents represents an an importantimportant part ofof thethe foundationfoundation ofof thethe locallocal group group theory theory involved involved inin thethe classification.classification. ForFor example the importance of p-groups ofof symplecticsymplectic typetype isis reflectedreflected inin thethe secondsecond casecase of of TheoremTheorem 48.3. 48.3. MoreMore generallygenerally thethe resultsresults ofof thisthis chapterchapter will will bebe usedused repeatedlyrepeatedly inin chapterschapters 10 10 through through 16.16.

ExercisesExercises for chapter 8 8 1.1. LetLet q q be be a a prime prime and and A A anan elementaryelementary abelian abelian q-groupq-group acting acting on on aa q'-group G.G. ProveProve GG == (CG(B): (CG(B): IA:!A: BI =q). (Hint: (Hint: Use Use 18.7 18.7 to to reduce reduce to to the the case case GG aa p-group.p-group. Then Then use use Exercise Exercise 4.1 4.1 andand 23.1.)23.1 .) 2.2. Let Let G G = S Model. Mod,. ,, n >2 3,3, withwith 7Lp"-1Z,.-I S = X X = = (x) (x) 2, = Dzn , n > 2, Qzn Q2.,n23, , n > 3, or O~SD~~,~>~.L~~Z~~-LZX=(X)<]G SDzn , n > 4. Let 7L2--1 X = (x) 4 G andand yy (=-E G G - - X with y anan involution if G isis dihedraldihedral oror semidihedralsemidihedral andand yy ofof orderorder 4 4 if if GG is is quaternion. quaternion. Prove Prove (1)(1) GO)G(') = = (D(G) @(G) == (x2) (x2) =7L2S Z2n-2. -2. (2)(2) EitherEither G G is is dihedral dihedral of of order order 4 4 or or Z(G) Z(G) = = (x2 (x2'-') -2) isis of of order order 2. 2. (3)(3) GG is isof of classclassn n - 1. 1. (4)(4) XX is is the the unique unique cyclic cyclic subgroup subgroup of of GG ofof indexindex p,p, unless unless G G is is dihedral dihedral ofof orderorder 4 4 oror quaternionquaternion ofof orderorder 8. 8. (5) GG - - X X is is the the union union of of two two conjugacy conjugacy classes classes of of GG withwith representativesrepresentatives yy andand yx.yx. Each Each membermember of GG -- X X is is an an involution involution if if GG is is dihedral, dihedral, eacheach isis ofof orderorder 4 if GG isis quaternion,quaternion, whilewhile if GG isis semidihedralsemidihedral then yy isis of of orderorder 2 2 andand xyxy ofof orderorder 4. 4. (6) GG hashas two two maximalmaximal subgroupssubgroups distinct from X. If G isis dihedraldihedral of orderorder atat leastleast 8,8, bothboth areare dihedral.dihedral. If GG isis quaternionquaternion ofof orderorder atat leastleast 16,16, both are quaternion. If G is semidihedral then one is dihedral and thethe otherother quaternion.quaternion. (7)(7) QuaternionQuaternion groups groups have have a a unique unique involution. involution. 4. LetLet G G be be a a p-group p-group with with no no noncyclic noncyclic normal normal abelian abelian subgroups.subgroups. ProveProve GG is cyclic, quaternion,quaternion, semidihedral, semidihedral, or or dihedral, dihedral, and and in in the the last last case case I GI GI I > 8.8. 116 p-groups

If H isis aa p-groupp-group withwith just one subgroup of orderorder p, prove H is cyclic or quaternion. 5. LetLet GG bebe aa nonabelian nonabelian p-groupp-group ofof symplecticsymplectic typetype andand exponentexponent p oror 4.4. Set Z = Z(G), G,G, == G/Z, G/Z, A A = = Aut(G), Aut(G), and and A* A* = Out(G). Out(G). Prove (1) Inn(G)Inn(G) = = CA((3). cA(G). (2) CA(Z)*CA(Z)* = Sp(O) sp(G) and and A*A* isis the group of all similarities of some symplectic form on G if p isis odd.odd. (3) IfIf pp == 2 2 thenthen GG -E D', Dn, Di-1D"-' Q, Q, or or Z4 Z4 * *D', Dn, and and A* A* - EOZ 0L(2), (2), OZn(2), 0,(2), or SP2n(2),Sp2,(2), respectively. 6. Let GG be aa 2-group containing anan involution involution x xwith with CG(x) CG(x)E - E4. E. Then G is dihedral or semidihedral. 7. LetLet pp bebe an an odd odd prime.prime. ProveProve (1) Up Up toto isomorphismisomorphism therethere isis a unique extraspecial group E of order p3 and exponent pp. . (2) Au~A~~(E)(Z(E))E Aut(Z(E)). (3) UpUp toto isomorphismisomorphism therethere is a unique central product ofof nn copies ofof EE with identified centers. 8. LetLet AA be be aa 7r'-group nl-group actingacting onon aa 7r-groupn-group G. Prove (1) GG == [G, [G, A]CG(A), A]CG(A), and (2) [G,[G, A]A1 _= [G, [G, A, A, A].A]. 9. LetLet rr bebe a a prime, prime, A A an an elementary elementary abelian abelian r-groupr-group actingacting onon aa solvable solvable r'-grouprl-group G,G, DD <5 A,A, andand BB aa noncyclic noncyclic subgroupsubgroup of A.A. ProveProve [G,[G, D]Dl = ([CG([CG(b),(b), Dl:D]: b E B')B#). . 10. Let AA be aa p'-groupp'-group withwith aa uniqueunique minimalminimal normal subgroup B, assume A acts faithfully onon aa nontrivialnontrivial p-groupp-group P, and assume A is faithful on no proper subgroup ofof P. Prove Prove that either (1) PP is is elementary elementary abelian abelian andand AA is irreducible onon P, or (2) PP == [P, [P, B] B] is is special, special, [B, [B, Z(P)] Z(P)] = = 1, 1, and and A A is is irreducible irreducible on P/Z(P). If [A, Z(P)] = = 1 1and and AP AP possesses possesses a a faithfulfaithful irreducible representation over some field, thenthen P is extraspecial. 11. Let p bebe anan oddodd prime and G a p-groupp-group with m(G) > 3. Prove GG hashas aa normal abelian subgroupsubgroup of of p-rankp-rank atat least 4.4. (Hint: LetLet G be a coun-counterexample, V an elementaryelementary abelian normal subgroup ofof G ofof max-maximal rank,rank, HH == CG(V), CG(V), and and Ep4 EP4 - AE

Change of fieldfield ofof aa linearlinear representationrepresentation

Let n:jr: G -+-> GL(V,GL(V, F) be be anan FG-representation, EE a subfield ofof F,F, and KK an extension fieldfield of of F.F. ThenThen V isis alsoalso aa vectorvector space space over over E E with with GE(V, GL(V, F) F) 5< GL(V, E), soso 7rn alsoalso definesdefines anan EG-representation.EG-representation. Further, by aa tensoringtensoring process discussed in sectionsection 25, nn inducesinduces a a KG-representation KG-representation 7rKnK on a K-K- space VK. This chapterchapter investigatesinvestigates the relationship among these representations. ItIt will often be very useful toto extend FF toto KK byby passingpassing fromfrom 7rn toto n7r K. For example several results at the end of chapterchapter 9 areare establishedestablished inin thisthis way. nTr is is said said toto bebe absolutelyabsolutely irreducible ifif 7rn Kis is irreducible irreducible forfor eacheach extensionextension K ofof F,F, and and F F is is said said to to be be a asplitting splitting field jield forfor G G if if every every irreducible irreducible FG-FG- representation isis absolutely irreducible.irreducible. ItIt developsdevelops in in sectionsection 2525 that n7r isis absolutely irreducible preciselyprecisely whenwhen FF = EndFG(V) EndFc(V) and in section 2727 thatthat ifif G is finite then a splitting field is obtainedobtained by adjoining a suitable root of unity to F. It'sIt's particularlyparticularly nice nice to to work work overover aa splittingsplitting field. field. For example in section 27 it isis shownshown that,that, overover aa splitting splitting field,field, the the irredycible irreducible representations representations ofof thethe directdirect product of groupsgroups areare just the tensor products of irreducible representations of of the factors. Section 26 investigates representations over finitefinite fields,fields, wherewhere changechange ofof field goes very smoothly. Lemma 26.6 summarizes many of the relationships involved. Section 27 introduces the minimal polynomialpolynomial ofof aa linearlinear transformation.transformation. Semisimple and unipotentunipotent elementselements areare discusseddiscussed andand it it is is shownshown that that if if F F is perfect then each member g of GL(V) admitsadmits aa Jordan decomposition;decomposition; that that isis g can be written uniquely asas the commuting product of aa semisimplesemisimple element and a unipotent element.

25 Tensor products In this section G isis aa group,group, FF is is a a field, field, and and VV a a finite finite dimensional dimensional vector vector space overover F. Let (Vi:(V1: 0 0 5 < ii (< m)m) bebe vectorvector spaces over F, and and denotedenote by by L(V1,L(Vl, ...... , ,Vm; Vm; Vo) the the setset ofof all maps a: Vl Vl x . . - x V.Vm -,-+ VoVo suchsuch that that for each i,i, 11 <5 i

-(25.1)(25.1) Tensor products exist and are unique up toto isomorphism.isomorphism.

Proof.Prooj SeeSee for for example example page page 408408 inin LangLang [La].[La].

BecauseBecause of 25.1 there isis aa uniqueunique tensortensor productproduct ofof V1,Vl , ...., . . Vm , V, whichwhich isis dede- noted byby VlVl ®@ ... - . @® V,Vm or or ®m 1&. V. Write Write vl v1 @ ® n . ®- @ vm v, forfor thethe imageimage of (vi,(vl, ...,. . . , vm)v,) underunder thethe mapmap (denoted by by nr above)above) associated associated to the tensortensor product. TheThe elementselements vl vl ®@ ... . .® . @Vm, v, V1, vi EE V1.Vi, are called fundamentalfundamental tensors.tensors. It is easy toto verify fromfrom thethe universaluniversal propertyproperty that:that:

(25.2) VlVl ®@ .. -. . ®@ VmV, is is generatedgenerated asas anan F-spaceF-space by the fundamentalfundamental tensors.tensors.

Here are some elementary properties of the tensor product; they can be found for example inin Lang, Chapter 16.16.

(25.3)(25.3) Let (Vi:(&: 11 5< i <5 m)m) bebe F-spaces.F-spaces. ThenThen (1) V1vl@v2~v2@Vl. ® V2 = V2 ® V1. (2) (Vl(Vi®V2)®V3V1®(V20V3) @ V2) @ v3 Vl @ (V2 @ V3). (3) ((DUEI(e,,, U) ®@ VV % (DUEI(U@,,,(U @ ® V) V) for for any any directdirect sumsum e,,,®UEI U U ofof F- spaces.spaces. (4) LetLet XiXi bebe aa basisbasis of Vi,Vi, i = 1, 1,2. 2. Then Then XXI 1 @®X2 X2 == {X {XI 1 ®x2:8x2:~~ Xi EE X1 Xi} } is a basis forfor Vl Vl ®@ V2. V2. (5) Let alai EEndF(V1).E EndF(Vi). Then Then there there exists exists a uniquea unique map map a1 @al ®. -. ®@ an a, EE EndF(VlEnd~(V1@... ®...(9 @V,)with(vl Vm)With (VI @..-@v,)(al@...@a,) ®...®vm)(a1(9 ...®am) = = vial vial @...@v,a,®...®vmam. . (6) Forvl,u1Forvi,ui E Q,i = 1,1,2,anda 2, and aE E F:

(VI + ul) ® V2 = (Vi (9 V2) + (ul ® V2), and VIVl 8® (V2 (212 + u2) u2) == (VI (211 (9€3 V2)212) ++ (v](211 €30 u2), a(vl (9 V2) = avl ® V2 = V1 ® avg. Tensor products 119

If 7r1:ni: G G -+--). GL(Vj), GL(V; ),1 15 < i i < m,m, are FG-representations, then byby 25.3.525.3.5 therethere is anan FG-representationFG-representation 7r1 nl 8.®. .. ..@ ® nmzr,,, ofof G onon V1Vl ®@ .. . .@® V,, Vm defineddefined byby g(nlg(n1 ®.8.--@nm)=gnl(9 7r,,,) = gJr1@...@gnm,forg ®... ®g7r,,,, for gt EG.nl G. nl @...@n,isthetensor ® ®7rm is the tensor product ofof the the representations representations 7r1, nl, ... . ., .7r,,, , n,. . A special case of these constructions is of particular interest.interest. LetLet KK be an extension field ofof F. Then K isis aa vectorvector spacespace over F, so so thethe tensortensor product K ®@ U cancan be formed for any F-space U. Let X andand BB bebe bases for U and K over F, respectively. respectively. By 25.3 each member of K ®@ UU cancan bebe writtenwritten uniquelyuniquely as F-(b,x)EBxxC(b,x)EBxXab.x(b ab,x(b @ (9 x),withab,x with ab,x E E F. F. ASAs ab,x(b ab,x(b @ (9 X) x) == ((ab,x)b) @® Xx with (ab,x(ab,,)b )b EE K, K, itit followsfollows thatthat eachmembereach member of KK 8® U is of thethe formform ~,,,(c,ExEx(c® @(9 x), cxc, Et K.K. Indeed Indeed it it turnsturns outout KK ®@ UU can can be be mademade intointo a vector space K ®@ F U = UKlJK overover K by defining scalar multiplication via:

a1:(cx®x))_(acx ®x)a, cx E K, X E X. (XEX XEX TheseThese remarksremarks areare summarizedsummarized in thethe following following lemma;lemma; seesee chapterchapter 16,16, section 3 in Lang [La][La] forfor example.example.

(25.4) LetLet KK be be an an extensionextension fieldfield of F and and XX aa basis for a vector space U over F.F.ThenuK Then UK = K KmF OF U Ui~avectorspaceoverKwith1 is a vector space over K with 1 @X® X = { (1 1 @x:x® x: x E X)X} a basis for UK.uK. I"i It will be useful to have thethe followingfollowing well knownknown propertyproperty of thisthis construction,construction, which can be found onon pagepage 419419 inin LangLang [La].[La].

(25.5) If L > KK >> F F is is a atower tower of of fields fields and and UU anan F-spaceF-space then then LL ®F UU Z L ®K@K (K (K @FOF U).U)

Notice that, for gg Et EndF(U),EndF(U), 11 ® @ gg Et EndK(UK),E~~K(u~), where 11 8® g is thethe mapmap defined in 25.3.5 with respect toto thethe identity mapmap 1 on K. That is 11 ®@ g: a ®@ x H++ aa ® @ xg. xg. In In this this way way EndF(U) EndF(U) is is identified identified with with a a subalgebra subalgebra of of EndK(UK). ~nd~(~~). Further if jr:n: G ----+ GL(U) is anan FG-representationFG-representation then we obtain a KG- KG- representation nK:.rrK:G G -+ --± GL(U~) GL(UK)defined defined by by n 7rK = = 1 1 @ ®7r, n, where where 11 isis the the trivialtrivial representation ofof GG on K. Equivalently (1(1 @(9 ~)(~n~) x)(girK)= = 110 @ xg7r xgn forfor each x E X, gg EE G.G. ObserveObserve thatthat if UU isis finite finite dimensionaldimensional then Mx(gn)Mx(gir) = M(i®x)(gJrK)M(l@x,(gnK> Recall the definitiondefinition ofof envelopingenveloping algebraalgebra inin sectionsection 12.12.

(25.6) Let 7r:n: G G -+--- EndF(V) = E E be be an an FG-representation, FG-representation, A A thethe envelopingenveloping algebra of n7r inin E, and K anan extensionextension of F. Then Then thethe envelopingenveloping algebra of n7rK inin ~ndKEndK(VK)(vK) is is isomorphicisomorphic toto AK asas aa K-space. 120 Change ofjieldof field ofof a a linearlinear representation

ProojProof. WeWe may regardregard EE and EndK(vK)Endx(V') == EK E~ asas the the rings rings Fnxn FnXn andand KnxnKnXn, respectively, withwith EE thethe setset ofof matricesmatrices inin EKEx whose entries areare inin F.F. Now

A ag(g7r):ag EF} soAK = bg(g7r):bg c K . gEG gEG }

But, as matrices, gng7r and gnKg7r K are are the the same, same, so, so, as as G G isis aa group,group, AKAx is the subalgebra ofof EEx generatedgenerated by G7rx.GnK. ThatThat is Ax isis thethe envelopingenveloping algebra of x 7rnK.

Let G1G1 andand G2G2 bebe groupsgroups and and 7ri ni anan FGi-representation.FGi-representation. Denote Denote by by 7r1 nl ®7r2 8 n2 the tensor productproduct ititl 1 @® it2n2 where itini isis thethe representationrepresentation of of G G 1 1 xx G2G2 with with : ftiri restricted to Gi equalequal toto 7rini and itini restrictedrestricted to G3_iG34 trivial.trivial. This is aa smallsmall abuse of notation which will hopefully cause no problem.problem. TheThe conventionconvention isis used in the proof of the next lemma. Notice that if G1G 1 E = GzG2 E= G G then then G G is is diagonally diagonally embedded embedded inin GG1 1 x G2G2 via the map g H (g, (g, g), g), and and if if we we identify identify G G with with this this diagonal diagonal subgroup subgroup via via thethe isomorphism, thenthen thethe tensortensor productproduct representationrepresentation 7r, nl ®8 7r2: n2: GG -+-> GL(ViGL(Vl 8 V2) of of GG isis thethe restrictionrestriction ofof thethe tensortensor productproduct representationrepresentation ofof GIG1 xx G2 to this diagonal subgroup.

(25.7) Let K be a Galois extension ofof F,F, Fr thethe Galois group of K over F, andand V an FG-module. Then (1) ThereThere is auniquea unique F(rF(r x x G)-representation G)-representation on on VKVK withwith (y, (y, g):g): a a @® vv HH ay@vgforeacha~K,v~V,yay 0 vg for each aK, vV,y E~,~EG.F, g G. (2) IfIf WW isis a KG-submodule of VK with with Wr Wr = W, W, thenthen WW == UK uK forfor somesome FG-submodule U of V.

Proof.Prooj TheThe representation representation in in (1) (1) is is just just thethe tensortensor productproduct representation a ®8 fi B of r xx G G where where aa is is the the action action ofof rr on on K, K, and and $ ,f3 is is the the representation representation ofof GG on V.V. Assume WW isis asas inin (2), and extend aa basisbasis ZZ == (zi: 11 i< ii _(< m) for WW inside of of X' X' U U Z Z to to aa basis basis Y Y= = (zl, (zl,...... , zm, x;,,,xm+1 1...... , xn)x:) of Vx,vK, where where xfxi == 11 (9 @ xi, and XX = {xi:{xi: 11 5m(aijxl ++ wi), wi), forfor wi EE WW and aijaij E K. Let y eE F.r. ThenThen MY = xi - 1:(aij)Yx' = (aij - aijy)x + wi. i>m j>m

But byby hypothesishypothesis wi wiy y Ee W,W, so, so, as as Y Y is is a abasis basis for for vK, Vx,aij aij = = aij aijy y forfor allall i, i, j. j. Tensor products 121

Hence ai,aij E Fix(r)Fix([') = F. F. As As X' X' isis a a basis basis for for VK, vK, (wi: (w,: 1 5< i <_( m}m) is is aa basisbasis for W and we have shown w,wi = 11 ® @ vi,vi, wherewhere

vi=xi V. j>m

Thus W = UK uK where U is the subspace ofof VV generatedgenerated by by (vi: (vi: 1 1 _( < ii (< m).

An irreducible FG-module VV isis absolutelyabsolutely irreducibleirreducible ifif VKvK remains remains irre-irreducible for each extensionextension KK ofof F.

(25.8) Let VV bebe anan irreducibleirreducible FG-module.FG-module. ThenThen VV is is absolutely absolutely irreducible irreducible if and only ifif F == EndFG(V). EndFG(V).

Proof. AssumeAssume F F = = EndFG(V). EndFG(V). Then, Then, by by 12.16,12.16, E == EndF(V) EndF(V) is is the the envelop- envelop- ing algebra for G on V. So, if K isis anan extensionextension of F andand A'A' thethe envelopingenveloping algebra ofof G in E'E' == EndK(VK), ~nd~(V~), then,then, byby 25.425.4 andand 25.6,25.6, dimK(A')dimK(Af) =_ dimF(A) == n2n2 = dimK(E'), dimK(E1), soso A'A' == E. E'. In In particular particular A', A', andand hence hence alsoalso G,G, is irreducible onon VK.vK. Conversely assumeassume V is absolutely absolutely irreducible.irreducible. Then VKvK is irreducibleirreducible where K is thethe algebraicalgebraic closureclosure of of F.F. By 12.17, E' == EndK(VK) ~ndK(v~) is thethe enveloping algebraalgebra for for G G in in E', E', so,so, byby 25.6,25.6, n2 n2 = = dimK(Ef)dimK(E') = dimF(A). dimF(A). LetLet 'r D == EndFG(V). EndFG(V). By 12.16, AA 2= DmxmDmXm wherewhere m m == dimD(V).dimD(V). ThenThen n n == dimF(V) == mk,mk, k = dimF(D) dimF(D) andand dimF(A) == m2k.m2k. SoSo m2k2m2k2 = =n2 n2 = m2km2k and hencehence kk == 1. ThatThat is is F F == D.D.

If n:7r :G G + -> GL(V)GL(V) isis anan FG-representation,FG-representation, X isis abasisa basis for for V, V, and and a a EE Aut(F),Aut(F), then nu:7r°: GG +- GL(V) GL(V) is is the the FG-representation FG-representation withwith Mx(g7r°)Mx(gnU) = Mx(g7r)°. Mx(gn)". Here ifif AA == (aij) E FnxnF""" then A°A" = (a (a;). ). NoticeNotice thatthat ifif n°it" is is the the represen-representation defined withwith respectrespect toto aa different basisbasis ofof V, then it"ir° isis equivalentequivalent to 7r°nu by aa remarkremark after after 13.1.13.1. SoSo 7r°nu is is independent independent of X,X, upup to to equiva-equivalence. I'llI'll sometimessometimes write V°V" forfor VV regardedregarded asas anan FG-moduleFG-module withwith respectrespect to 7r°.nu. Recall the character of an FG-representation 7rn isis thethe functionfunction X:X: G G -+ -± F defined by X~(g) (g) = Tr(g7r Tr(gn). ). Let V be an FG-moduleFG-module and k a field with F <5 k k < 5 EndFG(V). EndF&). Then the action of k on V makes V into a k-space.k-space. FurtherFurther thatthat k-spacek-space structurestructure extendsextends the F-space structurestructure and is preserved byby G, so we can regard VV asas aa kG-kG- module. Similarly ifif K is a subfield ofof F then V is certainly a K-space and G preserves that K-structure. So V is also a KG-module. 122 Change ofofjeld field ofof aa linearlinear representation

(25.9) Let a: G G -+-+ GL(V)GL(V) be be an an irreducibleirreducible FG-representationFG-representation such thatthat KK = EndFG(V) isis aa field, andand letlet p,8 bebe thethe representationrepresentation ofof GG on V regarded asas aa KG-module, andand Xx the character of P.p. ThenThen (1) K K == F[X], F[x], where where F[X]F[x] is is the the F-subalgebra F-subalgebra ofof KK generated generated byby thethe elements Xx (g),(g), gg E G. (2) Assume LL is a a normalnormal extension ofof KK and a EE Gal(K/F)#.Gal(K/F)#. ThenThen L ®KBK VV" ° is not LLG-isomorphic G-isomorphic to to L L®K BK V V.

Proof. Let Let X X bebe a a basis basis for for V V over over K, K, m m = = I X1,1x1, andand MM the the enveloping enveloping algebra algebra of a inin EndF(V).EndF(V). ByBy definitiondefinition gaga = g18gp forfor each each g g E E G G andand MM isis thethe F- subalgebra generatedgenerated by by Ga. Ga. ThusThus MM consists ofof thethe elementselements EU,(~~),Eag(ga), aga, EE F. ByBy 12.16,12.16, the mapmap E>ag(ga) a, (ga) i-+ H >agMx(g,8) E a, Mx(g/?) is is an an isomorphism isomorphism of M with the ring KmxmKmXm ofof allall m m by by m m matrices matrices over over K. K. Hence Hence for for each each x x Ec K therethere is A E M with Tr(A)Tr(A) = = x,x, andand AA == E,,,EgEG a,Mx(gj?)agMx(gf) forfor somesome a, ag E c F.F. Now x == EgEG CgEG agX(g)a,x(g) EE F[X].F[X] SoSO (1)(I) is is established. established. Assume the hypothesis of (2)(2) and let EE bebe thethe fixed fixed field field ofof or.a. As As Cra ## 1,1, E 0# K. K. As As K K = = EndFG(V) EndEG(V) wewe may may assumeassume E == F. F. Let Let y y be be an an extension extension of ora toto LL andand U = L L ®K gK V.V. Then LL ®KBK V°V" == UY UY andand the the character character of of GG onon UU isis stillstill the character Xx of p.P. Thus if U is LG-isomorphicLG-isomorphic to to UYUY then Xx = XY, x Y, soSO Xx (g)(g) is is contained contained in in the the fixed fixed field field k ofk ofy. y. But But k nk Kn K= = F, F, so, so, by by (I), (1), K K = = F,F, a contradiction.

(25.10)(25.10) Let V V bebe anan irreducibleirreducible FG-module, k a Galois extension ofof F,F, andand K == EndFG(V). EndFG(V). ThenThen (1)(1) VKVK == ®aEA eaEA WaWa forfor some some irreducibleirreducible kG-modulekG-module W and some A C Gal(k/F) == F r with with rr = = ANr(W), ANr(W), where where Nr(W)Nr(W) == {y {y E E Fr : :WY WY G= W).W). (2) LetLet U U bebe an an irreducible irreducible kG-module. kG-module. ThenThen VV isis anan FG-submoduleFG-submodule ofof UU (regarded as an FG-module) precisely when U is kG-isomorphic toto W"W° for some aor EE r.r. (3) IfIf kk <5 K K then then A A = = I' rand and W° W" is iskG-isomorphic kG-isomorphic to to V V for for some some or a EE F.r. (4) IfIf KK isis a a Galois Galois extension extension ofof FF then then A A is is a a set set of of cosetcoset representatives representatives forfor Nr(W)Nr (W) inin 1.I'.

Proof. LetLet Fr == Gal(k/F). Gal(k/F). By By 25.7, 25.7, Fr x x G G is is represented represented onon VK.VK. Let Let WW be anan irreducible kG-submodulekG-submodule of of vk Vkand and M M = = (Wy:(Wy: yy E r).F). If M 0# VkVk then, then, byby 25.7.2, MM = Uk,uk, forfor some FG-submoduleFG-submodule U Uof of V. V. As As 0 0# 0 MM ## Vk,Vk, 0 ## U U # #V contradictingV contradicting the the irreducibility irreducibility of of V.V. HenceHence Vkvk = M M and and thenthen (1) follows from 12.5. AlsoAlso Vkvk isis generated asas an F-module by thethe copiescopies (a(aV:a V: a EE V)k#) of of V, V, so so Vk vk is is a homogeneous a homogeneous FG-module FG-module andand hencehence eacheach summandsummand W°Wa is thethe sumsum ofof copiescopies of V,V, asas anan FG-module.FG-module. ThisThis givesgives half of (2).(2). Representations over finite fields 123

Assume U isis anan irreducibleirreducible kG-module and V an FG-submodule of U. Let X be an F-basisF-basis ofof V.V. AsAs UU is is irreducible irreducible andand VV anan FG-submoduleFG-submodule of U, X generates U asas aa kG-module.kG-module. AlsoAlso 11 (9 B XX isis aa basis basis forfor Vkvk overover k,k, soSO wewe cancan define: a:Vk --> U

ax(1 x) H axx ax E k. XEX XEX Then a is a surjective kG-homomorphism,kG-homomorphism, soso Vk/ker(a) ==" U U as as a a kG-module. kG-module. Hence (1) and 12.5 implyimply UU ="= W°W" forfor somesome oa EE A.A. SoSo (2)(2) holds. In (3) we may regardregard VV asas anan irreducibleirreducible kG-module, kG-module, so, so, by by (2), (2), V V % - W°W" for some aQ E A. Then,Then, by by (I), (1), IAl 1Al = = dim~(V)/ dimF(V)/dimk(V)dimk(V) = _ Ik:Ik : FIFl = Il?IFI 1 so A = F l? andand (3) (3) isis established.established. To prove (4) let L be the extension generated by k and K and assumeassume KK isis Galois over F. ThenThen LL isis GaloisGalois overover K,K, and, and, by by (3),(3), 25.5,25.5, andand 25.3.3:25.3.3: V L- L ®K V K= ® L ®K U° °eGal(K/F) with U KG-isomorphicKG-isomorphic toto V. Now L ®KBK U° U" isis anan irreducible irreducible LG-module LG-module for each a EE Gal(K/F),Gal(K/F), since, since, by by 25.8, 25.8, V V is is absolutely absolutely irreducible irreducible as as a a KG- KG- module. FurtherFurther if if a o ## rs then then LL ®KBK U° U" 9 L ®KBK UTU` byby 25.9.2.25.9.2. On the other hand if a andand b areare distinct members ofof A withwith WaW° S= Wbwb thenthen LL ®kBk Wa Ll 2 L ®kWbBkwb SOso some irreducible occurs inin VL''withvL%ith multiplicitymultiplicity greatergreater thanthan 1,1, a contradiction. This completescompletes thethe proofproof ofof (4).(4).

A splitting jieldfield for a finite group G is a field F withwith thethe propertyproperty thatthat everyevery irreducible FG-representation isis absolutelyabsolutely irreducible.irreducible. NoticeNotice thatthat byby 25.8 an irreducible FG-moduleFG-module VV isis absolutely irreducibleirreducible precisely precisely when when F F = EndFc(V).EndFG(V). Hence, by 12.17:12.17:

(25.11) If FF is is algebraically algebraically closedclosed thenthen FF is is a a splitting splitting field field for for each each finite finite group.

It will turn out in section 27 that if G isis aa finitefinite group then a splitting field for G is obtained by adjoining a suitable root ofof unity toto GF(p)GF(p) for any primeprime p.

26 RepresentationsRepresentations over over finitefinite fieldsfields The hypotheses of section 25 are continued in this section. In addition assume F isis ofof finitefinite order. order. The following observations makemake thingsthings gogo particularlyparticularly smoothlysmoothly when when FF is finite. 124 Change ofofjield field ofof aa linearlinear representation

(26.1) (1)(1) EachEach finite finite dimensional dimensional division division algebra algebra over over FF is is a a finite finite field, field, and and hence a Galois extension of F. (2) IfIf VV is is an an irreducible irreducible FG-module FG-module then then EndFG(V) EndFG(V) isis aa finitefinite GaloisGalois ex-extension of F.

The first remark follows from the well knownknown factsfacts thatthat finitefinite divisiondivision rings are fields, and that every finite field is Galois over each of its subfields.subfields. The second remark is a consequenceconsequence of the firstfirst and the hypothesis that V isis ofof finitefinite dimension over F.

(26.2) Let V V bebe an an irreducibleirreducible FG-module,FG-module, k a finitefinite extension of F, and and Pr == Gal(k/F). Gal(k/F). Then Then (1) Vkvk = = ®QEA eaGA W°WO for for somesome irreducibleirreducible kG-modulekG-module W and anyany setset AA ofof coset representatives forfor Nr(W)Nr(W) in r.P. (2) LetLet UU be be an an irreducible irreducible kG-module. kG-module. ThenThen VV isis anan FG-submoduleFG-submodule ofof UU precisely when U is kG-isomorphic toto WuW° for some aa EE F.r.

Proof. ThisThis is is a a direct direct consequence consequence of of 26.126.1 andand 25.10.25.10.

Let jr:n: G -*-+ GL(V)GL(V) be be an an FG-representation FG-representation and and KK aa subfieldsubfield of F. We say nn cancan be written over K if there exists an F-basis XX ofof VV suchsuch thatthat eacheach entry of MX(gr)isinKforeachMx(gn) is in K for each g g EE G.G.

(26.3)(26.3) LetLet jr:n: G G ---, GL(V) GL(V) be be an an irreducible irreducible FG-representation, FG-representation, K K aa subfield subfield ofof F,F, and and (a)(a) = = Gal(F/K). Gal(F/K). Then Then the the following following are are equivalent: equivalent: (1) rrn cancan bebe writtenwritten overover K. (2)(2) VV = = F F ®KU @K U for for some some irreducible irreducible KG-submodule KG-submodule UU ofof V.V. (3)(3) VV is is FG-isomorphic FG-isomorphic toto V°.V".

Proof. TheThe equivalenceequivalence of (1) andand (2)(2) is trivialtrivial asas is thethe implicationimplication (1) implies (3). Assume (3) and let UU bebe anan irreducibleirreducible KG-submoduleKG-submodule of V. V. ThenThen Fr == Gal(F/K) Gal(F/K) = = NF(V) Nr(V) as as f' r = = (a) (m) and and V V = ZV°. V". Hence, Hence, by by 26.2, 26.2, OFuF = V. V. That is (3)(3) implies (2).(2).

(26.4)(26.4) LetLet V V be be anan irreducibleirreducible FG-module, FG-module, KK aa subfield subfield of F, U U anan irreducible irreducible KG-submodule of of V, V, and and E E = = NF(U).NF(U). ThenThen VV == F ®EBE U.

Proof. LetLet UuF F == F BEU®E U andand rh == Gal(F/E). ByBy 26.2,26.2, UFuF = =®QEA eaEA VQVa, where A is a set of coset representatives for for A A = = Nr(V)NF(V) inin r.F. Let L bebe the fixedfixed field of A andand WW an an irreducible irreducible LG-submoduleLG-submodule ofof V; as VV is is Representationsepresentations overoverjnite finite jeldsfields 125 a homogeneous EG-moduleEG-module we we may may assume assume U U I < W. If LL # F then,then, byby induction on on IF IF : : E E1, 1, L LgE ®EU U2 = W,W, while,while, by by 26.3, 26.3, V V Z = FF ®LgL W,W, SoSo VVZ = F FgE ®E U.U.ThuswemaytakeL= Thus we may take L = F.ButthenAF. But then A == 1soI so dimE(U) = dimF(UF) = I F : EI dimF(V) = dimE(V), so U = V. V. HenceHence F == NF(U) NF(U) = = E, E, and and the the lemma lemma holds. holds.

(26.5) LetLet VV be be anan irreducibleirreducible FG-module. FG-module. Then Then the the following following areare equivalent:equivalent: (1) VV can can bebe writtenwritten overover no proper subfield ofof F.F. (2) VV is is anan irreducible irreducible KG-module KG-module for each subfieldsubfield KK ofof F. (3) NAut(F)(V)NAU~(F)(V) == 1.1.

Proof.ProoJ ThisThis follows follows from from 26.3 26.3 andand 26.4.26.4.

An FG-module VV is condensed if V is absolutely irreducible and can bebe writtenwritten over no proper subfield of F.

Theorem 26.6. Let p bebe aa prime, prime, FpFp the the fieldfield ofof orderorder p,p, FFp p its its algebraicalgebraic closure, A the set of finite subfieldssubfields of of E,, Pp, andand aIl the set of pairspairs (F,(F, V) where F EE AA andand VV is is an an (isomorphism (isomorphism type type of of an)an) irreducible irreducible finite finite dimensional dimensional FG-module. DefineDefine a a relation relation,, r on on 0 1 byby (F,(F, V)r(K,V)T(K, U) ifif FF <5 KK and and VV isis anan FG-submodule ofof U. Let --- be be the the equivalence equivaleqce relation on 1 generatedgenerated by by T.r. Then (1) (F,(F, V)T(K, V)r(K, U) U) if if andand onlyonly if F <(. K K and and U U is is a a summand summand of of KK OF@,P V-V. (2) Let AA bebe anan equivalenceequivalence classclass of of --. ^-. Then, Then, for for each each F F E E A, A, AF AF == ((F,{(F, V)V) E E A) A} is is nonemptynonempty and and Aut(F) Aut(F) isis transitive onon AFAF. InIn particularparticular lAFpAFp 1 I= = 1 and the map A HI+ AFpAFp is is aa bijection bijection betweenbetween the setset ofof equiva-equivalence classes of -- and the isomorphismisomorphism classes of irreducibleirreducible finitefinite dimen-dimensional FpGFpG modules.modules. (3) IfIf (F, V),V), (K, U) E 1 withwith F <5 KK thenthen (F, V)V) --- (K, (K, U) U) if if and and onlyonly ifif (F, V°)/(K,Vu)r(K, U), U), for for somesome aa eE Aut(F).Aut(F). (4) In In eacheach equivalenceequivalence class A of ^--- there exists aa uniqueunique FF EE AA suchsuch thatthat the members of AF are condensed. Indeed forfor (Fp(Fp,, V) E A, FF == EndFpG(V) EndFpG(V) and (F, V)V) EE AA with with VV a a condensed condensed FG-module. FG-module.

Proof. PartPart (1)(I) follows from 26.2.26.2. Let Let (F, (F, V)V) EE a.Q. IfIf EE < FF then then therethere isis anan irreducible EG-module VEVE of V, and we saw during thethe proof ofof 25.1025.10 thatthat VV is a homogeneous EG-module, soso VEVE is determined up to isomorphism. Write VpV, for VFp.VF,. Let (K,(K, U) E a.0. Claim Claim (F, V)V) -- (K, U)U) ifif andand onlyonly ifif VpVp = Up. Up. The The suffi-suffi- ciency of Vp = UpUp isis immediateimmediate from the definition of --; to prove necessity 126 Change offieldof field ofof a a linearlinear representationrepresentation it suffices toto taketake (F, (F, V)r(K,V)/(K, U) U) and and to to showshow VpVp =2 Up.Up. ButBut this follows from the lastlast paragraph. Let A be an equivalence class class of of -- - and (Fp,(F,, V)V) EE A.A. ByBy 26.2,26.2, for each F EE A, VF == (DaEA eaEAWa,Wa for for some some set set A A of of coset coset representatives representatives for NANA"~(F)(W) t(F)(W) in Aut(F). ByBy (1)(1) and and thethe claim,

AF = {(F, {(F, W°): Wa):a a EE Aut(F)}.Aut(F)}.

That is (2) holds. To proveprove (3) (3) observe observe that that if if(K, (K, U) U) E EC2 0and and F F 5 < KK thenthen (F,(F, UF)r(K,UF)/(K, U).U). Then, byby (2),(2), (F, (F, W)W) --- (K,(K, U) U) if if andand only if W"W° = UFUF for some aa EE Aut(F). So (3) holds. Let F == EndFPG(V). EndFp&'). Then (F, V) EE AA and,and, byby 25.8,25.8, VV isis anan absolutelyabsolutely irreducible FG-module. By 25.10.3, 25.10.4, and 26.5, VV cancan bebe writtenwritten overover no proper subfield ofof F, so so VV isis condensed.condensed. Finally suppose (K, U) isis anotheranother condensedcondensed membermember of A.A. ToTo completecomplete the proof ofof (4)(4) wewe must must show show K K = = F. Let k be the subfield ofof F,F. generatedgenerated by K andand F.F. Use Use 26.526.5 andand thethe factfact thatthat VV isis condensedcondensed as an FG-module to conclude V 25t V"V° forfor aa EE Aut(F)#;Aut(F)#; then then byby 26.2, F ®GF(p) V = ® VQ. a EAut(F)

Then by 25.3 and 25.5, k ®GF(p) V = k ®GF(p) (F ®GF(p) V) = k OF ( ® V y 1 yEAu[(F) /// ® (k OFV'). yEAut(F)

As VV isis absolutelyabsolutely irreducibleirreducible as as anan FG-module,FG-module, k k OF @F V VY Y is irreducible forfor each y. HenceHence kk ®FPgFp V has exactly IFIF :: Fp Fp II irreducible summands. But, by symmetry between K andand F,F, k®FP kBFp V V also also has has (K IK :: Fp F,J I irreducibleirreducible summands, sosoK=F. K = F.

Theorem 26.6 defines an equivalence relation on the finite dimensional representations of G over finite fields ofof characteristiccharacteristic p.p. This equivalence relationrelation has the property that each class contains aa representative over each finite field of characteristic p. Hence we cancan thinkthink ofof suchsuch aa representation representation asas writtenwritten over any finite fieldfield ofof characteristiccharacteristic p.p. However thethe lemmalemma suggestssuggests thatthat toto each class there is associated aa field F overover which the representation is best written: namely the unique field over which the representation is condensed. I willwill referrefer toto FF as as the the field field ofof definitiondefinition of thethe representation. FF can alsoalso be described as theth smallest field over which the representation is absolutely Minimal polynomials 127

irreducible. Equivalently F == EndF,G(V EndF,,c(V), ), where V is thethe uniqueunique F,G-moduleFpG-module in the class.

27 MinimalMinimal polynomialspolynomials In section 27 F isis aa field,field, V is a finite dimensionaldimensional vectorvector spacespace over over F, F, andand G is a group. Suppose for the moment that A is a finite dimensional algebraalgebra overover FF and F[x] isis aa polynomial polynomial ring. For a EE A and f (x)(x) _= Ym Cy=o o b;x`bixi EE F[x],F[x], define define f (a)(a) = CrZ0o biaibra` E A. Then thethe mapmap a,:aa: FF[x] [x] + A suchsuch that that a,: aa: f fH H f (a), is an F-algebraF-algebra homomorphism.homomorphism. TheThe assumptionassumption thatthat AA isis finitefinite dimensionaldimensional forces ker(a,)ker(aa) #54 0 0 since since F[x]F[x] is of infiniteinfinite dimension. AsAs F[x]F[x] isis aa prin-principal ideal domain (PID), ker(aa)ker(a,) isis aa principalprincipal ideal,ideal, andand indeedindeed there is a uniqueunique monic polynomial f,(x)fa(x) with ker(a,)ker(aa) == (f,).(fa). ByBy definitiondefinition fa isis the minimal polynomial ofof a. ByBy constructionconstruction fafa dividesdivides eacheach polynomial polynomial which annihilates a.a. Further fa isis monicmonic ofof degreedegree atat mostmost dimF(A),dimF(A), sincesince dim~(A)dimF(A) >> dim~(F[xIl(f,))dimF(F[x]/(fa)) = deg(fa). deg(f,). Applying these observations toto the n2-dimensional F-algebraF-algebra EndF(V), it follows thatthat eacheach gg EE EndF(V)EndF(V) hashas aa minimal minimal polynomial polynomial min(g) min(g) = = min(g,min(g, F,F, V). Observe that if n: A A + B is an F-algebra isomorphism then then a a andand anan have the same minimal polynomial.polynomial. In In particularparticular if if X X isis aa basis forfor V then MX:Mx: EndF(V)EndF(V) +-+ Fn"nFnXn is such an isomorphism, soso min(g)min(g) == min(Mx(g)). If K isis anan extensionextension ofof F,F, then then Ml®X(1 MIBx(l (9 @ g) = MX(g), Mx(g), soso thethe minimalminimal poly-polynomial of 1 ®@ g over K dividesdivides the minimal polynomial of g over F. IndeedIndeed an easy applicationapplication of rationalrational canonicalcanonical formform showsshows thethe twotwo minimalminimal poly-poly- I nomials are equal. (Reduce to the case where V is a cyclic FG-module. Then there isis aa basisbasis XX == (xi:(x;: 11 (< ii <( n) forfor V in which xlgxjg == xi+lxj+1 for for i i << n and rz=o xng == -- ~yziEn o ai+lxi,aj+lxj, wherewhere f f (x) (x) = = xnx" ++ ~:iia,x`alxi is is thethe minimalminimal polynopolyno- mial for g over F. Then f (1 (9@ g)g) = 0 and if h(x) E K[x]K[x] properlyproperly divides f thenh(1then h(l (9@ g) # 0Oas as [1(1 @xi:® x1:11 5< i (deg(h)+< deg(h) + 1}1) is linearly independent and 1 @®xj+1 xi+l = (1 (1 (9 @ xl)g`.)x~)~'.) ThusThus we we havehave shown:shown:

(27.1) The minimal polynomial of a linearlinear transformation isis unchanged by extension of of thethe basebase field.field. ThatThat is,is, ifif KK is is an an extension extension of of F F and and g gE EEndF EndF(V), ( V ), thenthenmin(g, min(g, F, V)V) ==min(l min(1 (9@g, g, K,K, VK). vK).

Let g EE EndF(V) and a EE F.F. We We saysay aa isis aa characteristiccharacteristic valuevalue for g if there exists v E V# with vg = av.av. We We callcall vv aa characteristic vectorvector for a.

(27.2) Let g E EndF(V) and a EE F.F. Then Then aa isis a a characteristic characteristic valuevalue for g ifif and only if a is aa root of the minimal polynomial of g. 128 Change ofofjield field of aa linear representation

Proof. LetLet ff (x)(x) = = min(g).min(g). If If a a isis not not a a root root of of f f then (f,(f, xx -- a) = 11 soso there existexist r, r, s s EE F[x] F[x] withwith f f rr ++ (x - a)s a)s = = 1. 1. Now Now ifif vv EE V#V' with vg == av then vv == vv1 1 == v(v(f f (g)r(g)(g)r(g) ++ (g(g -- aI)s(g)) == 0, 0, since since ff (g)(g) ==0 0 == v(g -- aI).a I). This contradiction shows characteristiccharacteristic valuesvalues of of g g areare roots roots of of f. f. Conversely if if a a is is a aroot root off, of f, then then f f= = (x(x - a)h, a)h, for for somesome h E F[x]. IfIf a is not acharacteristica characteristic value value of of g gthen then ker(g ker(g - - aa I) I) == 0 so (g-a(g -a I)-1I)-' exists.exists. Hence, Hence, as 0 = ff (g)(g) == (g (g -- aI)h(g), aI)h(g), we we also also have have h(g)h(g) == 0. 0. But But thenthen f dividesdivides h, a contradiction. g EE EndF(V) isis semisimplesemisimple if the minimal polynomial of g hashas nono repeatedrepeated roots.

(27.3) LetLet g EE EndF(V)EndF(V) and and assume assume min(g) min(g) splitssplits over F. Then g is semisimplesemisimple if and only if g isis diagonalizable.diagonalizable.

Proof. ThisThis isis ExerciseExercise 9.2.9.2.

(27.4) LetLet SS bebe aa finite subset of commuting elements of EndF(V),EndF (V), and assume the minimal polynomial of each number of S splits over F. Then Then therethere existsexists a basis X of VV suchsuch thatthat forfor eacheach s EE SS the the following following hold: hold: (1) MX(s)Mx(s) isis lowerlower triangular. (2) TheThe entriesentries onon the main diagonal of Mx(s) areare the eigenvalues of min(s). (3) IfsIf s is is semisimple semisimple then then Mx(s)Mx(s) isis diagonal.diagonal.

Proof. InductInduct onon n ++ IS I SI. 1. If If n n == 1 the result isis trivial,trivial, soso take take n n >> 1. If some g EE SS isis aa scalar scalar transformation transformation thenthen by by inductioninduction on on BSI, IS/, the result holds for S - {g}, (g}, and and thenthen alsoalso forfor S.S. SoSo nono membermember ofof SS isis aa scalarscalar transformation. transformation. Let g EE S.S. ByBy 27.2,27.2, gg possesses possesses a a characteristic characteristic value value a1.al. LetLet V1Vl bebe thethe eigenspace of ala1 for g. Then S C_C(g) C(g) 5< N(V1).N(Vl). As g is not a scalar, VlV1 # 0 V, so, by induction onon nn therethere is is a a basis basis X X1 1 for for V1Vl with MX,Mx, (SIv,)(SI v) as claimed in the lemma. In particular therethere isis aa 1-dimensional1-dimensional subspace subspace U U of of V1Vl fixedfixed by S. If g isis semisimple,semisimple, then, then, byby 27.3,27.3, V V = = ®im-t1 Vi, where Vi isis thethe eigenspaceeigenspace of the characteristic valuevalue aiai of g. Now,Now, choosing a basis Xi for V,Vi asas inin thethe last paragraph, we see thatthat XX = U` UyZl 1 Xi Xi is is a abasis basis with with the the desired desired properties. properties. So we can assume no member of S is semisimple. Thus (3) is established.established. Finally S acts on V/V/U U = V V and,and, by by inductioninduction onon n, therethere is a basis X for VV with MX(SIv)Mz(SIv) triangular. triangular. Now Now pick pick X X = = (xi: (xi: 1 15 < i i5 < n) n) withwith UU == (xi)(XI) andand (fi:(ii: 1 1 <5 ii <5 n) n) = = X; X; thenthen Mx(s) Mx(s) is is triangular triangular for for each s E S. g EE EndF(V) isis nilpotent if if g"g' == 0 0 for for some some positivepositive integer m, and g is unipotent ifif g g == I ++ h h for for some some nilpotentnilpotent h E EndF(V). MinimalMinimal polynomialspolynomials 129

(27.5)(27.5) LetLet gg EE EndF(V)EndF(V) andand n == dimF(V). dim~(V). ThenThen (1)(1) TheThe following following are are equivalent: equivalent: (i)(i) g gis is nilpotent, nilpotent, unipotent, unipotent, respectively. respectively. (ii)(ii) min(g)min(g) = = x', xm, (x (x - 1)',- l)m, respectively,respectively, for for some some positive positive integer integer m. m. (iii)(iii) There There exists exists a a basis basis X X for for V V such such that that MX(g) Mx(g) is is lower lower triangular triangular andand allall entriesent'ries onon thethe mainmain diagonaldiagonal are are 0,0, 1, 1, respectively. respectively. (2)(2) Unipotent Unipotent elements elements are are of of determinant determinant 1, 1, and and hence hence nonsingular. nonsingular. (3)(3) IfIf gg is is nilpotent nilpotent andand semisimplesemisimple then g = 0. 0. (4)(4) IfIf g g is is unipotent unipotent andand semisimplesemisimple then g = I. I. (5)(5) LetLet char(F)char(F) == p p > > 0. 0. Then Then gP" gP" = = 1 1if ifg gis isunipotent. unipotent. Conversely Conversely ifif gP=gP"' 1= for 1 for some some positive positive integer integer m, m, then then g g is is unipotent. unipotent. (6)(6) IfIf char(F)char(F) = = 0 0and and g gE EGL(V) GL(V) is is of of finite finite order order then then g g is is semisimple. semisimple. (7)(7) IfIf char(F) char(F) = =p >p 0> and 0 andg g E GL(V) E GL(V) is of is finiteof finite order order m m then then g gis is semisim- semisim- pleple ifif andand only if (m, p) = 1. 1.

Proof.Proof. AllAll partsparts ofof thethe lemma lemma are are reasonably reasonably straightforward,straightforward, but I'll makemake aa couplecouple ofof remarksremarks anyway.anyway. IfIf gg EE GL(V)GL(V) isis of of finitefinite order order m m the the minimal minimal polynomialpolynomial of of g divides xmx'" --1, 1, which which hashas no multiple rootsroots ifif (m, (m, char(F)) char(F)) = 1. 1. HenceHence (6)(6) andand halfhalf ofof (7)(7) hold. hold. PartsParts (4)(4) andand (5)(5) implyimply thethe remainingremaining halfhalf ofof (7),(7), sincesince any any powerpower ofof aa semisimple semisimple element element is is semisimple. semisimple.

RecallRecall a field FF is perfect if if char(F) char(F) = = 0 0 oror char(F) char(F) = = pp > 0 and F = FP, FP, wherewhere FPFP is is thethe imageimage ofof FF under under the the p-power p-power map map aa i-+ H aP. up. ForFor exampleexample finitefinite fields fields are are perfect asas are algebraically closedclosed fields. fields. We need the following elementaryelementary fact,fact, whichwhich appearsappears forfor exampleexample as the CorollaryCorollary on page 190190 ofof LangLang [La].[La].

(27.6)(27.6) IfIf FF is is perfect perfect then then everyevery polynomialpolynomial in F[x]F [x] is is separable.separable.

(27.7)(27.7) IfIf FF is is perfectperfect andand a EE EndF(V),EndF(V), then then there exists ff EE F[x]F [x] and and a a posi- posi- tivetive integerinteger e e with with (f, (f, f f') ') = 11 (where(where f'f' is is thethe derivative off) of f) andand f fe(a) e(a) = 0.0. If ,BB E EndF(V) withwith ff (,B) (B) = 0 0 then then ,B B isis semisimple.semisimple.

Proof.Proof. Let Let min(a) min(a) = = flm ny!Jei 1 fee' with with fif; irreducible. irreducible. Let Let mm ee=max(e;:l = max{el: l ii

The proof of the following lemma comes from page 7171 ofof [Ch[Ch 2].21.

(27.8) Let F bebe perfectperfect andand a EE EndF(V).EndF(V). ThenThen therethere exist j3, yY E EndF(V) with a == ,B j3 + y, y, ,Bj3 semisimple, and y nilpotent.nilpotent. Further ,Bj3 = = t(a) forfor somesome t(x) E F[x].

Proof. ChooseChoose f andand e e as as in in the the last last lemma. lemma. ThenThen therethere exist h, h1hl E F[x] with 1 == f'h ++ f fh hl.1. Define Define an F-algebraF-algebra homomorphism4: homomorphism o: F[x] F[x] + -+ F[x]F[x]by by (go)(x)(g@)(x) = g(x g(x -- f (x)h(x)).f (x)h(x)). Observe Observe go g@ _ (g-g'(g-g'f f h) h)mod mod f 2,f 2, so so inparticular in particular ff4 O = f -- f'f f h'fh = f= - ff (1- f- (1f h - 1) fhl) = 0 =mod Omod f 2. f Then 2. Then proceeding proceeding by by induc-induction on mm andand usingusing the the fact fact that that @ 0 is is a ahomomorphism, homomorphism, f @mf 0' = = 0 modmod f f2-. 2m. Choose mm with with 2m 2' > e.e. Then: Then: (27.8.1) fo'f 4" =_ = 0 0 mod mod ffe, ', so (f¢')(a)(f @")(a) = = 0. 0. Next, for for g(x) g(x) = = xi Tj aixi a,x' E F[x],F[xl,

gok = (aixi)ok= >ai(xgk)` = g(xok), as q5k@k is is a a homomorphism, homomorphism, so

(27.8.2) g@kgok = g(xok)g(~~k)for each gg E F[x]. Also x4 xo == xx -- ff hh = x x modmod f,f, so proceeding by induction on k, and using the fact that @ is a homomorphismhomomorphism withwith ff @0 = 0 0 modmod ff, , wewe conclude: (27.8.3) x@~xqk = x x mod f.f. We cancan nownow completecomplete the the proof proof of of the the lemma. lemma. Let Let t t= = x@", x0"', j3,B = = t(a),t(a), and y = a a -- P. j3. Then Then .f f (fl) (j3) == .f f (t)(a)(t)(a) == .f f (xom)(a)(x@")(a) = (fom)(a)(f @")(a) == 0, 0, by by 27.8.227.8.2 and 27.8.1, respectively. Hence, by 27.7, ,Bj3 is semisimple. Also, byby 27.8.3,27.8.3, (X(x - t) t) = = 0 0 mod mod ff, , so ye = (a(a -- ,B)e j3le = (x(x -- t)e(a) t)e(a) = = 0. 0. Thus Thus y y is is nilpotent. nilpotent.

(27.9) Let F bebe perfect and a!a EE GL(V).GL(V). Then (1) There exist a,,as, a,a EE GL(V) GL(V) withwith as,a,, semisimple,semisimple, a,a, unipotent,unipotent, andand a = asauasa = auas. (2) If a = cpµ = = µ pc with with , c,p pc GL(V),E GL(V), ( semisimple, and pµ unipotent,unipotent, then=then ( =as a, and and /-t p == au. a,. (3) ThereThereexistpolynomialst(x), exist polynomialst(x), v(x) v(x) E EF[x] F[x] witha, withal = = t(a) t(a)andau anda, == v(a).

Proof. ByBy 27.8, aa = +j3 y,+ withy, with P, j3,y cy EndF(V),E EndF(V), B j3semisimple, semisimple, y y nilpotent, nilpotent, and j3 = t(a), t(a), for for some some t(x)t(x) E E F[x].F[x]. As As ,Bj3 = t(a),t(a), $j3 EE Z(C(a) Z(C(cf) fl fl EndF(V)).EndF(V)). Let F bebe thethe algebraicalgebraic closure ofof F. ByBy 27.427.4 and 27.5 there isis a basis X of F withwith MX(,B)Mx(j3) diagonaldiagonal and and Mx(y)Mx(y) strictly lower triangular. FromFrom thisthis itit is evident that det(,B)det(j3) == det(er).det(a). Thus, as a isis nonsingular, nonsingular, so so is, is P.j3. Minimal polynomials 131

Let a,as = ,Bj3 andand a, a.= = I I -I- + j3-'P-1 y. Asp-'As j3-' andand y ycommute commute and and y y is is nilpotent, nilpotent, j3-',B-1y y isis nilpotent,nilpotent, so a,a isis unipotent. unipotent. ByBy constructionconstruction a a= = a,a, asa == auas.So (1) and (3) hold. Suppose 6 and pµ areare asas inin (2).(2). AsAs ,Bj3 E Z(C(a)), 6 and µp commutecommute with with ,B j3 and a,.a,,. ByBy 27.4,27.4, j3,B and and 6 can be simultaneously diagonalized over F, and hence j3-'6,B-1 is diagonalizable over F, so,8-1 so j3-'6 is semisimple by 27.3. Similarly a,p-' is unipotent. Finally as Pa, = a = cp, j3-'6 = a&-' is both semisimple and unipotent, so,so, byby 27.5.4,27.5.4, ,Bj3 = 6 and a, a = µ.p. as andand aa, are are called called the the semisimple semisimple partpart of a andand thethe unipotentunipotent partpart of a,a, respectively, and and the the decomposition decomposition a =a = a,a, asa == auasis called thethe JordanJordan decomposition of a. As As aa consequenceconsequence of of 27.527.5 andand 27.927.9 wewe have:have:

(27.10) LetLet aa E GL(V) bebe ofof finitefinite order.order. Then Then as as and and a, a are powers ofof a.a. If char(F) = 0 then aa = as,a,, while ifif char(F)char(F) = = pp > 0 then laslIas _= aI lalp~ p, andand laulIaul = Ialp.Ialp.

(27.11) Let F bebe perfect and aa cE EndF(V),EndF(V), b b a a characteristiccharacteristic value ofof aa inin F, U the eigenspace ofof bb for aa in V,V, andand KK anan extensionextension of F. Then Then UKuK is the eigenspace of b forfor 11 ®€3 a inin VK.vK. /.

Proof. This This is is essentially essentially an an applicationapplication of of JordanJordan Form;Form; I sketch a proof. Recall thethe mapmap ff H f(a) f (a) is isan an F-algebra F-algebra representation representation ofof F[x]F[x] on on VV withwith kernel (M(x)), where M(x) = min(a). rnin(a). Let Let M(x) M(x) = = Fl'= niZl 1 pi(x)elpi(x)" bebe thethe prime factorization ofof M.M. From the theory of modulesmodules over a principalprincipal ideal domain (cf. Theorems 3 and 6 on pages 390390 andand 397397 ofof LangLang [La]) [La]) wewe knowknow V = ®i-1 @;=, V(i),V(i), where where V V(i)(i) = =ker(pi ker(~i(a)~l). (a)ej ). IndeedIndeed as as the the polynomials polynomials p; pi(x)'' (x )e' are relatively prime the same holds in VK, SOso as V(i)K~(i)~ is containedcontained in thethe kernel (VK)(i)(vK)(i) of p{(x)eipi(x)" onon vK,VK,we we concludeconclude ~(i)~V(i)K == (VK(i)).(vK(i)). Thus without loss MM == (x - b)e. b)'. Again from the theory of modules over a PID, V = ®l=1 @:=, Vi,Vi, wherewhere V,Vi = v,F[x]viF[x] isis aa cycliccyclic modulemodule for F[x]F[x] with with annihilatorannihilator (x(x -- b)ei, b)" ,e e = = e1 el >2 e2 ez >L - . L> e5e, >3 1, 1, and and thethe invariantsinvariants elei areare uniquelyuniquely determined.determined. As VKvK = viKVK is is aa modulemodule withwith invariants ei,e1, wewe concludeconclude thethe eiel are also the invariants of F[x] on on VK,vK, andand itit remainsremains toto observeobserve thatthat UU == ®;=1 U;,Ui, with with UU Ui = U U fl n V, Vi of dimension 1.1.

(27.12) Let 7r:n: GG +-+ GL(V) be an FG-representation and K an extension of F. ThenThen dimK dimK(Cv~(gnK)) (CVK (g7rK))= dimF dimF(Cv(gn)) (Cv (g7r )) for for each each g g Ec G. 132 Change ofjieldof field ofof aa linearlinear representation

Proof. ThisThis isis aa directdirect consequenceconsequence of 27.11.27.1 1.

Recall the definition of a splitting field in section 25.

(27.13) Let G be a finite groupgroup ofof exponentexponent m m and and n: r:G G +-- GL(V)GL(V) an FG-FG- representation. Let Let k k = = mm if if char(F) char(F) == 0 andand kk == m,~m p,if if char(F) char(F) = = pp > 0.0. Let n = dimF(V). dimF(V). ThenThen (1) LetLet Xx be thethe charactercharacter ofof nr and and gg EE G.G. ThenThen X(g)~(g) isis a a sumsum ofof nn kthkth roots of unity. (2) FF is is a a splitting splitting field field for for G G if if FF is is finite finite and and contains contains a a primitive primitive kth kth rootroot of unity.

Proof. LetLet F F be be the the algebraic algebraic closure closure of of F.F. By By 27.4 27.4 there there is is a a basis basis XX for for VF V' such that Mx(g) isis triangular.triangular. By 27.10, g = gsg gsgu with with gs g, semisimple semisimple and and 1gsI Igsl dividing k, withwith g,g, unipotent, and with g,gs andand g,g powerspowers ofof g.g. ThusThus thethe entries on the main diagonal of Mx(g,) are 1 and so the entries of Mx(g) areare thethe samesame as those of Mx(gs).MX(gs). InIn particular particular if if a a isis suchsuch anan entry entry then then ak ak == 1 as (gs)k == 1.1. So (1) holds. Assume FF is is finite, finite, F F contains contains co, o, aa primitive kthkth root of 1, V is an irreducible FG-module, andand let K == EndFG(V). EndFG(V). ByBy 26.1.2,26.1.2, KK isis aa finitefinite field extending F and hence containing w.co. Let Let + r be thethe charactercharacter of V regardedregarded asas aa KG-module.KG-module. The argument of (1) shows +(g)Vr(g) E E F F for for each each g g Ee G. As i/r(g)+(g) E F forfor eacheach g eE G,G, we we concludeconclude from 25.9.1 thatthat KK = F.F. Thus Thus F F is is a a splitting splitting field field for G by 25.8, completing thethe proof.proof.

Recall that given representations cr a and BP of groupsgroups G andand H,H, respectively, respectively, therethere is a representationrepresentation a a @I®P B of GG xx H.H. The The definition definition ofof acr ®@I fB appears appears justjust before before thethe statementstatement ofof LemmaLemma 25.7.25.7. ThisThis representationrepresentation appears in the statementstatement ofof thethe next twotwo lemmas.lemmas.

(27.14)(27.14) LetLet GG <5 GL(V GL(V), ), MM = = CGL(V)(G), CcL(v)(G), and assume V is aa homogeneoushomogeneous FG-module. Write I == Irr(G, Irr(G, V, V, F)F) for for the the set set of of irreducibleirreducible FG-submodules of Vv and choose %Vi Ee I,1, 1 5< ii <5 m, m, with with VV = = ®ml Vi.6. LetLet K K = =EndFG EndFG(V1) (VI ) be a field and A = HomFG(VI, HomFG(V1, V). Then (1)(1) ThereThere exists YY = = (ai:(ai: 11 5< ii <5 m)m) cE A A withwith ViaiVIai == Vi andand a1 = 1.1. (2)(2) AA isis aa KG-moduleKG-module andand YY isis aa K-basisK-basis forfor A.A. YY inducesinduces a unique K- spacespace structurestructure on VV extendingextending the the F-structureF-structure such such that that ai cri E E HomKG(VI, Hom~G(v1, Vi)Vi) for each i.i. This structurestructure is preserved by G.G. (3)(3) TheThe map n: M -+ GL(A; GL(A; K) K) defined defined by by xir:,8xn: B H H 18x, Bx, xx eE M, M, P B E E A, A, isis an isomorphism. M preserves thethe K-space structure on V. Minimal polynomialspolynomials 133

(4) The The mapmap 7/r:+: A# A# + ± I defineddefined by 1:,8+: ,B HH Vl,8 V1,B is is a a surjection surjection and and definesdefines a bijection 0: S(A) SG(V) BH(b:beB) between thethe set S(A) of all K-subspaces of A and the set SG(V) of all FG- submodules of V. 04 isis aa permutationpermutation equivalenceequivalence of the actions of M on S(A) and SG(V). (5) TheThe mapmap B:8: MM x G -++ GL(V, GL(V, K)K) defined defined by

(x, g)8: uv HH vxguxg is a K(M xx G)-representationG)-representation whose whose imageimage isis GMGM andand whichwhich isis equivalentequivalent to the tensor product of thethe representationsrepresentations ofof MM onon A andand GG on Vl over K.K.

Proof. AsAs VV is is homogeneous homogeneous therethere existsexists anan isomorphismisomorphism al: VlVl +-+ ViVi of FG-modules. Composing ai with the inclusion ViVi Ec VV we may regard aiai as a member of A. ChooseChoose alar = 1. 1. Then Then (1)(1) holds.holds. As K == EndFG(VI) EndFG(Vl) and ai isis anan FG-isomorphism, K K = = EndFG EndFG(X) (Vi) for each i and aiai is also aa KG-isomorphism.KG-isomorphism. Thus we have a unique scalar multiplication ofof KK on Vi extending thatthat ofof FF such that ai eE HomKG(V1, HomKG(Vl, Vi), soso therethere isis aa K-spaceK-space structure on V extending that on F. ItIt isis defineddefined by:

m m F a(uai)= a(uai), u e Vi, a E K. i=1 i=1 Next AA == HomFG H0mFG(V1, (V1, ®m 1 Vi)vi) = ®m j HomFGHOmFG(Vi, (Vi ,Vi) Vi) is is isomorphic toto Km as an F-space. F-space. SimilarlySimilarly HomKG(V1,HomKG(Vl,V) V) is is isomorphicisomorphic to to KmKm asas aa K- space and is an F-subspace ofof A,A, soso AA = = HomKG(V1, HomKG(Vl, V)V) isis alsoalso aa K-space. Also, as Y is a K-linearly independent subset of A of order m, Y isis aa K-basis for A, so (2) holds. Evidently, forfor ,BP E A and x cE M,M, the the compositioncomposition ,8x,Bx is also in A, A, soso thethe map nr inin (3)(3) isis aa well-definedwell-defined KM-representation. If If six aix == aiai then, as G is irreducible onon Vi,Vi, Vi Vi 5 < Cv(x).Cv(x). HenceHence n r is faithful. LetLet (uj: (vj: 1 1 ( < jj < d)d) be be aa KK-basis -basis forfor V1.Vl . Then X = (vjai: 1 < j < d, l < i < m) is a K-basis forfor V.V. AnAn elementelement ofof thethe general linear group GL(A, K)K) on on AA (re-(regarded as a K-space) maymay be regarded asas an m by m matrix (aij)(aid) withwith respectrespect toto the basis Y ofof A.A. Given Given such such an an element element define define x xE eM M by by UjffiX vjai x = = xkEk vjaikak.ujaikffk. Then x cE M M with with x xn 7r = (aid), (aij), and xx preserves thethe K-structureK-structure onon V.V. HenceHence n r is an isomorphismisomorphism andand (3)(3) holds.holds. 134 Change of fieldId ofof aa linearlinear representation

Evidently *maps A# into I and the induced map 0 takes S(A) into SG(V) and preserves inclusion. It is also clear that 0q5 isis anan injection fromfrom thethe setset S1(A)Sl (A) of 1I-dimensional -dimensional subspaces subspaces ofof AA into into I.I. LetLet W E I andand letlet 7ri:ni: W +-k V, Vi bebe thethe ii th projection. 7rini isis trivialtrivial oror anan isomorphismisomorphism by Schur's Lemma, andand therethere existsexists an isomorphism B:f: VlVl + W. ThenThen ai ai = = Bnia;'ftiai 1 EE K andand a = >2aiaiCapi E E AA with a$a* == Via Vla == W, W, so so 0: q5: S1(A) Sl(A) -k + I Iis is a abijection. bijection. For For eacheach B,B, DD EE S(A),S(A), (B ++ D)O D)@ == BO Bq5 + + DO; Dq5; from from this this remark remark and and its its predecessor predecessor it it is is notnot difficultdifficult to complete the proof of (4).(4). Finally, by (3), M preserves the K-space structure on V. Hence the map 08 in (5) isis indeedindeed aa well-definedwell-defined K(MK(M x G)-representationG)-representation whose whose imageimage isis GM. The map ujaivjai H++ vj uj ® €9 ai ai induces induces an an equivalence equivalence of of 08 with with thethe tensortensor productproduct representationrepresentation (x, g): ai ® vi ra aix ® vj g of MM xx G G on on A A ® €9 V1, Vl, so so (5) (5) holds. holds.

(27.15) Let Gi, ii == 1, 1,2, 2, be be groups, groups, F F a asplitting splitting field field for for G1 G 1 andand G2,G2, andand AiAi a collection of representatives for for the equivalence classes of finite dimensional irreducible FF Gi Gi -representations. Then the map (7r1,(771 ,772) 7r2) t+ H n1ni ®7r2€9 772 is a bijec- tiontion between AA 1 x L2A2 andand thethe setset of of equivalenceequivalence classesclasses of of finitefinite dimensionaldimensional irreducible F(GlF(G1 xx G2)-modules.G2)-modules.

Proof. Let7riLet ni c E Li Ai with with module module Vi. Vi. By By 27.14.4 27.14.4 and and 27.14.5 27.14.5 there there is is aa bijection between the F(G1F(Gl x x G2)-submodules G2)-submodules of of V1 Vl ®€9 V2V2 andand thethe FG2-submodulesFG2-submodules of V2,V2, so,SO, as 7r2772 isis irreducible, so isis 7r1771 €3®7r2.772. ConverselyConversely letlet n:7r: G1G1 x x G2 +-* GL(V) be an irreducible FG-representation. By By Clifford'sClifford's Theorem,Theorem, 12.13,12.13, VV isis aa homogeneous FGi-module,FGi-module, so, so, by by 27.14.5, 27.14.5, 7r IT isis equivalent equivalent to to 7ri ni ®7r2€9 for some 7rini E Ai. Indeed 7rini is determined up to equivalence by the equivalenceequivalence class of irreducible FGi-submodules ofof V.V. SoSo the the lemmalemma holds. holds.

(27.16) LetLet G G bebe aa finite group and n:7r :G G + -- GL(V)GL(V) anan irreducibleirreducible FG-represen-FG-representation.tation. Then Z(G7r)Z(Gn) = (z)(z) isis aa cyclic cyclic groupgroup of of orderorder relativelyrelatively prime to thethe characteristiccharacteristic of F and,and, ifif FF contains contains a a primitive primitive IzIth 1~1th rootroot ofof unityunity w,w, then then zz actsacts onon VV byby scalarscalar multiplication multiplication via via a a power power wkwk of of ww with (Iz1,(lzl, k)k) = 1.1.

Proof.Proof. By By 12.15, 12.15,Z(Gn) Z(G7r) is is cyclic, cyclic, saysay Z(G7r)Z(Gn) == (z). (z). By By Exercise Exercise 4.3, 4.3, nn == JzI lzl isis relatively prime to char(F). So So we we cancan assume assume w o is is aa primitiveprimitive nthnth rootroot of of 11 inin F. Now Now z satisfies thethe polynomial polynomial f f (x)(x) == x"xn - - 1 1so so its its minimal minimal polynomial polynomial divides f andand hence hence has has roots roots powers powers of of w.w. So So by by 27.227.2 wkwk is is aa characteristiccharacteristic value for z for some 00 << k < n, and then by Clifford's Theorem z acts by scalar multiplication viavia wkwk on on V.V. Thus Thus n n = = 1z lz I 1 _= I wklok 1, soso (k,(k, n)n) = 1. 1. Minimal polynomialspolynomials 135

(27.17) Let V be anan irreducibleirreducible FG-module. Assume G is finite and a semidirect productproduct ofof H 2a_ G G byby X ofof primeprime orderorder p.p. AssumeAssume dim(V)dim(V) # p(dim(Cv(X))). Then V isis aa homogeneoushomogeneous FH-module andand ifif F is finitefinite of order prime to p thenthen VV isis anan irreducibleirreducible FH-module.

Proof. By By Clifford's Clifford's Theorem, Theorem, 12.13, 12.13, V V is is the the directdirect sum sum ofof the homogeneous components (Vi:(V,: 1 1 5 < ii

(27.18) LetLet p andand q be primesprimes withwith q q >> p, G G a group ofof orderorder pq, pq, XX E Syl,(G),Sylp(G), and V a faithfulfaithful FG-moduleFG-module with with (pq, (pq, char(F)) char(F)) == 1 and CV(X)Cv(X) = 0. 0. Then G is cyclic.

Proof. ExtendingExtending FF if if necessary,necessary, wewe maymay assume with 27.12 that F contains a primitive qthqth root of 1.1. By Exercise 2.5, G hashas aa normalnormal SylowSylow q-group H. AsAs (pq,(pq, char(F)) char(F)) = = 1, 1, V V is is the the direct direct sum sum of of irreducible irreducible FG-modules FG-modules by Maschke'sMaschke's Theorem, soso H isis faithful faithful onon oneone ofof thesethese irreducibles,irreducibles, andand hence we may assume V is an irreducible FG-module. So, by 27.17, V is a homogeneous FH-module. Hence 27.16 says H actsacts by scalar multiplication on V, so H <( Z(G). Z(G). Thus Thus GG == HX HX is is cyclic. cyclic. 136 Change ofjieldof field ofof aa linearlinear representation

Remarks. TheThe classical classical theory theory of of linearlinear representations representations of of finitefinite groupsgroups con- con- siders representations over the complex numbers where things gogo relativelyrelatively smoothly. Unfortunately many questions about finite groups requirerequire consideration of representations over fields of prime characteristic, particularly finite fields. ForFor exampleexample we'vewe've seen that the study of representationsrepresentations in thethe categorycategory of groups of aa groupgroup GG onon anan elementaryelementary abelianabelian p-groupp-group E is equivalent to the study of GF(p)G-representations on E regardedregarded as a GF(p)-space. Represen- tation theory over such lessless well behaved fields requiresrequires thethe kindkind ofof techniquestechniques introduced inin this chapter. Lemma 27.18 providesprovides one application of these techniques, and we will encounter others inin section 36. Little use is made in this book of the Jordan decomposition studiedstudied in in sectionsection 27. It isis howeverhowever fundamentalfundamental to the study of groupsgroups of LieLie typetype asas linearlinear groupsgroups or algebraic groups.

Exercises for chapter 99 1. LetLet U,U, V,V, andand WW be FG-modules.FG-modules. (1) DefineDefine 0:4: L(U, V;V; W) ->+ HomF((U,HOmF((U, HomF(V,HOmF(V, W))) by v(u(a4)))V(u(a!@)) =_ (u, v)av)a! for for uu EE U,U, V v EE V,V, andand aa! EE L(U, L(U, V; V; W). W). ProveProve 04 is is anan isomor-isomorphism of F-spaces. (2) Prove Prove 04 isis anan isomorphismisomorphism of L(U, V; F) with with HomF(U,HOmF(U, V*),V*), where V* is the dual of V.V. (3) GG preservespreserves ff EE L(U,L(U, VV; ; F)F) if f (ug, vg) = ff (u, (u, v)v) forfor eacheach g E G. Prove G preserves ff ifif andand only if f0f 4 E E HomFG(U, HomFG(U, V*). V*). (4) LetLet 0 6 bebe an an automorphism automorphism of of F F of of order order at at most most 2 2and and LG LG(V, (V, V v') B) the set of sesquilinear forms on V with respect to 06 which are preserved by G. Assume VV is an irreducible FG-module. ProveProve LG(V,LG(V, V9) v') # 0 if and only if V is isomorphicisomorphic to (Vs)*(Ve)* as an FG-module, in which case each member of LG(V, V ve)#9)# is is nondegenerate.nondegenerate. If V is absolutely irreducible prove the members of of LG(V,LG(V, V v') B) areare similar,similar, ifif 60 = 11 each each member is symmetric or each is skew symmetric, and and if 10116 1 == 2 some member is hermitian symmetric. 2. ProveProve LemmaLemma 27.3. 27.3. (Hint:(Hint: Use Use the the theorytheory of of modulesmodules over over aa PIDPID asas inin thethe proof of 27.11.)27.11 .) 3. Let Let n1nl andand n2n2 bebe FG-representations,FG-representations, Xixi thethe charactercharacter of nl,ni, andand Xx thethe character of 71nl 80 72. n2. ProveProve Xx == X1 ~1x2; X2,that that is, is, forfor eacheach g E G, X(g)~(g) =_ x1(g)x2(g>.X1(g)X2(g) 4. LetLet G G be be a a finite finite groupgroup andand Xx the character of a complex G-representation. Prove X(g) == X*(g)x*(g) = X(g-1) x(~-') forfor eacheach gg EE G,G, where where X* X* isis thethe charactercharacter of the dual representationrepresentation andand X(g)ft) is is the the complex complex conjugate conjugate of of X(g).~(g). 5. (Spectral(Spectral Theorem)Theorem) Let VV bebe aa finitefinite dimensionaldimensional vectorvector spacespace over the complex numbers andand ff aa positive positive definitedefinite unitary form on V. Then for Minimal polynomials 137

each g EE O(V,O(V, f) there there exists exists an an orthonormalorthonormal basis for (V, f) consisting consisting of characteristic vectors forfor g. In particular every elementelement ofof O(V,O(V, f)f) isis semisimple. 6. LetLet (Vi, (K, fi fi), ), ii == 1, 1,2, 2, be be 2-dimensional 2-dimensional symplectic symplectic spacesspaces overover a field F and let VV == V1 Vl ®@ V2.V2. Let O(V1)A(Vi) bebe thethe groupgroup ofof similarities gg of K; thatthat is gg EE GL(VV)GL(K) with with fi(ng, fi(xg, yg)yg) = X(g) h(g)fi(n, fi(x, y) forfor allall x,n, yy EE Vi,K, andand somesome X(g)h(g) EE F'.P. ProveProve (1)(1) ThereThere existsexists aa uniqueunique nondegenerate nondegenerate symmetricsymmetric bilinear bilinear form form f f = f,fl ®@ f2 f2 onon V V such such thatthat f(vl ® v2, U1 ® u2) = fi(vi, ui).f2(v2, u2), ui, vi E V. (2) ThereThere is aa uniqueunique quadratic formform QQ on V associatedassociated toto ff withwith Q(vl ®v2)@ v2) == 0 0 for for all all vivi EE Vi.V,. (3) (V,(V, Q) Q) is is a a 4-dimensional 4-dimensional hyperbolic hyperbolic orthogonalorthogonal space. space. (4) LetLet DiA, == O(Vi, A(Vi, fi),fi), GiGi = = O(Vi, O(V,, fi), f,), and and itn the the tensor tensor product product rep-representation of A = O1 Al Xx A2A2 onon VV (cf. (cf. the the convention convention beforebefore 25.7).25.7). ProveA7rProve An 5< O(V,A(V, Q)with(gi,Q) with (gl, g2)ir g2)n E E O(V, O(V, Q) Q) if if andand only ififk(gl) h(g1) = )1(g2)-1.~(~2)-'.ker(n) ker(ir) == (@I, h-'{(,LI,-1I):.l I): h E F#}.F'}. (5)(5) LetLet a:a: (V1, (Vl, fl) ->+ (V2, (V2, f2) bebe anan isometry.isometry. ProveProve therethere isis a a uniqueunique t EE GL(V)GL(V) withwith (u(u (&@ va)t va)t = vv ®@ ua.ua. Prove Prove tt isis aa transvectiontransvection or reflection inin O(V, Q), (O17r)`(Aln)' == O2ir,A2n, and (GI7r)`(Gin)' = G27r. G2n. (6) O(V,A(V, Q)Q) = = (An)(t).(on)(t) (7)(7) Q(V,O(V, Q) Q) = = (G1G2)ir (G1G2)n = Z SL2(F) SL2(F) * *SL2(F), SL2(F), unless unless IFSI Fl = 2. 2. 7. IfIf itn isis an an irreducible irreducible FG-representation FG-representation and and Or a E Aut(F), thenthen 7r°nu is is anan irreducible FG-representation. IfIf xX isis thethe character ofof nit thenthen X°xu isis thethe character of nu,7r°, where where xU(g)X°(g) = = (X(~(g))" (g))' for g E G. 8. LetLet VV be be an an n-dimensional n-dimensional vector vector spacespace overover aa fieldfield FF of of primeprime charac-charac- teristicteristic p andand xn anan elementelement ofof orderorder pp inin GL(V). GL(V). Assume Assume nn >> p. p. Prove Prove dim(Cv(x))dim(Cv(n)) >> 1. 1. 9. Let Let VV be be aa finite finite dimensionaldimensional vector space overover aa fieldfield F, F, ff aa nontrivialnontrivial sesquilinearsesquilinear form on VV withwith respectrespect toto anan automorphismautomorphism 96 ofof finitefinite orderorder m, and G = O(V, O(V, f). f ). Assume Assume G G is is irreducible irreducible on on V. V. ProveProve thatthat either (1)(1) VV is is FG-isomorphic FG-isomorphic to to Ve V' andand G G preserves preserves a a nondegenerate nondegenerate bilinear form on V, V, or (2) mrn isis eveneven andand VV is is FG-isomorphic FG-isomorphic to to V02 V" butbut notnot toto VB. v'. FurtherFurther V = F F ®K mK U U and and G G preserves preserves a a nondegenerate nondegenerate hermitian hermitian symmetricsymmetric form on U,U, wherewhere KK isis the the fixed fixed fieldfield ofof 92o2 andand UU isis anan irreducibleirreducible KG-submodule of V.V. (Hint: Use Exercise 9.1 and the fact that (V*)* is is FG-isomorphicFG-isomorphic to V.) 10. Let itn bebe an an irreducible irreducible CG-representation CG-representation andand ora a 1-dimensionalI-dimensional CG- representation.representation. Prove itn ®@ ora is an irreducible CG-representation. 10

Presentations ofof groups

A groupgroup FF isis free free with with freefree generating generating setset XX ifif it it possesses possesses the the following following universal property:property: eacheach functionfunction a:a: X + H ofof XX intointo aa groupgroup H extendsextends uniquely to a homomorphism of F intointo H.H. We We findfind inin sectionsection 2828 thatthat forfor eacheach cardinal C there exists (up to isomorphism) a unique freefree groupgroup F withwith freefree generating set of cardinality C. Less precisely: F is is thethe largestlargest groupgroup generatedgenerated by X. If W is a set of words in the alphabet XXU U X-',X-1, it develops that there is also aa largest group G generated byby XX withwith ww = 11 inin G forfor each w EE W.W. ThisThis is the group Grp(X : W) generatedgenerated byby XX subjectsubject toto thethe relationsrelations w w == 11 for w EE W.W. In section 29 we investigateinvestigate Grp(X:Grp(X: W) whenwhen XX = {x1, {XI, ...... , ,x, x,) } isis finitefinite and W consists of the wordswords (xi xj)'"ii = 1, 1, for for suitablesuitable integralintegral matricesmatrices (mid). (mij). Such a group is calledcalled aa CoxeterCoxeter group.group. ForFor exampleexample finitefinite symmetric groups are Coxeter groups. We find thatthat CoxeterCoxeter groups admit aa representationrepresentation n:ir: G + O(V, Q)Q) wherewhere (V, (V, Q)Q) is is an an orthogonal orthogonal space space over over the reals and XnXir consists ofof reflections. If G is finite (V, Q)Q) turnsturns out to be Euclidean space. Finite Coxeter groups are investigated via this representation in sectionsection 30,30, whichwhich developsdevelops the elementary theorytheory of root systems.systems. The theory of Coxeter groups will be used extensively inin chapterchapter 14 toto studystudy the classical groups from aa geometricgeometric pointpoint ofof view.view.

28 FreeFree groupsgroups An object G in an algebraicalgebraic category AA is saidsaid to bebefree free with freefree generating set X if X is a subset of G and, whenever HH is an object inin AA andand a:a: X + H is a function from X into H, there, there,exists exists a unique morphism B:P: G + H of GG into H extendingextending a. This This sectionsection discussesdiscusses freefree groups. But first recallrecall that a monoid isis a set G together with an associative binary operation on G possessing an identity 1.1. Here's anan example of a monoid. Let X be a set. A word in X isis aa finitefinite sequencesequence xlx2~1x2...... X,X, with xi in X; n is the length of the word. The empty sequence is allowed and denoted by 1. Let M be the set of words in X andand definedefine the productproduct of two words xlx1 ...... x"x, andand ylyi . . . y,ym to to be be the the word word xl xl ...... x,yl.x"yl .... .Ym y, ofof lengthlength n n ++ m. Observe thatthat MM isis a monoid with identity the empty sequence 1.1. IndeedIndeed

(28.1) M isis aa freefree monoid with free generating set X. Free groups 139

For if HH is is a a monoid monoid andand a:a: X X + H is is aa functionfunction thenthen a cancan bebe extendedextended toto aa morphismmorphism /3: fi: M M + H defineddefined by (xl(xl ...... x7f)p x,)fi = xla xla ... . .x7za. .x,a. /3fi isis wellwell defined as each word has a unique representation as a product of members of X.X. EvidentlyEvidently /3 fi is the unique extension of a. Indeed Indeed in in generalgeneral inin any any algebraicalgebraic categorycategory if X isis aa generatinggenerating set for an object G and a: X + H isis aa functionfunction thenthen therethere isis atat most one extension fifi of a toto aa morphism of G intointo H.H. This This is is because if f'fi' is is another another extension extension thenthen KK == {g {g E E G: G: gj3 gfi == g/3'} gfi'} isis aa subobjectsubobject of GG containingcontaining X.X. Next assume X = Y Y UU Y-'Y-1 withwith YY fl fl Y-' Y-1 = = 00 andand y i-+H y-1y-l isis aa bijection of Y with Y-'.Y-1. SetSet (y-l)-l(y-1)-1 == yy forfor each y E Y; thus xx HH x-1X-l isis a a permutation permutation of XX ofof orderorder 2.2. DefineDefine twotwo wordswords u andand ww toto bebe adjacentadjacent if if therethere exist exist words words a, bb EE MM andand xx EE XX such such thatthat u == axx-lb axx-'b andand w w = = ab, ab, or or vice vice versa. versa. Thus adjacencyadjacency isis aa reflexivereflexive andand symmetricsymmetric relation. relation. Define Define anan equivalenceequivalence relation -- on M byby uu -- w if there existsexists a a sequence sequence u u= = ul, u1, ...... , u,, u = w ofof wordswords such that u,ui andand u;+lui+l areare adjacentadjacent for each i,i, 1 5< i < n.n. That That isis --- is is the the transitive transitive extension extension ofof thethe adjacency adjacency relation.relation. Write w forfor thethe equivalence classclass of of aa wordword w w under under -- - andand letlet FF be be thethe set set of of equivalenceequivalence classes.

(28.2)(28.2) If If u,u, v,v, w EE MM withwith uu -- v then uwuw -- - vwvw and wuwu --- wv.wv.

Proof. ThereThere is a sequence uu == u1,u 1, ...... , ,u, u, =+= v vof of words words with with u, ui adjacentadjacent to to ui+1.ui+l. Observe U~Wuiw = wiwi isis adjacentadjacent to w;+1wi+l andand uwuw = w1,wl, ...... , ,w w, = =VW, vw, SO so UW "VW.- VW.

Now define a product on F byby uvii3 = = ii-D. iZ. By 28.228.2 this product isis wellwell defined.defined. Further the product ofof thethe equivalenceequivalence classesclasses ofof the elements xnx;', 1, ...... , ,xi xcl 1 is is anan inverse for XIX1 ...... Xn,X,, so F isis aa group.group. HenceHence

(28.3)(28.3) FF is is a a group group and and w w i-+ H w W isis a a surjective surjective monoid monoid homomorphismhomomorphism of M onto F.

(28.4)(28.4) FF is is a a free free group group with with free free generating generating set set Y. P.

Proof. ObserveObserve first first thatthat YY generates F. This This followsfollows fromfrom 28.3 together with the fact thatthat X generates MM andand XX= = Y y UU Y-1. P-'. Now let H be a group andand a: a: EY +- H H a function.a function. Define Define /B: fi: XX + H by y/3yfi == ya andand y-lfiy-l Q = (ya)-1 (ya)-' forfor y y E E Y. Y. As As M M is is a a free free monoid monoid on on X, X, /3 fi extends to a morphismmorphismy:M y: M + H.H.Define6:F Define S: F + HHbyW6 by 08 = =wy.Imustshow6iswell wy. I must show S is well defined; thatthat isis ifif u u --- vv thenthen uyu y == vy. v y It . Itsuffices suffices to to assume assume u u isis adjacentadjacent to v, saysayu=axx-'bandv=ab.Thenuy u = axx-1b and v = ab. Then uy=(axx-'b)y = (axx-1b)y =ayxfi(xfi)-'by = ayx/B(xp)-l by =ayby, = ayby, 140 Presentations ofof groups as desired. EvidentlyEvidently 6b is is a a homomorphism homomorphism extending extending a. a.As As Y f generatesgenerates F,F, an earlier remark shows 6b is the unique extension ofof a.a.

Lemma 24.8 shows thatthat forfor each set S there exists a free free group group withwith free generating set S. The universal property implies:implies:

(28.5) UpUp toto isomorphismisomorphism there there existsexists aa unique free group with free generating set of cardinality C for each cardinal C.

If W C2 MM isis aa setset of words in X, write Grp(Y :: W) W) for for the the group group FIN, F/N, where N == (WF) (wF) isis the the normal normal subgroup subgroup ofof FF generated generated byby thethe subsetsubset Ww of F. Grp(Y :: W) W) is is the the group group generatedgenerated by by Y Y subject subject to to the the relations relations w w = = 1 forfor w cE W.W. ThatThat is is Grp(Y Grp(Y :: W) W) isis thethe largest group generated by the set Y in which w = 1 1 for for eacheach w cE W.W. ToTo be more preciseprecise Grp(Y Grp(Y : W): W) = = FIN F/N = GG withwith Y and W identified with (yN:(jiN: y cE Y)Y) and (wN:w(w- N: w E c W), respectively. As As Y Y generates generates F, F, Y Y generates generates G. G. As As w wE E N, N, w w == 11 inin GG forfor each w EE W.W. SoSo GG isis generatedgenerated byby YY and each of the words inin W isis trivial.trivial. I'llI'll also saysay thethe relationrelation ww == 11 isis satisfiedsatisJied in G to indicate thatthat ww == 11 in G.G. G isis the largest group with these propertiesproperties inin thethe followingfollowing sense:sense:

(28.6) Let a:a: Y 4- Ya Ya be be a functiona function of of Y Y onto onto a a set set Ya,Ya, HH a a group group generated generated bybyYa,andWasetofwordsw Ya, and W a set of words w = y8, y:' ...... yFinY~Y-lynn in Y UY-lwithwa with wa == (y,(yla)" a)" ... (y,a)"(yna)8n == 1 in H for each w E W.W. (That(That is H isis generated by Ya and satisfies the relations ww = 11 for for ww E E W.) W.) Then Then a aextends extends uniquely uniquely toto aa surjective surjective homomorphism of Grp(Y : W) onto H.

Proof. LetLet F F be be the the free free group group on on Y.Y. Then there existsexists aa uniqueunique homomorphism homomorphism j?:8: FF -4 H Hof of F Fonto onto H H extending extending a.a. Let Let NN == (WF) (WF) and and b:6: vv ++H vN vN thethe natural map map of of F F onto onto G G = = FIN.F/N. ThenThen N = ker(b). ker(6). For ww == xl ...... xn x, EE WW with xi xt E E X, X, 1 1 = = wawa == xla ...... x,a xna == xlj?xl/' ...... x,j? xn _= (XI,(XI,...... xn)j? xnV = = wj?, 4, so w E ker(j?).ker(b). ThusThus NN 5< ker(8),ker(j?), asas NN isis the the smallest smallest normalnormal subgroupsubgroup of F containing W. ButBut as as N N (< ker(j?), ker(8), j? induces a a homomorphism homomorphism y :y: G G 4 - HH withwith j? ==6y.Asj?issurjectivesois by. Asis surjective so isy.Also,fory y. Also, for Ey Y,c Y,ya ya = =yj? y = = yayyby == yy.

A presentation forfor aa group GG is a set YY of generators of G together with aa set W of words inin YY UU Y-'Y-l such that thethe relationrelation w w == 11 is is satisfiedsatisfied in G for each w EE WW andand thethe homomorphismhomomorphism ofof Grp(YGrp(Y : W)W) ontoonto GG described inin lemma 28.6 isis an isomorphism. I'llI'll summarize this setup with thethe statement G = Grp(Y Grp(Y :: W). W). Every groupgroup has atat leastleast oneone presentation; namely: Coxeter groups 141

(28.7)(28.7) ForFor eacheach groupgroup G,G, G = Grp(G: G~~(G:X~(X~)-' xy(xy)-1 = 1, 1, x, X, y y EE G)G) is aa presentation for G.G.

Proof.Proof. LetLet gg H++ k gbe be a bijectiona bijection of of G G with with a a set set G, G, let let F F be be the the free free group group on G, let WW be the set ofof wordswords ,?YE)-',x, xy(xy)-1, x, y yE E G,G, andand let let N N == (WF).(WF). Evidently GG satisfiessatisfies the relations defined byby WW soso the the map map g g ++ H gg extendsextends toto a homomorphismhomomorphisma a of F onto G with NN <_( ker(a). It remains toto showshow N N == ker(a).ker(a). Assume otherwiseotherwise and and let let v v = = Zl1 . . . Zn bebe a word in ker(a)ker(a) - N N of of minimal minimal lengthlength n.n. AsAs 11 E E N, N, n n > > 0.0. If If n n = = 1 thenthen xl x1 = = 1 1 and and v v= = jll a = 1.1x1flx12;' 1 EE W, contrary to the choice of v.v. HenceHence nn > 22 so v = x1x2u ZIZzu forfor somesome word u of lengthlength n - 2. 2. Now Now ww = xlx2(~)-'E W C N, sosow-1v w-'v = .1x2mu u EE ker(a) -- N.N. As (xix2)u(m)u is ofof lengthlength at most n - 1, 1, the the choice choice of of vv of of minimal minimal length length isis contradicted.

Here's aa slightlyslightly moremore nontrivialnontrivial example. The dihedral group of order 2n is defined to be the semidirect productproduct of of a a cyclic cyclic group group X X = = (x) of order n by a group Y = (y) (y) of of orderorder 2,2, withwith respect to the automorphism xYxy == x-'.x-1. TheThe case where n == oc co and and x x is is the the infinite infinite cyclic cyclic group group is is also also allowed. allowed. Denote Denote the dihedraldihedral groupgroup ofof orderorder 2n2n by by D2n.Dzn. DihedralDihedral 2-groups2-groups havehave alreadyalready beenbeen discussed in the chapter on p-groups. 4

(28.8) D2n = Grp(x, y : xn = y2 = 1 = xyx).

If n == oo co the the relation relation xn xn = = 1 1 is is to to be be ignored. ignored. TheThe proofproof ofof 28.828.8 isis easy.easy. Let D=XY=D2nandD = XY = Dzn and

By 28.6 therethere isis aa homomorphism homomorphism a aof of G G onto onto D D with with xa xa = = x x andand Ya ya == y.y. Then n == Ix 1x1 I divides 1x1, I I, so, so, as as 3" xn = = 1, 1, 1x1 1 I= = n n andand a: a: ji X 4 - XX isis anan isomorphism, where where X X= = (2).(x). Similarly,Similarly, setting setting Y Y= =(y), (y),a: a: Y4 - YY isis anan isomorphism. 1 = XXY33Y so so ZY V = (3)-'. Thus XX I!

29 CoxeterCoxeter groupsgroups Define a Coxeter matrix of size n toto bebe anan nn byby nn symmetricsymmetric matrix matrix with with 1 1 ss onon the main diagonal andand integersintegers ofof sizesize atat leastleast 22 offoff thethe main diagonal.diagonal. ToTo each Coxeter matrixmatrix MM = (mid) (mij) ofof rankrank nn therethere isis associatedassociated aa CoxeterCoxeter diagram: this diagram consists ofof nn nodes,nodes, indexedindexed by by integers integers 1 1 5 < ii _(< n,n, togethertogether with an edge ofof weightweight mij mid - - 2 2 joining joining distinct distinct nodes nodes i and i and j, j,1 _(1

willwill bebe mostmost concernedconcerned withwith CoxeterCoxeter matrices matrices with with thethe following following diagrams: diagrams:

1 2 n-1 n

An o-o

1 2 n-2 n-1 n Cn o-o o-®

n-1 1 2 n-2 Dn o-o 0 n

ThusThus aa diagramdiagram ofof typetype AnA, defines a Coxeter matrix of size nn with mijmid == 3 if Jili - j jI I == 1 1and and mid mij == 2 2 if if Ii li - - j Ij >1 >1. 1.Similarly Similarly a adiagram diagram of of type type Cn C, definesdefines aamatrixwithmij matrix with mid =3ifli- = 3 if Ji - jlj I == 1landi, and i, jj <1.mij=2ifli-j]> 1. AA CoxeterCoxeter systemsystem with Coxeter matrixmatrix M = (mid) (mij) of size n is aa pair (G,(G, S)S) wherewhere G is a group, S = (si: (si: 11 <_( i i< ( n) n) a a family family of of elements elements of of G,G, and and

GG ==Grp(S:(~~s~)~'j Grp(S: (sisj)m,i= = 1,1, 11 _(i < i In,< n, 15 1 < j <(n). n). GG isis aa Coxeter Coxeter group group ifif therethere existsexists aa familyfamily SS such such thatthat (G,(G, S)S) is is a a Coxeter Coxeter system.system. InIn thethe remainderremainder of of thisthis sectionsection let let (G,(G, S) S) be be aa Coxeter Coxeter system system with with matrixmatrix MM == (m1, (mij) .) of of sizesize n. n. Let Let S S == (si: 1 5< i <5 n). n). Notice Notice

s2= (SiSi)mii = 1.

(29.1)(29.1) LetLet TT be be the the set set of of conjugatesconjugates of of membersmembers ofof SS underunder GG and and forfor eacheach wordword rr == r1 rl .... . rn. rm in in the the alphabet alphabet S S and and each each t t E E T T define define N(r, t) = (i: t = ri < i < m}. Then,Then, if if r1 rl...... rm rn == rlri . ...rirk inin G,G, we we have IN(r,t)l(N(r, t)j =- IN(r', IN(rl, t)jt)l mod mod 22 for for eacheacht t EE T.T.

Proof.Proof. Let Let A A be be thethe setset product (f(±1}1) x TT andand for s EE S define snsn cE Sym(A)Sym(A) byby (E,(E, t)snt)sn = (EE(s, (~6(s,t), t), tS), t''),where where 6(s, 8(s,t) t) = _ -1 -1 ififs s = t t and and 8(s,6(s, t) = +1 +1 if if ss # # t. t.Observe Observe that that sn sn is is an an involution. involution. I'llI'll showshow (sinsjn)mii(sinsjn)"lj = 1 1 for for allall i, j; hence,hence, byby 28.6,28.6, nn extends extends toto aa permutation permutation representation nn of GG onon A.A. InIn particularparticular nn is is a a homomorphism homomorphism so, so, ifif gg == r1 rl .... . rm. rm c E G G and and r1 ri EE S,S, then then

tri...r;_1) (E, t)gn = (e, t)rln ... rnn = E r i , t8) i=1 Coxeter groups 143

t''...ri-') Further 3(ri,6(ri,tr'...'i-1) = --1 1 exactly when i E N(r,N(r, t). So (e,(E, t)gnt)g7r == (E(-l)IN(r,t)I,(~(-l)I~(',')l, tg)t8) and hence IN(r,IN(r, t) t)l I mod 2 depends only on g and not on r.

It remains to show (s,(si7rsj7r)m"jnsjn)"'l = = 1;1; equivalentlyequivalently a a = (si(s, ns,rsj7r)m'j n)"ll fixesfixes each t(s'si)»") (e,(E, t)t) E A.A. NowNow (E,(e, t)at)a = (Es, (~6,t(S~S~)""l) = (EE,(~6, t),t), as (s,~~)"~](sisj)m'j == 1,1, wherewhere of course 8 = 8(si, t)E(sj, ts')....

So we must show 86 == 1. 1. But But 8(si, 6(si, t(S,sjt('1'1)~) )k)_= -1 - 1for for some some 0 0 <5 kk << mij, m,j, preciseprecise- ly when t t = = (si (sisj)2ksi. s j )2ksi . Also 8(sj, 6(sj, t(S'Sj~(S~SJ)~'~) )ks')= --1 1precisely precisely when when t =t =(si (sisj)2k+1~i. sj )2k+1 si. Further, by Exercise 10.1, (si,(s,, sj) isis eithereither dihedral dihedral of of orderorder 2m, where m divides mij, or of order at mostmost 2.2. HenceHence if if t t $20 ((sisj))si then all terms inin 68 areare ++1, 1, while if t EE ((sisj))si((sisj))si exactlyexactly (2mij)/IsisjI(2mij)/lsisj I termsterms are -1,- 1, as as t t= = (sisj)dIs"'jIsi (S~S~)~~'~'Jlsi for 0 <5 d d < <(2mij (2mIJ)/lsisjl. )/ Isis j 1.

(29.2) TheThe membersmembers ofof SS areare involutions.involutions.

Proof. InIn thethe proofproof ofof 29.129.1 aa homomorphismhomomorphism 7rn of G into Sym(A) was con- structed for which sns7r waswas anan involutioninvolution forfor eacheach s s E S. So 2 = Isn Isn I I dividesdivides Is!.is (. But But of of course, course, as as (G, (G, S) S) isis aa CoxeterCoxeter system,system, s? s? == 1.1.

If H isis aa groupgroup withwith generating set R then the length of hh E H withwith respectrespect to R is the minimal lengthlength ofof aa wordword w w inin the the abhabet alphabet R R U U R-' R-1 such such that that w w = = h in H. Denote Denote thisthis lengthlength byby 1(h)l(h) = lR(h). l~(h).

(29.3) Let g cE GG and and rr = = r1 rl .... .rm . rm a a word word inin thethe alphabetalphabet S with gg = rr inin G.G. Define

r7(r)~(r) = II{t It E T:T: IN(r,IN(r, t) t)l I == 1 1mod mod 2}1 211 in the notation of 29.1. Then ~(r)1](r)= = l(g).l(g).

Proof. ByBy 29.1,29.1, if r' == g g then then 1](r) ~(r) =.1](r'),= .q(rl), whilewhile byby definitiondefinition of l(g) there is r'r' = riri ...... rirk with kk == l(g) and r' == g g in in G. G. So, So, without without loss, mm == l(g).l(g). So evidently ~(r) 1](r)5 < m m = = l(g). l(g). If If ~(r) 1](r)< < m m there there are are i, j,i, 1j, 51

(29.4) LetLet HH bebe a a group group generated generated byby aa setset RR ofof involutions.involutions. Then (H, R) is is aa Coxeter system precisely when the following Exchange Condition is satisfied: Exchange Condition: Condition: If riIf Eri ER, R, 0 50

Proof. SupposeSuppose firstfirst that (H, S)S) is aa CoxeterCoxeter system and letlet ri,rt, h, satisfy the hypothesishypothesis of the Exchange Condition. Then,Then, settingsetting r r = rorirorl ...... rn, r,, ri(r)~(r) <5 i '-'.__Y° l(roh)l(roh) << nn ++ 1 1 by 29.3, soso there there are are i, i,j, j,0 0 5 < i << jj <5 nn with with r;r,?-"..ro == rrj-'.'-roJ andand hence rir, ...... rj_1rj-1 = rt+i ri+l .... . rj..rj. By By 29.3, 29.3, ri(ri v(rl...... r,) r,) == 1(h) l(h) = = n, n, so so ii == 0. 0. Thus thethe ExchangeExchange ConditionCondition is is satisfied.satisfied. ConverselyConversely suppose (H, R)R) satisfies satisfies the ExchangeExchange Condition.Condition. LetLet a!: a: RR +-+ XX be a function intointo aa groupgroup XX such such that, that, for for each each r, r, s sE E R, R, (rasa)lrsl (rasa)I's' == 1. 1. It It will will sufficesuffice to show a!a extendsextends toto aa homomorphism of H intointo X.X. Let h EE H, n n == l(h), l(h), andand r1 rl ...... rnr, == h h = = s1 sl .... . sn. s, withwith ri,ri, si si cE R.R. Claim Claim r1arla...... r,a rna = slasla!...... s,asna andand {rt:(ri: 1 1 5 < ii 5< n]n) = {s,:1(si:1 5< i <5 n).n]. Assume not and pick aa counterexamplecounterexample withwith n n minimal. minimal. Then Then l(sl l (sih) h) = = n n - -1 1 << l(h),l(h), so,so, by by thethe ExchangeExchange Condition, Condition, s1r1 slrl ...... rk-1 rk-1 == r1 rl ... . .rk .rk for for some some k. k. Hence Hence s1r1slrl ...... rk-lrk+l.. rk-lrk+l ... .r, rn = hh = =s1 sl... . sn . .s, so so r1 rl.. ... rk_1rk+1 .rk-lrk+l ...... rnrn == S2 s2.. ... Sn..s,. Thus,Thus, byby minimalityminimality of n,n, r1a rla ...... rk_1ark+la rk-lark+la...... rna rna = seasp.. ... sna. s,a andand {S2,($2,...... , ,Sn) s,] == {ri:{ri: ii #0 k).k]. AlsoAlso if k << nn then, then, byby minimality ofof n, n, slurla!. siaria... . . rk_lark-la == r1a r1a ... . .rka .rka andand {Si, Isl, r1, rl , ...,. . . rk_11, rk-1) == jr, (rl ..., . . rk),. , rk], whichwhich combined combined with with thethe lastlast setset of equalities establishes thethe claim.claim. So So k k = = n, slrlsir, ...... rn-i r,-1 == h, h, and and {ri{rl,, ...... , ,rn_i) rnvl) == {s2, (s2, ...,. . sn. , s,).) . SimilarlySimilarly rise rlsl .... . sn_1.s,-1 == h handand {s1, {sl, .... . ,. ,Sn_1) s,-1) = {r2{r2...... , ,rn r,).). In particular jr,{rl...... rn) , r,) _ ={s1 (sl ...... sn . , s,),), establishing establishing half half the the claim. claim. ReplacingReplacing r1rl ...... ,rn, r,, andand si,sl, ...,. . . ,sn, s,, byby si,sl, r1, rl, ...... , , rn-ir,-1 andand r1,rl, sl,si, ...., . . , sn_i,s,-1, andand continuingcontinuing in this this manner, manner, wewe obtainobtain (sire)n/2=(rise)n/2(slr1)"I2 =(rls1)"I2 oror rl(~~r~)("-')/~ri(Siri)(n-1)/2= = ~l(rls~)(~-~)/~, Si(risi)(n-1)/2,withwith equalityequality of images under a failingfailing inin thethe respectiverespective case. ItIt followsfollows thatthat (slrl)"(sire)n == 1,1, soso thethe orderorder mm ofof s1r1slrl in HH dividesdivides n. But by hypothesis thethe orderorder ofof slarlaslarla divides divides m,m, so so equality equality ofof imagesimages underunder a doesdoes hold, hold, a a contradiction. contradiction. SoSo thethe claimclaim isis established.established. Since Since CoxeterCoxeter systemssystems satisfysatisfy the the ExchangeExchange ConditionCondition wewe cancan record:

(29.5)(29.5) Let g cE GG with with 1(g)L(g) == m m and and r1, ri, ttti cE SS with with r1rl ...... rn rm = = t1 tl ...... to tm == g. g. ThenThen (ri:{rt:11 5

NowNow back to thethe proofproof ofof 29.4.29.4. Define Define a: H ->.-+ XX by haha == r1a rla! ...... rna,ma!, forfor hh = = r1 rl ...... rn r, withwith nn == 1(h) l(h) and and r,ri EE R.R. TheThe claim claim showsshows a toto be be wellwell defined.defined. Let'sLet's seesee next next that that (rh)a (rh)a = = raharuha for for r rc ER. R. If If l(rh) l(rh) = =1(h) l(h) ++ 1 1 thisthis isis clear,clear, soso assumeassume not.not. Then,Then, byby thethe ExchangeExchange Condition,Condition, rr1rrl ...... rk-1 rkPl == r1rl ...... rk rk for some kk 5 < n.n. ByBy thethe claim, claim, rurla raria ...... rk-lark-1a = r1arla ...... rka.rka. Also Also rhrh = = r1 rl ...... rk-irk+1 rk-lrk+l .. .. . rn r,, isis ofof lengthlength at most n -- 1. 1. As As 1(h)l(h) = n, n, we we concludeconclude l(rh)l(rh) = = n n -- 1. 1. So So (rh)a(rh)a! = = r1a rla!...... (rk-i)a(rk+1)a (~~-~)a(r~+~)a...... ,,a rna= = rurla!.raria ... . .rnarna == raha,raha!, establishing establishing the the second second claim. claim. ItIt remainsremains to show gahagaha == (gh)a (gh)a forfor g, g, h h c EH. H. Assume Assume not not and and choose choose a a countercounter example with 11(g) (g) minimal.minimal. ByBy the the last last paragraph, paragraph, 1 (g)1(g)> > 1,1, so so g g = = rk, Coxeter groups 145145 rr EE R, R, k k EE HH with with l(k) = l(g)l(g) - 1. 1. ThenThen (gh)a(gh)a == ra(kh)a = rakaharakaha = = gahaguha by by minimality minimality of of l(g), l(g), completing completing the the proof. proof.

LetLet VV bebe an an n-dimensional n-dimensional vector vector space space overover thethe realsreals RR withwith basisbasis XX = = (xi:1(x,: 1 <5 i

(29.6)(29.6) (1) Q(xi)Q(x~) == 1/2. 112. (2)(2) (xi,(xi, xj) <5 00 for for ii 0# j, j, with with (xi, (xi, xj) xj) = = 0 0if if and and only only if if miimij = = 2. 2.

Proof.Proof. mii = 1,1, soso Q(xi)Q(xi) = (xi,(xi, xi)/2 == - -cos(h)/2 cos(n)/2 = = 1/2. 112. Similarly Similarly ifif ii ## j jthen then mil mij > > 2, 2, so so (xi, (xi, xj) xj) = -cos(7r/mid) cos(n/mij) <5 0 0 with with equality equality if if andand only only ifif milmij == 2. 2.

ByBy 29.629.6 and and 22.6.2 22.6.2 there thereis is aauniquereflection unique reflection riri onon VV withwithcenter center (xi)(xi); ; moreover vrivri == v v -- 2(v, 2(v, xi)xi xi)xi for for vv cE V.V.

(29.7)(29.7) For i 0# j, j, rirj rirj is isof of order order mil, mij (ri,, (ri ,rf) rj ) = ED2,,,,i, D2m,j, andand ifif milmij >> 2 2then then I (ri,(ri, rj) rj) is is irreducible irreducible on on (xi,(xi, xj).xi). '

Proof.Proof. LetLet UU = = (xi, (xi, xj), xj), D D = = (ri, (ri, rj), rj), m m= =mid, mij, and and 9 8= =7r/m. nlm. Observe Observe that, that, forfor a,a, b b cE III, R, 2Q(axl2Q(axl ++ bx2) bx2) = a2 a2 - - 2ab 2ab cos cos 98 + b2 b2 = (a(a -- b b cos cos0)2 812+ + b2(sin b2(sin8)2 9)2 >2 0,0, withwith equality precisely when a = b b = = 0. 0. Thus Thus Q Q is is a apositive positive definite definite quadratic quadratic formform onon U,U, soso in in particularparticular UU is is a a nondegenerate nondegenerate subspace subspace of VV andand hencehence VV = U U ® @ U1.u'. ButBut U1U' <5 xkxk <5 Cv(rk) CV(rk) forfor k = i i andand j,j, soso U1U' <5 Cv(D). Cv(D). HenceHence DD isis faithful faithful onon U.U. AsAs Q is positivedefiniteonpositive definite on U, U,(xk, (xk,xk) xk) = 1,= and(xi,1,and(xi, xj) xj) = _ - -cos9, cos 8, (U, (U, Q) Q) is is isometricisometric to to 2-dimensional2-dimensional EuclideanEuclidean space space R2 R2 with with thethe standardstandard innerinner product andand withxiwith xi = (1, (1,O) 0) andandx, xj == (cos(7r (cos(n -0),-@), sin(7r sin(n -0)) -8)) inin the the standard standardcoordinate coordinate system.system. Thus rir, andand rjr, areare the the reflections reflections onon R2R2 through through thethe verticalvertical axisaxis and thethe axis determined by n/27r/2-0, - 8, respectively. respectively. Hence Hence ri rir, rj isis the the rotationrotation through thethe angle -27r/m,-2n/m, andand therefore therefore is is of of orderorder m m as as desired.desired. Thus the firstfirst claim of 29.729.7 is establishedestablished andand thethe secondsecond is aa consequenceconsequence of thethe firstfirst andand Exercise 10.1.10.1. (Xk)(xk) and and xkxk fl U areare thethe onlyonly nontrivialnontrivial proper subspacessubspaces of U fixed by by rk rk and and if ifm m > > 2 2 then, then, by by 29.6.2, 29.6.2, (x,) (xi) # 0 xf x flf? U,U, soso DD isis irreducibleirreducible on U.U. 146 Presentations ofof groups

(29.8) LetLet WW == (ri:(ri: 1 <( i i< (n) n) be be the the subgroup subgroup of of O(V, O(V, Q) Q) generated generated by the reflections (ri:(ri: 1 1 5< ii <( n). n). Then Then there there exists exists a asurjective surjective homomorphism homomorphism a: G -++ W W withwith siasia == ri ri for for each each i. i. In In particular particular a a isis an an RG-representation RG-representation which identifies S with a set of reflections inin O(V,O(V, Q).

Proof. ThisThis isis immediateimmediate fromfrom 29.729.7 and 28.6.

(29.9) (1) S is of order n. (2) For each ii # j, IsisJsisj j I = = mi mij j and and (si, (si, s sj) j) = D2m;jDZmij.

Proof. TheThe map a: (si, (si, ssj) j) -++ (ri, (ri, r rj) j) induced induced byby thethe mapmap ofof 29.829.8 isis aa sur-surjective homomorphism.homomorphism. By By Exercise Exercise 10.1 10.1 and and 29.7, 29.7, (ri (ri,, rj) r j)=Grp(ri =Grp(ri, , rj: r: r?r? = rr: = =(ri (rirj)m~l) r j )mii) so,SO, asas (si,(si, ssj) j) satisfiessatisfies these relations, 28.6 says there is a homomorphism pY of (ri, rj) ontoonto (si, sj) with rkprkp = Sk.sk. ThenThen p /3 = = a-' a-1 so aa is an isomorphism and 29.7 implies (2).(2). As r,ri # rj forfor ii # j, (1)(1) holds.holds.

Let A0 == {1, {I, ... . ., . n} , n) be be the the set set of of nodes nodes of of thethe CoxeterCoxeter diagram of (G, S). TheThe graph of the diagramdiagram isis thethe graphgraph on on A0 obtained byby joiningjoining ii to j ifif thethe edgeedge between i and j inin thethe CoxeterCoxeter diagram diagram is is ofof weightweight atat leastleast 1,1, oror equivalentlyequivalently ififmij>3. mij 2 3.

(29.10) Let (Ok:(Ak: 1 5< k <5 r) r) bebe the the connected connected components components of of thethe graphgraph of the Coxeter diagramdiagram A0 of (G, S)S) and and let Gk = (Si: (si: i EE Ok).Ak). ThenThen (1) G G = G1G1 x . . x G, isis thethe direct product ofof the the subgroups subgroups Gk, Gk,1 1 5 < k k 5< r. (2) V isis thethe orthogonalorthogonal direct sum of the subspaces Vk == [Gka,[Gka, V],V], 11 (<- k 5< r.

Proof. If ii andand jj are are in in distinct distinct componentscomponents ofof A0 thenthen Isis/sisj/ j I= = mijmij = 2,2, so so [si, sj] sj] = 1. 1. Thus Thus GG isis thethe centralcentral product ofof thethe subgroupssubgroups Gk, Gk,1 1( < k k 5< r, and hence there is a surjective homomorphismhomomorphism pP of GIG1 x . . - x GrG, = D D ontoonto G with sip == si si for for each each i. i.Conversely Conversely S S satisfies satisfies the the CoxeterCoxeter relationsrelations in D, so by 28.6 there isis a homomorphism y y ofof GG ontoonto DD with si y y = si.si. ThenThen y == P-1, p-', soso /3 ,9 isis anan isomorphismisomorphism and (1) holds. Similarly,Similarly, by by 29.6.2, 29.6.2, xj xj EE xl,x; , so, ifif jj EE DaA, andand i EE Ab,Ab, then

Va=([V,rk]:kEA.)=(xk:kEDa)

Because of 29.10 it does little harm to assume the graph of thethe CoxeterCoxeter diagramdiagram of (G, S)S) is connected. In that event (G, S)S) is said to be an irreducible CoxeterCoxeter system. Coxeter groups 147

(29.11) AssumeAssume (G,(G, S)S) is is anan irreducibleirreducible Coxeter Coxeter system.system. ThenThen (1) G G acts acts absolutely absolutely irreducibly irreducibly on on V/ V/ Vv'. -L. (2) IfIf WW isis finite finite thenthen (V,(V, Q)Q) is is nondegenerate.nondegenerate.

Proof. LetLet UU bebe aa properproper RG-submodule ofof V.V. For For siSi E E S,S, [V, [V,si] si] = = (xi) is of dimension 1 soso either either xi xi EE UU oror UU (< CvCv(si) (si) = xi xl. . Thus Thus eithereither therethere exists i EE 0A withwith xixi EE UU oror U U <5 niEAXi niEAx; = =V1, v', and and I assumeI assume the the former. former. ClaimClaim xj EE UU forfor each jj EE 0,A, so so that V = (xj:(xj: jj EE 0) A) < _(U, U, contradictingcontradicting U U proper. proper. As the graph of A0 isis connectedconnected it suffices toto proveprove xj xj E U for midmij > 2.2. But,But, for such j, (si,(si, sj) is is irreducibleirreducible onon (xi, xj) xj) byby 29.7,29.7, soso asas xixi E U and U isis G-invariant, xjxj E U. I've shownshown G G isis irreducibleirreducible on on Vv == V/V v/v'. 1. F F= =EndRG EndRc@) (V) acts acts on on [ V[v, , si si] ] = (xi), so, forfor aa E F, axiaxi == bxi bxi for for some some b b EE R',R#, and and hence, hence, asas GG isis irreducibleirreducible on Vv andand centralizescentralizes a, a actsacts asas aa scalarscalar transformationtransformation via b on V.v. That is F == R. R. So, So, by by 25.8, 25.8, G G is is absolutely absolutely irreducibleirreducible on V.v. Suppose WW isis finite.finite. Then, by Maschke's Theorem,Theorem, V V == V V' -L CB Z for some RG-submoduleEf G-submodule ZZ ofof V. vV is EfRG-isomorphic G-isomorphic to to ZZ and [V,[v, si] # 1, so [Z, si] si] # 1. HenceHence (xi) (xi)= _ [Z,[Z, si] si] 5 < Z,Z, soso V V = = (xi:1 1 ( < i i (< n) <5 Z.Z. ThusThus V'V1 = 00 and (2) holds.

(29.12) Assume (G,(G, S)S) is is anan irreducibleirreducible CoxeterCoxeter system and W is finite. Then (V, Q) is is isometricisometric toto n-dimensionaln-dimensional Euclidean space under the usual inner product.

Proof. LetLet hh bebe thethe bilinearbilinear formform onon VV whichwhich makesmakes X intointo anan orthonormalorthonormal basis and define g:g: VV xx V -++ R R by by g(u,g(u, v) v) = = F_wEW CWEw ~(uw,h(uw, vw).VW). It is straightforward to check that g is a symmetric bilinear formform onon V preserved byby G. As the quadratic formform ofof hh is positive definite,definite, so so is is the the form form PP of g, so (V,(V, g) is nondegenerate. But, by 29.11, V V isis anan absolutelyabsolutely irreducibleirreducible RG-module,RG-module, so, by Exercise 9.1,9.1, PP == aQ aQ for for some some a aE Efl8#. R#. As As PP is is positive positive definite definite and Q(xi) = 1/2112 > > 0, 0,a a> 0,> so0, Qso isQ positiveis positive definite. definite. By By 19.9 19.9 therethere isis a basisbasis Y Y = = (yi: 1 1 ( 00 so, so, adjusting adjusting by by aa suitable suitable scalar, scalar, wewe cancan taketake Q(yi) == 1, 1, since since every every positive positive member member of of ER is is a a square square in in R. R. Thus Thus Y Y isis anan or-orthonormal basis for (V, Q),Q), soso (V,(V, Q)Q) isis EuclideanEuclidean spacespace under under thethe usualusual innerinner product.

Let0={1,...,n}andforJCOletSj=(SJ:jLet A = {I,. . . , n} and for J c A let Sj = (SJ:j EJ)andGj=(Sj).TheE J} and Gj = (Sj).The subgroups GJGj and and theirtheir conjugatesconjugates under G areare the parabolic subgroupssubgroups ofof thethe Coxeter system (G, S).S). 148 Presentations ofof groups

(29.13) LetLet J,J, K C2 AA andand g E Gj. ThenThen (1) IfIfl(g)=mandg=si, l (g) = m and g = si, ...... Si,,,si,,,withsi,~Sthenik~Jforeachl~k~m. with Si, E S then ik E J for each 1 < k< in. (2) (Gj,(Gj, Sj) SJ) is is a a Coxeter Coxeter system system with CoxeterCoxeter matrix matrix Mj Mj == (Mid),(mij), i, i, j j E J. (3) (Gd,(GJ, GK) GK) = = GJUK GJUK and GJGi fln GK'% == GJnKGJ~K. (4) IfIfGJ=&then Gj = GK then J=J = K.K.

Proof. LetLet gg = sats,, ...... saks,, with ak EE JJ and and k k minimal minimal subject subject toto thisthis con-constraint. ClaimClaim kk == m. By induction onon k, k, l(s,,g) l(sa,g) == k - 1, 1, so so by by thethe ExchangeExchange Condition eithereither kk == l(g) oror sa,s,, ...... sa,_, s,,-, = saZ s,, ...... s,,sar for some t, inin which which case g == sae s, .... .sa,_,sa,+,sak, .s ,,-, s,,+,s,,, contrary to minimality of k. HenceHence (1)(1) followsfollows from 29.5. Next (1) says thethe ExchangeExchange Condition Condition is is satisfied satisfied by by (Gj, (Gj, SJ), Sj), soso (Gj,(Gj, SJ)Sj) is a Coxeter system byby 29.4.29.4. 29.929.9 sayssays Mj Mj is is the the Coxeter Coxeter matrix matrix of of (Gj, (Gj, SJ).Sj). The first remark in (3) and the inclusioninclusion GJnKGJnK 5< Gj fl fl GK GK are trivial. Part (1) gives thethe inclusioninclusion GJGj fln GK

30 RootRoot systemssystems In this section V is a finitefinite dimensionaldimensional Euclidean space over a fieldfield FF equal to the reals oror thethe rationals.rationals. That isis V is an n-dimensional space over F togethertogether with a quadratic form QQ suchsuch thatthat (V,(V, Q) Q) possesses possesses an an orthonormalorthonormal basis.basis. Hence QQ isis positivepositive definite. Let ((, , )) bebe thethe bilinearbilinear formform defineddefined by Q.Q. ForFor v EE V#V# therethere exists aa uniqueunique reflectionreflection with center (v)(v) by by 22.6.2;22.6.2; denotedenote thisthis reflection by r,,.r,. A root system is is a a finite finite subset subset C E of of V# V# invariant invariant under under W(C) W (E) = = (r, (r :: vv EE C)E) andsuchthatI(v)and such that I (v) nCI fl E 5I < 2foreachv 2 for each vE E C.Observethatifv E. Observe that if vE E C E then-v= then -v = vr,vr EE E, C, so, so, as as ((v) I(v) fl n EICl Weyl group of E.C. Here's oneone wayway to obtain root systems:systems:

(30.1) Let G be aa finitefinite subgroupsubgroup of O(V, Q)Q) generatedgenerated by aa G-invariantG-invariant setset R of reflections. Let Let C E consistconsist ofof those those v v EE V#V# with with Q(v) Q(v) == 1/2112 andand (v)(v) thethe center of some membermember ofof R.R. ThenThen CE is a rootroot systemsystem andand GG == W(E).W(C).

Proof. ForFor v E E,C, (v) (v) is is the center ofof some some r rE E R, R, so so r r= = r,.r,,. Thus Thus G G = = (R) == W(E),W(C), so it remains toto showshow CE is a rootroot system.system. If If u u == av EE EC then, then, byby defi-definition ofof C,E, 1121/2 = Q(u)Q(u) == a2Q(v) a2~(v) == a2/2, a2/2, so so a a = = ±1. f 1. Thus Thus I E/C fln (v) (v)I I <5 2.2. Root systems 149

Also if g EE GG then,then, asas RR isis G-invariant,G-invariant, rg EE RR andand (vg)(vg) is is the the center center ofof rgrg with Q(vg) = Q(v)Q(v) == 1/2, 112, so so vg vg EE E.C. Hence Hence GG == W(E) W(C) acts acts on on E.C. Finally Finally I1x1 E J= = 21RI 21R1 5 < 21GI 2G I<< oo.GO. SoSo EC isis aaroot root system.system.

By 29.12 andand 30.1,30.1, every every finitefinite CoxeterCoxeter groupgroup cancan bebe representedrepresented as as the the Weyl Weyl group of some root system. We'll see later in this section that on the one hand thisthis representation is is faithfulfaithful and and onon thethe otherother that that thethe Weyl Weyl groupgroup of of eacheach rootroot system isis aa finitefinite CoxeterCoxeter group.group. ThusThus thethe finitefinite CoxeterCoxeter groupsgroups areare preciselyprecisely the Weyl groups of root systems.systems. For the remainder of thisthis sectionsection let let CE bebe aa rootroot systemsystem andand WW == W(E)W(C) itsits Weyl group. The elements of EC willwill bebe calledcalled roots.roots.

(30.2)(30.2) (1) (E)(C) isis a anondegenerate nondegenerate subspace subspace of of VV andand WW centralizes CL,El, soso W W isis faithful on (E).(C). (2)(2) TheThe permutation permutation representation representation of of WW onon EC isis faithful. faithful. (3)(3) WW is is finite. finite.

Proof. Let U = (E). (C). ThenThen HH == CW(U) Cw(U) = = CW(E) Cw(C) and and W/HW/H is is faithful faithful onon E,C, soso W/HW/H is is finite. finite. ThusThus toto prove (2) and (3)(3) itit sufficessuffices to to show show H H = = 1.1. As QQ isis positivepositive definite, UU isis nondegenerate.nondegenerate. Thus Thus V V = = U U @ ® u'.U'. But,But, for v E E,C, rr, centralizes centralizes vlvL Q 2 U1, uI, so so W W = = (r,,: (r,: v v E E E) C) centralizes centralizes U1. uI. Hence Hence H centralizescentralizes V,V, soso HH == 1, 1, completing completing the the proof groof ofof thethe lemma.lemma.

An ordering ofof VV isis aa totaltotal orderingordering ofof VV preserved preserved byby additionaddition andand multipli-multiplication by positivepositive scalars; scalars; that that is is if ifu, u, v, v, w wE E V V with with u u> > v, v, and and 0 0 < < aa E F, then u + w >> v v + + w w and and au au > 2 av. av. I leaveI leave the the following following lemma lemma as as an an exercise.exercise.

(30.3) (1)(1) IfIf <5 is is an an orderingordering on V and V+ = {v {v EE V:V: v >> 01 0) thenthen (i)(i) V+V+ isis closedclosed underunder additionaddition andand multiplicationmultiplication by by positivepositive scalars,scalars, andand (ii)(ii) for eacheach vv EE v#,V#, I{v, -v) -v) n V+) V+I = 1. 1. (2)(2) IfIf S S CV# 2 V# satisfies satisfies (i) (i) and and (ii) (ii) of of (1) (1) then then there there exists exists a a uniqueunique orderingordering ofVwithof V with SS=V+. = V+. (3) IfIf TT C 2 V# V# satisfiessatisfies (1.i) and I{v,I {v, -v) -v) nn TTI I 5< 11 for each v EE V#,v#, thenthen T C5 V+V+ forfor some some orderingordering ofof V.V.

A subset P of EC is aapositive positive system system if ifP P = = C+E+ = E C n n V+ V+ for for some some orderingordering of V.V. AA subsetsubset Tr n of EC isis aa simplesimple systemsystem if Trn isis linearlylinearly independentindependent and each v E EC cancan bebe writtenwritten v = ExE7r Ex,, a,xaxx such that eithereither 00 5< a,ax E F forfor all x EE 7rn orO>a,or 0 > ax EE FF forallxfor all x en.E r. 150 Presentations of groups

(30.4) Each simplesimple system isis containedcontained inin aa unique positive system and each positive system contains aa uniqueunique simplesimple system.system.

Proof. IfIf nn isis a a simple simple system system let let TT consistconsist ofof those elements FxEREx,, a,xaxx E V#V# with a,ax >> 0. 0. By By 30.3.3, 30.3.3, TT C E V+ V+ for for some some ordering ordering of of V.V. ByBy definitiondefinition of simple systemssystems andand 30.3.1,30.3.1, PP = T T n n E X is is the the unique unique positivepositive system containing n. Let Let SS bebe the set of those v E P suchsuch that, if v = EyEY EYE CyycY y forfor somesome YY EC P and 0 < cy E F, thenthen Y = {v}.{v}. EvidentlyEvidently S S E c n.n. I'll show nn cE S S whichwhich willwill prove n == S S is is the the uniqueunique simplesimple systemsystem in P. So assume vv EE nn and v = EyEYCyy C,, cYy forfor somesome Y Y C PP and 0 < Cycy E F.F. Now,Now,fory for Y E E Y,yY, Y = = ~,,,b,,xforsomeO ExEK byxx for some 05

(30.5) If n isis aa simplesimple systemsystem and x andand yy areare distinctdistinct members ofof n,n, thenthen (x,y)<0.(x, Y)5 0.

Indeed, returningreturning toto thethe proofproof ofof 30.4,30.4, let let n n bebe aa subsetsubset ofof P minimal subject to P beingbeing contained in thethe nonnegative linearlinear span span of of n. n. I'llI'll showshow nn satisfies 30.5 and thenthen useuse thisthis factfact toto showshow nn is linearlylinearly independentindependent and hence a simple system. This will complete the proof of 30.4. Suppose (x, (x, y) y) > > 0. 0. By By 22.6.2, 22.6.2, z = z =yr, yrx = =y -2(x,y-2(x, y)x/(x, y)x/(x, x),x), so so z z= = y y-cx -cx with cc >> 0. Now zz == C,,,ten atatt with with ata, > 00 if z Ec P andand atat 5< 0 ifif zz Ec -P.-P. If z cE P,P, consider consider the the equation: equation:

(1 -- ay)y aY)y = C att ++ (ax (a, + c)x.c)x. t#x,yt54x,y The right handhand sideside is is in in V+, V+, so, so, as as y y E c V+, 1 -- aya, >> 0. 0. But But then then yy isis inin thethe nonnegative span span of of n n- - (y),(y}, contradicting the the minimality minimality of of n. n. If Ifz zE c -P-P consider: (ax(a, ++ C)XOx = C (-at)t(-at) ++ (1 (1 -- ay)y aY)y t#x,yt54x,y and apply the same argument for aa contradiction. So n satisfies 30.5 and it remainsremains toto showshow nn is linearlylinearly independent.independent. SupposeSuppose O=CXEITa,xandleta0 = EXEn axx and let a=(x = (x En:a,c n: ax >>0}and/3 0) and,8 ==(y {y cEn:a, n: ay <

then z EE V+,V+, a acontradiction. contradiction. So So nn is is linearly linearly independent independent and and thethe proofproof isis complete.

For the remainder of this section let n bebe aa simplesimple systemsystem and PP thethe positive positive system containing nn. .

(30.6)(30.6) For each v E P therethere is xx E n withwith (v,(v, x) > 0.0.

Proof.Proof. WriteWrite v = EXEx,,E, axxa,x with a,ax 2> 00 andand observe observe 0 0 < < (v, (v,v) v) = = Ex>x axa, (v, x).

(30.7)(30.7) For eacheach x x EE n, n, xr,xrx = = -x-x andand rxr, acts onon P P -- (x).(x}.

Proof.Proof. Let r == rx. r,. ByBy definitiondefinition of of r, r, xr xr == -x.-x. LetLet V v EE PP - -{x}, {x}, soso that that vv = EYE, EYE, ayyayy with ayay 2> 0. AsAs (x}(x)== (x) (x) flfl P,P, aZ a, > 00 for some z z EE n n -- (x).(x}. Now yr = > ayyr ay(y - 2(y, x)x) ayy + bx

Y Y Y54x for some b E F. AsAs aZa, > 00 itit follows that vryr EE PP from from the the definitiondefinition of simple system.

(30.8)(30.8) WW isis transitive transitive on on simplesimple systems systems and and onon positivepositive systems.systems.

Proof. ByBy 30.3,30.3, W W permutespermutes positivepositive systems,systems, while it is evidentevident from the , definitions that WW permutespermutes simplesimple systems.systems. ByBy 30.430.4 itit sufficessuffices toto showshow WW isis transitivetransitive onon positive systems. Assume not and let P andand RR bebe positivepositive systems systems inin different orbits ofof W, and subjectsubject toto thisthis constraintconstraint with with IP ( Pfl fl (-R)I (-R)J = nn minimal. AsAs RR # P,P, n n > > 0. 0. Let Let n nbe be the the simple simple system system in in P.P. By By 30.4 30.4 there there isis x EE nn fl fl (-R). (-R). By By 30.7, 30.7, IIPr, Prx n fl (-R)I (-R)j = nn - -1, 1, so, so, by by minimality minimality of n, RR EE (Prx)W(Pr,) W == PW, P W, a a contradiction. contradiction.

For Vv E E,X, V v == EZE7[ E,,, a,xaxx for uniqueunique a,ax E F; define define the height of v to be h(v) = = ExEn Ex,, ax. a,. EvidentlyEvidently thethe heightheight functionfunction h dependsdepends on n. Notice Notice forfor v E P that h(v)h(v) >> 0 andand h(-v) h(-v) = -h(v)-h(v) << 0. 0. Also Also h(x)h(x) = = 1 1 for for x EE n.

(30.9) (1) (1) h(v)h(v) >> 1l for eacheach v v E E P P - - n. (2) WW == (rx: (r,:x x EE 7r).n). (3) EachEach member member ofof EX isis conjugateconjugate toto an element of nn under W.

Proof. Let G = (rx: (r,: x EE n)n) and and vv EE P.P. Pick Pick Uu EPE P fl fl vG vG such such thatthat h(u) is minimal. ByBy 30.6 30.6 there there is is y yE E n n withwith (u, (u,y) y) > > 0.0. ThenThen ur, ury == u - cy cy with with 152 Presentations ofof groups c = 2(u, y) > Oandifu0 andif u 0# yy then,then, byby 30.7, 30.7,ury ur, E EPnvG PnvG and andh(ury) h(ur,) = = h(u)- h(u)- c << h(u),h(u), aa contradiction.contradiction. SoSo uu == yy EE jr.n. In In particularparticular if v 60 n then h(v) > h(y) == 1, 1, so so (1) (1) holds. holds. FurtherFurther it it followsfollows that each member of ZE isis conjugate to an element ofof nTr under under G, G, so so that that (3) (3)holds. holds. Finally Finally v v= = ygyg forfor somesome g g EE G, so r,rU = = r,,ryg = = (r,),(ry)g E E G G and and hence hence W W = = (r,:(r,,: v vE E P) P) == G.

For wW E E WW let let N(w) N(w) = = IPwI Pw fl n(-P)I. (-P)I. Then N(l)N(I)=O= 0 and, and, by by 30.7,30.7, N(r,)N(rx) = 11 for x EE Tr.n. LetLet R = Jr,,: {r, :x x EE Tr}n] and 1l = lR 1~ thethe lengthlength functionfunction with with respectrespect toto R; that is, for w EE W,W, l(w) isis thethe minimalminimal lengthlength of aa word u inin thethe membersmembers of R such that u = w. w. It It will will develop develop shortlyshortly that the functions 1l and N agree on W.

(30.10) Let wW EE WW andand xX E E n. r. Then (1) N(rxw)N(r,w) = N(w) N(w) ++ 1 1 ifif xw >> 0,0, and and (2) N(rw)N(r,w) == N(w) N(w) -- 1 1 if if xwxw << 0.0.

Proof. PrxwPr,w == ({-x} ((-XI UU (P(P - -{x}))w (x}))w == {-xw} (-xw} U U (Pw (Pw - -{xw}) {xw}) byby 30.7, 30.7, soso (Pw n(-n (-P))P)) UU {-xw}(-xw] if xwxw >> 0, 0, Pr,wPrxw n(-n (-P)P) = (Pw fln (-P))(-P)) - {xw)(xw] if xw << 0.0.

(30.11) Let Let ri ri E E R,O R, 0 5 < ii <5 nn andand w == r1 rl ...... r,r,, EE WW withwith N(w) = n n andand N(row) <5 n.n. ThenThen there existsexists k,k, 11 5< k 5< n,n, such thatthat rorlrorl ...... rk_1 rk-1 = r1rl ...... rk. rk.

Proof. By 30.10, xw xw < 0,0, where where x EE Trn withwith ro == rx. r,. LetLet k k bebe minimal minimal subject toto xr1xrl ...... rk rk << 0. Then, by 30.7, y = xr1 xrl ...... rk_1 rk-1 EE Tr n and rk = ry. r,. Thus r o...rk - Yxrl,... = Y1 ... Yk. r;l".rk-lp - r,, ... ,,-,rA_ I= - r, Yy = = rk, rk, which which can can bebe rewrittenrewritten asas r0r1rorl . . . rk-1rk_1 = rl . . . rk.

(30.12) 1(w)l(w) = = N(w)N(w) forfor eacheach w EE W.W.

Proof. I'veI've alreadyalready observed observed that that 1l(1) (1) == 0 = N N(1). (1). Then 30.10 and induction on 1(w)l(w) implies implies l(w)1(w)2 > N(w).N(w). Suppose thatthat thethe lemmalemma is is falsefalse and and pick pick u u == rorl ...... rn r, EE WW withwith 1(u)l(u) = n n ++ 1 1 > > N(u), N(u), andand subject subject to to this this constraint constraint with with n minimal. Let ww = r1rl ...... rn r,. . As 1(u)l(u) = nn ++ 1, 1, 1(w) l(w) = = n, n, and, and, by by minimalityminimality of l(u),1(u),N(w) N(w) = = 1(w).l(w). As As N(row) N(row) = = N(u) << n n + + 1, 1, there there exists exists kk withwith rorlror1 ...... rk-1 rk_1 == rlr1 ...... rk rk byby 30.11.30.11. Thus Thus u u = = ro..ro ... .rk-irk.. rk_1rk ... .rn rn = = rlr1 ...... rzrk+l r rk+l

. . . rnr, = r1 rl .. . . rk-lrk+lrk-lrk+1 .. . . rn is of length at most nn - 1, 1, contrary contrary to to thethe choicechoice of U.u.

(30.13) (W,(W, R)R) is is aa CoxeterCoxeter system.system. 152 Presentations of groups Root systems 153 c = 2(u, y) > Oandifu : y then, by 30.7,ury E PnvG andh(ury) = h(u)- Proof. This is immediate from 29.4, 30.11, and 30.12. c < h(u), a contradiction. So u = y E r. In particular if v 0 r then h(v) > h(y) = 1, so (1) holds. Further it follows that each member of E is conjugate (30.14) W is regular on the positive systems and on the simple systems. to an element of 7r under G, so that (3) holds. Finally v = yg for some g E G, so rU = ryg = (ry)g E G and hence W = (ru: v E P) = G. Proof. By 30.8 and 30.4 it suffices to show NW(P) = 1. But NW(P) = {w E W: N(w) = 0) so 30.12 completes the proof. For W E W let N(w) = I Pw n (-P)I. Then N(l)=0 and, by 30.7, N(rx) =1 i for x E r. Let R = Jr,,: x E r) and l = lR the length function with respect to Let D consist of those v E V such that (v, x) > 0 for all x E Jr. R; that is, for w E W, l(w) is the minimal length of a word u in the members of R such that u = w. It will develop shortly that the functions l and N agree on W. (30.15) (1) vW n D is nonempty for each v E V. (2) (d, u) > O for each d E D and u E P. (30.10) Let W E W and X E r. Then (1) N(r,,w) = N(w) + 1 if xw > 0, and Proof. Pick Z E v W maximal with respect to the ordering defining P. Then, (2) N(row) = N(w) -1 if xw < 0. for x E Jr, z > zrx = z - 2(z, x)x, so (z, x) > 0 as x > 0. That is z E vW n D. Let d E D and u E P. Then u = XEn axx with a, > 0 so Proof. Prxw = ({-x) U (P - (x}))w = {-xw) U (Pw - {xw)) by 30.7, so (d, u) = >x ax(d, x) > 0. (Pw n (-P)) u {-xw}ifxw > 0, Prxw n (-P) = (30.16) Let d E D. Then Cw(d) is the Weyl group of the root systems E n dl (Pw n (-P)) - {xw) ifxw < 0. and has simple system 7r n dl. (30.11) Let ri E R, 0 < i < n and w = rl... r E W with N(w) = n and Proof. U = (ry: x E Tr n d') < Cw(d), so assume w E Cw(d) - U and sub- N(row) < n. Then there exists k, l < k < n, such that rori... rk-t = ri . .. rk. ject to this constraint with n = 1(w) minimal. w : 1 so n > 0. Then N(w) = n > 0 so there is x E zr with xw < 0. Now, by 30.15.2, 0 > (xw, d) = Proof. By 30.10, xw < 0, where x E Tr with ro = rx. Let k be minimal (x, dw-1) = (x, d), so x E dl by another application of 30.15.2. Thus rx E U subject to xr1... rk < 0. Then, by 30.7, y = xr1 ... rk-1 E 7r and rk = ry. Thus rk and, by 30.10, l(rxw) < n, so, by minimality of n, rxw E U. But then w E U. r'' = r r y , which can be rewritten as r r rk_ p xr ...i = y = k 0l rk So U = Cw(d). Evidently E n dl is a root system and an easy calculation using 30.15.2 shows Ir n dl is a simple system. (30.12) 1(w) = N(w) for each w E W.

(30.17) Let S C V. Then Cw(S) is the Weyl group of the root system E n S'. Proof. I've already observed that 1(1) = 0 = N(1). Then 30.10 and induction on 1(w) implies 1(w) > N(w). Suppose that the lemma is false and pick u = rori ... rn E W with 1(u) = n + 1 > N(u), and subject to this constraint with Proof. Let U = (rn: v EE n Sl). It suffices to show U = Cw(S). Cw(S) = n minimal. Let w = rl ... rn. As l(u) = n + 1, 1(w) = n, and, by minimality Cw((S)) = Cw(So) where So is a basis for (S), so without loss S is finite. of 1(u), N(w) = 1(w). As N(row) = N(u) < n + 1, there exists k with Replacing S by a suitable conjugate under W and appealing to 30.15, we may rorl ... rk_1 = rl ... rk by 30.11. Thus u = ro ... rk_lrk ... rn = ri ... rkrk+l take d E Dn S. Let G = Cw(d) and E0 = E n dl. Then, by 30.16, G = W(Eo) ... rn = ri ... rk_irk+l ... rn is of length at most n - 1, contrary to the choice and Cw(S) = CG(S), while, by induction on the order of E, CG(S) = U- of U. (30.18) Let (G, S) be a Coxeter system with G finite. Then the representation (30.13) (W, R) is a Coxeter system. a of 29.8 is faithful and G is the Weyl group of a root system. 154 Presentations ofof groups

Proof.Proof. ByBy 29.1229.12 andand 30.1,30.1, GaGa is is the the Weyl Weyl group group ofof aa root root system. system. By By 29.729.7 andand 30.13,30.13, Ga is is a a Coxeter Coxeter group group ofof typetype M,M, so so I /GalGal _ = I GIGI. 1. Hence a isis anan isomorphism.isomorphism.

(30.19)(30.19) Let H bebe thethe symmetricsymmetric groupgroup onon SZ52 = {1,(1, ...... , , n), (e) =Z 12,Z2, and and GG == (e)wroH (e)wrnH thethe wreath wreath product product of of (e) (e) by by H. H. Choose Choose notation notation so so that that CH(e)CH(e) ' isis thethe stabilizerstabilizer inin HH of n EE Q.52. LetLet G1G1 == H, H, S1 S1 == {(1, ((1,2), 2), (2, (2,3), 3), ...... , , (n -- 1,1, n)), G2G2 = G, G, S2 S2 = = S1 S1 UU (e),(e), S3 S3 == S1 S1 U U {(n{(n - 1, 1, n)e), n)e), and and G3G3 = = (S3). (S3). ThenThen (Gi,(Gi, Si) Si) is is aa CoxeterCoxeter systemsystem of type An-A,-1,1, Cn,C,, Dn,D,, for i = 1, 1,2,3, 2, 3, respectively. respectively. FurtherFurther G3G3 is of index 2 in G2 and is the semidirect productproduct ofof E2"-1E2'-, by S.S,.

Proof.Proof. LetLet V V be be n-dimensional n-dimensional Euclidean Euclidean spacespace withwith orthonormalorthonormal basic XX = (xi:(xi: 1 <( i i <5 n), n), ei ei the the reflection reflection with centercenter (xi), (xi), and and E E = = (ei: 11 5< i <( n). n). RepresentRepresent H onon VV via xixih h = Xih xih forfor hh E H. Then GG == (H, E) E) < 5 O(V) O(V) and and GG isis thethe wreathwreath product (e)(e) w rnrn H,H, wherewhere ee = en. en. Hence G isis the semidirect product ofof EE -Z E2,1, E2", by by HH =Z Sn, S,, andand G3 G3 is is thethe semidirectsemidirect product ofof DD = EE fln G3 G3 byby HH where where DD == (ei(eiej: ej: 2 2( < i i _(< jj <( n) n) =Z E2-,E2"-1.. The The transposition (i, j) isis thethe reflectionreflection with center (xij),(xih),where where xij xi j= = xi xi -xi. - xj. Also Gi == (Si) (Si) and and Si Si is is aa setset ofof reflections,reflections, so, byby 30.1,30.1, GiGi = W(Ei) W(Ci) where where Ei Ci is is the the root root system system consistingconsisting ofof thethe Gi-conjugatesGi-conjugates of of x12xl2 if i == 1 1 or or 3, 3, and and {x12, 1x12, x,)xn) if i = 2. 2. Thus Thus E1={±xxij:1

ForFor JJ CJ r n let let Vj Vj _ =(J) (J) andand Wj Wj = (rj:= (rj: j E j J).E J).RecallRecall the the subgroups subgroups WJ Wj and and theirtheir conjugates conjugates under under WW are are called called parabolic parabolic subgroups subgroups of of W. W.

(30.20)(30.20) LetLet 0 0 #0 JJ CJ r.n. Then Then (1)(1) EjCJ = =E flC vifl isVJ ais root a root system system with with simple simple system system J Jand and Weyl Weyl group group WrWJ. . (2)(2) WiWJ = = Cw(VJ cW(vJ'). ).

Proof.Proof. WiWj < (Cw(VjL) cw(vJ') == U Uand, and, by by 30.17, 30.17, U Uis isthe the Weyl Weyl group group of of the the root root systemsystem Ej.TrC J.n isis linearly linearly independentindependent and J spansspans Vj,Vj, soso JJ is is a a basis basis of of Vj.Vj. Root systems 155

Hence each member of EjCJ isis a a linear linear combination combination of of thethe membersmembers of of J,J, so, so, as as 7r n is a simplesimple system, system, J J is is a a simple simple system system for for El. Ej. Thus Thus U U= = (rj: (rj:j j Ec- J) J) == Wj.Wl.

Remarks. TheThe discussiondiscussion of CoxeterCoxeter systems in section 29 follows that of Bourbaki [Bo] and Suzuki [Su]. The presentation of root systems given herehere draws heavily on the appendix of SteinbergSteinberg [St].[St]. Coxeter groups and root systems play an important role inin branchesbranches ofof math-math- ematics other than finite group theory, most particularly inin the study of LieLie algebras, Lie groups, and algebraic groups. We will find inin chapterschapters 14 and 16 that they are crucial to thethe studystudy of thethe finitefinite groupsgroups ofof Lie type.type.

Exercises for chapter 1010 1. ProveProve D2nD2, =Grp (x, y: x2 = y2 Y2 = = (xy)n (~y)~ == 1). 1). ProveProve every every group group generated generated by a pair of distinct involutionsinvolutions is aa dihedral group.group. 2. ProveProve lemmalemma 30.3. 3. LetLet EC be be a a root root system, system, 7r n a simple system for C,E, PP the the positivepositive systemsystem ofof n,7r, RR == (ra:(r,: a EE 7r),n), and W = (R) (R) thethe Weyl Weyl groupgroup of E.C. ProveProve (1) ThereThere exists a unique wowo E E W W with with Pwo Pwo = = -P.-P. (2) wo wo isis thethe uniqueunique elementelement of WW ofof maximalmaximal length in the alphabet R. Further 1(wo)1 (wo) = I I P I I and wo is an involution. (3) nworwo = -7t-n andand Rw0RwO = R.R. 4. LetLet E C be be an an irreducible irreducible root root system, system, 7r n a simple system forfor C,E, P thethe positivepositive system forfor n, ,r,J J Cc n, E Cj, j, the subsetsubset of of C, E, spanned spanned by by J, J, and and $ lU = = P P - - Cj.Ej. Prove (C) (E)= _ ($).(i). 5. LetLet EC be be a a root root system, system, P P a a positive positive systemsystem forfor E,C, andand ww EE W. W. Prove Prove forfor each aa E P thatthat PwPw containscontains exactly one one of of a a and and -a. -a. 6. AssumeAssume thethe hypothesishypothesis ofof ExerciseExercise 10.3 10.3 andand letlet wowo be the element of WW of maximal lengthlength inin thethe alphabetalphabet R. R. Define Define a arelation relation 5 < on on W W by by u u 5 < ww if w = xu xu with with 1(w) l(w) == 1(x) l(x) ++ 1(u). l(u). ProveProve (1) < 5 is is a a partial partial order order onon W.W. (2) wo wo isis thethe unique maximal element of W. That isis ww 5< wowo for all w EE W.W. (3) For For rr = = ra r, E ER R and and W w E E W W the the following following are are equivalent: equivalent: (a) rwrw <5 w.w. (b) l(rw)Qrw) < 5 1(w). Qw). (c) aw << 0. 0. (d) w = rl ...... r,rn with rr = r1rl and l(w) = n. (4) IfIf uu <5 ww and and rr EE RR withwith l(ur) <5 1(u) l(u) and l(wr)l(wr) < 5 1(w)l(w) then then urur 5< wr. (Hint: ToTo proveprove (2) (2)let let wo wo # 0 w EE WW andand useuse ExerciseExercise 10.3.2,10.3.2, 30.10,30.10, and 30.12 toto show therethere exists exists r r EE RR withwith l(w)1(w) < < l(rw).)l(rw).) 11

The generalizedgeneralized FittingFitting subgroup

We've seen that the composition factors of a finite group control the structure of the group in part,part, butbut thatthat controlcontrol isis farfar fromfrom complete.complete. SectionSection 331 1 introducesintroduces a tool for studying finite groups via compositioncomposition factorsfactors 'near`near thethe bottom'bottom' ofof the group. The generalized FittingFitting subgroupsubgroup F*(G)F*(G) of a finite group G isis aa characteristic subgroup ofof G generated by the smallsmall normalnormal subgroupssubgroups of GG and with the propertyproperty thatthat CG(F*(G))CG(F*(G)) 5< F*(G). ThisThis lastlast property supplies a representation of GG asas aa subgroupsubgroup of of Aut(F*(G)) Aut(F*(G))with with kernel kernel Z(F*(G)). Z(F*(G)). G cancan be effectively investigated via this representationrepresentation because F*(G) isis aa relativelyrelatively uncomplicated group whose embeddingembedding inin GG is is particularlyparticularly well well behaved.behaved. It turns out that F*(G)F*(G) is is a a central central product product of of the the groups groups OP(G), O,(G), p p r= E 7r(G),n(G), with a subgroup E(G) of of G.G. To To definedefine E(G)E(G) requires requires somesome terminology.terminology. AA central extensionextension of a group X isis aa groupgroup YY togethertogether withwith aa surjectivesurjective homomorphism of Y onto X whose kernel is in the center of Y.Y. The groupgroup YY willwill also be said to be a central extension of X. A group L is quasisimple if L is perfect and the central extension of a simple group. The components ofof G are its subnormal quasisimplequasisimple subgroups,subgroups, andand E(G)E(G) is the subgroup of G generated by the components of G. It develops that E(G) is is aa centralcentral product product ofof thethe components of G. Recall that ifif p is aa primeprime thenthen a p-local subgroup of G isis thethe normalizernormalizer in G of aa nontrivial p-subgroup ofof G.G. The The locallocal theorytheory ofof groupsgroups investigatesinvestigates finite groups fromfrom thethe pointpoint of view ofof p-locals.p-locals. A question of great interest in this theory is the relationship between the generalized Fitting subgroup of GG and that of its local subgroups. Section 31 contains various resultsresults about such relationships. InIn thethe final chapter we'll get some idea of how such results are used to classify the finite simple groups. If F*(G) isis aa p-group,p-group, itit can can bebe particularlyparticularly difficultdifficult toto analyze the structure of G. OneOne tooltool forfor dealingdealing withwith suchsuch groupsgroups isis thethe ThompsonThompson factorization.factorization. Lemma 32.5 shows that, if G is solvable and the Thompson factorization fails, fails, then the structure of G isis ratherrather restricted.restricted. This result will be used used inin laterlater chapters to prove the ThompsonThompson Normal p-Complement Theorem and the Solvable Signalizer Functor Theorem. The Normal p-Complement TheoremTheorem will be used in turn to establish the nilpotence of Frobenius kernels.kernels. Finally the importanceimportance of componentscomponents focusesfocuses attentionattention on on quasisimplequasisimple groups. In section 33 wewe findfind therethere is is a alargest largest perfect perfect centralcentral extension extension The generalized Fitting subgroup 157

7r:n: G (? +G G ofof eacheach perfect perfect groupgroup G.G. G (? is is the the universal universal covering covering groupgroup ofof GG andand ker(7r) ker(n) isis thethe Schur multiplier ofof G.G. InIn particular ifif GG isis simplesimple thenthen G(? isis thethe largest largest quasisimple quasisimple group group with with G G as as a a homomorphic homomorphic image image and and the the Schur Schur multipliermultiplier of of G G is is the the center center of of G. (?. As As an an illustration illustration of of this this theory, theory, the the covering covering groupsgroups andand SchurSchur multipliers multipliers of of the the finite finite alternating alternating groups groups are are determined. determined. TheThe Schur Schur multiplier multiplier can can be be defined defined forfor nonperfect nonperfect groups groups using using an an alternatealternate definitiondefinition requiringrequiring homologicalhomological algebra. algebra. The The presentation presentation givengiven herehere isis group group theoretictheoretic and and restricted restricted to to perfect perfect groups; groups; it it follows follows Steinberg Steinberg [St]. [St].

3131 TheThe generalized generalized Fitting Fitting subgroup subgroup InIn thisthis sectionsection GG isis aa finitefinite group. AA groupgroup XX isis quasisimplequasisimple if if X X = = x(') V andand X/Z(X)X/Z(X) isis simple. simple.

(31.1)(31.1) LetLet X X be be a a group group such such that that X/Z(X) X/Z(X) isis a a nonabelian nonabelian simple simple group. group. Then Then XX == X(') x(')z(x) Z(X) andand XMx(') isis quasisimple. quasisimple.

Proof.Proof. LetLet Y Y = = X x(') M and and X* X* = =X/Z(X). X/Z(X). NowNow Y* Y* <9 XX* * andand XX* * isis simplesimple soso Y*Y* ==1 1 or or X*. X*. In In the the latter latter case case X X = = YZ(X) YZ(X) andand in in the the former former X* X* is is abelian, abelian, contrarycontrary toto hypothesis. hypothesis. SoSo X X == YZ(X). YZ(X). ThusThus X/ X/Y(') Y(1) is abelian so Y = YO).~('1. FurtherFurther Y/Z(Y) E= X*X* isis simple,simple, soso Y Y is is quasisimple. quasisimple.

(31.2)(31.2) Let XX bebe aa quasisimplequasisimple groupgroup and H H <99 < X. ThenThen either H = X X oror HH <5 Z(X). Z(X).

Proof.Proof. IfIf HH Z(X),$ Z(X), X X= =HZ(X HZ(X), ), soso thatthat X/HX/H isis abelian abelian andand hencehence XX == XMx(') <5 H.H.

TheThe componentscomponents ofof aa group XX areare its subnormalsubnormal quasisimple subgroups.subgroups. WriteWrite Comp(X)Comp(X) forfor thethe set of components of X. Set E(X) = (Comp(X)). (Comp(X)).

(31.3)(31.3) IfIf HH <9 <9 X X then then Comp(H)Comp(H) = Comp(X) Comp(X) n H.

(31.4)(31.4) Let L EE Comp(G) andand H <9 < 9 G. G. Then Then either either LL E E Comp(H) Comp(H) oror [L,[L, HIH] ==1. 1.

Proof. LetLet GG bebe a a minimal minimal counterexample. counterexample. If L = GG thethe lemmalemma holdsholds byby 31.2,31.2, so by 7.2 wewe maymay take take X X = = (Lc)(L~) ## G. G. Similarly Similarly if if HH == G G the the lemma lemma isis trivial,trivial, soso taketake Y Y == (Ho)(H~)f G.G. XX nn Y Y < 9 X,X, so, so, by by minimalityminimality ofof G,G, either either L E Comp(X n Y) or [L, xX nn Y] Y] = = 1. 1. In In the the first first case case L L EE Comp(Y) Comp(Y) by by 31.3 31.3 158 The generalizedgeneralized Fitting subgroup

and then,then, asas HH <5 YY << G, G, the the lemma lemma holds holds by by minimalityminimality of of G.G. InIn thethe secondsecond [Y, L, L, LI L] I <[Y [YnX,L]=1,so,by8.9, n x, LI = 1, SO,by 8.9, [Y,L]=1.[Y, LI = 1.

(31.5) Distinct components of G commute.commute.

Proof. ThisThis isis aa directdirect consequenceconsequence ofof 31.231.2 and 31.4.3 1.4.

(31.6) Let L E Comp(G) and H anan L-invariantL-invariant subgroup of G. Then (1) Either L E Comp(H) oror [L,[L, HIH] ==1. 1. (2) IfIf HH isis solvable solvable then [L,[L, H]HI == 1. 1. (3) IfIf RR <_( G G then then either either LL EE Comp([R,Comp([R, L])L]) or [R,[R, L] = 1.1.

Proof. Part (1) follows fromfrom 31.431.4 appliedapplied to to LH LH in the role of G. Then (1) implies (2).(2). IfIf RR <5 GG then then [L, [L, R] R] is is L-invariant L-invariant byby 8.5.6,8.5.6, soso byby (1)(1) eithereither L E Comp([L, R]) or [R,[R, L, L, L] L] = = 1. 1. In In thethe latterlatter casecase [R,[R, L] L] = = 1 1 by by 8.9.8.9.

(31.7) LetLet EE == E(G),E(G), Z = Z(E), and E* == E/Z. ThenThen (1) ZZ == (Z(L): (Z(L): L L EE Comp(G)).Comp(G)). (2) E*E* is is the the direct direct product product ofof thethe groupsgroups (L*:(L*: L EE Comp(G)).Comp(G)). (3) EE is is a a central central product product of of itsits components. components.

The Fitting subgroup ofof G, denoted by F(G), isis thethe largestlargest nilpotentnilpotent normalnormal subgroup of G. O,,.(G)O,(G) denotesdenotes thethe largestlargest solvablesolvable normalnormal subgroupsubgroup of G.G.

(31.8) Let G be a finite group. ThenThen F(G) isis thethe directdirect product of the groups (OP(G):(Op(G): PP E dG)).n(G)).

Proof. SeeSee 9.11.9.11.

(31.9) O.(CG(F(G))) = Z(F(G)).

Proof. Let Z = Z(F(G)).Z(F(G)). G*G* = G/Z, G/Z, and and H H = = O,,(CG(F(G))). O,(CG(F(G))). AssumeAssume H*#H*= 1 and let X* be a minimalminimal normal subgroup ofof H*.H*. ThenThen X* isis a p- group for some primeprime p,p, so X = PZ, PZ, where where P P E E SyIP(X). Sylp(X). XX centralizescentralizes Z,2, so P

(31.10) If G is solvable thenthen CG(F(G))CG(F(G)) 5< F(G).

Define thethe socle of G toto bebe thethe subgroupsubgroup generatedgenerated by allall minimalminimal normal subgroups of G, and write Soc(G) for the socle of G. TheThe generalizedgeneralized Fitting Fitting subgroupsubgroup 159

(31.11)(31.11) Let Z == Z(F(G)), Z(F(G)), G* G* = = G/Z, G/Z, and and S* S* = Soc(CG(F(G))*).= Soc(CG(F(G)>*). ThenThen E(G)E(G) = = SM S('! andand SS = = E(G)Z. E(G)Z.

Proof.Proof. Let Let H H = =CG(F(G)). CG(F(G)). By By 31.9,31.9, O,0(H*)O,(H*) =1.= 1. So, So, by by 8.28.2 and and 8.3, 8.3, each each minimalminimal normalnormal subgroupsubgroup ofof H*H* is is the the direct direct product product of of nonabeliannonabelian simple simple groupsgroups and and by by 31.331.3 these factors are components of H*.H*. ThusThus S*S* 5< E(H*).E(H*). Let Let K*K*beacomponentof~*.~~31.1, be a component of H*. By 31.1, K ==~(')~with~(')~uasisimple.~y31.3, K(' )Z with KM quasisimple. By 31.3, K('K(') E E Comp(G), Comp(G), so so S S < I E(G)Z.E(G)Z. By By 31.6.2, 31.6.2, E(G) E(G) < 5 H. H. Let Let L L E E Comp(G), Comp(G), and and M=M = (LH). (LH). Then Then M* M* is is a aminimal minimal normal normal subgroup subgroup of of H*H* by by 31.4,31.4, soso MM <( S. S. ThusThus SS == E(G)Z. E(G)Z. By By 31.1, 3 1.1, E(G) E(G) = = Ski). s(').

DefineDefine the generalized Fitting subgroupsubgroup ofof GG toto be be F*(G)F*(G) == F(G)E(G).

(31.12)(31.12) F*(G)F*(G) is is a a central central product of F(G) withwith E(G).E(G).

Proof.Proof. See See 31.6.2. 31.6.2.

(31.13) CG(F*(G)) < F*(G).

Proof.Proof. LetLet HH = = CG(F*(G)), CG(F*(G)), KK = CG(F(G)), CG(F(G)), Z = Z(F(G)), andand G*G* = G/Z. G/Z. ThenThen H*H* < 9 K*,K*, soso ifif H*H* 01# 1then then 10 1 #H* H* n nSoc(K*), Soc(K*), and and then, then, by by 31.11, 31.11, HH nn E(G) E(G) 0 # Z, Z, a acontradiction. contradiction.

I RecallRecall Op',E(G)O,J,~(G) is defined byby Op,,E(G)/Op,(G)Op',E(G)/OP,(G) == E(G/OP,(G)).E(G/O,,(G)).

(31.14)(31.14) LetLet pp bebe aa p-subgroup p-subgroup of of G.G. ThenThen (1)(1) Op',E(NG(P))O~~,E(NG(P>> <5 CG(OP(G)), CG(O~(G)), andand (2)(2) ifif PP < 5 Op(G) O,(G) thenthen OP(F*(NG(p))) OP(F*(NG(p))) == OP(F*(G)). Op(F*(G)).

Proof.Proof. Let Let X X = =OP,,E(NG(p)). 0,1,~(N~(p)). ToTo prove prove (1), (I), it it suffices suffices to to show show X X centralizes centralizes RR == COp(G)(P) Co,(G)(P) by the AA x BB Lemma.Lemma. But [R, Op,(X)]OP,(X)] (< R nn OP-(X) Op!(X) = 1 1 andand [R,[R, X]XI <5 OP,(X) Op,(X) by by 31.6.2,31.6.2, soso [R,[R, X]XI == 1 1by by coprime coprime action, action, 18.7. 18.7. SoSo taketake P

(31.15) If If GG is is solvablesolvable and and P P is is a ap-subgroup p-subgroup of of G G then then Op, O,I(NG(P)) (NG (P)) < 5 Op, O,!(G). (G).

Proof. PassingPassing toto G/Op,(G)G/O,l(G) andand appealing appealing to to coprime coprime action, action, 18.7,18.7, wewe maymay take Op,(G)O,f(G) = 1, 1, and and itit remainsremains toto showshow Op'(NG(P))O,~(NG(P)) == 1. 1. This This follows follows fromfrom 31.10 and 31.14.1.

(31.16) Let F*(G) = Op(G) O,(G)forsomeprime for some prime p.Then p. Then F*(NG(P)) F*(NG(P))= = O,(NG(P))Op(NG(P)) for each p-subgroup P of G.G.

Proof. ThisThis followsfollows fromfrom 31.133 1.13 andand 31.14.1.3 1.14.1.

The Schreier Conjecture says says Out(L) isis solvablesolvable for each finite simple group L.

(31.17) Let O,t(G)Op,(G) == 11 and P aa p-subgroupp-subgroup ofof G.G. ThenThen (1) Op',E(NG(P))O,J,~(NG(P)) fixesfixes eacheach componentcomponent of G. (2) If eacheach component component ofof GG satisfies satisfies the the Schreier Schreier conjecture, conjecture, thenthen Op',E(NG(P))°°Opf,~(N~(P>)" L< E(G).

Proof. Let HH = NG(P),NG(P), X == Op,(H), O,!(H), H*H* = H/X, H/X, and and Y Y == Op',E(H)°°. O,I,~(H)*. Let K <( X X or or KK <( H H with with K* K* E EComp(H*), Comp(H*), and and subject subject to to these these constraints constraints pick K minimal subject toto movingmoving a a component component of of G. G. Let Let P P (< Po E Sylp(CG(K)). As K satisfiessatisfies the same hypothesis with respect to Po, we maymay taketake P = Po. Po. In particular, by by 31.14, 31.14, Op(G) Op(G) 5 < P.P. Let RR E Syl,(HSylp(H nfl E(G)). Suppose firstfirst KK 5< X.X. Then K isis aa q-groupq-group forfor somesome prime q, andand byby co-coprime action, 18.7, there exists an R-invariant SylowSylow q-groupq-group Q of Op,(H).O,f(H). Replacing KK by a suitable conjugate,conjugate, we we may may assume assume K K ( < Q. Q. By By 24.4, 24.4, Q Q == [R, Q]CQ(R). NowNow [R, [R, Q]Q] 5< [E(G), Q]Q] <( E(G), E(G), so so [R, [R, Q] Q] fixes fixes eacheach com-component ofof G.G. HenceHence we we may may take take [K, [K, R] R] = =1. 1. Thus,Thus, by by choice choice of of P, P, R R 5< P. So P fln E(G) E(G) E E Sylp(E(G)) Syl,(E(G)) byby Exercise Exercise 3.2. 3.2. As As Op,(G)O,/(G) = 1, 1, p cE 7r n(L) (L) forfor each LL E Comp(G), soso 11 # # P flfl LL EE Sylp(L)Syl,(L) by 6.4. ThenThen PP nfl L $ Z(L), so L = [E(G), P P fl n L] L] is is K-invariant. K-invariant. So K $ X. ByBy 31.4,31.4, eithereither K*K* <5 [K*,[K*, R*] or [K*,[K*, R*] == 1. In the latter case by coprime action, 18.7, K =O,f(K)CK(R)=Op-(K)CK(R) so, by minimalityminimality of K,K, [K, R] == 1. 1. But But then then an an argument argument in in thethe lastlast paragraph supplies a contradiction. In the former,former, KK (< X[K,X [K, R]R]:< 5 XE(G),XE(G), soso K fixesfixes each component ofof G. Let L bebe aa componentcomponent of G.G. II havehave shownshown YY 5< N(L). If LL satisfiessatisfies thethe Schreier conjecture, then Autr(L)Auty(L) <5 Inn(L) Inn(L) as as Y Y == YO°, Y*, soso YY <( LC(L).LC(L). Hence, under the hypothesis ofof (2), Y 5< E(G)C(E(G))E(G)C(E(G)) and then, by 31.14,31.14, YY (< E(G)C(F*(G)) < 5 F*(G), F*(G), so so Y Y <( F*(G)O° F*(G)* == E(G). TheThe generalizedgeneralized FittingFitting subgroupsubgroup 161161

GG is is balanced balanced for for the the prime prime p p if if O O,l(CG(X)) p, (CG M) << Op- Opj(G) (G) for each X of orderorder pp inin G. G.

(31.18)(31.18) LetLet Op- Op~(G) (G) == 1, 1, x x of of order order pin p in G, G, L L E E Comp(G), Comp(G), and and Y Y = = Op'(CG Op~(CG(~)). (x)). ThenThen (1)(1) IfIf LL ## [L, [L, x] x] then then [L, [L, Y] Y] = = 1 1and and either either L L E E Comp(CG(x)) Comp(CG(x)) or L # LxLx andand C[L,x](x)('CcL,xl(~)(l) == K K E E Comp(CG(x)) Comp(CG(x)) withwith K K a ahomomorphic homomorphic image image of of L.L. (2)(2) IfIf L L = = [L, [L, x] x] andAuty(x)L(L) andA~t~(,)~(L) isis balancedfortheprime balancedfortheprime p, p, then then [Y, [Y, L]L] = = 1. 1.

Proof.Proof. AssumeAssume L ## [L, [L, x]. x]. Then Then eithereither [L,[L , x] = 1 or L ## Lx. Lx .In In the the first first casecase LL E E Comp(C(x)),Comp(C(x)), soso [L,[L, Y] Y] = 1. 1. InIn thethe secondsecond let M = [L, [L, x] x] andand M*M* = M/Z(M).M/Z(M). By By 8.9, 8.9, [M, [M, L] L] # # 1, 1,so, so, by by 31.4, 31.4, L LE EComp(M), Comp(M), and and we we conclude conclude MM = (0)) (L(")) is isthe the central central product product of of thethe groupsgroups (Lx': 0 0 5< ii << p) p) from from 31.5. 31.5. Hence,Hence, by by ExerciseExercise 3.5, 3.5, KK = = CM(x)(l) cM(x)(') isis aa homomorphichomomorphic imageimage of L.L. AsAs LL isis quasisimple,quasisimple, so so is its homomorphic image K. Also MM 1< <1 G, G, so so K K <1 < 1 CG CG (x),(x), andand hencehence K EE Comp(C(x)).Comp(C(x)). SoSo [K,[K, Y]Y] == 1. 1. Y Y acts acts on on L* L* by by 31.17, 31.17, so, so, asas [K*,[K*,Y] Y] == 1, l,[L*,Y] [L*, Y] == 1 1by by Exercise Exercise 3.5.3.5. ThenThen [L,[L,Y] Y] = 11 by by 31.6,31.6, completingcompleting the the proof proof ofof (1).(1). SoSo assumeassume L == [L, [L, x], x], let let U U = =Y(x)L, Y (x) L, and and let let rr: n:U U--> + AutG(L) AutG(L) be be thethe conjugationconjugation map. LetLet P == Op(U)(x) O,(U)(x) soso that that P P E E Sylp(ker(7r)(x)). Syl,(ker(n)(x)). ByBy 31.14,31.14, [P,[P, Y] Y] == 1, 1, so, so, asas Cu(P)CU(P) < 5 Cv(x), CU(x), YY <5 Op,(Cv(P)) OPf(CU(P)) == Op,(Nv(P)). Op~(NU(P)). Then,Then, asas PP E E Sylp(ker(7r)(x)), Syl,(ker(n)(x)), NUn((x7r))Nuk((xn)) = Nu(P)7r Nu(P)n by by a aFrattini Frattini Argument, Argument, soso YrrYn < Op'Op~(Cur(xn)). (Cu. (x 7r)).Hence, Hence, if ifAutu(L) Autu (L) is is balanced balanced for for thethe primeprime p, thenthen [Y,[Y, L] = 1, 1, so so that that (2) (2) holds. holds.

(31.19)(31.19) LetLet Op-(G)O,<(G) == 1 1 and and assume assume AutH(L) AutH(L) is is balanced balanced forfor thethe primeprime pp forfor eacheach LL EE Comp(G)Comp(G) andand each H <( G G with with L L < 1 H. H. Then Then G G is is balanced balanced for for the the primeprime p.p.

Proof.Proof. Let Let X X be be a a subgroup subgroup of of GG ofof orderorder p andand YY == Op'(CG(X)). Op,(CG(X)). II must must showshow YY = 1. 1. By By 31.18 3 1.18 and and the the hypothesis hypothesis on on thethe componentscomponents of G,G, [Y,[Y, E(G)] = = 1. 1. ByBy 31.14,31.14, [Y,[Y, O,(G)]Op(G)] ==1. 1. So Y 5< CG(F*(G)) == Z(F(G)), so, so, as as Op,(G)O,,(G) = 1, 1, Y=1.Y=l.

Here'sHere's technicaltechnical lemma lemma to to bebe usedused inin chapterchapter 15. 15.

(31.20)(31.20) LetLet AA bebe anan elementaryelementary abelian r-group actingacting on a solvable r'-group GG andand letlet aa EE A. Let p E nr cC 7r n(G), (G), pkp" = rr' n' UU {p},{p], and P an an A-invariantA-invariant p- subgroupsubgroup ofof G.G. 162 The generalizedgeneralized Fitting subgroup

(1) SupposeSuppose PP <5 Op(K) O,(K) forfor some some a-invariant a-invariant 7r-subgroup n-subgroup K ofof G suchsuch that CK(a) is aa Hall 7r-subgroupn-subgroup of CG(a). Then P 1 OPr(N)for for each a-invariant subgroup NN of G with NN == (P, CN(a)). CN(a)). (2) Assume AA is noncyclic andand letlet A bebe thethe set set ofof hyperplaneshyperplanes of A. AssumeAssume f foro r each BB EE A thatthat Cp(B)Cp(B) , P)). (3) IfIf BB isis a a noncyclic noncyclic subgroupsubgroup of CA(P) and P <5 O,r((CG O,((CG((a, ((a, b)), P)) for each bb EE B',B#, then then P P 5< On((CG(a),0, ((CG(a), P)).

Proof.ProoJ LetLet GG be be a a minimal minimal counterexample counterexample toto (1),(I), (2),(2), oror (3).(3). WithoutWithout loss, A = (a) inin (1)(1) andand AA = (B, (B, a) a) in in (3).(3). InIn eacheach casecase itit sufficessuffices to assume G = (P, CG(a)) CG(a)) andand prove P <5 Q, O,(G), (G), wherewhere aa = = p" p" in in (1) (1) and and a a = = 7r n inin (2) (2) and and (3).(3). Let H bebe a minimal A-invariant normalnormal subgroupsubgroup ofof G G and and G* G* = = GIH. G/H. Then P* < 5 O,,(G*) O,(G*) == S* S* by by minimalityminimality of GG andand coprimecoprime action 18.7.4, andand HH is a q-group for some prime q. Hence if P <5 O,,(HP) O,(HP) wewe areare done,done, soso thisthis isis not the case, and inin particularparticular qq V$ a.a. Let II=CG(a). = CG(a). Then CH(a)CH(a)<]I. < I. We show [P,[P, CH(a)]=l.CH(a)] = 1. Then Then G G == (I, P)P) < 5 N(CH(a)), N(CH(a)), so, so, by by minimalityminimality ofof H,H, CH(a)CH(a)= (aor 1 or H.H. In In the the latterlatter case [P, H]HI == 1, 1, contradicting contradicting P $ O,(PH). Thus CH(a) = 1.1. Now to verify that [P, CH(a)] = =1. 1. In In (1),(I), asas qq EEn 7r and CH(a) <]< I, CH(a)CH(a) isis containedcontained in each each Hall Hall 7r-groupn-group of CG(a)CG(a) by by Hall'sHall's TheoremTheorem 18.5, 18.5, andand hence in K. So [CH(a), PIP] 5

32 ThompsonThompson factorizationfactorization In this section p isis aa prime and G isis aa finite group. DenoteDenote byby M(G)d(G) thethe setset ofof elementary abelian p-subgroupsp-subgroups ofof GG ofof p-rankp-rank mp(G).m p(G). Set Set J(G) J(G) = _ (d(G)).(. ((G)). Thompson factorizationfactorization 163

We call J(G) the the Thompson Thompson subgroup of G. Of course J(G) depends depends onon thethe choice of p.

(32.1)(32.1) (1) J(G)J(G) char(G). char(G). (2)(2) IfIf A isis in in d(G) a (G) andand A A 5 < H H 5 < G G then then d(H) ,((H) G c d(G).,((G). (3)(3) LetLet PP E E Sylp(G).Syl,(G). Then J(P) = = J(Q) J(Q) for for each each p-subgroup p-subgroup QQ ofof GG con-containing J(P).J(P).

If VV isis aa GF(p)G-moduleGF(p)G-module define define GJ'(G, P(G, V) V) to to consist consist of of thethe nontrivial nontrivial elemen- elemen- tarytary abelianabelian p-subgroupsp-subgroups AA ofof GG suchsuch thatthat

m(A) + m(Cv(A)) >1 m(B) m(B) + m(Cv(B)) m(Cv(B)) for each B <5 A.A. NoticeNotice that if BB isis aa nontrivialnontrivial subgroup of A E P(G,°J'(G, V) for which thisthis inequality inequality is is an an equality, equality, then then B Bis isin inGJ'(G, P(G, V). V). A Asubset subset _1:1P P of P(G,P(G, V) V) isis stablestable if if GG permutespermutes °J' P via via conjugation conjugation and, and, whenever whenever A A isis inin GJ'P and BB isis aa nontrivial.nontrivial. subgroupsubgroup ofof AA withwith

m(A) + m(Cv(A)) = m(B) + m(Cv(B)), then B isis inin °J'9 AsAs aa finalfinal remark note that ifif A is in °J'(G,P(G, V) thenthen m(A) 2> m(Vm(V/Cv(A)); /Cv (A)); to see thisthis justjust taketake BB ==1 1 in in the inequality defining membership in °J'(G,P(G, V). V).

(32.2) LetLet V V be be a a normal elementary abelian p-subgroup ofof G,G, G* G* = =GI G/ CG(V CG(V), ), and GJ'P == {A*: AA EE d(G),1(G) and and A* A* # 11.1). ThenThen PGJ' is is a a stable stable subset subset of of P(G*,7)(G*, V).V).

Proof.Prooj LetLet AA be in a(G)d(G) andand CA(V) CA(V) < 5 B B < 5 A. A. A0 A. == ACv(A) ACV(A) isis an an elemen- elemen- tary abelian p-group, so,so, asas AA EE d(G),, (G), m(Ao)m(Ao) 5< m(A).m(A). HenceHence AA == A0 A. since A 5< A0.Ao. Thus CV(A)Cv(A) = A flil V.V. SimilarlySimilarly m(BCv(B)) < 5 m(A).m(A). AsAs CB(V)CB(V)= CA(V), we have m(BCv(B)) = m(B*) + m(CA(V)) + m(Cv(B)) - m(A fl v)

while

m(BCv(B)) < m(A) = m(A*) + m(CA(V))

so as Cv(A)Cv (A) == A flil VV itit follows that

m(A*) + m(Cv(A)) > m(B*) + m(Cv(B)) with equality only if m(BCV(B))m(BCv(B)) = m(A). m(A). ThusThus ifif A*A* # 1 then A* EE P(G,°J'(G, V) while if m(B*) ++ m(Cv(B))m(Cv(B)) = m(A*)m(A*) + m(Cv(A)) then m(A) = m(BCv(B)),m(BCv(B)), so B0Bo == BCv(B) isis in ,1(G)d(G) and and hence hence B* B* = = Bo B,* is is in in GJ'(G*, P(G*, V).V). 164 The generalizedgeneralized Fitting subgroup

(32.3) LetLet G be aa solvablesolvable group, V a faithfulfaithful GF(p)G-module, andand PP a Sylow p-group of G.G. Assume Assume Op(G) Op(G) == 1, 1,GJ' Pis is a astable stable subset subset of of -OP(G, P(G, V) V) and H=(P)#1.Thenp(3,H=H1 H=Then p<3,H=HI x . xx H,,,.-.x V=[V,H]H,, V=[V,H]@CV(H) ®CV(H) with [V, HI H] = [V, [V, Hl]HI] (D @ .. . . @®[V, [V, H,,],H,], GG permutes (Hi:(H,:11 5

Proo$Proof. LetLet rr consist consist of of thethe minimalminimal members ofof P,J', pick A EE r, I', andand set set U U == Cv(A). IfIf BB isis a ahyperplane hyperplane of of A A then then m(A) m(A)+m(U)+m(U) L > m(B)+m(Cy-(B))m(B)+m(Cv(B)) withwith B EE °J'P oror BB == 1 1 in in case case of of equality.equality. SoSo by minimality of A eithereither /A/Al I== p oror the inequality is strict, and in that event, as m(A)m(A) = = m(B)m(B) + 1,1, itit followsfollows that U = CV(B). Cv(B). Suppose ]A/Al I>> p p andand let let K K = = [F(G),[F(G), A]. As As Op(G)Op(G)= = 1, F(G)F(G) is a pf-groupp'-group and A is faithful on F(G) byby 31.10,31.10, soso KK # 1. By Maschke's Theorem, V = [V, K] @ ®CV(K), Cv(K), and, asas KK #0 1,1,0 0 # [V, K]. AA acts on K andandhence hence on [V, K], so C[V,K](A)CIV,Kl(A)# 0. Thus U $ Cv(K). OnOn thethe otherother hand,hand, byby ExerciseExercise 8.1, K = (CK(B): (CK(B): IA:BI/A : BI == p) p) and andby by thethelastparagraph, last paragraph, U U=CV(B),soCK(B) = Cv(B),so CK(B) 5< N(U). Thus Thus K <5 N(U), so, so, byby ExerciseExercise 3.6, K = [K, A] <5 CG(U),CG(U), contra-contradicting U $ Cv(K). So JAI/A/= = p p for each AAEI'. E r. As As AA E E P(G,P(G, V), V), m(V/CV(A))m(V/Cv(A)) (m(A) m(V/CV(L)). IfIf 00 #o CW(L)Cw(L) then A and B cen- tralize CW(L),Cw(L), W/Cw(L),W/CW(L), and V/W,V/W, SOso L centralizes Cw(L),CW(L), W/Cw(L),W/CW(L), and V/ W, which is impossible byby ExerciseExercise 3.13.1 sincesince LL isis notnot a p-group. There- fore W = [V, A] @® [V, B] is of rankrank 22 andand V == W @® CV(L).Cv(L). In particular L is faithful on W so L 5< GL(W) 2= GL2(p).GL2(p). Now A fixes only [V, A] amongstamongst thethe set 08 of p ++ 1 1 points points ofof WW and and hence hence is is transitive transitive onon thethe remainingremaining pp points.points. So as B movesmoves [V,[V, A], L isis transitivetransitive on 0.8. ThusThus asas LL containscontains thethe groupgroup AA of transvections with center [V, A], L contains all transvections in GL(W), so, by 13.7, L = SL(W) =E SL2(p). SL2(p). InIn particular,particular, as G is solvable, p <( 3 by 13.8.13.8. Next letlet YY = O"(L)Op(L) and and MM == NG(Y). NG(Y). NoticeNotice MM actsacts onon WW andand CV(L)CV(L) as W == [V, [V, Y]Y] and and Cv(L)Cv(L) = = Cv Cv(Y). (Y). Suppose Suppose L # Lo€Lo E 52S2 withwith YoYo= = OP(L0)OP(Lo) 5< M. Then eithereither WW = = [V, [V, Yo]Yo] oror [V,[V, Yo]Yo] 5< Cv(Y),CV(Y), since Yo isis irreducible on [V, Yo]. If If WW = = [V, Yo] then then CV(Y) Cv(Y) == CV(Yo),Cv(Yo), soso L L == SL(W) = Lo, Lo, contrary toto thethe choicechoice ofof Lo.Lo. ThusThus [V,[V,Yo] Vol 5 < Cv(Y),CV(Y),so so W W 5 < Cy-(L)Cv(L) and, by Exercise 3.6, [L,[L, Lo] _(< CG(W)CG(W) i'lfl CG(CV(L))CG(Cv(L)) == CG(V) = 11 Let A be a maximalmaximal set of commutingcommuting members of S252 and and D D == (A). By the last paragraph, D is the directdirect productproduct ofof (I: (I: II EE A)A) and VV = [V, [V, D]Dl ®@ CvCv(D) (D) with [V, DlD] =_ ®IEO[V, @,,,[V, I].I]. So So it it remains remains toto showshow S252 = = AA andand JJ = (-opfl (Pn P) <5 D.D. Thompson factorizationfactorization 165

If S252 # 0A then,then, byby Exercise 11.4,11.4, there there is is L L E E 52 c- - 0A withwith YY ==OP(L) OJ'(L) (< NF(G)(Op(D)).NF(G)(OP(D)).Thus Thus Y Y permutes permutes Z E = = ([V,([V, I]:I]: II E A)0) and,and, as m([V, Y]) == 2, it follows that Yf actsacts onon eacheach member of E,Z, andand hencehence alsoalso on each member ofof A.0. But But now by the next-to-lastnext-to-last paragraphparagraph andand maximality maximality of of A, A, I, L EE A, 0, contrarycontrary to the choice of L. So S252 = = A. A. Finally Finally assume assume J J$ D D and and let let E E E E °J' P with with EE minimal minimal subjectsubject to END.E $ D. Let Let Z=(WE).Z = (wE). AsAs E E is is abelian, abelian, CE(W)=CE(Z) CE(W)= CE(Z) andand E/NE(L)E/NE(L) is regularregular on WE.WE. Let SS bebe a a setset ofof cosetcoset representatives representatives for NE(L) in E.E. Then Z = ®SES esEs WsWs so Cz(E)CZ(E) == (USES (C,,, ws: w E Cw(N~(L)))CW(NE(L))]is is of of rankrank Es == m(CW/NE(L)).m(Cw/NE(L)). Therefore Therefore m(Z/Cz(E)) m(Z/CZ(E))= =m(Z) m(Z)- -E s= = 2pa2pa -- s,E, wherewhere aa = m(E/NE(L)). AlsoAlso m(NE(W)/CE(W)) m(NE(W)/CE(W)) == 3 6 < ( 1 1 with with s E = = 1 1 in in case case of of equality.equality. Further E s (< m(W) == 2.2. Thus ifif a a > 00 thenthen 2pa -- sE >> a a+ +S. 6. Finally, Finally, ifif E # NE(L), we conclude m(Cv(CE(Z))) - m(Cv(E)) > m(Z) - m(Cz(E)) = 2pa - s > a + S = m(E) - m(CE(Z)), contradicting E EE ?7P(G, (G, V). V). Therefore E actsacts on on each each membermember ofof A.A. Hence, Hence, as as L L= =o~'(GL(w)), OP'(GL(W)), E 5= = 1. 1. Now Now m(E)

This completes the proof of the lemma. Notice that, as D = H H < G,G, S2 52 isis thethe set of all subgroups of G generated by a pair of noncommuting members of F.r .

(32.4) Let F*(G) = Op(G), Op(G), PP E E Sylp(G), ZZ = c21(Z(P)), Q1(Z(P)), VV = (ZG) (zG) and G* =_ G/CG(V).G/CG(V ). Then V is elementary abelian andand Op(G*)Op(G*) == 1.1.

Proof. F*(G)F*(G)=Op(G) = Op(G) <( P P < I C(Z),C(Z), so,so, by by 31.13, 31.13, Z Z< 5S21(Z(Op(G))) 521(Z(Op(G)))= = Vo.VO. So, as VoVo L]< G, V = (ZG)(zG) <( Vo Vo and and hence,hence, asas VoVo is elementary abelian, so is V. Let K* ==Op(G*). Op(G*). By 6.4, Q = PP fl fl K K E E Sylp(K),Sylp(K), so,so, asas K*K* isis aa p-group, K = CG(V)Q 5< CG(Z).CG(Z).Hence, Hence, as as K KL]

Lemma 32.4 supplies aa tooltool forfor analyzinganalyzing groups groups G G with with F*(G) F*(G) == Op(G). Namely, as V is elementaryelementary abelian, we can regard V asas aa vectorvector spacespace overover GF(p) and and the the representationrepresentation ofof G onon VV byby conjugationconjugation makes V intointo a a faithful GF(p)G*-module. These observations are used inin conjunctionconjunction withwith 166 The generalizedgeneralized Fitting subgroup lemmas 32.2 and 32.3 in thethe proof of thethe next twotwo lemmas.lemmas. BothBoth areare versionsversions of Thompson Factorization.

(32.5) (Thompson(Thompson Factorization) Factorization) Let Let G G be be a a solvable solvable group group with with F(G) F(G) == Op(G), let PP E Sylp(G),Sylp(G), Z = 01(Z(P)), nl(Z(P)), V V = = (ZG), (zG), andand G* G* = = GI G/CG(V). CG(V ). Then either (1) G = NG(J(P))CG(Z), NG(J(P))CG(Z),or (2) pp <5 3, 3, J(G)* J(G)* is is the the direct direct product product of of copies copies of of SL2(p)SL2(p) permuted permuted by G, and J(P)* E E Sylp(J(G)*).Sylp(J(G)*).

Proof. If J(P)* # 1 then, by 32.2 and 32.4, G*,G*, V,V, d(G)*sl(G)* satisfiessatisfies the hy- ' pothesis of 32.3, andand hencehence (2)(2) holdsholds by by 32.3. 32.3. So So assume assume J(P) J(P) <5 DD = CG(V)- CG(V). By 32.1 andand a a Frattini Frattini Argument, Argument, G G = = NG(J(P))D NG(J(P))D 5< NG(J(P))CG(Z).

(32.6) (Thompson(Thompson Factorization) Factorization) Let Let G G be be aa solvablesolvable groupgroup withwith F(G)F(G) =_ Op(G). Let p bebe odd and if p == 3 3 assume assume G G has has abelian abelian Sylow Sylow 2-subgroups. 2-subgroups. Then G == NG(J(P))CG(S21(Z(P)))NG(J(P))CG(Q1(Z(P))) forfor PP E E Sylp(G).Syl,(G).

Proof. ThisThis follows follows fromfrom 32.532.5 andand thethe observation that SL2(3) hashas nonabeliannonabelian Sylow 2-groups.

33 CentralCentral extensionsextensions A central extension of of aa groupgroup GG isis a pair (H, n)ir) where H isis aa groupgroup andand n:ir: H + G is a surjective homomorphismhomomorphism with with ker(n) ker(ir) 5 < Z(H). HH isis alsoalso saidsaid to be a centralcentral extensionextension of G. Notice that the quasisimple groups are precisely the perfect central extensions of thethe simplesimple groups.groups. A morphism a:a: (GI,(G1, nl)r-+Ti) (G2,+ (G2, ir2) n2) of ofcentral central extensions extensions of of GG isis aa groupgroup homomorphism a: G1 G1 + G2 with nlnt ==a7r2. an2. A central extension (e,(G, n)lr) of G is universal ifif for each central extensionextension (H,(H, a)a) of GG therethere existsexists aa uniqueunique morphism a:a: (e, (G, n)Jr) +=-* (H, a)a) of of centralcentral extensions. extensions.

(33.1) UpUp toto isomorphism there there isis atat mostmost one universal central extension of a group G.

Proof. IfIf (Gi,(Gi, iri), xi), ii == 1, 1,2, 2, are are universal universal central central extensions extensions ofof GG thenthen therethere exist morphisms of central extensions ai:ai: (G;,(Gi, ,r,)ni) + (G3_1,(G34, n3-i).7r3-0. As As aia3-i aia3-i and 1 are morphismsmorphisms of (Gi(Gi, , 7ri) ni) to (Gi, nini), ), the uniqueness of such a morphism says ala2ala2 == 1 1 = = a2a1. ~~2~x1. ThusThus ai ai is is an an isomorphism. isomorphism. Central extensionsextensions 167

(33.2)(33.2) IfIf (G,(6, ir)n) isis aa universal universal centralcentral extension ofof G then bothboth 6G and G are perfect.

Proof.Proof. Let H = G6 xx (G/&))(6/6(')) andand define define a: H + G by (x, y)a y)a = xir. xn. Then Then (H,(H, a)a) isis a a central central extension extension ofof GG and and a1:ai: (G, (6, ir)n) + (H, a), ii == 1, 1,2, 2, are are mor-morphisms, wherewhere xal xa1= = (x, 1) 1) andand xa2 xa2 == (x,(x, x&)).~6")). So,So, by by the the uniqueness uniqueness of suchsuch a morphism, a,aI = a2 a2 andand hence G6 == 0(1). 6"). ThusThus G 6 is is perfect, perfect, so, so, by by 8.8.2,8.8.2, GG == Gir 6n is is perfect. perfect.

(33.3)(33.3) Let G bebe perfectperfect andand (H,(H, .7r)n) a central extension ofof G. Then HH = ker(ir)HO)ker(n)HC1) with H(')HO perfect. perfect.

Proof.Proof. By By 8.8, 8.8, H(1)7r H(')n = = (Hir)(1) (Hn)(') == G(1), G('), so,so, as as G G is is perfect, perfect, H(1)7r H(')n = G. G. Hence Hence HH == ker(.7r)H(1). ker(n)HC'). AsAs (H, (H, n) ir) isis acentralextension,a central extension, ker(n) ker(,r)5 < Z(H), Z(H), soso H/H(~)H/H(2) = Z(H/H(2))(H(')/H(2))z(H/H(~))(H(')/H(~)) isis abelian, abelian, and and hencehence HOH(') = = H(2) HC2) by by 8.8.4.8.8.4. ThusThus HOH(') isis perfect.perfect.

(33.4)(33.4) GG possesses possesses a a universal universal central central extension extension if if andand onlyonly ifif GG is is perfect. perfect.

Proof. ByBy 33.2, 33.2, if if G G possesses possesses a a universal universal central central extension extension thenthen GG is is perfect. perfect. Conversely assume G isis perfect.perfect. LetLet g g F+H g g be be aa bijectionbijection of G with a set (?a and let F be the free groupgroup onon (?.G. Let Let r r be thethe setset ofof words words X xy(xy)-1,jilxy)-', x,x, y EE G,G, and let M be the normal subgroup of F generated by F.r. Next let A be the set of words [w,[w, i], Z], w w E E P, r, Z z EE G,G, andand let N be the normal subgroup of F generated by A. AsAs MM a4 F, N = [M, [M, flF] :4 FF and and M/N MIN < 5 Z(F/N) Z(F/N) byby 8.5.2. 8.5.2. ThenThen by by 28.628.6 and 28.7 there is a unique homomorphismhomomorphism n: ir: FINF/N + G, with (XN)n(xN)7r ==x x forfor all x E G,G, and and indeedindeed MIN M/N = ker(ir). ker(n). Therefore (FIN,(F/N, ir)n) isis a a central central extension extension of G. Let (H,(H, a) be be a acentral central extension extension ofof G.G. For For X x E E G,G, let let h(x)h(x) EE HH withwith h(x)orh(x)a = = x. Then, for x,x, y, y, zz EE G, w = h(~)h(~)h(x~)-'h(x)h(y)h(xy)-1 E ker(a)ker(a) < 5 CH(h(z)),C~(h(z)), so [w,[w, h(z)]h(z)] == 1. HenceHence byby 28.628.6 there there exists exists a aunique unique homomorphism homomorphism a: FIN+F/N H with (XN)a(xN)a = h(x) foreachxfor each x E E G. G.Noticea: Notice a:(FIN, (F/N,n) ir)+ - (H, a) is aa morphism. Now let 6G == (FIN)('). ByBy 33.3,33.3, FIN=FIN = ker(7r)O ker(n)6 and and G6 is is perfect. perfect. HenceHence (G,(6, ir)n) isis alsoalso aa centralcentral extensionextension of G, and and a: (G,(6, ir)n) + (H, a) aa morphism. morphism. Suppose ,B:p: (6,(G, n)ir) + (H, a) isis a a second second morphism, morphism, and and definedefine y:y: G + H byby uy = ua(uf)-1, ua(up)-', for for u u E E G.6. Then aa aQ ==7r n =pa, =for, so GYGy c Z(H). Thus (uv)y(uv)y == (uv)a((uv)~)-'(uv)a((uv),B)-1 = uava(v,8)-1(u$)-1uava(v,!?)-' (up)-' = uavy(up)-' uavy(u,8)-1 = ua(u,8)-lvyua(up)-'v y = uyvy,uy vy, so so y y is is a ahomomorphism. homomorphism. MoreoverMoreover Gy6 isis abelian,abelian, so,so, asas G6 isis per-perfect, y isis trivial by 8.8.4. ThusThus a a == P.p. 168 The generalizedgeneralized Fitting subgroup

(e,(G, 7r)n) has been shown to be a universaluniversal central extension, soso thethe proofproof isis complete.

If G isis aa perfectperfect groupgroup andand (G,(e, 7r)n) itsits universaluniversal centralcentral extension, then G isis called the universal covering group ofof GG andand ker(7r)ker(n) the Schur multiplier of G. Notice that, by 33.2, G isis perfect. A perfect central extension oror covering ofof aa perfectperfect groupgroup GG is a central extension (H,(H, a) of G withwith H perfect.perfect.

(33.5) Let (H, a) be be aa centralcentral extensionextension of a group G, and and (K,(K, ,8) p) a perfect central extension of H. Then (K, Pa) pa) is is a a perfect perfect central central extension extension of of G.G.

Proof. ,8a:pa: K + G is thethe compositioncomposition of surjective homomorphisms andand hence a surjectivesurjective homomorphism. homomorphism. LetLet X x EE ker(pa)ker(,da) andand yy EE K.K. x,dxp E E ker(a) <5 Z(H), so so [x, y],8y]p = [x,d,[xp, yp]yd] == 1. ThusThus [x,[x, y]y] EE ker(p) ker(b) 5< Z(K).Z(K). Thus [ker(pa),[ker(da), K, K]Kl= = 1, so, byby 8.9,8.9, ker(pa) ker(8a) 5< Z(K).

(33.6) Let (H,(H, a) a) and and (K, (K, ,8) p) be be centralcentral extensions extensions of a group G with K perfect, and y: (H, a) - + (K, (K, ,8) p) a amorphism morphism of of centralcentral extensions.extensions. Then (H,(H, y) is aa central extension of K.

Proof. y: H 4 K isis aa homomorphism homomorphism with a == y,8. yp. TheThe latterlatter factfact impliesimplies ker(y) <<_ Z(H), Z(H), so so itremains it remains to to show show y y is is a a surjection. surjection. As As a == y,8 yp is is aa surjection, K = (Hy)ker(,8),(Hy)ker(p), so, asas ker(p) ker(b) 5 < Z(K), Z(K), HHy y 49 K K andand K/HyK/H y is is abelian. abelian. Hence Hence K = Hy as as KK isis perfect.perfect.

(33.7) Let Ge be the covering group ofof aa perfectperfect groupgroup GG andand letlet (H,(H, a)a) be a perfect central extension of e.G. Then a isis anan isomorphism.isomorphism.

Proof. LetLet ir:n: G e + G be thethe universaluniversal covering. ByBy 33.5,33.5, (H, (H, an) an) isis aa per-perfect central extension of G, soso byby thethe universaluniversal propertyproperty therethere isis aa morphismmorphism p:8: (G, 7r)n) + (H, an). Then Then ,8a ,!?an r =7r= n so so by by the the uniqueness uniqueness propertyproperty ofof universal extensions, pa,8a = 1. 1. Hence Hence ,8: p: Ge + H is an injection, while ,8,3 is aa surjectionsurjection by 33.6. ThusThus pfi isis anan isomorphismisomorphism and and as as pa ,8a = = 1, 1, a a = =,8-1 p-' is too.

(33.8) LetLet GG bebe perfect, (G,(e, 7r)n) the universaluniversal centralcentral extensionextension ofof G, G,and and (H, (H, a)a) a perfect central extension of G. Then (1) ThereThere existsexists a covering a: G (? + H withwith n r == aa.aa. (2) (G,(6, a)a) is is the the universal universal centralcentral extensionextension of H. (3) TheThe SchurSchur multipliermultiplier ofof HH isis aa subgroup subgroup ofof thethe SchurSchur multiplier of G. Central extensionsextensions 169

(4) IfIf Z(G)Z(G) =1 = 1then then Z(G) z((?) isis the the~churmulti~lier Schur multiplier ofof G, G, and and Z(H) Z(H) = = ker(a)ker(a) S ker(n)/ker(a)ker(7r)/ker(a) isis thethe quotientquotient of the Schur multiplier of G by the Schur multi-multiplier of H.

Proof. ByBy the the universal universal property property therethere existsexists a morphism a: (G,(e, 7r)n) + (H, a). Then n7r == aaas and, and, by by 33.6,33.6, a a is is a a covering. covering. Let Let (H, (H, ,B) B) bebe thethe universal universal cover-covering of H. ByBy thethe universaluniversal propertyproperty therethere isis aa morphismmorphism y:y: (H, (fi, ,B),!?) + (G,((?, a). By 33.6 and 33.7, y isis anan isomorphismisomorphism soso (2)(2) holds.holds. Now (3) andand (4)(4) areare straightforward.

(33.9) Let GG bebe aa group with G/Z(G) finite. finite. Then Then GMG(') is is finite.finite.

Proof. LetLetn n = IG/Z(G)I. IG/Z(G)I.Forz For Z E Z(G)Z(G)andg,h and g, h E G,[g,hz]G, [g, hz] == [g,[g,h] h] = G(1) [gz, h],h], soso thethe setset 0A of of commutators commutators is is of of order order at at most most n2. n2. I I claim each g EE G(') can bebe expressed expressed as as a aword word g g= = XI xl ...... x,x inin the the members members of A0 ofof length min 5< n3.n3. The claim together with the finiteness of 0A show show G(1) G(') isis finite.finite. It remains to establish the claim. Pick an expressionexpression forfor gg of minimalminimal lengthlength m. If mm > n3 then, as 1I A1 0 <5 n2, therethere isis somesome dd EE 0A with with I' r= =f i{i: : x1 x, == d]d} of orderk >> n. n.As~,x,+~ Asxix1+l =xa+1xx=x~+~x~+' `+' withx?+'withxx`+' E E A 0 wecanassumerwe can assume F = = {I,.[1, .... ., ,k}. k]. Hence it remains to show that do+1dn+' cancan bebe written as a product of n commutators, since then thethe minimalityminimality ofof mm willwill bebe contradicted.contradicted. Let Let d d == [x, y]. As (dx)n-'[x2, IG/Z(G)I == n, n, do dn E E Z(G),Z(G), so dn+'d"+1 == (dn)Xd(dn)xd = = (dn-')xdxd(dn-')Xdxd == (dx)n-' [x2, y] by 8.5.4. In particular do+1dn+' is a product of n commutators.

(33.10) LetLet GG bebe aa perfect perfect finitefinite group.group. ThenThen thethe universaluniversal covering group of , G and the Schur multiplier of GG areare finite.finite.

Proof. ThisThis is is a a direct direct consequence consequence of of 33.9.33.9.

(33.11) LetLet (H,(H, a) a) bebe a a perfect perfect central central extension extension of of aa finite group G, p a prime, and P E Syl,(H).Sylp(H). ThenThen P P n n ker(a) ker(a) 5 < @(P). ((P).

Proof. PassingPassing to H/(@(P)H/((D(P) nnker(a)) ker(a)) we maymay assume assume @(P) 1(P)n nker(a) ker(a) = 11 and it remains toto showshow X x == P n ker(a) == 1. 1. But But as as P/(DP/@(P) (P) = = P* P* is is elementary elementary abelian therethere isis aa complementcomplement Y* Y* to to X* X* in in P*. P*. Then Then P P = = X x YY so P splitssplits over X. Hence by Gaschiitz' Theorem, 10.4, H splits over X. Hence, as H isis perfect andand XX 5< Z(H),Z(H), XX = 1.1.

(33.12) LetLet GG bebe aa perfectperfect finitefinite group and M the Schur multiplier of G. Then .7r(M)n(M) S:c 7r(G).n(G). 170 The generalized Fitting subgroup

Proof. ThisThis is is a a consequence consequence of of 33.11.33.1 1.

(33.13) LetLet (H, (H, or) a) bebe aa perfect central central extension extension of of aa finitefinite group G with ker(aker(a) ) a p-group, let Go be a perfectperfect subgroup of G containingcontaining a SylowSylow p-groupp-group of G, let Ho == Q-1(Go), a-'(~o), andand ao a0 = =or, al~,: H0: HoHo + Go. Then (Ho, 00) ao) is a perfect central extension of Go with ker(ao) == ker(a).ker(o ). HenceHence aa SylowSylow p-groupp-group ofof thethe Schur multiplier ofof G is aa homomorphichomomorphic image of a SylowSylow p-group of of thethe SchurSchur multiplier of Go.

Proof. EvidentlyEvidently (Ho,(Ho, ao)00) is a central extension ofof GoGo withwith ker(a) ker(a) = ker(ao), ker(ao), so, by 33.3, HoHo = ker(a)~i')ker(a)Ho" withwith Hot)H:') perfect.perfect. AsAs GoGo contains contains a a Sylow- Sylow p- group of G,G, Ho containscontains a a Sylow Sylow p-group p-group P P ofof HH and,and, asas ker(a) ker(a) isis aa p-group,p-group, ker(a)ker(or) i < P. P. Then,Then, byby 33.11,33.1 1, ker(a) <( (D(P), @(P), so HoHo == @(P)H:').0 Thus P =-- P flil HoHo = t(P)(P@(P)(P fl nHo1)) H,'") byby the the modular modular property,property, 1.14, soso PP = P fln Hot) H:') 5< H'H~'') ) byby 23.1.23.1. ThusThus HoHo = PHo1)P H:') = Hot>, H:'), soso Ho isis perfect,perfect, completing the proof.

(33.14) If If GG is is aaperfect perfect finitefinite group with cyclic Sylow p-groups thenthen thethe SchurSchur multiplier ofof GG isis aa pl-group.p'-group.

Proof. LetLet (H,(H, a) a) be be a a perfect perfect central central extensionextension of G with ker(a) aa p-group;p-group; I must show ker(a)ker(or) = = 1. 1. Let Let P PE ESyl,(H), Sylp(H), so so that that Z Z= ker(a)=ker(a) ( < P P and and P/Z P/Z is cyclic.cyclic. By 33.133.11, 1, ZZ _(< @(P),4)(P), so PIP/@(P) 4)(P) is is cycliccyclic and hence, by 23.1, P is cyclic. At this point I appealappeal toto 39.1,39.1, which which says some h E HH inducesinduces a nontrivial pl-automorphismp'-automorphism onon P.P. But then, by 23.3, [S21(P),[Q1(P), h]h] # 0 1, so,so, as ZZ(Z(H), < Z(H), zZ n il S21(P)Q1(P)=l. = 1. As As PP isiscyclicand cyclic and ZZ( < PP thisthis sayssays ZZ= = 1. 1.

The section on the generalized Fitting group focused attention on quasisimple groups. Observe that the finite quasisimple groups are precisely the perfect central extensions ofof the finite simple groups. Hence, for each finite simple group G, the universal covering group of G isis thethe largestlargest quasisimplequasisimple groupgroup with G asas itsits simplesimple factor,factor, and the centercenter of anyany suchsuch quasisimplequasisimple group is a homomorphic imageimage ofof the Schur multiplier ofof G.G. Thus it is of particular interest to determine the covering groups and Schur multipliers of thethe finitefinite simple groups. This section closes with a description of the covering groups and Schur multipliers of the alternatingalternating groups.groups.

(33.15) LetLet G == A,,A, n > 5, eG the universal coveringcovering group group of of e, G, and and 2 2 == Z(O)~(e) thethe Schur multiplier of e.G. Represent G on X == (1, ...... ,,n). Then (1) 22 = G 716 z6 ifif n == 6 6 or or 7, 7, whilewhile 22 =712 E Z2 otherwise. otherwise. Central extensions 171

(2) LetLet i i be be a a 2-element 2-element in in G6 such such that that thethe imageimage it ofof i inin G6 isis anan involution. Then tf has 2k2k cyclescycles ofof lengthlength 2,2, if isis anan involutioninvolution ifif k is even, and i1 isis of order 4 if k is odd. (3) IfIf nn == 6 6or or 7 7then then 31+2 31+2 isis aa SylowSylow 3-group3-group of G.6.

This result is usually proved using homological algebra,algebra, butbut I'llI'll take a group theoretical approachapproach here. There are two parts to the proof. First show the order of the Schur multiplier of AnA, is at least 2 (or 6 ifif nn = 66 oror 7).7). SecondSecond showshow the multiplier has order at most 2 (or 6) and establish 33.15.2 and 33.15.3. Exercise 11.5 handles part one (unless n = 6 6 or or 7 7 where where thethe proofproof thatthat 33 dividesdivides thethe order of the multiplier is omitted). The second part is moremore difficultdifficult andand appearsappears below. Assume for the remainder ofof thethe section thatthat (G,(G, n)n) is a a perfect perfect centralcentral extension ofof eG =A,= An andand write write Sfor for the the image image Sn Sn of of S S E C G. G. Let Let Z Z == Z(G) and assume ZZ is a nontrivial p-groupp-group forfor somesome prime prime p. p. Let Let PP E Syl,(G).Sylp(G). It will suffice toto showshow pp <5 3,3, IZI IZI ==p, p, and and 33.15.2 33.15.2 and and 33.15.333.15.3 hold.hold. AssumeAssume otherwise and choose G to bebe aa countercounter exampleexample toto oneone ofof thesethese statementsstatements with n minimal. The idea of the proof is simple: exhibit a perfect subgroup H of G containing P,p, useuse thethe inductioninduction assumptionassumption to show the multiplier ofof H is a p'-group, andand hencehence obtainobtain a contradiction from 33.13. When p = 2, andand sometimes when p = 3, 3, thethe situationsituation is is more complicated but the same general idea works. I begin a series of reductions.

(33.16) n > 2p.

Proof. ByBy 33.14,33.14, Pp isis not not cyclic.cyclic.

(33.17) n > 5.

Proof. Assume nn = 5.5. By 33.16, pp = 2.2. Assume IZII Z =I = 2.P 2.P ( < HH <( G 6 withwith H =E A4, A4, so so Pp - EE4 E4 and and H H is istransitive transitive on on P#. P#. By By 33.11,33.1 1, (D(P)@(P) = Z,Z, so so byby 1.13 there is an element of orderorder 44 inin P. Hence, as H isis transitivetransitive onon P#,p#, every every element inin P - Z Z is is of of order order 4. 4. So,So, byby Exercise 8.4, P isis quaternion of order 8. Notice this givesgives 33.15.2 in this case, so it remains to show IIZI Z I( < 2. 2. Now Now p P == (8,(g, A) h) so P(')PM = (z) (z) where where zz == [g, [g, h]. h]. IfIf UU is is of of index index 22 inin ZZ we'vewe've seenseen thatthat P/P/U U -E Q8, Qg, so so z zV 4 U. U. Thus Thus Z/d)(Z) Z/@(Z) is iscyclic cyclic so so Z Z is is cyclic. cyclic. So, So, as as Z/ Z/U U is is the the unique subgroup of P/ UU ofof order 2, P also also hashas aa unique involution. Hence,Hence, byby Exercise 8.4, P is quaternion. So,So, as as I P:IP: Z(P)I Z(P) I= = 4,4, PP -E Q8, Qg, ThatThat is is I IZIZI = 2. 2. 172 The generalized Fitting subgroup

For S cC G or G writewrite Fix(S) andand M(S) forfor thethe setset ofof pointspoints in XX fixedfixed by SS and moved by S, respectively.

(33.18) LetLet Z3713 Z = Y 5< GG with IM(Y)IIM(~)I= = 3. 3. Then Then Y Y = = (y) (y) x x Z for somesome y of order 3, and if 8

Proof. YE <5 HH -Z A5 A5 with with IM(H)IIM(H)I == 5. 5. WithoutWithout lossloss pp = 3,3, so, by 33.17, HH = Z x HO).~('1. Thus Y == (y) x Z wherewhere yy EE H(').H(l). Also, if 8 5< n # 9 or 10, 10, thenthen 03'(cG(y))O3'(CG(Y))= _ (7) (y) x x K, k, kK -Z An_3, An-3, and, and, byby minimalityrninimality ofof n,n, KK == Z x K(l).K(').

(33.19) IfIf pp = 3 then nn > 6.6.

Proof. AssumeAssume pp = 3 3 andand n = 6. 6. Then E9 Z- PP = = (y, (9, g) g) with with y yand and g g moving moving 3 points. ByBy 33.1833.18 wewe maymay taketake y y andand gg ofof orderorder 3.3. LetLet RR = = (y,(y, g).g). ThenThen RR is of class at most 2 so, by 23.11,23.1 1, R is of exponent 3.3. ByBy 33.133.11, 1, RR == PP andand Z 5< @(P).t(P). PM P(') = = ([g, ([g, y]) y]) so, so, asas PP isis ofof exponent 3, 3, Z Z = = P(')P(l) = Z(P).Z(P). Thus Thus P isis extraspecialextraspecial of order 27 and exponent 3, so, by 23.13, P -Z 31+2 31+2 .. Hence all parts of 33.15 hold in this case.

I now consider two cases: Case I: nn is not a power ofof p; Case II:11: n is a power of p.p. InIn CaseCase I IP P is isnot not transitive transitive on on X Xso sothere there is ais partition a partition 171, P = = [X (XI, 1, X2)X2) of X withwith PP acting acting on on X XI 1 andand X2.X2. Let HH bebe the the subgroup subgroup acting acting on on X XI. 1. Then H == (a)Ho (ii)Ao withwith HoHo = = H1 HI Xx H2,Hz, where where Hi == Ox,_, Gx,-< -Z An; A, is is the the subgroup subgroup of G fixingfixing each point ofof X3+,X3_;, nini = IXi1,/Xi1, and (ii)(a)Hi Hi Z- Sn; S,, (unless(unless nIn 1 or or n2n2 is 1).1). In case 11,II, P isis transitivetransitive on X and there isis a partition P°J' ofof X into p subset Xi,Xi, 1 5< i <5 p,p, of of order order n/p,nlp, with with PP transitive transitive onon °J'.P. The The subgroup subgroup of G preserving P?P contains contains a asubgroup subgroup H H with with P P5 < H H andand HH Z- (An/P)wr (AnIp)wr Ap A, if p # 2, while H isis ofof indexindex 22 inin Sn12Sn12wr wr Z2 Z2 if if p p == 2. Notice that, by 33.16, n/pnip >2 p.P. Observe next that:

(33.20) One of the following holds:

(1) KK == Op'(H) OP'(H) is is perfect. perfect. (2) pp == 2 2 and, and, in in CaseCase I,I, n1nl > 11 << n2- n2. (3) pp == 3 3and and n1 nl or or n2 n2 is is 3 3 oror 44 inin CaseCase I.

Moreover, byby minimalityminimality of of n, n, ExercisesExercises 11.211.2 and and 11.3, 11.3, and and 33.13, 33.13, if if K k is perfect then p 5< 3, and if p = 3 3 andand CaseCase II holds,holds, thenthen nnl I or n2 is 6 or 7. In particular: Central extensionsextensions 173

(33.21) p <( 3 3 and and ifif pp == 3 3 and and Case Case I I holds holds then then n1nl oror n2n2 isis 3,3,4,6, 4, 6, or 7.

(33.22) p = 2.

ProoJProof. IfIf notnot pp == 3. 3. Suppose Suppose Case Case I Iholds. holds. Then Then n1 nl oror n2n2 isis 3,3,4,6, 4, 6, or 7; say n1.nl. Then P actsacts onon aa subsetsubset XiX/, of of X1X1 of of orderorder 3,3, soso replacingreplacing X1X1 by XiX/, we we may take nln1 == 3.3. ButBut nownow 33.18 33.18 implies implies K K = = (y)(y) xx ZZ x L if 8 _(< n #0 99 or 10,10, contradicting 33.133.11. 1. Hence Hence n n = =7, 7, 9, or 10. AsAs nn isis notnot aa powerpower ofof 3,3, nn #0 9,9, while if n = 1010 we could have chosenchosen nln1 = = 1, 1, n2 n2 = = 9. 9. So Son n = 7,7, wherewhere wewe choosechoosenl n1 = 1,1, so that K 2 A6 and,and, byby 33.1333.13 andand minimality minimality of of n, n, /Zl=IZ = 3 3 andand PP=31+2 2 3l+2. So Case III1 holds. Here Kk == (a)Ko (ii)KO wherewhere KoKO isis thethe directdirect productproduct ofof 33 copies of An/3rAn13,and and ii ais is of of order order 3 3 with with I M(ii)l=IM(a)I =n.n. Then Then ii a = bgJg for for somesome b E KO,and, and, byby minimalityminimality ofof n andand ExerciseExercise 11.2,11.2, K0 KO == Z Z x x K0(1) K;') if n > 9. If nn = 9 9 then,then, byby 33.18,33.18, K0KO == (Z, (Z, gi: g,: 1 1 <( i <(3) 3) withwith IM(gi)lI M(gi) == 3. FurtherFurther g,gi E z(03'(cG(cG(gi))))Z(O3'(CG(CG(gi)))) SO so KOK0 isis abelian. Finally there there is a 2-group2-group T7 = [T, [T, a] ii] <_( G withwith Ko KO == [K0, [KO, T],71, so, so, as as K0 KO isis abelian, abelian, 24.6 24.6 impliesimplies K0TKoT = Z Xx O~(KOT)O3(K0T) andand a a ofof coursecourse actsacts on on 03(KoT). O3(K0T).Let Let L L = = K;')Kol) ifif nn > 99 andand L = O3(K0T) O~(K~T) ifif nn = = 9. 9. We We can can choose choose b b E E L, L, so so that that aa isis of of orderorder 3. 3. But But nownow P == Z Z x x ((P ((P fl nL)(a)) L)(a)) contradicting contradicting 33.11. 33.1 1.

This leaves pp = 2. 2. The The proofproof herehere is similar to the casecase pp = 3. 3. IfIf nn isis oddodd then - Case I holdsholds withwith nln j == 1, so HH =2 An-1 A,-1 isis perfect perfect andand 33.13 and minimality of n complete thethe proof.proof. ThusThus n n is is even even and and H H = = Ho U withwith H0Ho == HOH0 = H H1 1 Xx Hz,H2, where, inin CaseCase I,I, HiHi 2= Ani,A,, ni is even, and U = (u)(ii) = 2 Z2, Z2, while, while, in in CaseCase II, 11, n is a power of 2. Hi 2 An/2An12 andand uU = (u,(ii, v) rir) =2 E4. E4. We We can can take take Ug Ug <( Ho Ho HI(1) for some gg EE G. G. Let Let ni ni == n/2 inin Case II.11. By minimality of of n, n, Hi Hi == Z * H,(') with ZiZi = H:')HH(1) nfl Z of order atat mostmost 2, 2, unless unless ni ni( < 4. 4. If If ni ni == 44 thenthen Hi I_( 4 set L~Li == H:').Hi(1) IfIfnl n j >4(nz> 4 < n2 thenthereisthen there is yi yi €Li ELI withwith IM(yi)l=4andI M(yi) =4 and yfyi == Y2 y2for for somesome a EE G. Notice ZiZi == (y?)(y:) SO,so, asas ZZ (< Z(G),Z(G), Z2 == ZfZl = Z1. Z1. Thus ZoZ0 = ZZ fln LoL0 = Z1 where LoL0 = L L1 1 L2. L2.This This of of course course also also holds holds if if ni ni << 4 for some i, since then ni == 2 2 and and Hi Hi = = 1. 1. L0 Lo char char Ho Ho so so U U acts acts on on L0, Lo, and and wewe maymay take Ug <( L0, Lo, so Z fln U9Ug (< Z1. Thus H ==Z Z ** ULoUb with Z fln UL0ULo=Z1, = Z1, so, by 33.11, Z = Z1 Z1 is of order 2. Thus 33.15.1 is established and it remains to establish 33.15.2. Let t be aa 2-element2-element in G with It an involution in G. As G = AnA, consists ofof even permutations on X, t has 2k cyclescycles of lengthlength 22 onon XX forfor some some positive positive integer k. ThenThen therethere isis aa partition partition XX = = Yo Yo UU YlYl U . . . U YkYk ofof X withwith Y0Yo = Fix(t), II Yi I I= = 44 forfor 11 (< ii (< k, and t acting onon Yi.Yi . We've seen that for i > 00 there 174 The generalizedgeneralized Fitting subgroup is Ki fixing XX -- Y; Yi pointwise pointwise withwith KiKi S- SL2(3) SL2(3) and Ki centralizing GyiGy.. Thus (Ki: 1 5< ii 5< k)k) == KlK1 * . - a* Kk isis aa centralcentral productproduct with identified centers and t = tltl . . . tk with ti E Ki. As Ki S SL2(3), titi isis of of order order 4 4 with with (ti2) (t?)= = 2,Z, SO sot t is of order 4 if and only if k is odd.odd. Hence the proof ofof 33.1533.15 isis atat lastlast complete.complete.

Remarks. I'mI'm not not sure sure who who shares shares credit credit for for thethe notion of `components'.'components'. Cer- tainly Bender [Be 1],11, Gorenstein and and Walter [GW],[GW], and Wielandt [Wi[Wi 1]11 shouldshould be included. I use Bender's notation of E(G) forfor thethe subgroup subgroup generated generated byby the components of a group G. The Gorenstein and Walter notation of L(G) isis also used in the literature to denote thisthis subgroup.subgroup. II believebelieve Bender [Be[Be 1] 11 waswas the first toto formally define the generalized Fitting subgroup. Thompson factorization was was introducedintroduced by by ThompsonThompson inin [Th[Th 1 11 ] and [Th 2],21, although 32.5 was presumably first proved byby Glauberman [Gl[G1 21.2]. There are analogues of Thompson factorization forfor arbitraryarbitrary finitefinite groupsgroups GG with F*(G) = = Op(G). Op(G). Such Such results results require require deepdeep knowledgeknowledge of the GF(p)-represen-GF(p)-representations of nearly simple groups. In particular one needs to knowknow thethe pairspairs (G,(G, V)V) with G a nearly simplesimple finitefinite groupgroup and and VV an an irreducible irreducible GF(p)G-module GF(p)G-module suchsuch that GJ'(G,P(G, V) isis nonempty. Generalized Thompson factorization plays an important role in the classification; see in particular the concluding remarks in section 48. Thompson factorizationfactorization forfor solvablesolvable groupsgroups isis usedused inin this book to establish the Thompson Normal p-Complementp-Complement Theorem, 39.5, and thethe nilpotence of Frobenius kernels (cf. 35.2435.24 andand 40.8).

Exercises for chapter 1111 1. LetLet r r bebe aaprime, prime, G a solvable rl-group,r'-group, and and AA an r-group acting acting on on G.G. Prove (1) IfIf rn C E 7r n(G) (G) and and H H is is aHall a Hall 7r n-subgroup -subgroup of G, thenthen Op(H)Op(H) <5 Oph(G)Oph(G) for eacheach p p EE n,r, where pnp" = r' n1 UU { {p}. p). (2) IfIf [A,[A, F(G)] = 1 1 then then [A,[A, G]GI = 1.1. (3) IfIf AA centralizescentralizes a Sylow p-groupp-group of G thenthen [A,[A, GIG] I< Op,(G).Opt(G). 2. Let Let (Gi (Gi : : i EE I) bebe perfect groupsgroups andand GiGi thethe universal covering group of Gi.Gi . Prove the universal covering groupgroup ofof thethe directdirect productproduct DD of the groups Gi, i E I, is the direct product of thethe coveringcovering groupsgroups G~,6j, i EE I, and and hence the Schur multiplier of G is the direct product of thethe SchurSchur multipliersmultipliers ofof the groupsgroups GiGi, , ii E I. 3. LetLet HH andand GG be be perfectperfect groupsgroups with universal coveringcovering groups groups H h and G,e, let H actact transitivelytransitively onon aa set I,I, and and letlet WW == G G wr1 wrl H. Prove the universal covering group of W is thethe semidirectsemidirect product product ofof aa centralcentral productproduct ofof I1 11I copies of G with H,I?, and and thethe SchurSchur multipliermultiplier ofof WW isis thethe directdirect productproduct of the multipliers of G and H. 4. Let G G bebe aa finitefinite group,group, p p a a prime, prime, Q!2 aa G-invariant G-invariant collectioncollection of p- subgroups ofof G,G, PP a p-subgroup of G, and A EC P n Q !2 withwith AA CE Np(O).Np(A). Central extensionsextensions 175

Prove thatthat eithereither AA = = PP nfl Q or therethere existsexists X X E E (P (P n fl Q) Q) - - A with X <( Np(O).NP(A). 5. LetLet UU bebe aa 2-dimensional2-dimensional vector space over CC withwith basisbasis XX == (x, y) andand define a, $, B, andand y y to to 1be be the elements of GL(V) such that ( Mx(a) = t0 f,Mx($) =(0 ), Mx(Y) _ (0 -Ol),

wherewhereiisinCwithi2 i is in C`withi2 = [email protected]@U,bethetensorproduct Let V= Ul ® .. ®U be the tensor product of n copies UJUi of U and let sksk in in Endc(V) Endc(V) be defined (cf. 25.3.5) as follows:

s2J =(1®...®1®((1®$)+(a(9 Y))®y(9 ...(, Y)l', 1

S2J-1 =(1®...®1®(a+$)®y(9 ...®Y)/I,1 1.1. (3) LetLet G G be be the the subgroupsubgroup of GL(V) generated by (sk:(sk: 11 5< k < 2n).2n). ProveProve G/(-I)G/ (- I) = EStn S2, with with -I -I E GM,E G('), and and GM G(') is isquasisimple quasisimple if if nn >> 2. 2. 6. LetLet GG be be a a nonabelian nonabelian simplesimple group group and and G thethe universal universal coveringcovering groupgroup of G. Prove Aut(G)Aut(G) S- Aut(G).~ut(c). 7. LetLet G G be be aa solvablesolvable group with F(G)F(G) = Op(G) Op(G) = = T, T, let let R R be be aa p-subgroupp-subgroup of GG containingcontaining T,T, let let (NG(C): (NG(C): 11 # C charchar R) (< H <5 G,G, letlet XX = J(ROp,F(G)), V V == (Q1(Z(R))x), (Q~(Z(R))'), andXR* and XR* =XR/CXR(V). = XR/CxR(V). Prove Prove eithereither (1) Op(H)Op(H) = T,T, oror (2) p <(3, 3, X*X* == X;Xi x .. .. x X*X,* with Xi*X = E SL2(p), SL2(p), R R fl n X X E E Sylp(X), Sylp(X), and NG(R) permutes {Xi:{X1: 1 1 ( < ii (< n).n]. 8. Let GG bebe aa finitefinite group,group, r aa prime,prime, nr aa setset ofof primesprimes with r V4 n, r, andand E,JE,.3-A

H ofof GxG, containedcontained in Gy.G,. Prove there exists a primeprime p such that for each x cE Fr andand yy cE Fx, r,, F*(Gx)F*(G,) andand F*(Gxy)F*(G,,) areare p-groupsp-groups andand eithereither ZxZ, or ZyZ, is a p-group. (Hint: Let x cE F,r, y y CE r,,Fx, zz EC r,, Fy, and and q q aa primeprime suchsuch thatthat Z,Zx isis notnot a q- group. For H <5 GG set set B(H) 0(H) = = Oq O,(H)E(H) (H)E(H) andand letlet Qx Q, == Gx,rz G,J~. . ProveProve Z,Zx gl<4 Gy G, andand GyZ.G,,. Then use 31.431.4 andand ExerciseExercise 11.911.9 toto showshow O(Gy),O(Gy), O(Gy,)9(GyZ) 5< Gx.G,. Conclude: (*) B(Gy) < 9(Gxy) < 9(Qy) < 9(Gy) Then (interchanging thethe rolesroles ofof xx and y if necessary) conclude Z,Zx is a p-group for some prime p.p. UsingUsing (*)(*) showshow 9(Zy) < 0(Qx) < O(Gx,y) = 9(Qy) and hence concludeconclude F*(Z,)F*(Zy) is a p-group. Finally useuse (*)(*) to show 0(G,)O(Gx) =_ 0(G,,)B(Gxy) == O(Gy) forfor eacheach primeprime q 54# p.p. This proof is due to P. Fan.) 11. (Thompson (Thompson [Th 31)3]) ProveProve there there exists a function ff fromfrom the positive integers into the positive integers such that, for each finite set X, each primitive permutation group G onon X, eacheach x cE X, X, andand eacheach nontrivialnontrivial orbitorbit YY of GxG, on X, either (a) IGxIlGxl 5< f(lYI19f(IYI), oror (b) F*(Gx)F*(G,) isis aa p-groupp-group forfor somesome prime p. Remark: Let y EE Y,Y, 9= (G,(Gx, , Gy),G,), and apply Exercise 11.1011.10 to the action of G on the geometrygeometry r(G,F(G, 9).). The Sims Conjecture says that a function f existsexists withwith IGxI

Linear representationsrepresentations ofof finitefinite groups

Chapter 12 considers FG-representationsFG-representations where where G G is is a a finitefinite group, group, F F isis a splitting fieldfield forfor G, and the characteristic characteristic ofof F doesdoes notnot divide divide thethe orderorder of G.G. UnderUnder these these hypotheses, hypotheses, FG-representationFG-representation theory theory goes goes particularly particularly smoothly. ForFor example Maschke's Theorem says each FG-representation is the sum of irreducibles, while, as F isis aa splittingsplitting fieldfield for G, each irreducible FG-representation is absolutely irreducible. Section 34 begins the analysis of the characters of such representations. We find that,that, ifif m is the number of conjugacyconjugacy classesclasses ofof G,G, thenthen G has exactly m irreducible characterscharacters (x,(Xi: : 1 1 5 < ii <_( m),m), andand thatthat thesethese characterscharacters formform aa basisbasis for the space of classclass functionsfunctions fromfrom GG intointo F. A result ofof BrauerBrauer (which(which is beyondbeyond thethe scopescope ofof thisthis book)book) showsshows that,that, under the hypothesis of the first paragraph,paragraph, the representation theory of G over F isis equivalentequivalent toto thethe theorytheory ofof GG overover C, C, soso sectionsection 3535 specializesspecializes toto thethe case F = C. C. TheThe charactercharacter tabletable ofof GG overover CCis is defined;defined; this is the m byby mm complex matrixmatrix (x,(g,)),(Xi(gj)), where where (gj: (gj: 1 1 5 < j <5 m) m) is is a aset set of of representatives representatives for thethe conjugacyconjugacy classesclasses of G.G. VariousVarious numericalnumerical relations on thethe charactercharacter table areare established;established; amongamong thesethese thethe orthogonalityorthogonality relationsrelations ofof lemmalemma 35.535.5 are most fundamental. The concepts of induced representations and induced characters are alsoalso discussed.discussed. These conceptsconcepts relate thethe representationsrepresentations and characters of subgroups of G to those of G.G. InducedInduced characterscharacters and relations among characters on the oneone handhand facilitatefacilitate thethe calculationcalculation ofof thethe charactercharacter table, and onon thethe otherother make make possiblepossible the the proofproof ofof deepdeep group group theoreticaltheoretical results. Specifically chapter 35 containscontains a proofproof of Burnside'sBurnside's paqb-Theorem,pagb-Theorem, which says a groupgroup whosewhose orderorder isis divisibledivisible byby justjust twotwo primesprimes isis solvable, solvable, and of Frobenius' TheoremTheorem onon thethe existence of Frobenius kernels in Frobenius groups. Section 36 applies some of the results in section section 3535 andand previousprevious chapters to analyze certain minimal groups and their representations. This section is inin thethe spiritspirit ofof thethe fundamentalfundamental paperpaper ofof HallHall andand HigmanHigman [HH],[HH], which showed how many group theoretic questions could be reducedreduced toto questionsquestions about the FG-representationsFG-representations of certaincertain minimalminimal groups,groups, particularlyparticularly extensions of elementaryelementary abelian abelian or extraspecial extraspecial p-groups by groupsgroups ofof primeprime order. 178 Linear representations ofof Jinitefinite groups

34 CharactersCharacters in in coprimecoprime characteristic In this section G is a finite group and F isis a splitting field forfor G with characteristic not dividing thethe order order of of G. G. Let Let R R = = FF[G] [GI be the group ringring ofof GG overover F.F.

(34.1) RR isis aa semisimplesemisimple ringring and (1) R is the direct productproduct ofof idealsideals (Ri: (Ri: 1 1 5 < ii <5 m)m) which which areare simplesimple asas rings. In particular Ri Rj Rj == 0 forfor ii # j jand and 1 1= = Em CyZl 1 e;ei, , where where ei ei =1 = Ri1R~. . (2) Ri is isomorphic to the ringring FniFn' Xnl"n' ofof allall nini byby nini matrices over F. (3) (Ri: 1 5< ii <5 m) m) is is the the set set of of homogeneous homogeneous components components ofof R,R, regardedregarded as a right module over itself. (4) Let Si consist of the matrices in Ri with 0 everywhere except in thethe firstfirst ' row. ThenThen (Si:(Si: 11 5< i <5 m) m) is is a a set set of of representativesrepresentatives for the equivalence classesclasses of simple right R-modules. (5) FF=EndR(Si) = EndR(Si) andand dimF(S()=ni.dimF(Si) = ni.

Proof. AsAs char(F) char(F) does does not not divide divide the the order order of of G,G, RRis is semisimple by Maschke'sMaschke's Theorem. Hence we can appeal to the standard theorems on semisimple rings (e.g. Lang [La], chapter 17) which yield 34.1, except that Ri is the ring DFD!' xni of matrices over the division ring Di = EndR(Ti) EndR(z) where Ti is a simple submodule of the ith homogeneous component Ri. But by hypothesis F isis aa splittingsplitting field forfor G, so F = EndR(T) EndR(T) forfor eacheach simplesimple R-moduleR-module TT byby 25.8,25.8, completingcompleting the proof.

Throughout thisthis sectionsection m,m, Ri,Ri, Si,Si, ni, ni, andand eiei willwill bebe as in 34.1. Notice that IG/IGI == dimF(R) = EYE1dimF(Ri) = ELl n?, which I record as:

(34.2) IGI=Ymin?.

I have already implicitly used thethe equivalenceequivalence between FG-modules and R-R- modules discussed inin section 12, and will continue to do soso withoutwithout furtherfurther comment. Let (Ci: 11 5< ii <5 m') m') be be the the conjugacy conjugacy classes classes of of G, G, gi g; E E Ci,Ci, and and definedefine zi E RR byby zizi == F-g,c.g.EgEc,g. ItIt develops develops in in thethe nextnext lemma that m == m'.

(34.3) (1)(1) TheThe numbernumber mm ofof equivalenceequivalence classes of irreducible FG-representations is equal to thethe numbernumber of conjugacy classes of G. (2) (zi:(zi: 11 5< i <5 m)m) andand (ei:(ei: 1 5< ii <5 m) m) areare basesbases for Z(R) over F.

Proof. RecallRecall ei ei =1= R,.1R~. AsAs R R is is the the direct direct product product ofof thethe idealsideals (Ri(Ri: : 1 5< i 5< m),

Z(R) = ®m@Y="=,(Ri) 1 Z(Ri) = = @y="=,ei,®miFei, withwith thethe lastlast equalityequality followingfollowing fromfrom thethe isomorphism Ri =Z Fn'Fni Xnixni and 13.4.1. Thus if I show (zi:(zi : 1 5< ii 5< m') = B isis Characters inin coprimecoprime characteristic 179 also a basis for Z(R), the proof will be complete. AsAs GG is a basis forfor R, B is linearly independent, so itit remainsremains toto showshow BB spansspans Z(R)Z(R) over F. Let z == CgEGL-gEG aggagg E R, a,ag E F.F. Then Zz E Z(R)Z(R) precisely when zh = z forfor each h E G, which holdsholds inin turnturn preciselyprecisely when when ag a, == a(gh) a(,h, for all g, h EE G. Thus z E Z(R)Z(R) ifif and only ifif a,ag = ai ai is is independent independent of of thethe choicechoice ofof gg EE Ci, oror equivalently, whenwhen z z = = CF_ aizi.aizi. SoSo indeedindeed BB spansspans Z(R)Z(R) over F.

A class functionfunction onon GG (over F)F) is a function fromfrom G into F which is constant on conjugacy classes.classes. DenoteDenote byby cl(G) cl(G) == cl(G, F)F) the the set set ofof all class functions on G and make cl(G) into an F-algebra by defining:

g(a -1- 0) = ga + g,8a, ,B E cl(G), g(aa) = a(ga)a E F, g E G, g(ao) = ga go. Evidently dimF(cl(G)) == m.m. By 14.8, ifif nr isis an an FG-representation FG-representation andand Xx itsits character, thenthen xX isis aa classclass function.function. LetLet n,ni be the representation ofof GG on Si by right multiplication and xiXi itsits character.character. Then Then (ni: (Tri: 1 15 < i i 5< m) is a setset ofof representatives for thethe equivalenceequivalence classesclasses ofof irreducibleirreducible FG-representations. FG-representations. By 14.8, equivalent representations have the same character,character, soso {xi:f Xi 1: 15 < i i5 < m)m) is the set of irreducibleirreducible characterscharacters of G over F; that that isis thethe set of all characters of irreducible FG-representations. The degree of a representation 7rn is just the dimension of its representation module. As this is also the trace of the d by d identity matrix which is in turn ~(l),X (1),it it follows follows that:

(34.4) Xixi (1) = ni isis thethe degreedegree of of yri.ni .

Each group G possessespossesses a so-calledso-called principal representationrepresentation of degreedegree 11 inin which each element of G actsacts asas thethe identityidentity onon thethe representationrepresentation module V. As dim(V)dim(V) = 1, 1, thethe principalprincipal representationrepresentation is certainlycertainly irreducible. By convention 7rlnl is taken to be the principal representation. Hence:

(34.5) SubjectSubject to to the the convention convention that that 7r1nl isis thethe principal principal representation,representation, xlXi (g) (g) = _ nl=1n1=1forallgEG. for allg~G.

Observe that there is a faithful representationrepresentation ofof thethe F-algebraF-algebra cl(G)cl(G) onon R defined byby (C (F agg)aa,g)a = F C ag(ga), a,(ga), for for a aE Ecl(G) cl(G) and and F C agg a,g E E R. R. Recall Recall also also that ni is just the restriction to G of the representation of of RR on Si and Xixi (r) (r) ==Tr(r7ri Tr(rni) ) for r EE R. Finally notice thatthat Xixi (r)(r) isis alsoalso thethe valuevalue ofof xiXi at at r r obtainedobtained fromfrom thethe representation of cl(G) on R. 180 Linear representations ofofjinite finite groups

(34.6) (1)(1) Rini=0=xi(Rj)foriRJ7i=0=Xi(Rj)for #i 0 j. j. (2) Xixi(Ri)=F (Ri) = F forforeachi. each i. (3) Xixi(eir) (ei r) = = Xi xi(r) (r) for eacheach ii andand eacheachr r EE R. R. In In particularparticular Xi xi(ei) (ei) = Xi xi(1) (1) = ni.

ProojProof. ByBy 34.1 RiRiRj Ri =0= 0 for for i i0 # j, j,so so (1) (1) holds. holds. As As ei ei = = 1R;, lRi, (1)(1) implies implies (3).(3). Part (2) is an easy exercise given the description of Si in 34.1.4.

(34.7) The irreducible characters formform aa basisbasis forfor cl(G)cl(G) overover FF.

ProojProof. AsAs dimF(cl(G)) =m= m = =number number of of irreducible irreducible characters, characters, it it sufficessuffices to show (xi:(Xi: 1 1 5< i <5 m) m) is is a alinearly linearly independent independent subset subset ofof cl(G).cl(G). ButBut thisthis isis . immediate from 34.6.1 andand 34.6.3.

The sum of FG-representations was was defineddefined in section 12.12. An immediate consequence of that definition is that XaX, ++ xgXp = = xa+gXa+p for for FG-representations FG-representations a andand $B andand theirtheir sum a +t- P, B, where where Xy X, denotesdenotes thethe charactercharacter of thethe representation y. DenoteDenote by char(G) thethe Z-submoduleZ-submodule of cl(G)cl(G) spannedspanned byby thethe irreducibleirreducible characters ofof G.G. char(G) is the set of generalized characterscharacters of G. As each FG-representation isis thethe sumsum ofof the the irreducible irreducible representations representations ni ni, , 1 1 5 < ii 5< m,m, (cf. 12.10), itit followsfollows fromfrom thethe precedingpreceding remarksremarks thatthat eacheach charactercharacter ofof GG isis aa nonnegative Z-linear combination of the irreducibleirreducible characters.characters. ThusThus charac-characters are generalized characters, although by 34.7 not all generalized characters are characters. Further if (mi :: 11 5 < ii 5< m) are nonnegative integers not all zero then >Cimi Xixi is thethe charactercharacter of thethe representationrepresentation Ci>i mimini. ni. Finally, Finally, by by ExerciseExercise 9.3, 9.3, Xa®p xa@g = xaxg,X«Xp, SO so the the productproduct ofof characterscharacters is is aa character. HenceHence char(G)char(G) isis a Z-Z- subalgebra of cl(G). These remarksremarks areare summarizedsummarized in: in:

(34.8) The Z-submodule char(G) of cl(G) spanned by the irreducible charac-characters is a Z-subalgebraZ-subalgebra of G. TheThe membersmembers of char(G) areare calledcalled generalizedgeneralized characters. The characters are precisely the nonnegative Z-linear span ofof thethe irreducible characters, and hence aa subsetsubset ofof thethe generalizedgeneralized characters. characters.

Representations and characters ofof degreedegree 11 areare saidsaid toto bebe linear.lineal:

(34.9) Let G bebe anan extraspecialextraspecial p-group p-group of of order order pl+2nand and Z Z= = Z(G) Z(G) =_ (z).(z). Then (1) GG has has per p2n linearlinear representations. representations. (2) GG has has p p - -1 1faithful faithful irreducible irreducible representations representations (P1, 41, ...... , ,(pp-1. Notation can be chosen so that zOizq5i acts via the scalar wknwi on the representation module Characters in characteristic 0 181

Vi ofof q5i,O , wherewhere ww isis some fixed primitiveprimitive pth pth root root of of 11 inin F. TheThe Otq5i are quasiequivalent for for 1 15 < i i5 < pp -- 1.1. (3) O; q5i isis ofof degreedegree p".pn. (4) G G has has exactly exactly penp2n +t- pp -- 1 1irreducible irreducible representations: representations: those those describeddescribed in (1) and (2).(2). (5) Let Let E E be be the the enveloping enveloping algebra algebra of of 0,q5i andand YY aa set of cosetcoset representativesrepresentatives for Z in G. Then E =E FP""p" FP"~P" and and YY isis a basis forfor E over F.

Proof. ByBy ExerciseExercise 12.1,12.1, GG hashas exactlyexactly IG/G(1)I 1 GIG(') I linear representations and each isis irreducible.irreducible. As As G G is is extraspecial extraspecial of oforder order pl+2nG(') G(' = = Z and I IG/ZIG/Z I =_ pp2",en, soSO (1)(1) holds. For xx EE G G - - Z,Z, xxG G == xZ, soSO G has m = p2n p2n - 1 1+ + p pconjugacy conjugacy classes. classes. As G hashas p2np2n irreducibleirreducible linear representations, this this leavesleaves exactlyexactly p p -- 11 nonlinear irreducible representations, by 34.3. Let q50 be such a representation. By Exercise 12.1,12.1, ZZ = = G(1) G(') $ ker(q5).ker((P). But But ZZ is the unique unique minimalminimal normalnormal subgroup of G, so ker((P)ker(q5)= = 1.1. ThatThat is is q50 isis faithful.faithful. ByBy 27.16,27.16, zq5zO == a((P)Ia(q5)I for some primitive pthpth root root of of unity unity a((P), a(@), saysay a((P)a(@) = w. w. By Exercise 8.5, 8.5, there there is is anan automorphism aa ofof GG of of order order p p - -1 1 regularregular on on z'. Z#. Let Let cPi Oj = = a'-'@, ai-1 0 1, 15 < i i< < p.p. Then

.(,-I) -1) PO-1) q5 = mj(8-')(i-1) zq5izo; = zJz' = w I

where za za = =z' zj and {j('-'):f j('-'): 11 5< ii <y Cy byby y,y, for for some some by by E E F, as as desired.desired.

35 CharactersCharacters inin characteristiccharacteristic 0 The hypothesis and notation of the last section are continuedcontinued inin thisthis section. InIn addition assumeassume FF = C.@. 182 Linear representationsrepresentations of ofJinite finite groups

(35.1) FG-representations 7r x and 0C#I areare equivalentequivalent ifif andand onlyonly ifif theythey havehave thethe same character.

Proof. ByBy 14.8, 14.8, equivalent equivalent representations representations havehave the same character.character. ConverselyConversely assume x7r andand C#I q havehave the the same same character character X. X. Now Now n 7r= = C > miximi7ri and and C#I 0 = _ >C ki7ri kixi for some nonnegative integers mi, ki. It suffices toto showshow mim, == ki forfor eacheach i.i. But C mi xiXi == xX == C ki xi,Xi, SO, so,by by 34.7,34.7, mim i= = kiki forfor eacheach i.

The regular representation ofof GG isis thethe representation of G by right multiplication on R.

(35.2) Let eiei == F(ai,g)g,C(ai,,)g, ai,g ai,, EE F. Then Then ai,gai,, = =X(eig-')/IGI ~(eig-')ll~l =ni~i(g-')ll~l,=niXi(g-')/IGI, where Xx is the character of the regular representation of G.

Proof. Observe X(eig-')X(eig-1) == X(F-h(ai,h)hg-1)~(E~(ai,h)hg-') == F-hEhai,hx(hg-').ai,hX(hg-1). But, But, by by Exercise 12.2, 12.2, X ~(x) (x) = 0 0 if x + 1 and X~(1) (1) == IGI IGI,, SOso X~(eig-')(ei g-1) = IGI IGI(ai,,), (ai,g), yield- ing the first equality inin thethe lemma.lemma. Next,Next, byby ExerciseExercise 12.2,12.2, xX = = Ey=lF-mi niniXi, Xi, SOso m - 1 X(eig-')=x(eig-')= Y.Cnjxj(eig njxj(eig 1) )= = nixi(eig-1) nixi(eigF1) = niXi(g-')niXi(gP1) j=1 with the last two equalities holdingholding byby 34.6.134.6.1 andand 34.6.3,34.6.3, respectively.respectively.

We now definedefine aa hermitianhermitian symmetricsymmetric sesquilinear sesquilinear form form ( (,, ) onon cl(G)cl(G) withwith respect to the complex conjugation map on @.C. (Recall (5.c denotes the complex conjugate of c in @.)C.) Namely for X, 98 EE cl(G)cl(G) definedefine (X'0)=(x()())g/IGI.

It is straightforward to check thatthat ((, , ) ) is is hermitian hermitian symmetricsymmetric and and sesquilinear.sesquilinear. Indeed the next lemma showsshows the formform isis nondegenerate.

(35.3) The irreducible characters form an orthonormal basis for the unitaryunitary space (cl(G), (( , )). That isis (Xi,(xi, Xj)xi) = Sip 6ij.

Proof. ByBy ExerciseExercise 9.4, 9.4, Xji(g) (g) = X~(g-') (g-1) for each character xX and each g E G.G. By 35.2: (xi(g)xi(g1)) Xi(ej)lnj = I , Xj) gEG Hence (Xi,(xi, X xi) j) == JijSi j by by 34.6.134.6.1 and 34.6.3. Characters inin characteristic 0 183

Recall (Ci(Ci: : 1 5< ii <5 m) m) are are the the conjugacy conjugacy classes classes of of G,G, gi gi EE Ci,Ci, andand by convention g1gl == 1. The charactercharacter table table of of G G (over (over C) C) is is the the m m by by m m matrix matrix (Xi (xi(gj)). (g j )). Thus the rows of thethe charactercharacter tabletable areare indexedindexed byby thethe irreducibleirreducible charac-characters of GG andand thethe columnscolumns byby thethe conjugacyconjugacy classesclasses of G.G. InIn particularparticular thethe character table is defined only up to aa permutationpermutation ofof thethe rowsrows andand columns,columns, except that by conventionconvention Xl~1 is always always thethe principalprincipal charactercharacter andand g1gl = 1. 1. Subject to this convention, the character table has each entry in the first row equal to 1, while the entriesentries in thethe firstfirst columncolumn areare thethe degreesdegrees ofof thethe irreducible irreducible CG-representations.

(35.4) LetLet A A be be the the character character table table of ofG andG and B the B matrixthe matrix (I Ci (ICiI g j 1 Xj(gi) (gi)/l / I G G 1). I). Then B == A-1.A-' .

Proof. (AB)ij(AB)ij == Ek xk Xi(gk)ICk1Xj(gk)/IGIxi(gk)lCklXj(gk)/lGI = (CgEGXI(g)Xj(g))/IGIxi(g)Xj(g))/lGI = (Xi, (xi, Xj)xi) = 8ij. Jij. Therefore Therefore ABAB == I Iis is the the identity identity matrix. matrix. So So A A is is nonsingular nonsingular andand B = A-1.A-l.

Observe that, byby 5.12,5.12, I ICilCi I == IG: CG(gi)l,CG(gi)I, SOso ICiICil/lGI I/IGI = ICG(gi)Il~G(gi)l-' in lemma 35.4.

(35.5) (Orthogonality(Orthogonality Relations) Let hk = ICkI lCk1 bebe the order of the kth conjugacy class of G. Then the character table ofof GG satisfiessatisfies thethe followingfollowing orthogo- orthogo-

; nality relations:

n 0 if is j, (1) hkXi(gk)XJ(gk) _ k=1 IGIifi = j,

n 0 if (2) Xi(gk)Xi(gi) = i=1 ICG(gk)I =IGI/hkifk=l,

n (3) IGI = n2, i=1

(4) niXi(gk)=0ifk > 1, i=1

n (5) >hkXi(gk)=0ifi > 1, k=1

n (6) > hk I Xi (gk) 12 = IGI k=1 184 Linear representationsrepresentations ofofJinite finite groups

Proof. Part (1) is aa restatementrestatement of 35.3, while part (2) is a restatementrestatement of 35.4. Parts (3) and (4) are the special cases of (2) with k =l== l =1 1 andand 11 = 1, 1, respectively (using Xi~~(1) (1)= = ni).ni). Similarly Similarly (5)(5) and and (6)(6) areare just just (1)(1) with with j j == 11 and j == i, i, respectively. respectively.

The orthogonality relations may be interpreted as follows. Part (1) sayssays thethe inner product of the rowsrows ofof thethe charactercharacter tabletable (weighted(weighted by thethe factorfactor hk)hk) is 00 oror II G 1,J, while while part part (2) (2) sayssays thethe innerinner productproduct of thethe columns (twisted by complex conjugation) is 0 oror IGI/hk.I G l/hk.

(35.6) Let h, g EE G. Then hh cE gGgG ifif andand onlyonly ifif Xixi(g) (g) = = xi(h)Xi (h)for for each each i, i, 11 5< i

Proof. IfIf hh E E gG gG thenthen Xixi(g) (g) == Xixi(h) (h) sincesince characters characters areare classclass functions.functions. Con-Con- versely assumeassume Xixi(g) (g) == Xixi(h) (h) for for eacheach i,i, butbut h $0 gG.gC. By Exercise 9.7, Xi ji is also an irreducible character, character, so so Xi ji(g) (g) = Xi ji(h). (h). Hence, Hence, by 35.5.2:

n n 0= EXi(g)X1(h)= EXj(g)Xj(g) =I CG(9)1 i_i i_i which is of course a contradiction.

At this pointpoint we'llwe'll need a few factsfacts aboutabout algebraicalgebraic integers.integers. RecallRecall anan algebraicalgebraic integer is an element of C whichwhich isis aa rootroot of aa monic polynomial inin 71[x].Z[x]. TheThe following factsfacts areare wellwell known and can be found forfor exampleexample inin chapter 9 of Lang [La].

(35.7) (1)(1) TheThe algebraic algebraic integers integers form form a a subring subring ofof C.C. (2) Z71 is is the the intersection intersection ofof thethe algebraic integers with'with Q. (3) JNorm(z)J]Norm(z)l >2 11 forfor eacheach algebraicalgebraic integer z #0 0. (4) AnAn elementelement cc cE C(C isis anan algebraicalgebraic integerinteger ifif andand onlyonly ifif therethere existsexists aa faithful 7/[c]-moduleZ[c]-module which isis finitelyfinitely generated generated as as a a 71-module. Z-module.

(35.8) ni dividesdivides n for each i.

Proof. LetLet a =n/ni. By By 35.2, 35.2, ei ei = =Eg1G(ai,g)g CgEG(ai,,)g withwith ai,g ai,, = = Xi(g)la. ;Xi(g)/a HenceHence aei = F-gcGCgEG ji(g)g.Xi(g)g. AS As e: e?=ei, = ei, also aei == CgGGXi(g)gei.;Xi(g)gei. By 27.13.1, X(g)~(g) is the sum of JGJ-thIGI-th rootsroots of unity forfor each character Xx and each g E G. Let M be the 71-submoduleZ-submodule ofof R generated by the elements

({ge,:g E G, { is a IGI-th/GI-th rootroot of 1). Characters in characteristic 0 185

Then M isis a a finitely finitely generated generated 71-moduleZ-module and urgei = {gaei = ChEGjii(h)Xi(h) {gheighei c EM,M, as as XZ(h) jii(h) isis thethe sum sum of of IGI-th/GI-th roots roots ofof unity.unity. HenceHence MM isis a a 71[a]- Z[a]- submodule ofof R,R, andand certainlycertainly M M is is faithful faithful as as Z[a] 7/[a] 5 < CC and R is a vector space overover C.C. Therefore aa is an algebraic integerinteger byby 35.7.4,35.7.4, andand thenthen aa isis an integer by 35.7.2.

Define the kernel of Xx to be

ker(X) = {g E G: X(g) = X(1)}.

(35.9) Let Xx be the character of a CG-representation a andand letlet gg EE G. Then (1) IX(g)/X(1)IIx(g)/x(l)I 5 < 11 3 > INorm(x(g)/x(1))1INorm(X(g)/X(1))I with with equality equality if if andand only ifif ga = = coI wI forfor somesome rootroot ofof unityunity to.w. (2) ker(X)kerO() = ker(a) 2< G.

Proof. By 27.13.1,27.13.1,X(g)= ~(g)= Y_'=, Cy=, wi,wi,wheren wheren = deg(a) = X(1)andcoiis ~(1)andq is aroot of unity. Thus IX(g)IIx(g)l I< CyZl IwiIlwil 5 < x(1)X(1) withwith equalityequality ifif and only if wi ==co w is independent of i. Similarly, if Norm(x)Norm(X) isis thethe norm of xX thenthen Norm(X) _ fl X °EE where EC is the set of embeddingsembeddings of of Q(x)Q(X) into into Q. Q. But But xu X° = = CiYi w°, w:, soso IX°I]xuI 5 xX (1) (1) andand hencehence INorm(X)I/n < [1 IX°I/n < 1 ° with equality if and only if coiwi == cow forfor allall i. Recall in this last case from the proof of 27.13 that asas CC isis algebraicallyalgebraically closed, closed, ga ga ==col. wI. SoSo (1) holds. Also if xX (g) (g) = = n n then then w co = =1 1 so so ga ga == I. Hence Hence (2)(2) holds.holds.

(35.10) LetLet zizi = =: Y-gECj CgEc, g, aiiaij = hi hixj(gi)/nj, Xj (gi)l nj, andand bilkbijk == 11(g, I{(g, h): h): gg EE Ci, hh EE Cj,gh=gk)I.ThenCj,gh=gk)l.Then Ein=1 (1) ziZi = CyZlaijej, (2) ailaij = =xj(zi)/nj, Xj (zi)l nj, (3) aitapailajl == Ek Ck bifkak1,bijkaklr (4) ailaij is is an an algebraic algebraic integer. integer.

Proof. AsAs R R = =ED' 1Rj, Rj, zi zi= = Y'i zijzij for suitable zijzip EE Rj. R. By 34.3,34.3, zizi EE Z(R), soso zij EE Z(Rj). NowNow RjRj = = Cnj Cnj xnj'"j soso Z(Rj) = = Cep Cej is is 1-dimensional I-dimensional and and consists of thethe scalarscalar matrices matrices inin Cnj CnlXn~. xnj. Thus zijzit = ci1 cijej ej forfor some cij in C, and xj(zi)Xj(Zi) == xj(CkXj(Y-k cikek)cikek)= = CkEk cikXj(ek)cikxj(ek) == cijnjCijn j byby 34.6.34.6. ThusThus to to proveprove (1)(1) 186 Linear representationsrepresentations of ofjinite finite groups

and (2) itit remainsremains to to show show ai aij j = =c; jCij. . But But ci Cijnj j n j == X xj(zi)j (zi) = = X xj(CgeCij (Y-gEc,g) =_ Y-gEC,xgECixj(g) Xj(g) = = hixj(gi),hi xj(gi), soSO the proof of (1) and (2) is complete. Next zizjzizj Ee Z(R), so, as (zi: 11 5< i <5 m)m) is is basis of Z(R) by 34.3.2, zizjziz j == Y-kxk CijkZk.CijkZk- AlsoAlso

z j zi =\gEEci g/ \gE h gE gh hECj and, in the sum on the right,right, thethe coefficientcoefficient of gk is bibijk. jk. As As zizjzizj is a linear combination of the zk,Zk, andand asas G is a basis for R,R, thethe coefficientcoefficient ofof xx EE CkCk in the sum on the right isis equalequal toto thatthat ofof gkgk andand thatthat coefficient coefficient isis Cijk.cijk. That is bijkbi jk == Cicijk. jk. Next Next

Zizj = t Zik ZilZjl \ k i k,I i as RkRlRkRI = = 0 for kk 0# 1.1. AlsoA~SO ZilZjlzilzjl =:= ailelajlelailelajlel = ailajieiailajlel Soso ZiZjzizj = Y-ixi ailajlel. ailajlei. OnOn thethe otherother hand hand zizj zizj = = Y-k xk bijkZkbijkzk == Y-k,l xk,l bijkaklelbijkaklei by the last paragraph and and (I), (1), so so xk Y-kbijkakl=ailaji,bijkakl = ailajl, as as (ei: (ei:1 1 ( < ii <5 m)m) is is aa basisbasis for Z(R). HenceHence (3) holds. Fix (i, 1) and set a = ai,i. ai,l. Then Then (*) as j,l = E bi jkak,i1 < j < n. k Consider thethe followingfollowing system system of of n n equations equations in in n nunknowns unknowns x x= = (XI, (xi,...... ,, x,):x,):

("") 0 = E bijkxk+ (bi jj - a)xj1 < j < n. k#j Observe (**)(**) hashas the the solution solution a a == (al,l,(a~,~,. . . , a,,l)a.,l) byby (*).(*). xl(g1)xi(gi) =nl= ni #0 0, so a1.lal,i = hiXi(gl)/ni hlxl(gl)/nl 0 #0. 0. So, So, as as (0, (0, ... . ., .0) , 0) is is also also a asolution solution to to (**), (**), the the matrixmatrix M of (**)(**) is ofof determinantdeterminant 0. Consider thethe matrixmatrix N N withwith entriesentries in in Z[x]l[x] obtained byby replacingreplacing aa by x inin M.M. (Observe(Observe thatthat byby definitiondefinition the bijksbijks are nonnegative integers.) integers.) Let Let f (x)f (x) = = det(N) det(N) E E Z[x]. l[x]. ThenThen ff (a)(a) == 00 andand ff is a monic polynomial, so a isis anan algebraicalgebraic integer. Thus (4) holds.

(35.11) Let (ni, hhj) j) = 1. 1. Then either (1) Xi(gj) = 0, or (2) Ixi(gj)I =ni, so gjker(xi) E Z(G/ker(Xi))

Proof. ByBy 27.13.1 27.13.1 and and 35.7.1, 35.7.1, a a= = Xi xi(gj) (g j) is is an an algebraicalgebraic integer, integer, asasis is b = ai aij j = hja/nih ja/ni byby 35.10.4.35.10.4. Assume aa 0# 00 andand letlet ff be be the the minimal minimal polynomial polynomial of a over Q.Q. Say ff (x) =_ xf=oyd_k=0akxk, akxk,ak ak E E Q. Q. Now Now the the field field extensions extensions Q(a) 0(a) and Q(b)0(b) areare equalequal as r = hj/ni is is in in Q. Q. So So the the minimal minimal polynomial polynomial f (x) of b over Q0 isis also of degreedegree d. d. As As b b is is a aroot root of of g(x) g(x) = = ~f=~ k=i akrd-kxkakrdPkxk, itit follows follows Characters in characteristic 0 187 that g = f. f. AsAs b bis is an an algebraic algebraic integer, integer, g g eE l[x],Z[x], so so akrd-k akrd-k is is an an integer. integer. AsAs r == hi hj/ni /ni with with (hj, (hj, ni) ni) == 1, 1, it itfollows follows that that nd-k ntVk dividesdivides ak ak forfor each each k. k. HenceHence d h(x) (akl rid k)xk E 7L[x]. k=1 But a/ni isis a a rootroot ofof h(x)h(x) and and h(x)h(x) is is monic, monic, soso a/nia/ni is is an an algebraic algebraic integer. integer. OnOn the other hand ]Norm(Norm (a/ni)l(a/ni)l <5 11 by by 35.9.1.35.9.1. Thus (NormINorm (a)[(a)I =ni=ni byby 35.7.3,35.7.3, so, by 35.9.1, gjgjni 7ri = = wl.wi. Therefore (2) holds, completing the proof.

(35.12)(35.12) Assume hihj == pe is a prime powerpower for for some some j j > 11 and and GG isis simple.simple. Then G isis ofof primeprime order.order.

Proof. We We may may assume assume G G is is notnot ofof prime order. Then,Then, asas GG isis simple, simple, Z(G) Z(G) ==1 1 and ker(xi)ker(Xi) ==1 1 for each i > 1. 1. So, So, by 35.11,35.1 1, for eacheach i,i, jj >> 1 1 either either Xi xi(gj) (gj) == 0 or p dividesdivides ni. ni. Next, Next, byby thethe orthogonalityorthogonality relations: relations: n 0niXi(gj)=1+pc i=1 for some algebraicalgebraic integer integer c. c. Hence Hence c c= = -pP1, -p-1, impossible asas -p-'-p-1 isis not not anan algebraic integer, by 35.7.2.

(35.13) (Burnside's(Burnside's pagb-Theorem).paqb-Theorem). Let I[GI GI == paqb,pagb, with p, q prime.prime. ThenThen G is solvable.solvable.

Proof. LetLet GG bebe aa minimalminimal counterexample.counterexample. If If1 1 # 0 HH 9a G then G/H and H are of order lessless thenthen GG andand bothboth areare {p, {p, q}-groups, q}-groups, so, so, byby minimalityminimality ofof G,G, each is solvable. But now 9.3.2 contradicts the choice of G. So GG isis simple.simple. LetLet QQ EE Sylq(G).Sylq (G). If If Q Q = =1 1 thenthen G is aa p-groupp-~oup andand hencehence GG isis solv- solv- able. So QQ 0# 1,1, soso inin particularparticular therethere is gjgj E E Z(Q)#.z(Q)#. Then QQ

(35.14) LetLet i/r@ EE cl(G). ThenThen (1) 1r@ = ELI(@,Xi")`Xi,~i)~i, andand (2) (Y',(@, Vf)@) = = yinEy=l(@y 1(Y', Xi)2. xi)'.

Proof. As the irreducible characters characters are are a abasis basis for for cl(G), cl(G), @ *= = xy=lEmt aiaiXi Xi forfor somesome complex complex numbers ai. ai. BYBy 35.3, (i,(@, Xj)xj) = (xiaiXi, xj) Xi) ==: xiai(xi, xj) =ai, SO (1) holds.Eiai(Xi,Xj)=ai,so(1)holds. SimilarlySimilarly (@, @) = riaiaj(Xi,Xj)=xiaiaj()(i, xi) = Y_ia?, Xi a?, so (2) holds.

WeWe next considerconsider inducedinduced representations.representations. LetLet HH 5< G,G, FF a a field, field, and and aa an an FG-FG- representation. It will be convenient to regard the image ha of h e H under 188 Linear representations ofofJinite finite groups a asas aa matrixmatrix ratherrather thanthan aa linearlinear transformation.transformation. Extend a toto GG byby definingdefining ga = = 0 0 forfor g EE G - H. H. Let Let X X == (x1: (xi : 1 <5 ii <5 n) n) be be a a set set of of cosetcoset representatives for H in G; hence n == IGI G : HI.H1. For For g g E E G G define define gaG gaG = = ((xigxil)a) ((xigx 1)a) to be the n by n matrix whose (i,(i, j)-th entryentry is the deg(a) by deg(a) matrixmatrix (xjgxj-1)a.(xigxjl)a. We cancan alsoalso regardregard gaGgaG as aa squaresquare matrixmatrix ofof sizesize deg(a)n deg(a)n overover F.F. If we take this point of view then:then:

(35.15) aGaG isis an an FG-representation FG-representation of of degree degree deg deg (a) (a)lG I G : :H1. HI.

Proof. Let u, u, vv E G,G, AA = = uaG,uaG,B B = = vaG, vac, and CC = (uv)aG. (uv)aG. WeWe mustmust showshow AB == C. Regarding A and B as n by n matrices with entries fromfrom FdxdF~~~ (where d = deg(a)) wewe have:have: (AB)ij = AikBkj = (xiuxk 1)a(xkvx,j 1)a. k k Most of the terms inin the sum on the right are 0, since (xiuxil)a(xi uxk 1)a = = 0 unless xiuxklxiuxk 1 E E H. H. But But xiuxkl xiuxkl E HH precisely whenwhen HxiuHxiu == Hxk, SOso (AB)ij = 0 unless there exists some k withwith HxiHxiu u == Hxk and Hxkv == Hxj.Hxi. ThisThis holds holds precisely whenwhen HxiuvHxi u v= = Hxj, Hxj, and and in in thatthat eventevent therethere isis a unique k withwith the property; namely kk is defineddefined by HxkHxk == Hxiu. Moreover inin this case (xi uxF1)a uxk 1)a . (xkvxil)a(xk vxJ 1)a = (xi (xi uuvxrl)a vxj 1)a = = C13 cij. . ~husThus (AB(AB)~~ )ij = C11 cij if if HxiHX~ uvu v == HxHxj j and (AB)1(AB)ij j == 0 otherwise. Finally Finally if if Hxi Hxi u uv v # HxjHx1 then xi uvxu vxj j1 1 $V H,H, so CiCij j = = (xiuvxil)a(xi uvxi 1)a = = 0 0 = = (AB)ij.(AB)i1. Hence Hence AB AB = = C, C, soso aGaG is a homomorphism, and the proof is complete. aGaG isis calledcalled the induced representation of a toto G.

(35.16) (1)(1) UpUp to to equivalence, equivalence, aG aG isis independent independent of of thethe choicechoice ofof cosetcoset representatives forfor HH in G. (2) IfIf aa and and ,8 B areare equivalentequivalent FH-representationsFH-representations thenthen aGaG andand pGBG are are equivalent FG-representations.

Proof. Let Y == (yi: 11 5< ii

(35.17)(35.17) Let xx bebe thethe charactercharacter ofof aa! andand extendextend xx toto G G by by defining defining x(g) ~(g) == 0 0 forfor g E GG -- H. H. ThenThen (1)(1) TheThe character character X XG G ofof thethe inducedinduced representation representation aG aG isis defined defined by by

xG(S) =Y,x(Sv)) IHI. (VEG (2) XG(g) =0 if g Hv for some v E G.

Proof XG(g) = Tr(aG) = Tr((gX'')a) == Ei=1 X(Sx' i). Moreover Proof. xG(g) ~r(a~)C:='=, 1 ~r((g"i-l)a) C;='=,x(~";'). Moreover xX(gxi-l) (gx; ') = 0 unless g E H",Hxi ,so so (2) (2) holds. holds. Finally Finally G G = = U;='=,U" 1 Hxi and, as x isis aa t ) classclass function, function, x(gxi x(gf ")I") = =x(gx+ X(g-l) ') for for h Eh H,E H, so EyEHX;so Z,,,, X(Syx(gy-I t) = I= H I HI X(Sx' I ~(g";'). HenceHence (1)(1) holds.holds.

XXG G is is called called the the inducedinduced character character of x toto G.G.

(35.18)(35.18) LetLet H <5 K K < 5 G G and and x xa charactera character of of H. H. Then Then XG xG == (xK)G (XK)G.

Proof.Proof. Let ~et 08 = = XK. xK. Then, hen, forfor gg EE G,G,

eG(S) =(9gv)/IK- /IKIIHI. vEG gEG UEK

FurtherFurther the map (v, u) H vu vu is is a a surjection surjection of of GG xx K K onto onto G G whose whose fibres fibres areare ofof orderorder IKI,IKI, soso OG(g)=(E,,,EGOG(g)= (C,,, X(Sw))/IHIx(gW))/IHI == XG(8) xG(g).

ForFor *$r EE cl(H) define define t/,G:$rG: G G -+-- FF by by 1/,G(g) $rG(g) = (EvEG(CVEG $r(gV))/I*(gv))/IHI, HI, where where asas usualusual *(x)$r(x) = = 0 0 for for xx EE GG -- H. H. If If ,/r + is is a acharacter character of of HH then, then, by by 35.17 35.17 , , $rGG isis aa charactercharacter of G,G, andand evidentlyevidently thethe map $r* HH ,/rG $rG preservespreserves addition addition andand scalarscalar multiplication,multiplication, so so asas the irreducible characters form a basis for cl(H) wewe conclude:conclude:

(35.19)(35.19) The mapmap i/r$r H ,/1G $rG isis a linearlinear transformation transformation ofof cl(G) intointo cl(H)cl(H) whichwhich mapsmaps characterscharacters to to characterscharacters andand generalizedgeneralized characterscharacters to to generalizedgeneralized characters.characters.

TheThe mapmap i/r$r HH 1116 $rG isis calledcalled thethe inductioninduction mapmap ofof cl(G)cl(G) intointo cl(H).cl(H). Notice that 35.1735.17 andand the the definition definition of of ,/1G $rG show:show:

(35.20)(35.20) LetLet *$r EE cl(H). cl(H). Then Then (1)(1) *G(g)=0forgEG-(UEGH"),and$rG(g)=O for g E G - (UVEGHV), and (2)(2) *G(1)=IG:HI,/r(1).$rG(l)= IG: Hl$r(l). 190 Linear representationsrepresentations of ofjinite finite groups

Recall that there are hermitianhermitian symmetricsymmetric forms forms ( (, , )H and (,( ,)G )G defined defined onon cl(H) and and cl(G),cl(G), respectively. respectively.

(35.21) (Frobenius(Frobenius Reciprocity Theorem) LetLet +1 EE cl(H) andand Xx E cl(G). Then (V',XIH)H(+, XIHIH = =(V'G,X)G (+G, X)G.

Proof. ('G, X)c = Y *G(g)X(g) IGI gEG

*(g"WO IGIIHI =CY-U

_ j(gv)X(gv) /IGHI v,gEG since jX isis aa classclass function.function. AsAs thethe mapmap gg F+i-, g"g" isis aa permutationpermutation ofof GG forfor eacheach v E G,G, itit followsfollows that that ($G, (1G, X)G X)G = = (CgeG(F-gEG+(g)j(g))/l *(g)X(g))/I H HI. I. As As +(g) fi(g) = = 0 for g E GG -- H, H, this this sum sum reducesreduces to (EhEH(CheH +(h)j(h))llHI1(h)2(h))/I HI == (v', (+, X XIHIH. I H)H

A subset T of G isis saidsaid toto be be a a TZ-set TI-set in in G G if ifT T n n Tg T9 E c{ 11) 1) for for each each g g E E G G- - NGNG(T>. (T ).

(35.22) LetLet TT be aa TZ-setTI-set inin G, HH == NG(T), andand +,r, 98 E cl(H)cl(H) with +1 andand 08 equal toto 00 on on H H -- T.T. ThenThen (1) *G(t)IlrG(t) _= fi(t) +(t) for for each each t tE E T#.T'. (2) IfIf 1r(1)+(l) == 0 0 thenthen (1,(+, O)H 19)~ = (V" (+G, BG)G- oG)~.

Proof.ProoJ Let g E G'.G#. Then Then +G(g) lG(g) = = (CUeGEVEG+(gU))/I i(g" ))/IHIHI with +(gU)*(g")=0= 0 unless unless g"gU EE T.T. Thus Thus fG(g)IlrG(g) == 0 0unless unless g gE E T" Tu for for some some u u E E G, G, in in which which case case Exercise Exercise 12.3 says TuT" is the unique conjugate of T containing g g whilewhile g"gU EE TT ifif andand onlyonly if v EE u-' u`H. H. In In particularparticular ifif gg EE T T then then +G(g) *G(g) = _ (ChEH (F-hEH+(gh))/l *(gh))/I H HII = = +(g). f(g) Thus (1) holds. Also,Also, asas +G/,.G is is a a class class function, function, +G(g) fG(g) =_ +(gu)*(g") if if g"gU EE T forfor some vv E G, and +G(g)G(g) = 0 otherwise. Assume +(l)*(1) = 0. 0. ThenThen f+G(l) G (l) = 0 0 byby 35.20.2,35.20.2, so

(V'G,OG)G =((g(g)) IGI gEG (,(gv)(gv)) /IGI gEA Characters in characteristic 0 191 where AA consistsconsists of of thosethose g g EE G#G' withwith 9'(g)gu(g) EE T forfor somesome v(g)v(g) EE G.G. Then, Then, byby Exercise 12.3,12.3,

(*G, OG)G = (*(t)e(t)I HI e)H ET# 1 as*is0onH-T#.

A Frobenius group is a transitive permutation groupgroup GG on a finite setset XX such that no member ofof G#G' fixesfixes more than one point of X andand somesome membermember ofof G#G' fixes at least one point of X.X. The The followingfollowing lemmalemma isis leftleft as as Exercise Exercise 12.4. 12.4.

(35.23) (1) If G is a Frobenius groupgroup onon aa setset X,X, xX E E X, X, and and H H = = G,, G, then H is a proper nontrivial subgroupsubgroup ofof T,T, HH is a TI-set inin G,G, andand HH = NG(H). Let K be the subset of GG consistingconsisting of of 11 togethertogether withwith the elementselements ofof GG fixingfixing nono points ofof X.X. Then Then I KIKI I= = I 1x1.X I. (2) Assume H isis aa proper nontrivial subgroup of a finite groupgroup G such thatthat H is is aa TI-setTI-set inin GG and and HH = = NG(H). NG(H). Then Then G G is is faithfully faithfully representedrepresented as aa Frobenius group by right multiplication on the coset space G/H.

Notice that, under the hypothesishypothesis of LemmaLemma 35.23.1,35.23.1, thethe representationrepresentation ofof GG on XX isis equivalentequivalent toto thethe representationrepresentation of GG byby rightright multiplicationmultiplication on thethe coset spacespace G/HG/H byby 5.8.5.8. HenceHence thethe permutationpermutation groupgroup theoretictheoretic hypotheseshypotheses of 35.23.1 areare equivalentequivalent to the group theoretic hypothesis of 35.23.2 by that lemma. I'll referrefer toto aa groupgroup GG satisfyingsatisfying either either hypothesishypothesis as as aa Frobenius groupgroup and call thethe subgroup H the Frobenius complementcomplement of G. Exercise 12.412.4 says H is determined up to conjugacy in G. The subset K of 35.23.1 can be described group theoretically by

K = G - (UH#). gEG K will be calledcalled the Frobenius kernel of G. The following important theorem of Frobenius shows K is a normal subgroup of G.

(35.24) (Frobenius'(Frobenius' Theorem)Theorem) The FrobeniusFrobenius kernel of a Frobenius group G is a normal subgroup of G.

Proof. LetLet HH and and K K bebe the the Frobenius Frobenius complement complement and and kernelkernel ofof G,G, respec- respec- tively. TheThe ideaidea isis toto produce produce aa character character xX of of G G with with K K == kerO();ker(X ); then then K K 9a G by 35.9.2. This will be achieved by applying 35.22 to the TI-set H of G.G. 192 Linear representationsrepresentations ofof$nite finite groups

Let BiOi,, 1 5 1. $i(l) =diOl(l)-Oi(l)=didi = 0. Hence, by 35.22.2 and the orthogonality relations:

(1/fiG>V`G)G = (Y'i, Vj)H = (diet - ei, dje1 - ej) =didj + 8ij.

Next, by 35.19,35.19, 11,: if = =E jCy=l 1 CJ cijxjX j for for suitable suitable integers integers ci cij. j . ByBy 35.1435.14 andand Frobenius reciprocity,reciprocity, cil ci1 = = ($?,(iiG, ~1)~X1)G= = ($i,(Vi, ~1 X1IH)H IH)H = = ($i, (fi, 81)Oi) == di. SoSO 2 ~ilc 1 == di. Also d;d? + + 11 == (($:, iG, $:) 1G) == Cy=l(~ij)2_j1(Cij)2= = di2d? + Ej'_2(cij)2,Cy=2(~ij) soSO $?1lf ; == di XI+X1 + EiXt(i) Ei Xt(i) for for some some irreducible irreducible character character xt(i) Xt(i)with with t(i) t(i)> > 1, andand some ~iEi = =±1. 51. Next Next didjdidj + + 8ij Sij =(riG = ($:, VfG)G$,?), =didj=didj + +EiEj(Xt(i), E~E,(x~(~), Xt(j)),xt(j)), soSO the map i HH t(i)t(i) is isan an injection, injection, and and without without lossloss we may take t(i)t(i) = i. i. Finally, Finally, byby 35.20.2, $:(I)ii/ (1) = Yr1(1)IG: $i(l)lG : HIHI == 0, so, as $:(l)iiiG(1) = = didi ++EiXi(l) ~ixi(1) and xi(1)Xi(1) =_ deg (xi)(rri) isis aa positivepositive integer,integer, it it follows follows that that ~i Ei = = --1 1 and and Xixi (1)(1) == di.di . Define xX == rk_1zfz1 didixi. Xi ThenThen Xx (1)(1) = _ rk=1~f=, d?d; == IHI 1 HI by by 35.5.3. 35.5.3. ByBy 35.20.1,35.20.1, Vf q (g) $7 (g)= 0 0 for g EE K',K#, so Xixi (g) (g) = = di --1/''iG (g) (g)= = di di. . Thus Thus x X(g) (g)= = EL1 Ei1 d?di2 _= ~(1)X(1)for for all all g g EE K,K, soso KK 5c ker(X).kero(). OnOn the other handhand eacheach member member of of G G - - K is conjugate toto anan elementelement ofof H',H#, soso toto showshow KK = ker(X) kero() (and(and complete the proof) itit remains to showshow X(h)~(h) = 0 0 for for hh EE H#. H'. ByBy 35.22, 35.22, V/iG(h) $?(h) = Y'i $i(h) (h) == di - 9 Oi(h), (h), soso Xi(h)xi(h) == di - Vlf $:(h) (h) = 9Oi(h). (h). Therefore ~(h) X(h)= = xi >i diOi(h) di9i(h) == 0 by 35.5.4.

(35.25) Let GG be a Frobenius group with FrobeniusFrobenius complementcomplement HH and kernel K. Then (1) KK <9 G G and and H H is is a acomplement complement to to K K in in G. G. ThusThus G G is is a a semidirectsemidirect product ofofKbyH. K by H. (2) H H actsacts semiregularly semiregularly on on K; K; that that is is CH(x) CH(x) == 1 1 for for eacheach xx EE K#,K', oror equiv-equivalently CKCK(y) (y)= = 11 for for each each yy EE H'.W. (3) K K isis a a regular regular normalnormal subgroupsubgroup of of GG inin its representation as a FrobeniusFrobenius group.

Proof. ByBy Frobenius'Frobenius' Theorem,Theorem, KK a9 G. G. By By definitiondefinition of H and K, H fln K = 1.1. By 35.23.1,35.23.1, lK/ IKI=IG:HI,= lG:HI, so, by 1.7, IKHI IKHI = =IKIIHI=IGI.Thus lKllHl= IGI. Thus G= G=HK. HK. Hence (1)(I) holds. Notice (1) and 15.10 imply (3), while (3) and 15.115.11 1 implyimply (2).

36 SomeSome special actions In this section representation theory developeddeveloped in previous sectionssections isis usedused toto derive various results on the representations of certain minimal groups, and these results are used in turn to prove a number of groupgroup theoretic lemmas. Some of these lemmas will be used in chapter 15 and one is used in 40.740.7 toto prove the nilpotence of Frobenius kernels.kernels. Some specialspecial actions 193

(36.1) LetLet p,p, q, q, and and r r bebe distinct distinct primes, primes, X X aa group of order r acting on an extra-extraspecial q-group QQ with with CQ CQ(X)= (X) = Z(Q),Z(Q ), andand VV aa faithful faithful GF(p)XQ-module GF(p)XQ-module such that Cv(X)=O.Cv(X)=0. ThenThen r=2"+1r =2"+ is1 isa Fermata Fermat prime, prime, q q =2, =2, and and Q Q is is of of width width n.n.

Proof. ReplacingReplacing VV byby anan irreducibleirreducible XQ-submoduleXQ-submodule of [V, Z(Q)],Z(Q)], we may assume V isis anan irreducibleirreducible XQ-module.XQ-module. Let FF be be a a splitting splitting field field for for XQ XQ over

GF(p); by by 27.1327.13 wewe maymay take F to to bebe finite.finite. Pass to VFvF = V V ®@GF(p)V.GF(p)V. By 26.2, F isis thethe directdirect sum sum of of cc Galois Galois conjugates conjugates of of anan irreducibleirreducible FXQ-module FXQ-module W.W. By 27.12, Cw(X)CW(X) == 0.0. In particular dimF(W) # r r dimF(CW(X)),dimF(Cw(X)), soso by 27.17, W isis an irreducibleirreducible FQ-module.FQ-module. ThereforeTherefore by by 34.9, 34.9, dimF(W) dimF(W) = = qe, where e is the width ofof Q, and forfor anyany setset YY ofof coset coset representatives representatives for for Z Z = = Z(Q)Z(Q) in Q, Y is a basisbasis forfor EE =EndF(W)=EndF(W) over F. AsAs CQ(X)CQ(X) == Z,Z, wewe maymay pick Y to be invariant under XX viavia conjugationconjugation andand wewe maymay pickpick 1 1 E E Y.Y. Next, XQ is a subgroup of E andand hencehence actsacts on EE viavia conjugation,conjugation, andand asas Y is a basis for E overover F,F, E E is is the the permutation permutation module module forfor thethe permutationpermutation representation of X on Y. As CQ(X) == Z, XX isis semiregularsemiregular on Y - {1}, {I}, so, by Exercise 4.6.1, CE(X)CE(X) is of dimension d=1+JYJ -1 =1+g2e-1 IYI - 1 q2e - 1 d=l+=l+~r r On the other hand we may assume F containscontains a a primitive primitive rth rth root root of of unity,unity, so so a a generator x of X can bebe diagonalizeddiagonalized on on W. W. Let Let ai, ai, 1 1 5 < ii 5< r,r, be be thethe rth rootsroots of 1, and mi the multiplicity of ai as an eigenvalue of x. As x is diagonalizable, CE(x) is isomorphic as anan F-algebraF-algebra to to the the direct direct product product of of algebras algebras Fm' Fml xm;""I, 1 -< i <5 r,r, sod so d = = Ei=1 xi=, m?.m?. Also Also qeqe = = dimF(W) dim~(W) = Eixi mi.mi. Thus Thus

r r q2e=CEm)2_Tm?+T 2mimj. i=1 i=1 i>j Therefore

T(mi-mj)2=(r-1)m?)- 2mimj i>j i i>

= r m2)-(mi)2=rd -q2e = r - 1.

i i Pick 1l inin thethe range range 1 1 5 < 1 l 5< r, and letlet ss = = s(1)s(l) == I{i : :mi mi = = ml}l.mi}I. If Ifs s == r then qe = dim(W) dim(W) == rmi, rml, impossible impossible as as r rand and q q are are distinct distinct primes. primes. So So there there areare at least twotwo multiplicities.multiplicities. NowNow sincesince therethere areare s(rs(r - s) s) differences differences of of thethe formform miml -mi-mj withwithmi mi#mj, #mi, (*) r - 1 = T(mi - Mj)2> s(r - s) i>j 194 Linear representationsrepresentations ofofjinite finite groups with equality only if there are exactly two multiplicities m1 ml and mkMk and and ImlIm1 - - Mkmkl l = = 1. Then

(**) r(s - 1) < s2 - 1 =(S - 1)(s + 1), sosoeithers=1 either s = 1 or orr r ( 1.1. Therefore qeqe == xiFt m, mi == r r- -1, 1,completing completing the the proof proof of of thethe lemma.lemma.

(36.2) LetLet p,p, q, q, and and rr bebe distinct distinct primes, primes, XX a a group group of of orderorder rr actingacting faithfullyfaithfully on a q-group Q,Q, andand VV aa faithfulfaithful GF(p)XQ-module. If q = 2 2 andand r isis aa FermatFermat prime assume QQ isis abelian.abelian. ThenThen CvCV(X) (X) # 0.

Proof. Assume Assume otherwiseotherwise and and choose choose a a counterexample counterexample with with m(V) m(V) and and I /XQlXQI minimal. Then XQ is irreducible on V by minimality of m(V), and,and, by mini-minimality of IXQJ,I XQ 1, XX centralizes every proper subgroup ofof Q. The latter remark and Exercise 8.10 imply Q isis elementaryelementary abelian or Q Q isis extraspecialextraspecial andand Z(Q)Z(Q)=CQ(X). = CQ(X). Moreover Moreover inin eithereither casecase X isis irreducibleirreducible on Q/(D(Q).Q/@(Q). Now by 36.1, QQ isis elementaryelementary abelian,abelian, and then Exercise 4.4 and 27.18 supply a contradiction.

(36.3) Let a bebe anan involutioninvolution actingacting onon aa solvablesolvable group GG ofof oddodd order,order, letlet p EE 7rn cg ,r(G),n(G), and and p" p" = = {p} {p} U,r'.U n'. Assume Assume K K is is an an a-invariant a-invariant subgroup subgroup of G such that CK(a) contains contains aa HallHall n-subgroupar-subgroup of of CG(a) CG(a)and and X X = = [X, a] is aa p-subgroup of of OFnOPn (K). ThenThen XX < ( O Opn pn (G).

Proof. Take G toto bebe a a minimal minimal counterexample. counterexample. Then OnOPx(G)r(G) = 11 and and itit remains to to showshow X X = 1.1. Let Let V V be be a aminimal minimal normal normal subgroupsubgroup of G(a) contained inin GG and G* = G/ G/V. V. By By minimality minimality ofof G,G, X* X* < ( O OPn(G*),pn (G*), so CX(V)Cx(V) (< Opn(G)Opn(G) == 1. 1. Also Also V V is is a aq-group q-group for for some some primeprime q,q, andand asas Op-(G)Op*(G) == 1,1, q E n7r - - (P).{p}. By coprime action, Exercise 6.2, X isis containedcontained inin anan a-invariant a-invariant Hall Hall 7r- n- subgroup H ofof K,K, and and as as CK CK (a)(a) contains contains a a Hall Hall 7r-subgroup n-subgroup ofof CG(a),%(a), CHCH(a) (a) is a Hall n-subgrouppr-subgroup of of &(a).CG(a). As As X X _(< Op.Opr(K), (K), X <( Op Op(H). (H). Thus Thus settingsetting Some specialspecial actions 195

KOKo = (Cxv(a), (&(a), X),X), X X < 5Opn(Ko) Opn (KO) byby 31.20.1. 3 1.20.1. Therefore Therefore (Ko, (KO, XV)XV) satisfiessatisfies the hypotheses of (K, G), soso GG == XV XV byby minimality minimality of of (G1. IGI. InIn particularparticular XX is irreducible on V andand GG isis aa pr-group.n-group. AsAs GG isis a n-group,pr-group, CG(a)CG(a) 5< KK andand X <5 Op(K),Op(K), so so [Cv(a), [Cv(a), X] XI <5 V V fl n Op(K) Op(K) = = 1. 1.Therefore, Therefore, as as X X is is faithful faithful and irreducible onon V, Cv(a) == 1. 1. But But now now 36.2 36.2 supplies supplies a acontradiction, contradiction, completing thethe proof ofof thethe lemma.lemma.

Remarks. TheThe representation representation theory theory in in sections sections 34 34 and and 35 35 isis basicbasic andand belongs in any introductory course on finite groups. The results inin sectionsection 3636 areare moremore technical. They are in the spirit of the fundamental paper of Hall and Higman [HH]. In particular lemma 36.1 is due toto ShultShult [Sh][Sh] althoughalthough thethe proofproof given given here is fromfrom CollinsCollins [Co] andand usesuses techniquestechniques ofof HallHall andand Higman Higman [HH].[HH]. LemmaLemma 36.3 will be used in the proof of thethe SolvableSolvable 2-Signalizer Functor Theorem in chapter 15. 15. InIn thethe firstfirst editionedition ofof thisthis text,text, sectionsection 3636 containedcontained strongerstronger results used inin thethe proofproof ofof thethe SolvableSolvable SignalizerSignalizer Functor Functor TheoremTheorem givengiven inin the first edition. These results have been omitted, since theythey areare unnecessaryunnecessary for 2-signalizers.

Exercises for chapter 1212 1. LetLet GG bebe aa finite finite groupgroup andand FF a a splitting splitting field field forfor GG whosewhose characteristiccharacteristic does not divide the order of G. Prove (1) A A charactercharacter X x ofof GG isis linearlinear if if andand onlyonly ifif Xx isis irreducibleirreducible andand G(1)G(') 5< ker(Xker(x). ). (2) IfIf GG == (g) (g) is is cyclic cyclic ofof orderorder n thenthen FF contains contains aa primitiveprimitive nth rootroot ofof unity w and the irreducibleirreducible characters characters of of G G are are xi, Xi, 1 15 < i i 5< n, withwith xiXi (gj) (gi) =_ a)ijwij . (3) GG hashas exactly exactly IG/GM)IG/G(')I linearlinear characters. characters. 2. LetLet 7r n be be thethe regularregular representation of a finite group G over a splitting field F whosewhose characteristiccharacteristic does not divide thethe orderorder ofof GG and let Xx be the character of 7r.n. Prove (1) 7rn is is the representationrepresentation inducedinduced by thethe regular permutation representation. (2) n7r = = Cy=lFml ni7ri,nixi, where (xi:(nri: 1 1 5 < i <5 m)m) areare the irreducible FG-represen-FG-representations andand nini == deg(7ri). deg(ni). (3) x(g)=OforgX(g)=0forgEG#andX(1)=IGI. ~G#andx(l)=IGI. 3. LetLet GG bebe aa finite group,group, TT gC G, AA= = {tg: {tg: tt EE T#,T', g E G),G}, and and H H= = NG(T). NG(T). Prove (1) TT is is a a TI-set TI-set inin GG ifif andand onlyonly if, if, for for each each t tE E T#,T', tG nfl T = tH andand CG(t)CG(~) 5< H. 196 Linear representationsrepresentations ofofJinite finite groups

(2) IfIf TT is is a aTI-set TI-set in in G G then then the the set set of of conjugates conjugates of of T T# # under GG partitionspartitions

A,0, and, and, for for each each t Et ET T',#, I ltGltG I= = IG IG :: H HIJ~~I. II tH I 4. (1) (1) Prove Prove lemma lemma 35.23. 35.23. (2) ProveProve a a finite finite group group has has atat most one faithful permutation representation as a Frobenius group, andand concludeconclude aa Frobenius groupgroup hashas atat mostmost oneone class of Frobenius complements.complements. (You (You may use the fact (proved in 40.8)40.8) that Frobenius kernels are solvable.) 5. LetLet GG =E S5 S5 be be thethe symmetric groupgroup on on X X = = (1,.it, .... ., ,5).5}. (1) Use Use 15.3 15.3 to to determine determine the the conjugacyconjugacy classesclasses of G. (2) As As in in ExerciseExercise 5.1.2,5.1.2, showshow GG has has 2-transitive 2-transitive permutationpermutation represen-representations of degree 5 andand 6,6, andand find find theirtheir permutationpermutation characterscharacters (cf. Exercise 4.5). (3) Find Find allall linearlinear characterscharacters of G. (4) DetermineDetermine thethe charactercharacter table of G. (Hint: Use (2)(2) andand ExerciseExercise 12.612.6 toto determinedetermine twotwo irreducibleirreducible characterscharacters of G.G. ThenThen useuse thethe nonprincipalnonprincipal linearlinear charactercharacter from (3)(3) andand ExercisesExercises 9.3 andand 9.109.10 toto produceproduce two two more more irreducible irreducible characters. characters. Finally, Finally, givengiven these characters and the linear characters, use the orthogonality relations to complete the table.) 6. LetLet 7rn bebe aa permutationpermutation representationrepresentation of the finite groupgroup GG on a set X, a the CG-representation inducedinduced byby n,r, andand Xx the character of aa (cf. Exercise 4.5). We say xX isis thethe permutationpermutation character ofof Jr.n. Prove (1) (X,(x, Xi) xl) is is thethe numbernumber of orbits of G on X. (2) IfIf 7rn isis transitivetransitive then (X,(x, X) is the permutationpermutation rankrank of G on X. (3) GG is is doubly doubly transitivetransitive onon X if and only ifif GG == x1Xi +xi+Xi forfor somesome ii > 1.1. GG is of permutationpermutation rankrank 33 if and only ifif G = Xix1 + XtX, + Xjxj forfor somesome i > jj > > 1. 1. (Hint: (Hint: See See Exercise Exercise 4.5.) 7. LetLet aa be be anan elementelement ofof primeprime order r actingacting on an r'-group G. Let p be be a prime, with pp == 2 if r isis aa Fermat Fermat prime,prime, letlet PP = = Op(G), Op(G), andand assumeassume Cp(a) ==1. 1. ProveProve [a, OP(G)]O"(G)] <5 CG(P).Cc(P). 8. LetLet a bebe anan elementelement of prime orderorder rr acting on on an an r'-group r'-group G. G. Let Let 11 #} X = [X, a]a] be a q-subgroup ofof GG with XX abelian ifif qq == 2,2, andand letlet pp be a prime distinct fromfrom q. q. Prove Prove [Op [O,(G), (G), X]XI = = (C[op(G),x1(a)X? (Cp,cc,,xl(a)x). 13

Transfer and fusionfusion

If G is a finitefinite group,group, H H (< G,G, andand a:a: HH + A is a homomorphismhomomorphism ofof H into an abelian group A, thenthen itit isis possiblepossible to to construct construct a a homomorphism homomorphism V: V: G G ->. -+ AA from a inin aa canonicalcanonical way. V is called the transfer of G into A via a. If If wewe cancan show therethere existsexists gg E E G G - - ker(V),ker(V), then,then, as as G/ker(V) G/ker(V) isis abelian,abelian, g g 40 G(')Gf1 the commutator group of G. In particular GG isis notnot nonabeliannonabelian simple.simple. It isis howeverhowever in generalgeneral difficult toto calculatecalculate g VV explicitlyexplicitly andand decidedecide whetherwhether g E ker(V).ker(V ). To To do do so so we we need need informationinformation about about thefusion the fusion of of gg inin H; that isis information about about gGgG nfl H. Hence chapter 1313 investigatesinvestigates bothboth thethe transfertransfer map and techniques for determining the fusion of elements in subgroups of G. Section 38 contains a proof of Alperin's Fusion Theorem, which says that p-local subgroups subgroups control the fusion of p-elements. To be somewhatsomewhat more precise, if P isis aa SylowSylow p-subgroupp-subgroup ofof GG thenthen wewe cancan determinedetermine whenwhen subsets of P arearefised fused inin GG (i.e.(i.e. conjugateconjugate in G) by inspecting thethe p-localsp-locals H of GG withwithPflHSylowinH. P n H Sylow in H. Section 39 investigatesinvestigates normal p-complements. A normalnormal p-complementp-complement for a finite group G is a normal Hall p'-subgroup ofof G.G. Various Various criteriacriteria forfor thethe existence of such objects are generated, The most powerful is the Thompson Normal p-Complement Theorem, Theorem, which which is is used used in in the the next next sectionsection to to establishestablish the nilpotence of Frobenius kernels. Section 39 also contains a proof of thethe Baer-Suzuki Theorem whichwhich sayssays aa p-subgroupp-subgroup XX of a finitefinite group G isis contained in Op(G)O,(G) if and only if (X, Xg) is a p-group forfor eacheach gg inin G.G. A group AA is said to act semiregularly on a group G ifif AA isis aa groupgroup ofof auto-automorphisms of G with &(a)CG(a) ==1 1 for each a cE A.A'. SuchSuch actions actions are are investigated investigated in sectionsection 40.

37 Transfer Let G be a finitefinite group, HH 5< G, and a: H + A a homomorphism of H intointo an abelian group A. Given a set X of coset representatives forfor H in G,G, definedefine V:G->.V:G+A Aby by

gV = fl((xg)(xg)_1)a xEX where xg denotes the unique member of X in the coset Hxg. WeWe say VV is the transfer of G into A via a. 198 Transfer andfisionand fusion

(37.1) The transfertransfer map VV isis independentindependent of the choice of the set X ofof cosetcoset representatives.

Proof. LetLet Y Y bebe a second set of coset representatives forfor HH in G. Then therethere isis a bijection xx ++H y(x)y(x) of of XX withwith YY suchsuch that y(x) = h(x)x for for somesome h(x)h(x) EE H. For y E YY andand gg EE G, G, writewrite ygyg for the member ofof Y in Hyg. ObserveObserve thatthat {y(x)g){y(x)g}= = Hy(x)gnY Hy(x)gfY = HxgfY HxgnY = = Hxgf1Y HxgnY = = {y(xg)}; {y(xg)};thatis that is y(x)g=y(x)g = y(xg). Also y(x)g = h(x)xg h(x)xg = h(x)xg(xg)-1(xg) h(x)xg(xg)-l(xg) = h(x)xg(xg)-1h(xg)-1y(xg)h(x)xg(xg)-'h(xg)-' y(xg) so that y(x)g(y(x)g)-'y(x)g(y(x)g)-1 == h(x)xg(xg)-lh(xg)-'.h(x)xg(xg)-1h(xg)-1

Therefore

fl(yg(yg)-1)a = fl(y(x)g(y(x)9)-1)a yEY xEX = fln h(x)a(xg(xg)-1)ah(xg)-1a h(x)a(~g(xg)-l)ah(xg)-~a = n(xg(xg)-l)a fl(xg(xg)_1)a xexxEX xexxEX with the last equality holding asas AA isis abelianabelian andand thethe mapmap xx i-±I+ xg isis aa permu-permutation of X. SoSo thethe lemmalemma holds.

(37.2) The transfer V is a group homomorphism ofof G into A.

Proof. Lets,Let s, t t EE G.G. Observe Observe x(st)x(Z) == (xs)1. (x5)t. HenceHence

(st)V = fl((xst)(xst)-1)a = fl(xs(xs)-1(xs)t(xst)-1)a xEX XEX _fl(xs(xs)-1)a((xs)t(xst)-1)a xEX

(fl(xs(x)_1)a) (fl(xt(xi)_1)a)=sVtV, xEX using the fact that A isis abelian andand thethe mapmap xx I+H x5xg is a permutation on X.

(37.3) Let (Hx1(~x~gj: gj:0 0 5 < j j < ni),nl ), 1 1 _( < ii 5< r,r, be the cycles of g EE G on thethe cosetcoset space G/H. PickPick XX == {x1gi: (xigj: 11 _(< i <_( r, 0 _(< j <_(nil. n1). Then (1) (g"1)x+`EHfor(gni)x;'~Hfor15isr. 1

Et-1nl=1G:HI. (2) C;=,ni=lG: HI. (3) gVg~ = n;,];=1((gn')x' ,ccgni)x;')~. )a. Transfer 199

Proof. PartsParts (1) (1) andand (2) (2) areare immediate immediate from from the the definitions. definitions. By By 37.137.1 we may calculate V with respect to this particular choice of coset representatives. By definition of X, (xigj)g= xigj+'(xigj)g=xigi-H1 = (xigj)g =(xig1)gforfor j j < ni-1.Alsoni - 1. A~SO (xig"j-1)g=xig"j(xigni-')g = xigni and (xigni-')g(xlg"i-1)g =xi. So SO (3)(3) holds.holds.

(37.4) Let G be a finitefinite group, group, p p a a prime, prime, H H< 5 G withwith (p,(p, IGI G: :HI) H p = = 1,1, KK < H with H/K abelian, abelian, and and gg aa p-elementp-element in H -- K K such such that that g" gma e eE gmK gmK forfor allall integers m, and allall a EE G, such such thatthat gmagma E H. Then g V4 G(l).G(').

Proof. Let A == HIH/K K andand a :: H + A the naturalnatural surjection. surjection. I'll I'll show show g gV V # # 1. Hence g V4 ker(V),ker(V), and and ofof course,course, asas GVGV 5< A, GV is abelian, so GMG(') 5< ker(V). Choose a set X of coset representatives forfor HH in G asas inin 37.3.37.3. By 37.3.1,

(g"(gni)x;l )x E H, so by hypothesis gnix;lg"xI ' E gnig"i K. K. Hence (g";x(gnix;l)a)a = g"gnia a = (gc~)~~.(g,)" . Hence, byby 37.3.3, gVgV == (ga)",(ga)", where nn == E;-1EL=, ni.ni. Finally, Finally, byby 37.3.2,37.3.2, n = IG :: HI,HI, soso byby hypothesis (p,(p, n)n) = = 1. Hence, as g is aa p-elementp-element andand gg V4 K, also gng" 4V K.K. ThusThus 11 ## g"agna =gV,= g V, asas desired. desired.

(37.5) Let GG bebe aa finitefinite group,group, p a a prime, prime, HH < 5 G G with with (p, (p, I IGG: :H1) HI) = = 1, 1, and and assume gGgG nn H = gH for allall p-elementsp-elements gg inin H.H. Then (oP(G)G('))(OP(G)G()) nn H = OP(H)HO).OP(H)H(').

Proof. LetLet Go Go = OP(G)G(l) O~(G)G(') and K = OP(H)HO). OP(H)H('). Then certainlycertainly KK 5< Go fln H, and each p'-element inin Go nn H isis inin OP(H)OP(H) <5 K,K, soso it it remains remains toto showshow each p-element g in Go n HH isis inin K.K. But But if if mm is is an an integer integer andand aa E E G withwith gma E HH then by hypothesis gma gma= gmh, forfor somesome h h EE H, H, so,so, as as H/K H/K is abelian, gma = = gmh EE gmgmK. K. It It follows follows from from 37.437.4 thatthat if g 4V KK then g V4 G(').G(l). Then, as all p-elements inin GoGo are are in in G(1),G('), also gg V4 Go, contrary contrary to the choice of g. SoSo thethe lemma holds.

Let H <5 G G and and S S an an H-invariant H-invariant subset subset of of G.G. Then Then HH is is said said to to control controlfusion fusion in S ifif sG n S=sH S = sH forfor each each s sE E S. S. For For example example one one of of thethe hypotheseshypotheses in the last lemma says that H controlscontrols fusion of its p-elements.p-elements. Let X CE H H < 5 G. G. Then Then X X isis said said to to be be weakly weakly closedclosed in H withwith respectrespect to GG ifXGnH={X}.if xGn H = {X).

(37.6) LetLet p be a prime,prime, T E Sylp(G),Sylp(G), W W 5< TT withwith WW weaklyweakly closed in T with respect to G, and DD = CG(W). CG(W). ThenThen NG(W)NG(W) controls fusion in D.

Proof. Let dd EE D andand gg E E G,G, with with d9dg EE D.D. ThenThen W,W, W9_'~8-' 5< C(d). By By Sy-Sy- low's TheoremTheorem therethere isis xx EE CG(d)CG(d) with UU = (W, (W, Wg-`x)wg-lX) a a p-group. Let 200 Transfer and fusion

(37.7) AssumeAssume T T isis an abelian SylowSylow p-group p-group of of G.G. ThenThen TT n O'OP(G) (G) =_ [T, Nc(T)I.NG(T)].

Proof. LetLet HH = = NG NG(T). (T). ByBy thethe Schur-ZassenhausSchur-Zassenhaus Theorem there is a com-complement XX toto TT inin H.H. ThenThen KK == X[T,X [T, XI X]

38 Alperin'sAlperin's FusionFusion TheoremTheorem In this section GG isis aa finitefinite group,group, pp isis a prime,prime, andand PP is some Sylow p-groupp-group of G. A p-subgroup XX ofof GG isis saidsaid to be a tame intersection of Sylow p-groups Q and R of G ifif XX = Q Q ni? RR and and NQ(X) NQ(X) and and NR(X)NR(X) are are SylowSylow p-groups of NG(XNG(~). ). The main result of this section is:

(38.1) (Alperin's(Alperin's Fusion Theorem) LetLet P E Sylp(G),Sylp(G), gg EE G, G, andand A, A, Ag Ag C C P.P. Then therethere existsexists QiQi EE Sylp(G), Sylp(G), 1 1 5 < ii 5< n, andand xi E Nc(PNG(P iln Qi) Qi) suchsuch that:

(1) gg=x1.. = x1...xn. .x,. (2) PP n n Qi Qi is is a a tame tame intersection intersection ofof PP and Qi forfor eacheach i, i, 1 1 ( < ii 5< n. (3) AA CE P Pi? n Q1 Ql and andAxl...xl Ax'x, Ec P iln Qi+i Qi+l for for 1 l

Alperin's Theorem will follow from Theorem 38.2 andand anan easy argument.argument. But first some definitions. For R, QQ E Sylp(G) write R +-+ QQ ifif therethere existexist SylowSylow p-groupsp-groups (Q1:1(Qi:1 5 i <5 n) of GG andand elements xi E NG(PNG(P i? n Qi)Qi) suchsuch that:

(1) PP n i? QiQi is is a a tame tame intersectionintersection ofof P and Qi for eacheach ii withwith 11 5 < ii 5< n. (2) PPnR(PnQland(PnR)"l...Xi n R < P n Q1 and (P n iPnQi+lforeachiwithlP n Qt+1 for each i with 1 (i < i

I'll alsoalso write R 54 Q Q when when it's it's necessary necessary to to emphasize emphasize the the role role ofof thethe elementelement x in (3), and saysay that that (Qi, (Qi,xi: xi: 1 1 5 < ii 5< n) accomplish R +-+ Q. Alperin's FusionFusion TheoremTheorem 201

Theorem 38.2. Q +- P P for for each each Q Q E E Sylp(G). Sylp(G).

The proof of 38.2 involves several reductions. Observe first that:

(38.3) P P.

Indeed P: P4 PP is is accomplished accomplished byby P,P, 1. 1.

(38.4) -*+ is is a atransitive transitive relation.relation.

Proof. Let (R1,(Ri, yi: yi: 1 1 < ii <( m) m) andand (Q;,(Qi,xi: x; : 1 1 (< ii <( n) n) accomplish accomplish SS+ -+ RR and R +-+ Q,Q, respectively.respectively. ThenThen R1,.R1, .... . , R,,Rm, el,. Q1, .... . ,, Q,,Q,,, and yl,yl,... . . ,. ym,, y,, x1,xl, ...... , ,xn x, accomplishaccomplish S -++ Q.Q.

(38.5) If If S S4P,Qx-+ 5 P, Qx + P, P, and and P P nil Q Q < ( P P n il S,S, then then QQ +-+ P.

Proof. ByBy 38.4 38.4 it it sufficessuffices to show Q +-+ Qx.QX. Let (Si,(S;, xi:x; :1 1 5 < ii 5< n)n) accomplishaccomplish S+S -+ P. Then Then (Si,(Si, x1:xi: 11 5< i <( n) n) alsoalso accomplishaccomplish QQ -++ Qx. QX.

(38.6) AssumeAssume R, R, Q Q E E Sylp(G) Sylp(G) with with R R+ A P Pand and P Pn n Q Q << R ni? Q.Q. AssumeAssume further,further,forallS~Syl,(G)withISnPI for all S E Sylp (G) with I S n P I > > IQnPI,thatS+I Q n P 1, that S -+P.Then P. ThenQ Q+ -+ P.

Proof. ByBy hypothesishypothesis therethere isis x x EE G G with with R R 5 4 P. Now Now P ni? QxQx = Rx Rx ni? QxQx = (Rn(R n Q)x,Q)",soIPn so I P n QxI=IRnQx I= I R n Ql Q >I> IPnP n Q1. Ql.Hence Hence Qx+Qx ± PPbyhypothesis. by hypothesis. Now apply 38.538.5 toto completecomplete the the proof.proof.

(38.7) Assume P ni? QQ isis aa tametame intersectionintersection of P and Q such that S +-+ P forfor all SS E Sylp(G)Sylp(G) with with IS IS i? nPI P) > > lQI Q n n PI. PI. Then Then Q Q + - P.

Proof. ByBy 38.3 wewe maymay assume assume Q Q # 0 P.P. ThusThus P P i?n Q < PoPo == Np(P i?n Q).Q). By hypothesis PoP0 andand QoQo = = NQ(PNQ(P i? n Q) are Sylow inin MM == NG(PNG(P i?n Q),Q), soso there is x EE M Withwith QoQg == Po.P0. Notice Q +-+ Qx isis accomplishedaccomplished by (Q,(Q, x). x). FurtherFurther PPi? n Q << P0 Po < 5 P Pi? n Qx, Qx, soso by by hypothesishypothesis Qx +± P. P. Therefore, Therefore, by by 38.4, Q +-+ P.

We are now inin a positionposition toto prove prove 38.2. 38.2. Pick Pick a a counterexample counterexample Q Q with with P P i? n Q of maximal order.order. By By 38.3, 38.3, P P # 0 Q,Q, soso P n QQ # P, andand hence PP i?n QQ < Np(P n Q).Q). LetLet SS E Sylp(G) withwith Np(P Np(P n n Q)Q) <5 Ns(PNs(P n n Q) Q) E ESylp(NG Sylp(NG (P(PnQ)).n Q)). As As P P n n Q Q < < Np(P Np(P n i? Q)Q) < 5P nP S, n S,it itfollows follows that that S S -* + P P by by maxima-maximality of P ni? Q. Q. Thus Thus there there is is xx EE G withwith S 5 P. 202 Transfer and fusionfusion

Evidently (P (P fln Q)XQ)x 5< QX.Qx. AlsoAlso P P nn QQ 5< S and SXSx= = P, P, soso (P (P fln Q)XQ)x 5 < P. Thus(PnQ)x 5< PnQx.If(PnQ)xPnQx.if(PnQ)x # PnQxthenPnQxthenlPnQl IPnQI << IPnQxl,so,lPnQxl,so, by maximality of I P P nn QQl, 1, QX Qx+ -+ P. P. ButBut thenthen Q +-+ P byby 38.5,38.5, contradictingcontradicting the choice of Q.Q. So(PnQ)x=PnQx.NextSo (P n Q)X = P n Qx.Next letT let TESylp(G)with E Sylp(G) with NQx(PnQx) Np(P n QX) <5 NT(PnNT(P n Qx) E Sylp(NG(P n n Qx)). AgainAgain P P n n QXQx << Np(PNQ=(P n n QX)Qx) ( < T, so P n Qx << T n Qx.Qx. HenceHence if T +- P Pthen, then, by by 38.6, 38.6, Qx QX -++ P, P, which which we we havehave alreadyalready observed toto bebe false. ThusThus we do not have T +-+ P, so,so, byby maximalitymaximality of IPnQI,PnQx=PnT.IPn QI,P n Q~=Pn T. By choiceofchoice of T,T, andasand as PP n QxQX = P P nn T, T, we wehave have NT NT(PnT) (PnT) E ESylp Sylp(N~(Pn (NG (P n T)). By choice ofof S, Ns(PNs(P nn Q)Q) EE Sylp(NG(Pn n Q)) Q)) so, so, asas (P(P n Q)xQ)X = P n QxQx = PP nn TT andand SxSx = = P, P, we we have have Np(P Np(P n T)n T) E ESylp(NG(P Sylp(N~(P nn T)). T)). Thus P nn TT is is a a tame tame intersection intersection of P andand T.T. But But now,now, by 38.7,38.7, T +-+ P, contrary to the last paragraph. This completes the proof ofof 38.2.38.2.

Now for the proof of Alperin'sAlperin's Fusion Fusion Theorem.Theorem. Assume the hypothesis of Alperin7sTheorem.Alperin'sTheorem.By38.2,By 38.2, pg-' Pg-' + -+ P. P.Let(Q1, Let (Qi, xi:x;: 1 1 5 < i i 5< n-1)n-1)accomplish accomplish Pg-'-+pg-' + P.P.AsA,AgCP,ACPnPC',so,bydefinition As A, Ag E P, A P n ~g-',SO, by definition of-+,Aof +, A c_E PPn n Pg-'pg-' <5 P nn Q1Ql andand AXl...XlAxt...xic- C(P (P nn P~-')X~...~IPg-')x,...xiP n< PQi+l n Qi+1 for for1 1

39 NormalNormal p-complements In this section p is aa prime andand GG isis aa finite finite group.group. AA normalnormal p-complement for G is a normal Hall p'-subgroup of G;G; thatthat is aa normalnormal p-complement is aa normal complement to a Sylow p-subgroup of G.

(39.1) (Burnside(Burnside Normal p-Complement Theorem) If a Sylow p-subgroupp-subgroup of G is in the center of its normalizer then G possesses a normal p-complement.

Proof. ThisThis is is immediateimmediate from from 37.7.37.7.

(39.2) If p isis thethe smallestsmallest prime divisor of thethe order ofof G and G has cycliccyclic Sylow p-groups, then G has a normal p-complement.

Proof. LetLet P P E E Sylp(G). Sylp(G). By By hypothesishypothesis P P isis cyclic. cyclic. AsAs PP isis abelianabelian and Sylow in G, AutG(P) is a p'-group, so, by 23.3, lAut~(P)lIAutG(P)I dividesdivides p p -- 1.1. Hence,Hence, as Normal p-complements 203 p isis thethe smallestsmallest primeprime divisordivisor of (G1,I G I, AutGAutG(P)(P) = 1.1. Equivalently, P isis inin thethe center of its normalizer, so 39.1 completes thethe proof.proof.

G is metacyclicmetacyclic if there exists aa normalnormal subgroupsubgroup H H ofof G such that H andand G/H are are cyclic. cyclic.

(39.3) Assume each Sylow group of G is cyclic. Then G is metacyclic.

Proof. Let Let p p be be the the smallest smallest prime prime divisor divisor of of I IG I. ByBy 39.2,39.2, GG hashas aa normalnormal p-p- complement H. By By induction on the order of G, H isis metacyclic.metacyclic. In particular H is solvable, so, as G/H is is cyclic, cyclic, GG isis solvable.solvable. Let K = F(G). F(G). K K is is nilpotent nilpotent with with cyclic cyclic Sylow Sylow groups,groups, soso KK isis cyclic. cyclic. Thus Aut(K) is abelian. However,However, by by 31.10, 31.10, K K = = CG(K), CG(K),so so Au~G(K) AutG(K)= = G/K G/K is abelian. As G/K hashas cyclic cyclic SylowSylow groups, G/K is is cyclic, cyclic, so so GG isis metacyclic.

A subgroup H ofof GG isis aa p-localp-local subgroupsubgroup if HH == NG(P) NG(P) for for somesome nontrivialnontrivial p-subgroup P ofof G.G.

(39.4) (Frobenius(Frobenius Normal p-Complement Theorem)Theorem) TheThe following are equivalent: (1) GG has has a a normal normal p-complement.p-complement. (2) Each Each p-localp-local subgroup subgroup of of GG hashas a a normal normal p-complement.p-complement. (3) AutG(P)AutG(P) isis aa p-groupp-group forfor eacheach p-subgroup P ofof G.G.

Proof. TheThe implicationsimplications (1)(1) impliesimplies (2)(2) andand (2)(2) impliesimplies (3)(3) areare easyeasy andand leftleft as exercises. Assume GG satisfiessatisfies (3)(3) but not (1),(I), and, subject to this constraint, choose G minimal.minimal. Observe first that G = OP(G). OP(G). For if not, by minimalityminimality of G,G, OP(G) OP(G) hashas aa normal p-complement, whichwhich is also a normal p-complement forfor G.G. Next let's see that a Sylow p-subgroupp-subgroup P of G controls fusion in P. For let g E G, a E P, and and agag EE P. ApplyApply Alperin'sAlperin's TheoremTheorem toto obtainobtain QiQi E E Sylp(G) and xix, E NG(PNG (P nn Q1)Qi) satisfyingsatisfying thethe conditionsconditions ofof thatthat theoremtheorem withwith AA == (a}.(a). In particular g =x1=XI ...... x,,. x,. ItIt will will suffice suffice to to show show ax,... xiE E aP aP for for eacheach 11 <5 i <5 n. n. Assume Assume otherwise otherwise and and letlet i ibe be a aminimal minimal counterexample. counterexample. Then b = axl...x,-l ax,...X; = LX( ax1...x'-1E E P P n Q1 Qi= U, U, x,xi E H = NG(U), P n H H EE Sylp(H),Sylp(H), and and aX1...Xi= (H/bxl. By hypothesis AutG(U) AutG(U) is is a ap-group, p-group, so so H H = = CG(U)(P CG(U)(P fl n H), H), as as PP fln H E Sylp(H). Thus, as b E U, bXlbx' ==b` bf for some t E PP n H, H, so, so, as as bb EE aP,up, aX1...Xzax,...x1= _ bXrbx' EEa1. up. So P controlscontrols fusionfusion in P. Hence Hence byby 37.537.5 (OP(G)G(1))(oP(G)G(')) n n PP ==OP(P)PO). OP(P)P('). But OP(P) ==1 1 and, and, asas G = OP(G), (OP(G)G('))(OP(G)G(l)) n n PP == P. Thus P = PM, P('), so,so, as p-groups areare solvable, P == 1. 1. But But thenthen GG isis itsits ownown normal p-complement. 204 Transfer andandfusion fusion

(39.5)(39.5) (Thompson(Thompson Normal p-Complement Theorem)Theorem) LetLet pp be odd and PP EE Sylp(G).Syl,(G). Assume NG(J(P)) andand CG(Q1(Z(P)))CG(Q1(Z(P))) have have normal normal p-complements.p-complements. ThenThen GG hashas aa normalnormal p-complement.p-complement.

Proof.Proof. AssumeAssume otherwiseotherwise and let G G bebe aa minimal minimal counterexample. counterexample. By thethe FrobeniusFrobenius p-Complementp-Complement TheoremTheorem there is a p-local subgroup H ofof GG whichwhich possessespossesses no normal p-complement. By Sylow's Theorem we take Q = P P fl n HH EE Sylp(H).Syl,(H). ChooseChoose H H withwith Q Q maximal maximal subject subject to to these these constraints. constraints. ClaimClaim PP == Q. Q. If If notnot QQ < < Np(Q), Np(Q), so so I NG(C)IpING(C)lp > > I 1QI Ql forfor eacheach C charchar Q.Q. Hence,Hence, byby maximalitymaximality ofof Q,Q, NG(C)NG(C) has has aa normalnormal p-complement.p-complement. So So NH(C) NH(C) alsoalso hashas aa normalnormal p-complement.p-complement. In particular thisthis holdsholds forfor C == J(Q) and and S21(Z(Q)),Ql (Z(Q)), so the hypotheses of the theorem hold in H. Hence, Hence, byby minimalityminimality ofof G,G, HH hashas a a normal normal p-complement, p-complement, contrary contrary toto thethe choice of H. SoSo PP EE Sylp(H).Syl,(H). Hence,Hence, asas HH hashas nono p-complement,p-complement, HH == G G by by minimalityminimality ofof G.G. Moreover Moreover if if Op, O,t(G) (G) # 11 then,then, by 18.718.7 and minimality of G, G/Op,(G)G/O,/ (G) hashas aa p-complement,p-complement, and and then then G G does does too. too. So So Op-(G) O,i(G) == 1. 1. AsAs GG = H H isis a a p-local, p-local, Op(G) O,(G) 0# 1. 1. Let Let G* G* = = G/Op(G). G/O,(G). ThenThen Op(G*) O,(G*) == 1 1 soso NG(J(P*))NG(J(P*)) andand CG(Q1(Z(P*))) CG(Q1(Z(P*))) are are proper proper subgroupssubgroups ofof GG containingcontaining P,P, andand hencehence byby minimalityminimality ofof GG havehave p-complements.p-complements. So, So, byby minimalityrninimality ofof G,G, G*G* has has aa p-complement.p-complement. Hence Hence G G = = O O,,,/,,(G). p,p,, p(G). Thus E(G) = = Op,p,, O,,,~,,(E(G)). p(E(G)). ButBut [E(G),[E(G), F(G)]F(G)] = = 1 1 so so Op(E(G))O,(E(G)) <( Z(E(G)), Z(E(G)), and and thus thus Op,p,(E(G)) O,,,t(E(G)) =Op= 0, (E(G))(E(G)) x x Op,(E(G)). O,!(E(G)). NowNow Op,(E(G))O,t(E(G)) <( Op,(G) O,t(G) == 1, 1, so so E(G)E(G) is is a a p-group. p-group. Hence,Hence, asas E(G) isis perfect,perfect, E(G) = = 1. 1. SoSo F*(G)F*(G) = = F(G)E(G) F(G)E(G) = = O O,(G). p(G). LetLet pp 0# r cr Eir(G) n(G) and and R RE ESyl,(G). Syl,(G). R R < (Op,p,(G), O,,,I(G), so,so, by by a aFrattini Frattini Argu- Argu- ment,ment, Op(G)NG(Z(R))Op(G)NG(Z(R)) containscontains aa SylowSylow p-groupp-group ofof GG which which wewe maymay taketake toto bebe P.P. In In particular particular P P actsacts on on Op(G)Z(R)O,(G)Z(R) soso PZ(R)PZ(R) = = K K is is a asubgroup subgroup ofof G.G. NowNow if K 0# G G then, then, by by minimality minimality of G, Z(R) = Op,(K) O,!(K) a5 K. K. But But thenthen Z(R)Z(R) <( CG(Op(G)) CG(Op(G)) <( Op(G) O,(G) byby 31.10,31.10, and hencehence Z(R) Z(R) == 1,1, aa contradiction.contradiction. SoSo G G == PZ(R). PZ(R). In In particular particular G G is is solvable solvable and and R R = = Z(R) Z(R) is is abelian. abelian. But But now now byby ThompsonThompson Factorization, Factorization, 32.6, 32.6, G G = =G G1 1G2 G2 wherewhere G GI 1 = NG NG(J(P)) (J (P)) andand G G2 2 = = CG(S21(Z(P)).CG(Ql(Z(P)). ByBy hypothesishypothesis Gi G, = Op,(G,)P.Opf(Gi)P. AsAs Op,(G,)O,t(Gi) <( CG(Op(G)) CG(Op(G)) ( OpO,(G), (G), Op-OPf(Gi)= (G,) = 1,l,soGi so G, = P.ThusP.ThusG=G1G2 G = G 1 G2 = P, P,contradictingthechoice contradicting the choice ofof GG as as a a counterexample. counterexample.

TheThe nextnext resultresult doesdoes notnot involveinvolve normalnormal p-complements,p-complements, butbut itit hashas thethe samesame flavor.flavor.

(39.6)(39.6) (Baer-Suzuki(Baer-Suzuki Theorem) Theorem) Let Let XX bebe aa p-subgroupp-subgroup of of G.G. Then Then either either X X <( Op(G)O,(G) oror therethere exists exists gg cE G G with with (X,(X, Xg) Xg) not not aa p-group. p-group. Semiregular action 205

Proof. Assume the theoremtheorem is is falsefalse andand letlet PP EE Syl,(G) Sylp(G) and and A A == XGxG fln P. (Xc)(xG)

40 SemiregularSemiregular action action InIn thisthis section GG isis aa nontrivialnontrivial finitefinite group and A is a nontrivial group of automorphisms ofof GG which which actsacts semiregularly semiregularly on on G. G. Recall Recall this this means means CG CG(a) (a) == 1 1 forfor eacheach aa E E A#. A'.

(40.1) 1IGI G I = 11 modmod CAI.IAl. In In particular particular (IAl,(IAA, IGI)= 1GI)=1. 1.

Proof. AsAs A A is is semiregular, semiregular, each each orbit orbit of of A A on on G# G' isis of of order order I jAl. A 1.

We will wish to apply coprime action, 18.6,18.6, to the representation of A onon G.G. Thus, until 40.7, assume eithereither GG or A is solvable.solvable. InIn 40.7 we prove G is nilpotent, atat whichwhich pointpoint wewe seesee thethe assumptionassumption waswas unnecessary.unnecessary.

(40.2)(40.2) AA is is semiregular semiregular onon eacheach A-invariantA-invariant subgroupsubgroup andand factorfactor groupgroup ofof G.G.

Proof. TheThe first first remark remark isis trivial trivial andand thethe secondsecond followsfollows fromfrom coprimecoprime action,action, 18.7.

(40.3)(40.3) ForFor eacheach pp EE ir(G), n(G), there there is is a aunique unique A-invariant A-invariant Sylow Sylow p-subgroup p-subgroup of G.G.

Proof. ByBy coprimecoprime action,action, 18.7,18.7, the set AA ofof A-invariantA-invariant Sylow p-groups is nonempty andand CG(A)CG(A) is transitive on A. SoSo as CG(A) = 1, 1, the lemma follows.

(40.4)(40.4) ForFor eacheach a EE A, the mapmap gg HH [g,[g, a]a] is is aa permutationpermutation of G.

Proof. [g,[g, a]a] =g-lga, = ,g-lga, So So [g, [g, a]a] _ =[h, [h, a]a] if ifand and only only if if hg-1 hg-' EE CG(a). CG(a). So, So, asas CG(a) = 1, 1, thethe commutatorcommutator map is anan injection,injection, andand hence also aa bijectionbijection asas GG isis finite.finite. 206 Transfer andandfision fusion

(40.5) If A is of even orderorder there there is is a a unique unique involution involution t int in A, A, gt g` = = g-l g-' forfor g E G, and G is abelian.

Proof. AsAs I IAlA I isis even even there there is is an an involution involution t t EE A. A. Let Let g g E E G. G. By By 40.4, 40.4, g g = = [h, t] for some h E G.G. ThusThus gtg` =_ (h-'ht)*(h-1h`)` = h-'hh-`h ==g-l. g-1. Therefore,Therefore, for for eacheach xx EE G, (g-lx_1)t xrxt = X-l,x-1, soSO xgxg = (xg)-`(xg)-' = (g-lx-l)t = g-`x-1g-tx-t = gx. SoSo GG isis abelian. abelian. Finally, ififs s isis anyany involutioninvolution inin A,A, then I've shownshown ss invertsinverts G,G, soso stst EE CA(G)CA(G) ==1. 1. Thus Thus tt is is unique.unique.

(40.6) Let p, q EE 7r(A).n(A). Then (1) IfIf pp is is odd odd then then Sylow Sylow p-groups p-groups of of AA are are cyclic. cyclic. (2) SylowSylow 2-groups2-groups ofof AA areare cycliccyclic or quaternion. (3) SubgroupsSubgroups ofof AA ofof order pq areare cyclic.cyclic. (4) SubgroupsSubgroups of of AA of of oddodd orderorder are are metacyclic.metacyclic.

Proof. ByBy 40.2 and 40.3 we may assume G is an r-group for some prime r, and indeed replacing G by G/c1(G)G/@(G) we we maymay assumeassume GG is is elementary elementary abelian. abelian. Observe that, for B <( A, A, (CG(b): (CG(b): bb E E B#)B') # G as CG(b) = 1 1 for each bb EE B#.B'. So, by Exercise 8.1, m,(A)mp(A)= = 1. 1. HenceHence ExerciseExercise 8.48.4 implies (1) and (2). Part (3) follows from 27.18, while (1)(1) and 39.339.3 implyimply (4).(4).

(40.7) GG isis nilpotent.nilpotent.

Proof. LetLet GG be be aa minimal minimal counterexample counterexample and a anan element of A of prime order. ByBy inductioninduction onon thethe orderorder ofof AA wewe maymay take take AA = = (a).(a). In particular A is solvable, so all lemmas in this sectionsection apply.apply. Suppose q EE 7r(G)n(G) andand 11 # Q is an A-invariant normal elementary abelian q-subgroup of G. By 40.2 and minimality ofof G,G, G/QG/Q isis nilpotentnilpotent soso RQ <9 GG for each Sylow r-groupr-group RR ofof G. In particular ifif rr == qq then R < G. But as G is aa countercounter example there exists r EE n(G)7r(G) withwith SylowSylow r-groupsr-groups of G not normal. By symmetry betweenbetween r r andand q,q, O,(G)Or(G) == 1. AsAs RQRQ 9 a G,G, CR(Q) <5 O,(G)Or(G) == 1,1, soso Z(R)Z(R) is faithful onon Q. By 40.3 we may choosechoose RR toto bebe a-a- invariant; thenthen a a isis faithfulfaithful onon Z(R).Z(R). But now,now, asas CQ(a)CQ(a)= =1, 1, 36.2 supplies aa contradiction. So FF(G) (G) == 1. 1. Now Now ifif HH is is a a proper proper A-invariant A-invariant normalnormal subgroupsubgroup of GG then,then, by minimality ofof G,G, HH is nilpotent, so,so, asas F(G)F(G) = 1, H = 1. 1. As G is not nilpotent, GG isis notnot aa 2-group2-group soso there is an odd prime p EE 7r(G).n(G). Let PP bebe an an A-invariant A-invariant Sylow Sylow p-group of G, G1G1== NG(J(P)) andand G2G2 = CG(n1(Z(P))).CG(S21(Z(P))). AsAs F(G)F(G) = 1, 1, G; Gi is is a a properproper A-invariantA-invariant subgroupsubgroup of G,G, soso G,Gi Semiregular action 207 is nilpotent by minimality ofof G.G. InIn particularparticular GiG; has a normalnormal p-complement,p-complement, so, by the ThompsonThompson Normal p-Complement Theorem,Theorem, GG hashas a normal p- complement. This is impossible as G possesses no nontrivial properproper A-invariant normal subgroups.

Recall from section 35 that a Frobenius groupgroup isis aa finite groupgroup HH which is the semidirect product of a nontrivial group K by a nontrivial groupgroup B with B semiregular on K. KK andand BB areare the Frobenius kernel and Frobenius complement of H, respectively. respectively. Notice that, by 40.7:

(40.8) FrobeniusFrobenius kernelskernels areare nilpotent.nilpotent.

Notice that the lemmaslemmas in this this sectionsection alsoalso givegive lotslots ofof informationinformation aboutabout Frobenius complements.

Remarks. TransferTransfer and and fusion fusion are are basic basic tools tools inin thethe study of finite groups. One can begin to see the power of these tools in somesome of thethe lemmaslemmas andand exercisesexercises in this chapter, but just barely.barely. The proof ofof thethe Baer-SuzukiBaer-Suzuki TheoremTheorem essentiallyessentially comescomes fromfrom AlperinAlperin and Lyons [AL]. Alperin's Fusion Theorem is (surprise)(surprise) due to Alperin [Al].[All. Thompson was the first to prove the nilpotence of Frobenius kernels in his thesis.

Exercises forfor chapterchapter 13 1. Let G be a finite group, pp aprime,a prime, HH <5 GG withwith (IG(IG: : HI,H1, p)=l,p) =1, and g a p-element in H. gg isis extremalextrernal in H ifif ICH(g)IICH(g)lp p == ICG(g)lp.ICc(g)1p (1) Represent G on G/H by by rightright multiplicationmultiplication and letlet Hx bebe aa fixedfixed point of g on G/H. ProveProve the orbit HxCG(g) of CG(g) onon G/HG/H is of order prime to p if and only if gX-'gx ' is extremalextremal inin H. (2) Assume gg isis ofof orderorder p, p, K K < HH withwith H/KH/K abelian, gg EE H H - - K, g is extremal in H, andand each H-conjugate ofof gg extremalextremal in HH isis contained contained in gK. Prove g 4 G(').Gel). 2. Let Let GG be be a a finite finite groupgroup withwith GG == 02 o~(G) (G) andand let let T T E E Sy12(G).Sy12(G). Assume T is dihedral, semidihedral, or Z2"l2n wrZ2,wr12, and and proveprove GG hashas oneone conjugacyconjugacy class of involutions. 3. LetLet GG bebe aa finitefinite groupgroup with G ==02 02(~). (G). Prove thatthat if if m2(G)m2(G) >> 2 then m2(CG(x)) > 22 for for each each involution involution xx inin G. G. (Hint: (Hint: Let Let T T E E Sy12(G). Sy12(G). Prove Prove there is E4E4 =Z U U < TT and and x xGG fl fl CT(U) CT(U) is is nonempty nonempty for for each each involution involution xx of G.) 208 Transfer and fusion

4. LetLet GG bebe aa finite finite group,group, pp a a prime, prime, andand assumeassume aa SylowSylow p-subgroup of GG is thethe modular groupgroup MpJZMp.,of of order order pn p" ## 8.8. Prove (1) OP(G)Op(G) hashas cycliccyclic SylowSylow p-subgroups. (2) IfIf pp == 2 2 then then GG has has a a normal normal 2-complement.2-complement. 5. LetLet G G be be a a finite finite group with quaternion Sylow 2-subgroups. Prove that either G has a normal 2-complement or or there there exists exists K K 9 < H H 5 < G G with with H/K H/K 2 sL2(3).SL2(3). 6. LetLet G G be be a a group group of of order order 60. 60. ProveProve eithereither GG isis solvablesolvable oror GG isis isomorphicisomorphic to the alternating group A5Ag of degree 5. (Hint:(Hint: LetLet T E Syl,(G).Sy12(G). Show Show T T 9a GG or G hashas aa normalnormal 2-complement2-complement or or JG (G : : NG(T)I Nc(T)( = = 5 5 and and kerNG(T)(G) ker~,(~,(G) = 1.) 1.) 14

The geometry of groupsgroups ofof Lie typetype

Chapters 4 and 77 introducedintroduced geometriesgeometries preservedpreserved by thethe classicalclassical groups.groups. Chapter 14 considers these geometries (and related geometries preserved by Coxeter groups) in detail, and uses the representations of the classical groups on their geometries to establish various group theoretical results.results. For example we'll seesee thatthat the finite classical groups L,(q),Ln(q), U,(q),Un(q), PSp,(q),PSpn(q), and PQE(q)PS2n(q) areare simple,simple, withwith aa fewfew exceptionsexceptions whenwhen nn and q are small.small. AlsoAlso Ln(F)L,(F) andand PSpn(F)PSp,(F) areare simplesimple for infinite fields F,F, as are U,(F)Un(F) and PQ,(F)PQn(F) under suitable restrictions onon F.F. If F isis finite of characteristic p, it will developdevelop that thethe stabilizer B of a maximal flag of thethe geometrygeometry ofof aa classicalclassical groupgroup GG over F isis thethe normalizernormalizer of a Sylow p-group of G, andand the subgroups of G containing B are precisely the stabilizers of flags fixed byby B. These subgroups and their conjugates are thethe parabolicparabolic subgroups of G. We say B is the BorelBore1 group of G. It also turns out that to each classical group G therethere isis associatedassociated aa Coxeter group called the Weyl groupgroup ofof G.G. The The WeylWeyl groupsgroups ofof thethe classicalclassical groupsgroups are of type A,,A, Cn,C,, oror Dn. D,. The The structurestructure of of GG isis controlledcontrolled to a large extent by that of its Weyl group: see for example lemma 43.7 and Exercise 14.6.14.6.

41 ComplexesComplexes Before beginning this section the reader may wish to reviewreview thethe discussiondiscussion of geometries in section 3. Section 41 is devoteddevoted to aa relatedrelated classclass ofof objects:objects: complexes. AA complexcomplex is is a a pair pair 6' -6' = = (r, (F, 6')-') where rF is a geometry over some finite indexindex setset II and -6'6' isis aa collection collection of of flags flags of of r F of of type type I. I. TheThe membersmembers ofof 6'67 areare calledcalled chambers.chambers. SubflagsSubflags of chamberschambers areare calledcalled simplices.simplices. Simplices of corank 1 are called walls. A complex is thin if eacheach wallwall isis containedcontained inin exactly two chambers. A complex is thick if each wall is contained in at least three chambers.chambers. Define the chamber graphgraph of 6'67 to bebe thethe graphgraph on 6'6' obtainedobtained byby joining chambers which have a common wall. A path in the chamber graph is called a gallery. AA complexcomplex is said to be connected if its chamber graph is connected.connected. A connected complex (F,(r, -6') 6') inin whichwhich everyevery flagflag ofof rF of rankrank 1 oror 2 isis aa simplexsimplex is called a chamber complex.complex. A morphism a: a:(r, (F, 6') f) -+ -* (A,(A, !%)9) of of complexescomplexes isis aa morphismmorphism r F+ - A of geometries withwith -6'a6'a contained inin !%.9. AA subcomplex subcomplex of (F,(r, -2)6') is a complex 210 The geometry of groups of Lie type

(A,(0, 5) GB) withwith 0 A a asubgeometry subgeometry ofof Fr andand _TGB =_ t' 6' fl rl oA the the set set of of chambers chambers contained in A. The notions of complex, building, and Tits system (which are thethe subjectsubject of this chapter)chapter) come from Tits Tits [Ti]. [Ti]. HoweverHowever the definition ofof a complex I've just just givengiven isis somewhatsomewhat less general than that of Tits [Ti] [Ti] in that under my definition there is a type function defined on simplices inherited from the associated geometry. HoweverHowever TitsTits showsshows howhow toto associateassociate typetype functionsfunctions toto Coxeter complexes andand buildings,buildings, soso inin thethe endend thethe classclass ofof objectobject considered is thethe same.same. The treatment of complexes, buildings, and Tits systemssystems givengiven herehere isis ex-ex- tracted from Tits [Ti] and Bourbaki [Bo], modulo remarks above. Let G G bebe aa group group andand -IF=F= (G;:(Gi: i i EE I)1) a familyfamily of subgroupssubgroups of G. TheThe coset geometry r(G,F(G, g)T) determineddetermined by by GG andand 9 is defined in section 3. Let -t'(G,6'(G, 9) bebe the the complex complex on on F(G, r(G, _,F) F) whosewhose chamberchamber setset (also(also denoteddenoted by -&(G, '(G, ST))F)) consistsconsists ofof the the flags flags SI,,, SI,,, x X E E G, G, where where SI,, SI,x = = (Gix: (G,x: i i EE I).fl.

(41.1) Let G be a group,group, F = (G1: (Gi: i E I)I) a family ofof subgroupssubgroups of of G, G, r F == r(G,F(G, F),ST), andand f3 -' = t'(G,6'(G, ST). F). Then Then (1) GG isis represented represented asas aa groupgroup ofof automorphisms of B-6 by rightright multiplica-multiplication on the cosets in F.r. (2) GG isis transitivetransitive onon thethe simplicessimplices of of 6' -t' of of type type J J forfor eacheach subsetsubset J J ofof I.I. In particular G isis transitivetransitive on on chambers.chambers. (3) G,IG is the stabilizer ofof thethe simplexsimplex Sj Sj ofof typetype J.J. (4) TheThe wallswalls of chambers of 6'-6 are the conjugates ofof Si,,Si', ii E 1,I, under G.G. SitSi, is contained in exactly IGp:I Gi,: GI (I chambers. (5) EveryEvery flagflag inin r ofof rankrank 11 or or 22 isis aa simplexsimplex of 6'.-t'.

The proof is straightforward; the notationnotation isis explainedexplained inin sectionsection 3. 3.

(41.2) Let C be a chamber in a chamber complex 6'-0 andand aa a morphism of B-6 fixing C. Then (1) aa fixes fixes eacheach simplexsimplex contained in C. (2) IfIf -06' is thinthin andand a a isis aa bijectionbijection on on 6', e, thenthen a a == 1.1.

Proof. aa is is a a morphism morphism of of geometries geometries so so it it preserves preserves type.type. HenceHence asas C contains a unique simplex of each type, (1)(1) holds. To prove (2) itit suffices toto showshow aa fixes each chamber adjacentadjacent toto C, since icf3 isis connected.connected. LetLet DD be such aa chamber;chamber; then then D D n fl C c = WW isis aa wallwall ofof C,C, so a fixesfixes WW by (1).(1). As -t'6' is thin, D andand CC areare thethe onlyonly chamberschambers containing W, so, so, asas WW = = Wa Wet= = Da Da rl flCa, Ca, (Ca, (Ca,Da) Da) == (C,(C, D). Now, asas CC == Ca, also D = Da.Da. Complexes 21121 1

Let 6' be a thin chamberchamber complex. A foldingfolding of 6'-t' is an idempotent morphism whose fibres on chambers are all of order 2.2.6' -t' is a Coxeter complex ifif forfor eacheach pair C, D ofof adjacentadjacent chamberschambers there exists a folding mapping C to D. In the remainder of this section assume 6' = (F,(r, -')6') isis a a Coxeter Coxeter complex.complex.

(41.3) Let 40 be a folding ofof 6'.t'. Then (1) 04 inducesinduces aa bijection bijection 4: 0:(6' (P - -6'4) eo) --+-> 6'4. eo. (2) 04 fixes fixes each each membermember of rr4 o andand i&?4. ' j.

Proof. PartPart (2)(2) is is just just aa restatement restatement of of thethe hypothesis hypothesis that that 4) 4 isis idempotent. idempotent. Let C E -64.-6'0. By By hypothesishypothesis therethere isis aa uniqueunique chamberchamber DD distinct from C with DoD@ = = C. C. By By (2),(2), D is not inin 6'4. eo.

For C, D in -'6' let let d(C, d(C, D) D) be be the the distance distance between between CC andand DD inin thethe chamberchamber graph.

(41.4) LetLet 4 0 bebe a a folding folding of of6', t', C C in in 6'4, eo, and and D D in in 6' t' - - 6'4.eo. Then Then (1) IfIf CC = Co,Co, ...... , , C,,C == D D is is a agallery, gallery, therethere existsexists 0 05 < ii < nn withwith Ci EE -6'06'4 and Ci+1Ci+l $6'4.eo (2) IfIf CC isis adjacentadjacent toto DD thenthen D@Do = C.C. (3) Let CiCf = CiCi if Ci isis notnot in in 6'4 eo andand Cf C = @-'(Ci)1(Ci) - - (Ci (Ci} } if Ci is in 6'4.i '. Then CiC; isis adjacent adjacent to to C1+1. C;+l.

Proof. PartPart (1)(1) isis clear.clear. Assume W = C C flrl D D is is a a wall. wall. ByBy 41.3,41.3,4 0 fixesfixes C andand W. Thus W = WO W@ c_E D4,Do, so,so, asas 6' is thin, D@Do = CC oror D.D. AsAs DD isis notnot inin 6'4,-00, Dip D@ = C, C, soso (2)(2) holds.holds. Part (3) is clear ifif neitherneither CiCi nornor Ci+lCi+1 is is in in 6'4, -00, so so let let Ci Ci be be in in 6'4. eo. LetLet U == Ci Ci fl rl Ci+1,Ci+l, vV == O-1(U) 4-'(u) fl rl C,C;, and and E E the the chamber chamber through through VV distinctdistinct from Ci'.ThenCf. Then U U=VcpcEO,so = V@E E4, so EO=C,orE@ = Ci or Ci+1.If Ci+1. If E@EO=Ci+1then = Ci+1 then EE=C+1 = Cf+, is adjacent toto Cf.Cl. If If E@ EO= = CiCi then,then, as as 4-'(ci) 0-1(Ci) == (Cf,(C, Ci},Ci}, we have EE = Ci. Hence VV == VOV@ = U U by by41.2.1, 41.2.1, SOso CfC == Ci+1. Ci+1. Therefore Ci+1@Ci+10 = = Ci # Ci+1,Ci+17 so Ci+lCi+1 $! e6'4, o, and and hence hence (Ci+i (Ci+1)' )' = Ci+i Ci+1. ThusThus C' C,! = = Ci+1 Ci+1 is is adjacent adjacent to Ci+1 == Ci+1Cf+l.

(41.5) Let CC andand C'C' bebe distinctdistinct adjacent adjacent chamberschambers and let 40 andand O'4' be foldingsfoldings with C'4C'O == C and Cep'C4' = = C'.C. ThenThen (1) IfIf D is in 6'4eo thenthen d(C',d(C1, D) = d(C, d(C, D) D) + 1. 1. (2) If D is not inin 6'4 eo then d(C1,d(C', D) D) == d(C,d(C, D)D) -- 1.1. (3) -e4l=eo' _ ee - eo.64.

Proof. AsAs CC isis adjacent toto C', C', Id(C,Id(C, D) D) - - d(C',d(C1, D)JD)I <_< 11 for each D E 6.P. 212 The geometry of groups of Lie type Complexes 213

Assume D is in-00 and C' = Co, ... , C, = D is a galleryoflength d(C', D). there exist foldings ¢ and 0' of 6 with CO = C' and C'O' = C. By 41.5, ¢ Then (Cio:0 < i < n) is a gallery between C = C'O and Do = D as 0 is and 0' are opposite, so by 41.6 they determine an involutory automorphism a morphism. By 41.4 there is 0i < n with CiO = C;+1 = Ci+10. Hence a(O, 0') of 6, called a reflection through C fl C. Let S denote the set of all C = CO, ... , C;_10, Ci+10, ... , CnO = D is a gallery of length at most reflections through C fl C' as C' varies over A(C), and let W be the subgroup n -1 from C to D. Therefore d(C, D) < n = d(C', D), so (1) follows from of Aut(-) generated by S. It will develop that (W, S) is a Coxeter system and the first paragraph of the proof. W = Aut(o). Similarly if D 0 -00 let C = Co, ... , Cn = D be a gallery of length d(C, D). For W E W let 1(w) be the length of w with respect to the generating set S. Define C' as in 41.4.3. By 41.4.3, C' = Co, ... , Cn = D is a gallery, while by 41.4.1 and 41.4.2 there is 0 < i < n with C; in ,°¢, C;+1 not in -6'0 and C;+1O = C. Hence C= C,+1 = Ci+i, so, as in the preceding paragraph, (41.7) Let w = sl... Sn E Wwith Si E S. Then (1) C, CSn, CSn_1Sn, ... , CS1... Sn is a gallery from C to Cw. d(C', D) < n and then (2) holds. (2) W is transitive on -0. Finally let D E'. If D is in 6'', then, by (1) applied to 0' and 0, D is not (3) d(C, Cw) = 1(w). in e o. So io' c_ -C - e o. If D is not in 6'then, by (2) applied to 0 and 0', D is in i/'. Thus i' - eo c io'. So (3) is established. Proof. Let S E S. By definition of S there exists C' E A(C) and foldings Foldings 0 and 0' are defined to be opposite if e o' = .6' - 60. through c fl C' with s = a(0, 0'), and, by 41.6, (C, C') is a cycle of s. Let w; = sn_i+1... sn. We've just seen C is adjacent to Csn_;+1 so Cw; _ CSn_i+l Wj_1 is adjacent to Cwi_,. Hence C, Cw 1, ... , Cwn = Cw is a gallery (41.6) Let 0 and 0' be opposite foldings and define a = a(O, 0'): r -* r by from C to Cw of length n, so d(C, Cw) < n. In particular d(C, Cw) < 1(w). va = v/ if v E ro' and va = vO' if v E ro. Then a is an automorphism of Conversely let C = CO, C1,..., C,n be a gallery of length d(C, Cm) = m. I'll - of order 2 whose orbits on -0 are the fibres of 0. show thereexistr, E S, 1 < i < m, such that Ck = CUk, where Uk = rm-k+1 ... rm. Notice that this establishes the transitivity of W on -0. Pro- Proof. By hypothesis 6' = 6' U -00. By definition of chamber complex, each ceed by induction on m; the case m = 1 has been handled in the first paragraph member of r is a simplex, so r = ro U ro'. By 41.3.2, 0 and 0' agree on of this proof, so take m > 1. By induction Ck = Cuk fork < m. Then Cm is r¢ fl ro', so a is well defined. By 41.3.1, a is bijective on '. If u and v are adjacent to Cm-1 = Cum_, so Cm(um_1)-1 is adjacent to C. Hence there is incident members of r, then as i' is a chamber complex there is C in -' with r1 E S with Cr, = Cm(um-1) 1, so Cum = Crlum-1 = Cm. u and v in C. Then ua, va c Ca c e, so ua * va. Hence a is morphism. By Finally let u = r1... rm and assume Cm = Cw. Then Cu = Cm = Cw so 41.4.1 there exist adjacent chambers C and D with C in io and D not in eo. uw-1 fixes C and hence, by 41.2, u = w. So 1(w) = 1(u) < m = d(C, Cw), By 41.4.2, DO = C and CO' = D. Then Ca 2 = Cq5'q5 = Do = C. Hence, by completing the proof. 41.2.2, a2 = 1. Thereforea-1= a is a morphism, so a is an automorphism of order 2. (41.8) (1) W = Aut(6'). (2) W is regular on e. The element a(0, rp') of 41.6 is called a reflection and is said to be a reflection (3)1 SI = 111; that is there exists a unique reflection through each wall of C. through c fl C' for each pair C, C' of adjacent chambers with C E eo and C' 0'q. Fix a chamber C in i. is defined over some finite index set I ; let m = I Proof. By 41.7, W is transitive on 6', so G = Aut(i') = WGc where GC is for the moment. Notice each of the m subsets of I of order m - 1 determines the stabilizer in G of C. But, by 41.2.2, Gc = 1, so (1) and (2) hold. For each a wall of C, and, as i' is thin, each wall X determines a unique chamber C' C' E A(C) there is S E S with cycle (C, C'), so BSI IA(C)I _ III. Further if with C' fl c = X. Let A(C) denote the set of these chambers; then A(C) is t is a member of S with cycle (C, C') then St E We =1, So S = t. the set of chambers distinct from C and adjacent to C in the chamber graph. Further A(C) is of order m. As -' is a Coxeter complex, for each C' in A(C) (41.9) (W, S) is a Coxeter system. Complexes 213 there existexist foldingsfoldings 4 0 and and 4' 0' of of fi? 6 with C4Co = C'C' andand C'O'C'4' = C. C. By By 41.5,41.5, 04 and 0'4' areare opposite, opposite, so so by by 41.6 41.6 theythey determine determine an an involutory involutory automorphismautomorphism a(0,a(4,4') 0') of of 6, fi?, called called a areflection reflection throughthrough C n C.C'. Let Let S S denotedenote thethe setset ofof all reflections throughthrough C n C' asas C'C' variesvaries overover A(C),A(C), andand letlet WW bebe thethe subgroupsubgroup of Aut(6)Aut(-') generated generated by by S.S. It It willwill developdevelop thatthat (W,(W, S) S) is a Coxeter system and W == Aut(o).Aut(fi?). For Ww EE WW letlet l(w)1(w)be be thethe lengthlength ofof ww with respect to the generating setset S.S.

(41.7) LetLet ww == sl ...... Sn s, EE WW with si EE S.S. Then (1) C, C, Csn, Cs,, CSn-IsCS,-~S,, n...... , Csl Csl ... . . sn. s, isis a a gallery gallery fromfrom CC toto Cw.Cw. (2) W W isis transitive transitive onon -0.fi?. (3) d(C, d(C, Cw)Cw) == 1(w). l(w).

Proof. LetLets S EE S.S .By By definition definition ofof SS therethere exists C' E A(C) and foldings 4,4' through Cc nf' C'C' withwith ss == a(0, a(4, 4'),40, and, and, by by 41.6, 41.6, (C, (C, C')C') is is a a cyclecycle of s. Let w,wi = Sn-i+l s,-i+l ...... sn. s,. We've justjust seen CC is adjacent toto CS,-,+~Csn_i+l soso Cw,Cwi ==

Csn-i+l wi -1 is adjacent to to CwCwiVl. i_ 1 .Hence C, Cwl,Cw 1,...... , Cw,Cwn = Cw Cw is is a agallery gallery from C to Cw of length n,n, soso d(C,d(C, Cw)Cw) 5< n. InIn particular d(C,d(C, Cw) < 5 1(w).l(w). Conversely letlet C C == Co, C1, C1, ...... , CbeC,,, be a agallerygallery of of lengthlength d(C, Cm)C,) = m.m. I'll show therethere exist rir, ES,E S,1 1 5 < i i 5< m,m, suchsuch that that CkCk = = CUk, Cuk, where where Ukuk= = rm_k+irm-k+l ...... r,.rm. NoticeNotice that that thisthis establishesestablishes the the transitivitytransitivity of of WW onon fi?.P. Pro-Pro- ceed by induction on m; the case m = 1 1 has has beenbeen handledhandled in in thethe firstfirst paragraph of this proof, so taketake mm > 1.1. By induction Ck == Cuk forkfor k << m. m. Then Then Cm C, isis adjacent to Cm-1CmP1 = Cum_1 CumPl so Cm(um_1)-1C,(U,-~)-' isis adjacent to C. HenceHence therethere isis r1rl E S with CrlCr, == Cm(um_1)-1, C,(U,-~)-', so Cum == Crium-1Crlu,,-1 == Cm. C,. Finally let u = r1rl ...... rm r, and assume C,Cm = = Cw.Cw. ThenThen CuCu = CmC, == Cw soso uw-1uw-' fixesfixes C andand hence, by 41.2, uu = w.w. SoSo 1(w)l(w) = = 1(u)l(u) 5< m = d(C, d(C, Cw), Cw), completing the proof.

(41.8) (1)(1) WW = Aut(6 Aut(8). ). (2) WW isis regularregular onon fi?.e. (3) I SI SI == II I11; I; that that is is there there exists exists aa unique unique reflectionreflection throughthrough each wall of C.C.

Proof. ByBy 41.7,41.7, WW isis transitivetransitive on 8,', so G = Aut(o)Aut(&) == WGc WGc wherewhere GeGc isis the stabilizer inin G of C. But, byby 41.2.2,41.2.2, GcGc = 1, 1, so so (1)(1) andand (2)(2) hold. For each C' EE 0(C)A(C) there there is is s s EE SS withwith cyclecycle (C,(C, C'),C'), so so ISIIS1 ?2 I IA(C)lA(C)I = III. 111. Further Further if t isis aa member ofof SS withwith cyclecycle (C, (C,C') C') then then st St E E Wc We = = 1,1, soso s s == t.

(41.9) (W,(W, S)S) is is aa CoxeterCoxeter system. system. 214 The geometrygeometry of groups of Lie type

Proof. It sufficessuffices to establish establish the exchange conditioncondition ofof 29.4.29.4. SoSo let w = slS1 ...... sns, E W withl(w)with l (w) ==n n andsands E E S S with withl(ws) l (w s) <5 ll(w). (w ). LetCk Let Ck = = C~n-k+lCSn_k+l ...... Sn forforl 1

--9=(C=Co,-LC7 = (C = Co, C1,C1,...... ,Cn=Cw) , Cn = Cw) is a gallery andand d(C,d(C, Cw)Cw) == l(w)1(w) 2> l(ws)l(ws) == d(C, Cws). Let 0@ bebe aa foldingfolding withCs@with Cso = C.Nowd(Cs, Cw) Cw) = = d(C, d(C, Cws) Cws) < 5d(C, d(C, Cw), Cw),so, so, by by41.5,Cw 41.5, Cw $ [email protected]. Then byby 41.441.4 applied applied to tog $? thereexiststhere exists i, i, 11 5 < i <5 nn withwith Ci in&@,in Co, Ci+1Ci+l not in &@,Co, andand Ci+1/.Ci+i@ == Ci. Ci. By By definition definition of of s,s, Cis Cis = = Ci+1, Ci+l, SO SO CSn-i+lCS,-~+~...... SnS S,S == CisCi s =Ci+l= Ci+1 == CS,-~.Csn_i .... .sn, sn,andhence,by41.8.2,sn-i+l.. and hence, by 41.8.2, Sn-i+1 . . S,SSnS == Sn_is,-i ..... Sn,. S,,SO SO the exchange condition isis verified.verified.

(41.10) Let (G,(G, R)R) bebe aa Coxeter Coxeter system, system, R R= = (ri: (r1: i Ei EI), I), Gi Gi = = (ri:(ri : j j# 0 i), and F = F(G, R) == (Gi:(Gi :i iE E I). I). Then Then &(G, P(G, 9) J) isis aa Coxeter Coxeter complex, complex, G == Aut(-'(G,Aut(&(G, 1F, F)), )), and R is the set of reflections through thethe wallswalls ofof the chamberchamber C = (Gi:i(Gi: i EE I).I}.

Proof. Adopt the notation ofof sectionsection 3.3. ByBy 29.13.3,29.13.3, Git Gi, = = (ri)(r1) (recall (recall i'= i'= I -- {i}) {i}) and and GI = 1.1. So,So, by 41.1.4,41.1.4, L3 _T = = B(G, -'(G, 9) J) is thin. AlsoAlso G G = = (R)(R) == (Gi,: i i E I),I), so,SO, by Exercise 14.3,14.3, _TL3 isis a chamber complex. By 41.1.2, G isis transitive onon g- - and,and, by by 41.1.3,41.1.3, GIGI is is the the stabilizer stabilizer ofof thethe chamberchamber C, so, as GI = 1, 1, G G is is regular regular onon -T.L3. InIn particularparticular G is faithful on g-9 so, by 441.1.1, 1.1.1, G is a subgroup of Aut(g).Aut(_9). Also G is transitive on 5B_q so itit remainsremains onlyonly toto show for each wall SitSi, ofof CC that therethere isis aa foldingfolding @0 through SitSi, withwith CC inin [email protected]. Let r = ri ri andand definedefine 0:@: G +-+ G by go = g if l(gr) = l(g) + 1 go = grif 1(gr) = 1(g) - 1.

As G,GI == 1 1we we can can identify identify G G withwith _TL3 via gg H CgCg and and regard regard 0@ as as a a function function from L3_T into into g. -9.Claim Claim that that if if D, D, D' D' EE g-9 andand DD nfl D'D' is a wall ofof typetype j'j' then DoD@ fln D'oDl@ isis alsoalso aa wall of typetype j'.j'. Now DD == Cg and D' = Crjg Crjg for for some some g E G, so wemustwe must show Cg@Cgo flfl C C(r;g)@ (r j g)o is a wallwall ofof type type j', j', so it sufficessuffices toto showshow (rjg)4(rig)@ ==rj(g4)foreachg rj(g@)for eachg EE G.G.Thisisclearifl(rjgr)-l(rjg)=l(gr)-1(g). This is clear if 2(rjgr) - 2(rjg) = 2(gr) - 2 (g). So, by symmetry betweenbetween rjg rjg and g, we may assume l(gr)l(gr) == l(g) -- 11 andand l(rjgr)l(rjgr)=l(rjg) = l(rjg) ++ 1.1. Thus g@ go=gr= gr andand (rjg)@(rjg)o=rjg.= rig. Let gg=sn...si = s, . . . sl with with l(g) == n n and and sk sk == rik. ri,. AsAs ll(gr) (gr) <5 l l(g),(g), the the ExchangeExchange Condition says smsn...... s1= sl = sm-1sii-1 ...... sir slr for for some some m. m. Thus Thus g g= =S, s,... .Sm+lSm-1 . .sm+lsm-1 .. . . slr,sir, soso without loss r = si, s1, HenceHence gog@ == gr gr = = Sn s, ...... S2. s2. Next rjrjgrgr = rjs,rjs, ...... s2 s2 isis ofof lengthlength at at most nn = 1(g)l(g) soso l(r;g)l(rjg) = l(rjgr) l(rjgr) - -1 <1 51(g). l(g). Then, Then, by by the the Exchange Exchange Con-Con- dition, r rjs,j Sn . . . sk+lSk+i == Sn s, ...... Sk sk for some k.k. IfIf kk #0 1 we may taketake s,sn = = r;. r. Buildings 215

But then rjgrr j gr == s,-1sn_ 1...... s2 s2 is is of of length length n -n -2 2 <

Gk'Gkl == (G{k,j}' (G{k,jY: ij EE k')k') for eacheach kk E I, ifif IIIIIl >> 2. 2. Hence Hence byby Exercise 14.8, thethe map map Gkg Gkg H H Gk(g0)Gk(g4) is a morphismmorphism ofof the the complex complex (r(G, (F(G, 9). (). -T). g). II alsoalso writewrite this map asas 4.0. Next observe thatthat forfor eacheach g g EE G,G, g42g02 = go.g4. So So 0 4 is is an an idempotent idempotent mor-morphism. Also ifif CgCg E g4-90 then 4-'(Cg)0-1(Cg) == (Cg, (Cg, Cgr} Cgr] so so 0 4 is is a afolding. folding. Finally Finally C and Cr areare the chambers through SitSi, andand Cr4 Cro = C C so so 0 4 is is a a folding folding throughthrough

Sij.Si'. By construction r isis thethe reflectionreflection through through Sip; $1; that is the fibres of 40 onon -9g are the orbits of rr on -9.g. So So RR isis thethe setset ofof reflectionreflection through the walls ofof C and hencehence Aut(g) Aut(-T) = = (R)(R) == G.

(41.11) The map (G, R) H '(G, &(G, -14-(G, F(G, R)) R)) isis aa bijectionbijection between the set ofof all Coxeter systems and the set of all Coxeter complexes (up toto isomorphism).isomorphism). F(G,4-(G, R)R) isis thethe familyfamily ofof maximalmaximal parabolicsparabolics defineddefined in 41.12 andand the inverse of the correspondence is is & 6 i-+H (Aut(o),(Aut(&), R)R) wherewhere RR isis thethe setset ofof reflectionsreflections through the walls ofof somesome fixedfixed chamberchamber of of -C.8.

Proof. IfIf (G,(G, R) R) is is a a Coxeter Coxeter systemsystem then by 41.10,41.10, (G,(G, R)pR)cp = = 8(G,-6(G, F(G, R)) is a Coxeter complex, whilewhile ifif &6 isis aa CoxeterCoxeter complex then by 41.8 and 41.9, &@6'* _= (Aut(o), (Aut(&), R) R) is isa Coxetera Coxeter system. system. By By ExerciseExercise 14.2.3,14.2.3, &@pPicp =Z -6', 8, whilewhile by 41.10, (G, R)p@R)cpi = = (G, (G, R), R), so so cp p and 1@ are are inverses inverses ofof eacheach otherother and the lemma holds.

42 BuildingsBuildings A building isis aa thickthick chamber chamber complex complex 93 0 =_ (F,(r, ') &) together together with with a aset set sad d of subcomplexes ofof 93,0, called called apartments, apartments, suchsuch that that the the following following axioms axioms areare satisfied: (B 1)1) The apartments are thin chamber complexes.complexes. (B2) EachEach pair of chambers ofof 930 isis contained contained inin anan apartment.apartment. (B3) IfIf A andand B are simplices ofof 93 0 containedcontained in apartments E and E', thenthen there exists an isomorphism of E withwith E'E' whichwhich is is the the identity identity on on A A U U B.B. In sections 13 and 22 a geometry r waswas associatedassociated toto eacheach classicalclassical groupgroup G (occasionally(occasionally subjectsubject to restrictionsrestrictions onon thethe field),field), andand itit waswas shownshown thatthat GG is flag transitivetransitive on r.F. Now B(r)-6'(F) isis a thickthick chamberchamber complexcomplex and inin ExerciseExercise 14.5 a set of apartmentsapartments sad is is defineddefined whichwhich admitsadmits thethe transitivetransitive actionaction ofof G, andand it isis shownshown that (6'(F),(&(r), s1)d) isis aa building building admittingadmitting G as as aa groupgroup ofof automorphisms. This representation is then used to study G in section 43. 216 The geometry of groups ofof Lie typetype

But before all that let's take a closer look at buildings. So for the rest of this section letlet B A = (F,(r,6') ') be a building withwith apartmentapartment set set sf. a.

(42.1) EachEach pairpair ofof apartmentsapartments isis isomorphic.isomorphic.

Proof. LetLet EXi, i = i =1, 2,1,2, be be apartments apartments and and pick pick chambers chambers CiCi E C,E. . By By axiomaxiom I B2, there is an apartment EC containingcontaining C1C1 and and C2,C2, and by axiomaxiom B3, E1C1% CE=E2. % C2.

Given aa pairpair (C,(E, C) with EC anan apartmentapartment and C a chamber in EC definedefine a map p = p(E, p(C, C): C): Fr - 4E byC byvp up = vo,= v4, wherewhere C, C,v are v are contained contained in in some some apartment apartment E'C' andand 0:4: E'--> C' + E C is is an an isomorphism isomorphism which which isis the identity on C. There are a number ofof pointspoints toto bebe made here. First by B2 there does indeed exist an apartment C'E' containing C and v, and by B3 the map 04 exists.exists. MoreoverMoreover if v, C are containedcontained inin anan apartment apartment 0 0 and and $: 1/f: 0 04 - EC isis an an isomorphism trivial on C then by B2B2 therethere existsexists anan isomorphismisomorphism a: a: C'E' + 0 trivial on C UU {v). Then (4-')a$(0-1)uVf = ,B,!? E Aut(C)Aut(E) isis trivialtrivial onon C,C, so, so, byby 41.2.2,41.2.2, ,B,!? = 1 1.. Thus a$a = 0,4, so v4v4 = = va$ vat = vv$.. What all thisthis showsshows isis that that p(C, p(E, C) isis independent of the choice ofof C'E' and 4,0, so so in particular p(C,p(E, C) C) isis well well defined.defined. As aa matter of fact it shows therethere existsexists aa unique unique isomorphism isomorphism 4: 0: C' E' +-> EC trivialtrivial on C. Let's seesee next that:

(42.2) p = p(E, p(C ,C) C) isis a a morphism morphism of B onto CE with p-'p-1(C) (c) = C. C .Moreover Moreover if E'C' isis anan apartmentapartment containing C then p: C'E' + EC isis thethe unique isomorphism of C'E' withwith EC trivialtrivial on C. In particular p is trivial onon C.E.

I've alreadyalready made the last observationobservation of 42.2. The remaining partsparts of the lemma are an easy consequence ofof thisthis observationobservation andand thethe buildingbuilding axioms.axioms. Given simplices A and B of .M93 define d(A,d(A, B) to be the minimal integer n such thatthat therethere exists exists a a gallery gallery 9 7 = = (Ci: 00 5< i <5 n)n) ofof AB with with A A containedcontained inin Co andand BB containedcontained inin C,.Cn. Then Then 9 C isis saidsaid toto bebe aa gallerygallery between between A A and and B B ofof length n.n .

(42.3) (Rainy'Day Lemma)Lemma) LetLet CE be an apartment, C a chamber of E,C, and and X a simplex ofof C.E. Then CE containscontains every gallerygallery of of B A between XX and CC of length d(X, C).

Proof. Let 9 == (Ce:0 (C,:0 <5 i i< 5 n) n) be be such such a a gallery gallery and assume 9' is is not not con-contained inin C.E. Then Then therethere exists anan i withwith C;C, notnot contained contained in EC andand Ci+iC,+1 contained inin C.E. Then wW = = Ci C, f1 n CZ+i C,+' isis aa wallwall ofof C,+1Ci+' so there is a chamber T 0# CZ+1C,+, of of C E with with W W E c T.T. Let p = p(E,p(C, T). T). AsAs pp isis a amorphism morphism trivialtrivial Buildings 217 on E,C, CipCip isis aa chamberchamber of EC containingcontaining W, so, as CE isis thin,thin, C;Cip p == T oror Ci+1.C,+1. AsAs p-l(~)p-1(T) == T, CipC,p = C,+1.Ci+1. But 9pqp is is a agallery gallery between between X = Xp Xp and Cp = CC ofof lengthlength n == d(X, d(X, C),C), so, so, asas C,Cip p == Ci+1 Ci+1 == Ci+lp, ci+l~, c COP,o p,---..., , Ci-1C1_1 p, p, Ci+1Ci+l p, p, . ..., . , C,,p is a gallery betweenbetween XX andand CC ofof lengthlength n n - - 1,1, contradicting nn == d(X, C). C).

(42.4) Let C and D bebe chambers,chambers, X a subsetsubset of C, and EC anan apartmentapartment containing C. Let pp = p(E,p(C, C) C) andand q 9 a agallery gallery of of length length d(X,d(X, D)D) between between X and D in BA. . Then Then d(Xp, Dp)Dp) = = d(X, d(X, D)D) and and qp 9p is is a a gallery gallery of length d(Xp,d(Xp, Dp) Dp) between XX = XpXp andand Dp.

ProoJProof. By B2, C UU D isis containedcontained in an apartment C'E' and, byby 42.3, 9C is contained inin C'.V. By 42.2, p:p: C'E' -+ E Cis isan an isomorphism isomorphism so so the the lemma lemma holds.

(42.5) Each apartmentapartment isis aa CoxeterCoxeter complex.complex.

ProoJProof. LetLet CC andand C' C' bebe adjacent adjacent chambers chambers in in anan apartmentapartment E.C. We We mustmust showshow there existexist oppositeopposite foldings foldings 4 0 and and 4' 0' of of C E through through B B = = C f1n C'C' withwith C'4 C'O == C and CO'C4' = C'. C'. For For the the first first time time thethe hypothesishypothesis that AB is is thick thick isis used. used. Namely as 2BB is thick there is a chamber C* distinct fromfrom CC and C' through B. Let C1E1 bebe anan apartmentapartment containing containing C C and and C*, C*, pl p,= = p(C1, p(E1, C), C), p:! P2 = = p(C,p(E, C'),C'), and and 04 = pipe: pl p2: EC + E.C. AsAs thethe composition ofof morphisms trivial on C, 04 isis aa mor-morphism trivial on C. Moreover Moreover applyingapplying 42.442.4 toto p,pl andand p2 pz wewe obtain:obtain:

(42.5.1) If 9 is is a a gallery gallery ofof lengthlength d(B, T) == n n between between BB andand aa chamberchamber T of E,C, thenthen d(B, TO) T4) == n n and and 9O 94 is is a agallery gallery of of length length n n between between BB andand TO. T4.

Next C'4C'O is aa chamber chamber containing containing B, B, so so C'4 C't = C C oror C'.C'. As As pl : CE + E1C1 is an isomorphism trivialtrivial onon C, C, C1plC'p1 = = C*. Then C'4C'O == C*C*p2 p2 #0 C'C' asas (p2)-1(C')(,3:!)-' (c')= C'. So C'q5C14 == C. Similarly therethere is is aa morphismmorphism 4' 0' of of CE trivialtrivial on on C',C', withwith C4'CO' == C',C', and and satisfying 42.5.1. Let D bebe a a chamberchamber inin EC andand (C,:0(Ci:0 <5 ii <5 n) n) == C 9 a agallery gallery ofof length length d(B, D) in in EC fromfrom BB toto D.D. NoticeNotice Co = CC oror C'.C'. ClaimClaim

(42.5.2) If Co = CC thenthen (i) 04 and and 0'04'4 are are trivial trivial onon D,D, andand (ii) D4'Do' #0 D.D. Assume otherwise and choose acounterexamplea counterexample withn with n minimal. minimal. As As C4' CO' = = C',C', C'OC'4 = C, C, andand 04 isis trivialtrivial on C, it follows thatthat n n >> 0. Now (Ci:(Ci:0 0 5 < ii < n) is 218 The geometrygeometry of groupsgroups of Lie type a gallery fromfrom BB toto CnV1Cii_1= -= E E of of length length d(B, d(B, E) E) = = nn -- 1, 1, soso (i)(i) andand (ii) hold for EE byby minimalityminimality of of n. n. Let Let A A= =D Dil flE. E. As As A A= =Aq5 Aq c c Dq5, Do, Dq5Do = DD or E. By 42.5.1,42.5.1, n n = = d(B,d(B, Dq5),DO), so,so, as as d(B, d(B, E) E) = = nn -- 1, 1, itit followsfollows that Dq5DO == D. Similarly, asas q50 andand 4'0' both satisfy 42.5.1,42.5.1, soso does does 4'4, 0'0, and thenthen asas 4'40'0 is trivial on E the same argument showsshows Dq5'q5DO'O == D. So (i) isis established. Notice Notice g' -9' = = Hq5'6/'0' isis gallerygallery ofof length length n n = = d(B,d(B, Dq5')Do') with Coq5'Coo' == CO'Cq5' = C', C', so, so, by by symmetrysymmetry between CC and C', 0'4' isis trivialtrivial onon Do'.Thus,Dq5'. Thus, if ifD D=Do',A=Aq'cDflEq',so= Dd', A = Aq5' c D n Eq5', so Eq5'Eq'=Dsince = D since Eq5'Eq'#E. # E. This isis impossibleimpossible as as n n- - 1 1 = = d(B, d(B, E)E) == d(B, Eq5')E/') while while d(B,d(B, D)D) = = n. n. This This completes the proof of 42.5.2. Now, byby 42.5.2,42.5.2, either either Co Co = = C C andand DD == Dq5,Do, oror CoCo = C',C', in in which which casecase . Coq5Coo == C and,and, applyingapplying 42.5.2 to gq5,qq5, wewe getget (Dq5)q5(DO)o == Do.Dq5. InIn eithereither casecase Dq5Dq = Dq52, Dq52, SOso q50 is idempotent. AlsoAlso if if DD == DoDq5 thenthen byby 42.5.2,42.5.2, CoCo ## C', so CoCo = = C,C, andand then then again again by by 42.5.2, 42.5.2, D #D Dq5'Do' and D = Dq'q.Dq5'4. Moreover Moreover if TT# # D D is is a a chamberchamber of of C E with with Tq5 To = = D D then then T T# 0 Tq5, To, so so T T= = Tq5' To' = =Too'. Tq5q5'. Thus T = Do' Dq5' andand so so {D, {D, Do'}D@'} isis the the fibre fibre ofof DD underunder 0.q5. HenceHence 0q5 is a folding of EC throughthrough B with COC'q5 == C, C, and and 0' q5' isis the the opposite opposite folding. folding. TheThe proof of 42.5 is complete.

43 BN-pairsBN-pairs andand TitsTits systems A Tits systemsystem is is aa quadruplequadruple (G, (G, B, B, N, N, S)S) suchsuch that that GG isis a group, B and N are subgroups ofof G, S is a finite collection ofof cosetscosets ofof BB ilfl N in N, andand thethe following axioms areare satisfied:satisfied:

(BNl)(BN1) G=(B,N)andH=BnNaN. G=(B,N)andH=BIN

By convention, w wB B = = nB andand B'BW == B'1, Bn, for for n n a a representative representative of the coset w = Hn EE W.W. As H

(43.1) Let 933 == (I', (r, t)B) be be a building a building with with apartment apartment set set .1 d and and assume assume GG isis a group of automorphisms of of -M93 transitive on

S2cZ={(C,E):CE = {(C,C): C E 8,6,EC EE d,W,CCE}.C C C}. BN-pairs andand TitsTits systems systems 219

Let (C, C)E) E 0,52, B B == Gc, Gc, and and NN == NG(E). NG(C). Then Then (1) TheThe representationrepresentation of N on EZ mapsmaps NN surjectivelysurjectively onto Aut(Z)Aut(E) with kernelkernel H = B B flil N.N. (2) (W,(W, S) S) is is a a Coxeter Coxeter system, system, where where W W = = N/H N/H and and S S is is the the setset of reflections inin WW throughthrough thethe wallswalls ofof C.C. (3) (G,(G, B, B, N, N, S) S) is is a a Tits Tits system. system.

Proof. ByBy 42.5, 42.5, E C is is a aCoxeter Coxeter complex, complex, while while byby hypothesishypothesis NN isis transitive transitive onon the chambers ofof E,C, so so (1) (1) follows follows from from 41.8. 41.8. ThenThen (2) (2) followsfollows from from (1)(1) andand 41.9. Lets E E S S and and ww EE W. Then C and Cs are adjacentadjacent andand EE == CC flil CsCs is is a a wallwall ofof C.C. EachEach b E BB fixesfixes E,E, soso EE EC_ Csb Csb and and Ew Ew CCsbw.Csbw. Let Let L+9 6' == (Ci: 00 5< i <5 n) n) be be a agallery gallery of of lengthlength d(C,d(C, Ew)Ew) fromfrom CC toto Ew,Ew, andand E'C' anan apart- apart- ment containing C and Csbw.Csbw. By By thethe RainyRainy DayDay Lemma,Lemma, 42.3, L+9 cE E Z flil V.C'. Let a E GG with (C,(C, Z')aE')a == (C, C).E). Then Ca = C,C, soso aa EE B. CiC1 andand C areare thethe only only chambers ofof CE or E'Z' containingcontaining CC fln CIC1 so,so, asas aa fixesfixes C,C, it it alsoalso fixesfixes C1. Proceeding by inductioninduction onon k,k, aa fixes CkCkfor for each each 0 0 5< k <5 n.n. ThusThus a a fixesfixes EwEw CC,,.C,. But EwEw C Csbw EC C'.V. So EwEw = = EwaEwa c C CsbwaCsbwa c C C'a E'a == E,C, soso CsbwaCsbwa == Cw Cw or or Csw. Csw. Hence Hence sbw sbw e EBWB BwB or or BswB,BswB, soso BN3BN3 isis estab- estab- lished.lished. As 393 is is thick, thick, there there exists exists a a chamberchamber DD throughthrough E distinctdistinct from C and Cs. Let 98 be anan apartmentapartment containingcontaining C C and and D, D, and and let let g gE eG G with with (C, (C, C)g E)g == (C, 0).8). Then g E BB and CC and CsgCsg are thethe chamberschambers through through E E in in Cg Eg == 8,0, so Csg == D # Cs.Cs. Thus sgs $ B, so BN4 is established and the proof is complete.

ItIt will developdevelop laterlater in thisthis sectionsection thatthat thethe converseconverse of 43.1 also holds; that isis each Tits systemsystem definesdefines aa building.building. Notice thatthat byby 43.1 andand Exercise 14.5 14.5 the the classicalclassical groups groups possess possess a a BN-pair. BN-pair. The BN-pair structure will be used to establish various facts about the classical groups. In the remainder of this section assume (G,(G, B, N, S) is a Tits systemsystem and letlet HH == B B fl il NN and and W W = = N/H. NIH. We We say say B B is is the the Borel Bore1 subgroup subgroup of G, H is is thethe CartanCartan subgroupsubgroup ofof G,G, and and WW is is the the Weyl Weyl group of G. Let 1l be the length functionfunction onon WW defineddefined byby thethe generating setset S.S.

(43.2) IfIf u,u, ww Ee W with BwBBwB == BuB then uu = w.

Proof. LetLet mm = 1(w) l(w) <5 1(u)l(u) and induct onon m.m. IfIf mm == 0 then w = 1,1, soso uu eE B flil N = H; H; that that is is u u = = 1 1(remember (remember the the convention convention on W).M. SoSo taketake m >> 0.0. Then ww = sv with ss EE SS andand 1(v)l(v) = mm - 1. 1. Now Now svBsvB c BuBBuB by by hypothesis, hypothesis, so vB Ec sBuBsBuB c2 BuB BuB U U BsuBBsuB by BN3. ThenThen BvBBvB == BuB oror BsuB,BsuB, so,so, byby 220 The geometry of groups of Lie type

induction onm,on m, v v = = uu orsu.or su. As As l(v) 1(v)= = m m - - 1 andmand m 5< l(u),1(u), v v # $ u, so vv = su and hence ww == sv == u u as as desired. desired.

(43.3) LetLet Ww E WWand and sS E S. (1) IfIf l(sw)l(sw) > 2 1(w) l(w) thenthen sBwsBw C2 BswB.BswB. (2) If If l(sw)l(sw) < 5 l(w) l(w) thenthen sBw sBw fl f' BwBBwB isis nonempty.nonempty.

Proof.Prooj FirstFirst the the proof proof ofof (1),(I), which which isis againagain by induction on m = 1(w). l(w). If mm == 00 then w = 1 1 and and the resultresult isis trivial, trivial, so so take take m m > > 0. 0. Then Then w w = = ur,ur, withwith rr E S and l(u)1(u) == mm - - 1. If l(su) < m - 1 1 thenthen l(sw)l(sw) = l(sur) < 5 l(su)l(su) ++ 1 1 << m,m, con-con- . trary toto hypothesis,hypothesis, soso Z(su) l(su) > m - 1 1= = 1(u). l(u). Then,Then, by by induction induction on on m,m, sBusBu Cs BsuB. HenceHence sBwsBw == sBursBur C2 BsuBr BsuBr C_ 2 BsuBBsuB UU BsurBBsurB by BN3. Also, by BN3, sBw C2 BwBBwB U BswB.BswB. If If BwBBwB = = BsuB BsuB then, then, by by 43.2, 43.2, w w = = su.su. Hence u = swsw soso mm -- 1 1= = 1(u) l(u) == l(sw) l(sw) > 21(w) l(w) == in, m, a acontradiction. contradiction. Similarly Similarly if BwB == BsurB thenthen ww == Sursur = sw, sw, a a contradiction. contradiction. ItIt followsfollows that sBwsBw C5 BswB, as desired. So (1) is established and it's on toto (2).(2). By By BN3BN3 andand BN4,BN4, BSBS flf' BsB is nonempty, soso sBsusBsu n fl BsBu BsBu is is nonempty nonempty for for each each u uE E W. W. Take Take u u = = sw.sw. Then l(su) = 1(w) l(w) >2 l(sw)l(sw) == 1(u), l(u), so, so, by by (1),(I), sBu sBu C 2 BsuB BsuB == BwB. BwB. Therefore,Therefore, asas sBsu nfl BsBu isis nonemptynonempty andand ww = su,su, (2) (2) isis established. established.

(43.4) Let ww = sisl ...... Sns, EE WW withwith Sisi E S. Then (1) IfIf n = 1(w) l(w) then siBsiB 2C (B,(B,B"-') B'_1) D2 BwB for for 1 1 5 < ii 5< n. (2) For each u E W, BwBuBBw BuB c5 UiEO Ui,, Bsi,Bsil ...... SiruB, siruB, where A consists of the sequences i = i il,l, ...... , ,i i,t withwith iij j EE {(1, 1, ...... , , n n} j and ili t < i2i2 < . . . < i,..i,. (3) IfIf Soso E SS with 1l(s0w) (sow) 5 < l(w) 1(w)= = n n then there exists exists 1 5< k 5< n withwith SOS1S0Sl ...... Sk-1Sk-1 == S1 ...... Sk. Sk.

ProojProof. PartPart (2) follows fromfrom BN3BN3 by inductioninduction onon n. NextNext partpart (1). Let wiwi=si =si ...... slws1w forfor 115 < i <5 n,n,n n ==l(w), 1(w), andand wowo = w.w. ByBy hypothesishypothesis l(wi+i)l(wi+l) < < 1(wi) l(wi) so,so, byby 43.3.2, si+iBwisi+lBwi fl f' BwiBBwiB is is nonempty. nonempty. Hence si+,B C BwiBwi-1B C (B, B"''.

-1 By inductioninduction on on i, i, sj sj E E (B, (B,B"-') B") == X X forfor each each jj <5 i, i, soso si+i si+l cE (B, (B, B-1_')BW~ ) 5< X. Hence ww == sl ...... sn s, cE X X andand then then BwBBw B C5 X. X. So So (1) (1) holds. holds. Similarly if l(s0w)l(sow) 5< 1(w)l(w) then, by 43.3.2, soBSOB 2 C Bw~w-'B BwBw-1B so, by (2),(2), BsOBBsoB = = ~xw-'~Bxw-1B for xX == si,Sil . ...sir. .sir and some i cE A.A. Hence,Hence, by 43.2, so = xw-'.xw-1. But l(xwP1)l(xw-1) > l(w-1)l(w-') - 1(x) l(x) > 2 n - r, r, so, so, as as l(so)l(so) = = 1,1, rr = nn - 1 1 and and x == S1 sl .... . Sk-1Sk+1 .sk-lsk+l .. . . SnS, for some 1 5< k <5 n.n. ThereforeTherefore (3)(3) holds.

(43.5) (W,(W, S)S) is is aa CoxeterCoxeter system.system. BN-pairs andand TitsTits systemssystems 221

Proof. ThisThis is is immediate immediate from from 43.4.343.4.3 andand 29.4.29.4.

(43.6) SS isis thethe setset ofof ww EE W*W' suchsuch thatthat BB U U BwBw B isis aa group.group.

Proof. ByBy BN3, BN3, B B U U Bs BsB B is is a a group group for fors S E S.S. Conversely if w = s1 sl ...... S"sn E W withwithsi Si E SandS and nn = l(w) l(w) > > 0 0then, then, by by 43.4.1, 43.4.1, Si si EE (B,(B, w), w), so so ifif BB UU BwB isis aa group then,then, byby 43.2,43.2, sisi == w for each i.

Let S = (Si:(si: i i EE I) I) andand for for J J 5C II let Pj = = (B, (B, sj: sj: j j EE J). J). TheThe conjugates conjugates of the subgroups Pj,Pj, J C5 I, I, are are called called parabolic parabolic subgroupssubgroups of G. Recall WjWj =_ (sj: j EE J)J) isis a aparabolic parabolic of of W. W.

(43.7) (1) PiPj = = BWjB. BWjB. In In particular particular GG == BNB. BNB. (2) The map J HH Pj Pj is isa bijectiona bijection of of the the power power set set ofof II with with thethe setset of all subgroups of G containing B. (3) PJUKPJUK = = (Pr,(PJ, PK)- PK). (4) PJnKPJ~K = PjPJ nf' PK.PK (5) IfIf g EE GG withwith BgB9 (< PJPj then then g EE PJ.Pj. In In particular NG(PJ)NG(Pj) = PiPj is is con- con- jugate in G to PK onlyonly if if J J = K.K. (6) wjWr=WnPj. = w n Pj.

Proof. To prove (1) it suffices toto show uBwuBw 2 C BWjB BWjB for for each each u, u, w w EE Wi.Wj. Proceeding byby inductioninduction on on l(u) l(u) it it suffices suffices to to show show this this for for u u= = sj, sj, j j EE J. But this is justjust BN3. So (1)(1) is established. LetB

Exercises 4.94.9 and 7.8 show that, if G isis aa classicalclassical groupgroup overover aa finitefinite fieldfield of characteristic p,p, then thethe CartanCartan groupgroup HH is a Hall p'-group of the BorelBore1 group B and B possesses a normal p-complement U U with with UU E E Syl Syl,(G). p (G). FurtherFurther B == NG NG(U), (U), soso the the parabolics are the subgroups of G containingcontaining the normalizernormalizer 222 The geometrygeometry of groups of Lie type of a Sylow p-group of G. Indeed Indeed itit cancan bebe shownshown thatthat the maximalmaximal parabolics P;',Pit, i EE I,I, are are the the maximal maximal subgroups subgroups ofof GG containingcontaining the Sylow p-group U, and further every p-localp-local is contained in some maximal parabolic (cf. section 47). Some more more notation. notation. For For i E i IE let I let Gi G; = =Pi, Pi, (recall (recall for for J c J IC that I that J' =J' =I I- - J)J) and let F =_ (G1: (Gi: i EE I)I) be be the the set set of of maximalmaximal parabolics containing B. Form the geometry r(G,F(G, _1F),F), the the subgeometry subgeometry Z E = = {Giw:{G;w: i i E I,I, w E W},W), and the complex G8A == -'(G, &(G, OT). 9). LetLet C C be be the the chamber chamber {G1: {Gi: ii E I);1); thenthen & = {Cg: gg E G)G} is the set of chamberschambers of of 93, 3, andand wewe cancan alsoalso regard EZ asas thethe complex (Z,(E, & fln E).Z). LetLet szld = {Eg:{Zg: g EE G},G), soso thatthat da is is a acollection collection ofof subcomplexes of of 93. A. Write B(G,3 (G, B,B, N,N, S)S) forfor thethe pairpair (A,(a, sa?).d). FinallyFinally let . UiUl = W',W~J, 6'8 = = (U1: (Ui : i EE I)I) and and formform thethe complex &(W,-'(W, d').8).

(43.8) (1)(1) TheThe map map G1 Gi w i-±H UiUU w wis isan an isomorphism isomorphism of ofthe the complex complex (Z (E, , Z En me) 8) with&(W,with '(W, 8).e). (2) GG isis representedrepresented as a group of automorphisms of of % 3 byby rightright multipli-multiplication; the kernel ofof thisthis representationrepresentation is is kerB kerB (G). (3) A(G,B(G, B, B, N, N, S) S) is is a a building. building. (4) G is transitive on on C2 S2 = = {(D, {(D,0): 9): D D E E',&, 90 EE s1,d, DCD 2 9} 0). (5) B == Gc Gc and and (BE)N (Bc)N = = NE(E) NG(Z) are are the the stabilizers stabilizers inin G G ofof CC andand E,Z, respectively. N' NE = = Aut(Z) Aut(E) S - W and H = NE. Nc.

Proof. By 29.13.3, WJnKWJnK = = Wj win n WK for for J, J, K K CC I.I. Thus UJUj = Wj,, Wj!, wherewhere UjUJ == njEJnj EJUi.Uj. InIn particular,particular, as (W,(W, S) is aa CoxeterCoxeter system,system, UIUl == WO WD = 1 1 and UitUp == W,Wi = (si)(q) isis ofof orderorder 2, so &(W,-'(W, 1)8) isis a athin thin chamber chamber complex complex by 41.1.4 and Exercise 14.3.14.3. Next 43.7.6 says the map 7r:n: GiwGi w Hi-+ U;Uiw w ofof (1)(1) isis wellwell defined,defined, afterafter whichwhich n7r isis evidentlyevidently aa bijectionbijection andand n-'7r-1 isis aa morphismmorphism ofof geometries. So to prove (1) it remains to show that ifif GkwGkw fl n G1 Gj v is nonempty then UkwUkw fln U1 Uj vv isis too.too. ButBut Gku fln Gj v isis nonemptynonempty if and only ifif 11 EE Gkuv-1Gj,G~uv-'G~, inin which which case case Ukw Ukw fl n UU Uj vv is nonempty by the following observation, whichwhich isis an easy consequence of 43.2 and 43.4.2:

(43.9) For each J,J, K K CE I Iand and each each w w E EW, W, (Gj (GJwGK) WGK) fln W w = = Uj UJWUK. WUK.

Hence (1) holds and therefore by thethe firstfirst paragraph ofof thisthis proof,proof, -0493 satisfies B1. Notice (2) is immediate from 41.1.1. Exercise 14.314.3 and 43.743.7 showshow -0493 is a chamber complex. To show -0493 is thick, by 41.1.4 and 43.7 wewe mustmust showshow IPi:BIJP,: BI >2foreachi> 2 for each i cE I.I.But,by43.6, But, by 43.6, PiPi == B BUBsiBsoIPi:BI U Bs, B so I P, : B I =2ifand=2 if and only if si E NG(B). Thus BN4 says 93-04 is thick. BN-pairs and TitsTits systemssystems 223

For xx E G and J Cc I Ilet let Cj,x Cj,, == {Gjx: {Gjx: j j E EJ} J) bebe the the simplex simplex ofof type J inin Cx. LetLet Cj,xCj,, andand CK,yCK,~ be simplices. Then there exist a, b E B and n E N with xy-'xy-1 == anb. anb. Now Now CK,y CK,~ == CK,by CK,by andand Cj,xCj, = = CJ,Rby Cj,nby are in Cby,Eby, soso 93 satisfies B2. Suppose Ci,x,Cj,,, CK,~CK,yC c C E nfl Cg.Eg. Then we can choose x, y E N and there exist n, m E NN with Cj,,Cj,x = CJ,ngCj,,, andand CK,~CK,y= = CK,~~. CK,mg.Notice Notice h =h = mgy-' mgy-1 E EGK. G. Define uu == nm-'nm-1 and and v = xy-1.xy-'. Then (Gj)uh == (~j)n~y-'(Gj)ngy-1 == (G~)X~-' (Gj)xy-1== (Gj)v, soso uu cE (Gj)v(GK).(G~)u(GK). So,SO, by 43.9, uu == rvs forfor somesome r EE GiGj n n NN andand s E GKnN.SetlGKfN. Setl = y-1 y-'shy.ThenGJxl shy. ThenGjxl == GjvylGj vyl = = GjvshyGj vshy = = GJr-'uhyGtr-1uhy == GangGJng = Gjx, and and GKylGKyl = GKShy GKshy == GKhy= = GKy,GK y,and and Cl El = = Cyl= Eyl = CshyEshy = ChyEhy = Emg Cmg= = Eg. Cg.So So El Cl = =Eg, Cg,(CJ,,)l (Cj,x)l = Cj,x,=CJ,,,and(C~,,)1 and (CK,y)l = =C~,~.Thatis CK,y. That is l1 induces an isomorphism of EC withwith EgCg trivial trivial on on Cj,x Cj,, andand CK,y.CK,~. ThereforeTherefore 933 satisfies B3. This completes the proof ofof (3).(3). As W is transitive on the chambers of -6(W,6(W, 6'&), ), (1) says N is transitivetransitive on fi?-' fln E.C. So, So, as as G G is is transitive transitive onon c1d by by construction,construction, (4) holds.holds. AlsoAlso NG(C) NG(E) == NNB(E)NNB(C) asas NB(E) NB(C) is is thethe stabilizerstabilizer of C in NG(E).NG(C). As 6'&is is thin, NB(C)NB(E) = BE BE by 41.2, so (5) holds.

(43.10) Let G*G* == G/kerB(G). G/kerB(G). Then Then (G*, (G*, B*, B*, N*, N*, S*) S*) is is aa TitsTits system.system.

Proof. ThisThis isis clear.clear.

Because of 43.10 it does little harm to assume kerB(G)kerB (G) = = 1. By 43.8 this has the effect of insuring that G isis faithfulfaithful on itsits building. Recall that the Coxeter system (W, S)S) is is irreducible if if the graph of its Coxeter diagram is connected.

(43.11) Assume (W, S) isis an irreducible CoxeterCoxeter systemsystem and and kerB(G) kerB(G) = = 1.1. Then (1) IfIfXgGthenG=XB,and X < G then G = XB, and (2) ifif GG is is perfect perfect andand BB isis solvable solvable then then GG isis simple.simple.

Proof. Let X <9 G. Then XBXB < < GG so,so, byby 43.7,43.7, XB XB = = Pj Pj forfor somesome J J EC I. LetLet Jo = {i{i E I: (Bsi(Bsi B)B) nfl Xx :# Q}. 4). IfIf ii EE JOJO thenthen si si Ec BsiBsi B B E c XB == Pj so so i E J by 43.7.6. Conversely, as as Pj Pj = XB, X intersects eacheach cosetcoset of of BB inin Pj non-nontrivially soso JJ CJO.c Jo. Thus J == JO. Jo. Claim JJ == I. I. If If notnot asas (W, S)S) isis irreducible irreducible there there exists exists i E i IE -I -J Jand and j j EE JJ with [si,[si, sj] # 1. Therefore, as (si,(si, sj) sj) is is dihedral, dihedral, l (sil(sisjsi) sjsi) > > l l(sjsi)(sjsi) > > l (sil(si), ), so, by 43.3.1, BsiBsiBsjBsiB Bsi Bsi B = = BsisjsiB.Bsi sjsi B. As As X X < GG andand JJ == JO Jo itit followsfollows thatthat sisisjsi sjsi EE WnW fl PJPj == Wj. Wj. But But then then si sisjsisjsi E E Wj Wj fl nW{i,J Wgi,jJ }= = WjW3 == (sj), contra- contra- dicting [si[si, , sj]sj] # : 1.1. 224 The geometry of groups ofof LieLie typetype

Thus II = J, J, so so XB XB = Pj PJ = = PI PI = G.= G.Hence Hence (1) (1) is is established established and and ofof coursecourse (1) implies (2).(2).

(43.12) Let F bebe aa fieldfield and 11 << n n andand integer.integer. ThenThen (1) SLn(F)SL,(F) isis quasisimplequasisimple and L,(F)Ln(F) is simple unlessunless (n,(n, I IFI)FI) =_ (2,(2,2) 2) oror (2,(2,3). 3). (2) Spn(F)Sp,(F) isis quasisimplequasisimple and PSp,(F)PSpn(F) is simple unless (n,(n, IIFI) FI) =_ (2, (2,2), 2), (2,(2,3), 3), or (4,(4,2). 2). (3) IfIf FF is is finite finite then then SUn(F)SU,(F) isis quasisimple quasisimple and and UU(F)U,(F) isis simplesimple unlessunless (n, IFI)IF11 == (2, (2,419 4), (2, (279) 9) or or (3, (3,4>. 4). (4) IfIf FF is is finite finite or or algebraicallyalgebraically closed, closed, and n >> 6, 6, then then Stn Qi(F) (F) is quasisimple and PS2n(F)PQi(F) is simple.

Proof. Let G == SLn(F), SL,(F), Spn(F),Sp,(F), SUn(F),SU,(F), oror 01Qi(F), (F), in in the the respective respective case. By Exercise 14.5, parts 1, 2, and 6, and 43.1, G possessespossesses aa BN-pairBN-pair (B, N). Observe next that B is solvable. This follows from Exercise 4.9 and Exercise 7.8. Indeed these exercises showshow BB is the semidirect product of a nilpotent group byby a solvable group.group. In In applying applying Exercise Exercise 7.8 7.8 observe observe n n- - 2m 2m 5< 2 as Z12' hashas nono singular points; thusthus O(Z',O(Z1, f) is is solvable solvable by by ExerciseExercise 7.2. Next, by Exercise 14.5.5, ker~(G)kerB(G) == Z(G). Further,Further, except in thethe specialspecial cases listed in the lemma, G isis perfect.perfect. This follows from 13.8,13.8, 22.3.4, 22.4, and Exercise 7.6.7.6. Finally,Finally, to to complete complete the the proof, proof, 43.1 43.11.2 1.2 says says G/Z(G) G/Z(G) is simple, since by Exercise 14.5.3 thethe Coxeter systemsystem ofof thethe WeylWeyl groupgroup ofof GG is irreducible.

Remarks. TheThe material material in in chapter chapter 14 14 comes comes fromfrom Tits [Ti] and Bourbaki [Bo]. Already in this chapter we begin to see the power of the Tits system building approach toto thethe study ofof groups of Lie type. The proof of the simplicity of various classical groups in 43.12 probably provides the best example. Further results are established in section 47, where groups with aa BN-pairBN-pair generated by root subgroups and satisfying a weakweak versionversion ofof thethe ChevalleyChevalley commutatorcommutator relations are investigated. These extra axioms facilitatefacilitate the proof of a number of interesting results. In [Ti], TitsTits classifiesclassifies all buildingsbuildings of rankrank at leastleast 3 withwith a finitefinite WeylWeyl group,group, and hence also all groups with a Tits system of rank at least 33 andand finitefinite Weyl group. The rank of a buildingbuilding is 111I I Iwhile while the the rank rank of of aTits a Tits system system (G, (G, B, B, N, N, S)S) is ISI.IS/.

Exercises forfor chapterchapter 14 1. If sfci == (A, (A, a) d) is ais complex a complex let let E C(d)(sxl) bebe the the geometry geometry whosewhose objects ofof type i areare the simplices of sa(d ofof typetype ii and whose edges are the simplices of dsl of of rankrank 2.2. If Fr isis aa geometrygeometry letlet B(r)6(I') bebe thethe complexcomplex (r,(I', &(r))-'(I')) BN-pairs and TitsTits systemssystems 225

where -6(I')&(r) is is the set of flags ofof rF of type I. ProveProve (1) TheThe inclusion inclusion map map 7rn is an injective morphism of C(d)E(sa') into into A;A; 7rn isis an isomorphism ifif and onlyonly ifif eacheach rankrank 22 flagflag ofof A is a simplex of d,sxl;7rn isis anan isomorphismisomorphism ofof-6(E(sy1)) &(C(d)) with difif and only ifif every flag of A is a simplex of d.c(. (2) 7rn is an isomorphism ofof C(&(r))E(6(I')) with with Fr ifif andand onlyonly ifif everyevery rankrank 2 flag of rr' isis containedcontained in a flag of type I. (3) ForForeachsimplex each simplex T T ofdletd, of a. let .(T = = {C{C - - TT: : T T 5< CC E d]s-I andand ETCT = E(AT,C(AT, aT).dT). ProveProve the the map map 7rT: nT: ETCT -+ + AT AT is is an an isomorphism isomorphism of of ATAT with ETCT for each simplex T ofof ,d 1 if if andand onlyonly ifif everyevery flag of A is aa simplex ofof d.s.. (4) Aut(F)Aut(r) = Aut(6(F)).Aut(&(r)). 2. Let Let i6B == (I', (r, -6) &) be be a aCoxeter Coxeter complexcomplex over I == (1, (1, ...... , ,n n], 1, letlet W = Aut(8),Aut(6), CC inin B,6, and forfor JJ EC II let J' == I I - - J Jand and Tj Tj the the subflag subflag of C ofof type J.J. For For i ic EI letI letri ber, bethe the reflection reflection through through the the wall wall Tip, T,I, let let R = {ri,{rl,...... , , r,], and RjRj = (ri:(r,:i i EE J').J']. ProveProve (1) IfIf J J C c I thenI then (ET, (Cz , -6T, gz) f) is is a CoxetercomplexCoxeter complex andand WT,WE == Aut(ET,,Aut(CTI, 6Tf ) (in the notation of Exercise 14.1). (2) WT,WE == (Rj)(Rj) and and ifif II # # J Jthen then (WT,, (WE, Rj) Rj) is is a aCoxeter Coxeter system.system. (3) LetLet TiT, == (vi (v,], 1, WiW, = WT,Wz, andand 9 = 9,-(W,9(W, R)R) == (Wi:(W,: ii E I).I). ProveProve the map viwI-Wiw iEI,wEW is an isomorphismisomorphism ofof 8-6 withwith 8(W, 6(W, 9). 3. LetLet GG bebe a group andand F97== (Gi (Gi : : i cE I) I) a afamily family of of subgroupssubgroups of of G.G. Prove the complex &(G,6(G, 9)is a chamberchamber complexcomplex if if and and only only if if G G = = (G,!:(G1': i i EE I),I), where i' i' == I -- {i {i 1.1. ProveProve B(G, 6(G, 9)) = =6(I'(G, B(r(G, 3T)) 9)) if andif and only only if if G G is is flagflag trasitive onon r(G,I'(G, 9).9-). 4. Let Let -6'& = (F, (r, -6) &) bebe aa Coxeter complex andand WW == Aut(8).Aut(6). ProveProve (1) LetLet A A and and B B be be simplices. simplices. Prove Prove A A U U B B is is a a simplex simplex if if andand only only if if AA U B is containedcontained inin r4F0 or r4'Fo' forfor eacheach pair 4,4'0, 0' of of oppositeopposite foldings. (2) LetLet Ai,A,, 11 <5 i i< 53, 3,be be simplices simplices such such that that Ai A, UU AjA, isis a a simplex simplex forfor each i, j. ThenThen U3=1 uLl Ai is a simplex. (3) Every Every flagflag of r'r isis aa simplex.simplex. (4) WW is is flag flag transitivetransitive on F.r. (5) ProveProve (in(in the notation of Exercise 14.1)14.1) that that (rfi,(IT,, BE) 6T) is is aa CoxeterCoxeter complex forfor each each J J cC I. (6) WW = Aut(I'). Aut(r). 5. LetLet FF be be a afield, field, V V a a finite finite dimensional dimensional vectorvector spacespace over F, and and assumeassume one of the following holds: (A) Fr isis the the projective projective geometrygeometry on V, Y == {(xi): 11 5< i <5 n)n] forfor somesome basis XX = (xi:(xi: 1 I:< i <5 m) m) of of V,V, m = dim(V), dim(V), and G = GL(V). GL(V). 226 The geometrygeometry of groupsgroups of Lie type

(C) (V,(V, f) is is a a symplectic symplectic or unitary spacespace and and G G = = O(V,O(V, f)f) or (V, Q) is anan orthogonalorthogonal space and G = O(V,O(V, Q). IfIf (V,(V, Q)Q) isis orthogonalorthogonal assume it is not hyperbolic. 0 < mm isis thethe WittWitt indexindex ofof thethe spacespace andand r itsits polarpolar geometry. XX == (xi:(xi: I1 <5 ii <5 2m) 2m) is is a a hyperbolic hyperbolic basisbasis for a maximal hyperbolichyperbolic subspacesubspace of of VV andand YY= = ((x):((x):x x E E X}.X). (D) (V,(V, Q)Q) isis aa 2m-dimensional2m-dimensional hyperbolichyperbolic orthogonal spacespace andand GG is the subgroup of O(V, Q)Q) preservingpreserving thethe equivalenceequivalence relationrelation ofof 22.8, rF is the oriflammeoriflamme geometrygeometry of of (V, (V, Q), Q), X X = = (xi: 11 5< ii 5< 2m) is aa hyperbolic basis of V,V, andand YY == ((x) ((x): : xx EE X XI. }. In each casecase let let 6% A == 8(r)-'(F) _= (F, (r, e) 8) bebe the the complex complex onon rr defined defined in Exercise 14.1, let EyCy be thethe subgeometrysubgeometry consisting of the objects of r generatedgenerated by by membersmembers ofof Y,Y, identify EyXy withwith thethe subcomplexsubcomplex (Ey,(Cy, -6 6 flil Ey),Xy), and and letlet .d' = ((Ey)g: ((Cy)g: g g EE G).G). Prove Prove (1) (A(B a)d) isis a abuilding. building. (2) GG isis transitivetransitive on SZ!2 = = ((C,((C, X):E): CC E -6,6, ZE EE sz'}.dl. (3) TheThe WeylWeyl group of AB is is of of type type A,,,, A,, C,,,,C,, orDmorD, in case A, C, and D, respectively. (4) Let C be a chamber inin Xy,Ey, BB == Gc, andand N == NG(Ey). NG(Xy). ProveProve G isis the semidirectsemidirect productproduct of of a a nilpotent nilpotent group group U U by by H H = = BB ilfl N, N, andand HH is abelian ifif A or D holds, or if (V, f) is is symplectic symplectic oror unitary;unitary, or ifif F isis finitefinite oror algebraicallyalgebraically closed. closed. (5) kerB(G)kerB(G) = = kerH(G) kerH(G) is is the the group group of of scalarscalar transformationstransformations of VV inin G.G. (6) SL(V)SL(V) and and Sp(V)Sp(V) are are transitive transitive onon Q!2 andand if F is is finitefinite or algebraically closed then S2(V,Q(V, Q)Q) and SU(V) are transitive on Q. 6. Let Let (G,(G, B, N, S) be a Tits systemsystem with with Weyl Weyl group group W W = = N/H N/H andand S S == (Si:(si : ii E I). Prove (1) ForFor each J, K CE I,I, the the mapmap

(PJ)w(PK)(PJ)W(~K) H (WJ)w(WK) (WJ)W(WK) is a bijection of the set of orbits of the parabolic WKWK on the coset space B/PJ with with the the orbits orbits of of the the parabolic parabolic WKWK on the coset space W/WJ.W/ Wj. (2) AssumeAssume thethe CoxeterCoxeter diagram A of WW is one of the following:

1 2 n-1 n An o--o...p_...p

1 2 n-2 n-1 n Cn o- o p------cz::=

Dn BN-pairs andand TitsTits systemssystems 227

Prove: (a) IfIf AA isis of of typetype AnA, then G is 2-transitive on G/Pl!,G/PI,, rankrank 33 on G/P2,G/P21 if n >> 2, 2, and and rank rank 66 on on P{t,n}'Ptl,ny ifif nn > 1.1. (c) IfIf AA isis ofof typetype CnC, and nn > 1,1, then then GG isis rank n + 11 on G/Pn,,G/ P,!, rankrank 3 on G/P1!,G/P1,, and rankrank 66 on on G/P2! G/P2 if nn > 2.2. (d) IfIf AA isis ofof typetype D,,Dn, nn 2> 4,4, thenthen G is rank 3 on G/Pl1G/Pl, andand rankrank 8 or 7 on G/G/P2, P2t for n = 4 4 oror nn >> 4, 4, respectively. respectively. 7. (1)(1) EveryEvery flag flag in in aa building building isis aa simplex.simplex. (2) IfIf (G,(G, B, B, N, N, S) S) is is a aTits Tits system system then then G G is is flag flag transitivetransitive on its building 989 (G, (G, B,B, N,N, S).S). (3) IfIf A98 == (F, (r, e) 8) isis a abuilding building then then Aut(A)Aut(98) = Aut(F). Aut(r). 8. Let Let (I',(r, -')8) bebe aa chamberchamber complex andand for for x x E E r F let let 8, ex == B ilfl r,. F. (1) IfIf I 11I I > 22 assumeassume for each x EE Fr thatthat (Fx,(r, ,8,) -ex) isis a connectedconnected complex. Assume @:0: 8 -.' HH -'B is is a a function function such that, forfor all all C, C, D D EE 8 i' andand all i EE I,I, if if CflDisC n D isawall a wall of of type type i'then i' then CO C@ flDoil D@ isis also also awalla wall of type P.i'. For xX E rF define xcrxa toto bebe thethe elementelement of of COC@ ofof thethe samesame typetype as x, where xx E C EE -e.8. ProveProve cr:a: (F,(r, 8)') -->+ (F,(r, 8)') isis aa well-definedwell-defined morphism. (2) AssumeAssume III11 I >> 2, 2, F r = = F(G, r(G, -12-), 9), and -'8 = = C(G, C(G, -12-) 9) forfor somesome groupgroup G and some family 9 9 == (G;: (Gi: i i EE I)I) of of subgroups. subgroups. AssumeAssume for each i E I thatthat G;,Git = = (Gy,j),: (G{i,,)~: jj E E i'). it). ProveProve (I'x, (r,, -fix) 8,) is aa connected connected complex. 9. LetLet TT == (G, (G, B, B, N, N, S) S) be be a aTitsTits system system with with finite WeylWeyl group group W W = = N/H.N/H. Let 983 _ =1-213 98 (T) (T) be be the the building building of of T, T, C C thethe chamberchamber ofof 98A fixedfixed by B,B, and EX _= (xw: (XW: xx E E C, C, W w EE W) the apartment ofof 98A stabilizedstabilized by N (cf. the discussion before 43.8). Let @ be a root system for W and nit aa simplesimple system forfor WW withwith SS = (r,,:(r,: cra EE 7r)n) (cf. (cf. 29.1229.12 and 30.1). LetLet 5< be the partial order on W defined in Exercise 10.6. Prove (1) If If u,u, ww EE WW withwith u <5 ww thenthen there exists aa gallerygallery (Ci: (C,: 0 0I

(Hints: Use 41.741.7 inin (1).(1). To To prove prove (2) (2) observe observe B Bfl nBSS BSWBW BW CE NBNB (Csw nfl Cw) andand use use thethe RainyRainy DayDay LemmaLemma andand (1)(1) to to proveprove NBNB (Csw fln Cw) < 5 BSW.BSW.Use Use (2) (2) toto proveprove (3)(3) andand (4) and use (4)(4) and and Exercise 10.610.6 to proveprove (5).)(3.) 10. (Richen(Richen [Ri])[Ri]) Assume the hypothesis of Exercise 14.9 andand assumeassume furtherfurther that T is saturated. LetLet aa E njr and s = ra r, EE S.S. For For Ww EE WW definedefine B,"Bw ==

B nfl BWOW-',Bw0w-' where, where wo w0 is is the the maximal maximal element element of of W. W. ForFor B,8 EE njr definedefine BgBp = Br, B,, . Prove

(1) IfIf w EE WW with awaw > 00 then then BaB, <5 BwBW-I. ' (2) IfIf Ww E W withwith aw aw < 0 then B,Ba nfl BW-IBw-' == H. (3) IfIf Ww E W withwith l(w) 1(w) 5 < l(ws) l(ws)then then B B = = (B(B flfl BS)(BB'')(Bfl fl BW). B'). (4) B B == Ba(B B,(B f1n BS) B" withwith BSBW f1B, Ba = = H. (5) BaB, # H. (6) ForFor w E W,W, B,Ba 5< BwB~-' ' ifif andand only ifif awaw > 0.0. (7) Let AA == {(Ba)w: {(B,)": a a EE n,7r, w w E E W). W). Prove Prove the the map map (B,)W (Ba)' H H awaw is a well-defined permutationpermutation equivalenceequivalence ofof thethe representationsrepresentations ofof W on A and (D.@. InIn particularparticular we we may may define define B, By = = (Ba)W(Ba)' forfor eacheach yy EE (D, @, where yy = aw.aw . (8) If If w w E E W W with with aw aw < <0 then0 then B", Bw = Ba(B,S,,,)S= B,(Bsw)S andand BaB, fln (B,r",)S (Bsw)S == H. (9) Let w0wo == rlr1 ...... r,r with with n n = = l(wo) l(wo) _= I(D+I I @+ 1 andand rlri = ra,, raL, a; ai EE jr.n. Let Let wi =riri-1= riri_1 ...... ri. rl. ProveProve BB = =BaiBa2wt Ba1Ba,,, ...B,,, ,,-, with Ba,w,-,fl n;=,+,j i+1 BaBaJwJ-,w = H.H. Further Further thethe map i H aiwi_1ai w,-1 isis aa bijection bijection of (1, . . . , njn) withwith 4)+.@+. (Hints: InIn (1)(1) useuse ExerciseExercise 10.6.410.6.4 toto show show w w-1 -' < woswos and then appeal to Exercise 14.9.4. In (2) use partsparts (2)(2) and and (5)(5) of of ExerciseExercise 14.914.9 toto conclude B~-~~Bw-'Sn fl BS BS _( < B and B fln Bw0BWO == H.H. In In (3)(3) useuse 43.3.143.3.1 toto get BB C BSBwBSBW andand then appeal to Exercise 14.9.3. Use (4) and BN4 to prove (5). In (7) use (6) to show for B,8 E n,Jr, (Ba)W(Ba)w = BpBg ifif andand onlyonly if aw == P. B. In In (8) (8) use use (1) (1) to to show show BaB, <5 Bw. B,. ThenThen useuse (4)(4) toto showshow Bw == Ba(BB,(B flfl BS) Bs) andand useuse Exercise 14.9.314.9.3 to show BswBiw 5< B.B. UseUse (8)(8) to prove (9).)(9).) 15

SignalizerSignalizerfunctors functors

Let r bebe aa prime,prime, GG aa finite finite group,group, andand AA anan abelianabelian r-subgroup of G. An A-signalizer functor functor onon GG is a a mapmap 08 fromfrom A#A' intointo thethe setset of of A-invariant A-invariant rl-subgroupsr'-subgroups of GG suchsuch that, that, for for each each a, a, b bE E A#, A', 0(a):!:-8(a) I: CG(a) andand 8(a)0(a) fln CG(b)I: <9(b). 8(b). The signalizer functorfunctor 80 is said to be complete complete if there is an A-invariant r'-subgrouprl-subgroup 0(G) 8(G) such such that that 0(a) 8(a) = = CB(G)(a) Ce(G)(a) forfor eacheach aa EE A#. A'. Notice that one way to construct an A-signalizerA-signalizer functor is to select some A-invariant rl-subgroupr'-subgroup X of G andand definedefine 0(a)8(a) = Cx(a) Cx(a) for for a a E E A#. A'. ByBy con-construction this signalizer functorfunctor isis complete. IfIf m(A)m(A) >2 3 3 it it turnsturns outout thatthat thisthis isis the only way to constructconstruct signalizersignalizer functors.functors. That is, ifif m(A) >2 33 thenthen everyevery A-signalizer functor is complete. This result is called the Signalizer Functor Theorem. It's oneone of the fundamental theorems in the classification of the finite simple groups. Unfortunately the proof of the Signalizer Functor Theorem is beyond the scope of this book. However, chapter 15 does contain a proof of a special case: the so-called Solvable 2-Signalizer Functor Theorem. It turnsturns outout that the SolvableSolvable SignalizerSignalizer FunctorFunctor TheoremTheorem sufficessuffices forfor manymany applicationsapplications of signalizer functors. - An A-signalizer functorfunctor 8 0 onon GG is said to be solvable if 8(a)0(a) isis solvablesolvable for each a EE A'.A#. WeWe saysay 80 isis solvablysolvably completecomplete if if 8 0 isis completecomplete andand 8(G)0(G) isis solvable. The main resultresult of chapterchapter 1515 is: is:

Solvable 2-SignalizerZSignalizer Functor Theorem. Theorem. Let Let A A be be an an abelian abelian 2-subgroup Zsubgroup of a finite group G with m(A) >2 3.3. Then each solvable A-signalizer functor on G is solvably complete.

Chapter 1616 contains a discussion of the Classification Theorem which illustrates how the Signalizer Functor Theorem isis used. ObserveObserve that the condition that m(A) >2 33 isis necessary inin thethe Signalizer FunctorFunctor TheoremTheorem byby ExerciseExercise 15.1. 15.1.

44 SolvableSolvable signalizer functors In section 44, r isis aa prime,prime, G isis aa finitefinite group,group, A is an abelian r-subgroup of G of r-rank atat leastleast 3,3, andand 08 is a solvable A-signalizer functor on G. Let Q(G)Q(G) denote denote thethe setset ofof solvablesolvable A-invariantA-invariant rl-subgroupsr'-subgroups X ofof GG withwith Cx(a) = X X fln 0(a)O(a) for each aa Ee A'.A#. For For A A 5 < H H 5< G,G, letlet Q(H)Q(H) == H fl Q(G) and 0(H) = (H fl 0(a): a E A#). 230230 SignalizerfunctorsSignalizer functors

IfIf 7rrr isis aa setset ofof primes primes then then &(H,Q(H, 7r)n) denotesdenotes thethe setset ofof n-groups7r-groups in in &(H).Q(H). Write Q*(H)&*(H) andand Q*(H,&*(H, 7r)n) for for the maximal members of Q(H)&(H) andand Q(H,Q(H, 7r),n), respec- respec- tively,tively, under the partial order of inclusion. For X E &(G),Q(G), write Qx(G,&x(G, 7r)n) forfor thethe set of H EE Q(G, n) withwith XX <( H, H, and and define define n(0)= U 7r(0(a)). aEA#

(44.1)(44.1) IfIf XX EE Q(G)&(G) andand Y Y is is an an A-invariant A-invariant subgroup subgroup of of XX thenthen Y Y E E Q(G).&(G).

(44.2)(44.2) IfIf X,X, Y Y E E Q(G)&(G) withwith X X < I NG Nc(Y) (Y) thenthen XYEXYE Q(G).&(GI.

Proof.Proof. ThisThis follows follows from from Exercise Exercise 6.1. 6.1.

(44.3)(44.3) LetLet HH EE Q(G)&(G) withwith HH a5 G, G, set set G* G* = = G/H, GIH, and and define define 9*(a*) O*(a*) == B(a)* O(a)* forfor a*a* EE A*#.A*'. Then (1)(1) 8*e* is is an an A*-signalizer A*-signalizer functor. functor. (2)(2) Q(G*) &(G*) == QH(G)* QH(G)* = ={X {X*: *: XX EE Q(G)&(G) andand HH < 5 X1. X}. (3)(3) Q(G*) &(G*) == Q(G)* &(G)* == {X*: {X*: XX E E Q(G)}.Q(G)}.

Proof.Proof. Let Let a, a, b bE E A#. A'. ToTo prove prove (1) (1) we we must must show show 9*(a*) O*(a*) flfl Cc*(b*)5 9*(b*).O*(b*). ByBy coprimecoprime action, action, 18.7, 18.7, CG- CG*(b*) (b*) == CG(b)*, CG(b)*, soSO it it suffices suffices to to show show CHOW C~s(,)(b) (b) <5 H9(b).HO(b). But But by by Exercise Exercise 6.1, 6.1, CHO(a)(b)=CH(b)CB(a)(b), CHsca)(b)= CH(b)Ce(a)(b), soSO asas CB(a)(b) Ce(a)(b) I<9(b), O(b), (1)(1) isis established.established. LetLet XX EE QH(G).QH(G). ThenThen XX isis a a solvable solvable A-invariant A-invariant r'-group,r'-group, soso itsits imageimage X*X* hashas thethe samesame properties. properties. Further Further for for a a E E A#, A', CxCx(a) (a)

(44.4)(44.4) ForFor eacheach 101 # B B < 5 A, A, 9(CG(B)) O(Cc(B)) Ef Q(G). &(GI.

Proof.Proof. Recall Recall 9(CG(B)) O(CG(B)) _= (CG(B) (CG(B) flfl 9(a): @(a): a aE E A#). A'). ButBut for for b b E E B#,B', CG(B)Cc(B) flfl 0(a)@(a)

(44.5) For each XX EE Q(G)&(G) andand Trn E:C.7r(9), n (0), CxCx(A) (A) 5 <0(CG(A)) O (CG(A)) and Cx(A)Cx (A) isis transitive on Q*(X,.7r).&*(X,n). Indeed, Q*(X,.7r)&*(X,n) is thethe setset of of A-invariant A-invariant Hall Hall .7r- n- subgroups of X.

Proof. LetLet T T be be the the set set of of A-invariant A-invariant yr-subgroups n-subgroups ofof XX and T* the set of maxi-maximal members ofof TT under under inclusion. inclusion. By By 44.1, 44.1, T T = =Q(X, &(X, .7r), n), so so T T** = = Q* &*(X, (X, .7r). n). By Exercise 6.2, T* is the set of A-invariant Hall n-subgroupsyr-subgroups of X and Cx(A)Cx (A) is transitive onon T*. Finally, as X E &(G),Q(G), wewe havehave Cx(A)Cx(A) 5 <0(a) O(a) forfor eacheach a EE A#,A', so CxCx(A) (A) <( 0(CG(A)),O (CG(A)), completing the proof.

(44.6) LetLet P p E E 7r n(O). (0). Then (1) For For each each a a E E A# A# therethere isis aa uniqueunique maximalmaximal B(CG(A))-invariantO(CG(A))-invariant membermember 0(a)A(a> ofof Q(6(a),&(O(a), p').PI). (2) AA is is an an A-signalizer A-signalizer functor.functor.

Proof. ByBy 44.4, 44.4, YY == B(CG(A)) O(CG(A))5 < XX = = O(a).9(a). LetLet R bebe thethe set set ofof Y-invariant Y-invariant members ofof Q(X,&(X, p') andand SS = = Q*(X, &*(X, p').p'). ThenThen for for R R E E R, R, R R <_ ( S E S,S, so as Y is transitive onon SS by 45.5, R is contained in each member ofof S.S. Therefore M = (R)(R) <5 S, S, so so M M is is the the unique unique maximalmaximal membermember of R byby 44.1.44.1. ThisThis estab-establishes (1).(1). Next forfor bb EE B',B#, A(a)0(a) flfl CG(b)CG(b) = = CM(b) 5 <9(b) O(b) asas BO is a signalizer functor. Also CM(b) is is anan A-invariantA-invariant pl-group,p'-group, so CM(b)5 < A(b),0(b), completingcompleting the proof of (2).(2).

, (44.7) (Transitivity Theorem)Theorem) O(CG(A))B(CG(A)) is is transitive transitive on on &*(G,Q*(G, p)p) for each Pp EE 7rn(@. (B).

Proof. TheThe proof proof isis left left asas ExerciseExercise 15.2.15.2.

(44.8) Let X E Q(G).&(G). Then (1) IfIf BB is is a a noncyclic noncyclic elementary elementary abelian abelian subgroupsubgroup ofof AA thenthen X = (B(C(D)) f1 X: I B: D) =r). (2) IfIf1#T

0(CG(T))O(CG(T)) = (0(CG(B)):(O(CG(B)): T <5 B <5 AA andand B isis noncyclic). (3) XX =O(X).=0(X).

Proof. IfIf 11 # T

X = (Cx(S): T/S isis cyclic).cyclic). 232 SignalizerSignalizerfunctors functors

Further ifif S #01 1 then then CX(S)Cx (S) <0(CG(S))5 8(CG(S)) byby 44.5,44.5, soso (1)(1) follows.follows. Similarly by 44.4, Y = =9(CG(T)) 8(CG(T))E E Q(G) Q(G) and and A A= = AITA/T is abelianabelian and acts on Y, so by Exercise 6.5, Y == (Cy(Cy(B):(B): A/BAlb isis cyclic)cyclic) with CyCy(B)(B) = 9(CG(B)). 8(CG(B)). IfIf TT is is noncyclic noncyclic then then (2)(2) isis trivial,trivial, so we may assume T is cyclic. Then as m(A) > 3, B isis noncyclic,noncyclic, soso (2) holds. For H 5< G, 9(H)8(H)

This about exhausts the results on solvablesolvable signalizer functors which can be established outside ofof an inductive setting. ToTo proceed furtherfurther wewe must work inside a minimal counterexample to the Solvable 2-Signalizer Functor Theorem. So in the remainder of chapterchapter 15 15 assumeassume (A, G, 9)8) isis aa counterexamplecounterexample to thethe Solvable 2-Signalizer Functor Theorem with (G(IG I + In(9)1 In (8)/ minimal.minimal. InIn particular now r = 2. 2. LetLet II == 9(CG 8(CG(A)). (A)).

(44.9) (1)(1) IfIf A 5< H << G G then then 9(H)8(H) E E Q(G).Q(G). (2) GG = = A(9(b): A(8(b): b b E E B#) B') forfor eacheach E4-subgroupE4-subgroup B B ofof A.A. (3) IfIf 101 # H H E E Q(G) Q(G) then then GG 0# NG(H). NG(H).

Proof.Proo$ InIn eacheach casecase wewe useuse thethe minimalityminimality ofof thethe counterexample.counterexample. Under the hypotheses ofof (I),(1), OH(a)9H(a)= = 8(a)9(a) nfl H is anan A-signalizerA-signalizer functorfunctor onon H,H, so (1) follows from from minimality minimality of of I G(GI. 1. Let B be a 4-subgroup of A and K = (9(b): (8(b): b cE B#).B'). If G 0# KA KA then by (1),(I), Y == 9(KA)8(KA) E Q(G). But nownow forfor aa EE A#, A', 9(a) = (9(a) fl 9(b): b E B#)

(44.10) Let nlr 2c n(9),n(8), MM EE Q*(G,Q*(G, 7r),n), andand 110 # X E Q(G) with XX 9< M. Then (1) MM isis a a Hall Hall n-subgroupn-subgroup ofof 9(NG(X)).8(NG(X)). (2) Op(M)0,(M)

Proof. AsAS X X <

(44.11) CA(P)CA(P) = 1 1 for for eacheach pp E7r(0)E n(8) andand eacheach PP EE Q*(G,Q*(G, p).

Proof. SupposeSuppose T = CA(P) CA(P) 0# 1. 1. Then Then byby 44.7,44.7, Q*(G,Q*(G, p) cc CG(T), Cc(T), so,so, byby Exercise 11.1.3, [H,[H, TIT] 5

[6(a), TIOp'(9(a)) < O(a) = CK(a) by paragraph one ofof thisthis proof.proof. RecallRecall that, that, by by 24.5, 24.5, [8(a), [9(a), T, T, TIT] = = [8(a),[9(a), TI,T], soSO [CK(a), Ti _ [9(a), T]. Then by 8.5.6, [CK(a), TIT]

[K, TiTI == ([CK(b), ([CK(b), Ti:TI: bb cE B#)B')

for each B E S. Therefore . 0(CG(B)) = n 9(b) bEB#b€B# acts on [K, T]. TI. Hence HenceB(CG(T)) 9 (CG(T )) acts acts onon [K, T]TI byby 44.8.2, soso by 44.2 and 44.4, H == [K, [K, T]9(CG(T)) T]~(CG(T)) EE Q(G). Q(G). ButBut by 24.4,24.4,8(a) 9(a) <9(CG(T))[9(a),58(C~(T))[8(a), T],TI, so, so, by by the second paragraph of this proof,proof, O(a)0(a) <5 H H forfor each each aa EE A#.A'. This contradicts 44.9.2.

(44.12)(44.12) ForForeachn each lr CEn(@, 7r(6), a cEA', A#, and M E Q*(G, n),7r), [a,[a, F(M)]#F(M)] 0 1.1.

Proof. AssumeAssume otherwise. otherwise. Then Then by by ExerciseExercise 11.1.2, 11.1.2, aa centralizescentralizes M. M. LetLetpEn(F(M))andO,(M)=X p c n (F(M)) and Op (M) = X 5P< P EE Q*(G,Q*(G,p).By44.11,[~, p). By 44.11, [a, PI#P] 0 1,1,so so by the Thompson A x BB Lemma,Lemma, [Np[Np(X), (X), a] 0 # 1. 1. But But Np Np (X) (X) is is containedcontained in in aa Hall n-subgroup ofof HH == 0(NG(X)), 8(NG(X)), so so by by 44.10.144.10.1 andand 44.5,44.5, Np(X)hNP(X)~ <5 M M forfor some h E CH(A). This This isis impossible,impossible, asas aa centralizescentralizes MM but [Np(xlh,[Np(X)h, a] #0 1.1.

For 7rn cn(8)C r(0) andand a E A',A#, letlet M(a,M(a, n)r) denote denote thosethose MME E Q*(G, n)r) such such that CM(a) E Q*(CG(a), n). Recall thatthat forfor Jrn a set of primes and p E n,7r, pKp" = 7r'U{p}.n' u {PI. 234 SignalizerfunctorsSignalizer functors

(44.13) LetLet jr n C- E 7r n(8),p cE 7r,n, a E A',A#, MM EE M(a, M(a, n), VV EE Q(Op(M)),Q(Op(M)), andand N EE Q(G) with N = (V,(V, CN(a)). Then V 5

Proo$Proof. AssumeAssume VV ¢ $ O OPT pn(N). (N). By By 331.20.2,1.20.2, therethere isis aa 4-subgroup4-subgroup BB of A with Cv(B) ¢$ Opn((CN(a), Cv(B))),Cv(B))), SOso replacing VV by Cv(B), we we maymay assumeassume [V, B] = 1. 1. NextNext by 31.20.3 therethere isis b c=E B' B#with with V V¢ $ OPn((C~((a,Op- ((CN((a, b)),b) ), V)), SOso we may assumeassume NN <0(b).5 8(b). However However as as Y Y = = Cm(a) CM(U) EE Q* Q*(Cc(a), (CG (a), n),7r), CYCy(b) (b) E Q*(Co(b)(a),Q*(Ce(b)(a),n). yr). Further Further CY(~)Cy(b) acts onon CoP(M)(b)CO,(M)(~) andand Cv(b)

For.7rFor n gc .7r(O)n(8)and and a a EE A', A#, let let U(a, U(a, n) 7r)consist consist ofof thethe thosethose U E Q(G,.7r)Q(G,n) such that [U,[U, a] # 1 and UU isis invariantinvariant underunder some some membermember ofof Q*(0(a),Q*(8(a), 7r).n). LetLet U*(a, 7r)n) be the minimalminimal membersmembers ofof U(a,U(a, 7r)n) underunder inclusion.inclusion. Given Given p p c E 1r, n, let V(p,.7r)V(p,n) consist of thosethose VV E Q(G, p) p) suchsuch that that VV _(OPr(H)for each each H E Qv(G).

(44.14)(44.14) ForFor each.7r each n cE: 7r n(8) (0) and a cE A#: A': (1) U(a,U(a, Ir)n) # 0. (2) For For U U E E U*(a,U*(a, pr),n), UU = [U, [U, a]a] is is aa p-groupp-group forfor somesome prime p. (3) Let Let Y Y EE Q*(O(a),Q*(O(a), n)7r) withwith YY 5 < Nc(U) NG(U) and and M M E E Q*,,(G, QJ (G, n).ir). Then UU <5 Op(M)O,(M) andand UU centralizes centralizes CF(M)(a). CF(~)(U). (4) UU E E V(p,V(p, 7r).n).

Proof.Proof. LetLet Y YE E Q*(O(a),Q*(O(a), n)7r) and and YY < XX EE Q*(G,Q*(G,n). n). By By 44.12,44.12, [a,[a, F(X)] F(X)] # 1,1, soso X E U(a,U(a, 7r),n), andand (1)(1) isis established.established. Let Let UU E E U*(a,.7r);U*(a,n); withoutwithout lossloss YY actsacts on U.U. By By ExerciseExercise 11.1.2,11.1.2, [Op(U), a]a] #* 1I for somesome prime p, andand byby 24.5,24.5, [Op(U),[O,(U), a, a] _= [Op(U), [O,(U), a], a], soso (2) (2) followsfollows from from minimalityminimality of U. Let M E QQTuy(G, I (G, 7r).n). ThenThen YY = CM(a), CM(a), soso UU is is CM(a)-invariant.CM(~)-invariant. Thus U <5 Op(M)O,(M) by 36.3.36.3. HenceHence UU centralizescentralizes OP(F(M)),Op(F(M)), whilewhile byby thethe Thompson Thompson Lemma andand minimalityminimality of of U, U, UU centralizes centralizes COP(M) Co,(M)(a). (a). This establishes (3). (3). Finally letlet H E Qu(G) and N=N = (CH(a), (CH(U), U).U). By By 44.13,44.13, UU

(44.15)(44.15) LetLet p p c E 7r n C7r E n(8), (0), BB a a hyperplane hyperplane of of A,A, P P E E Q* Q*(G, (G, p), p), and and for forb b E E B#,B', let M(b) E M(b, 7r).n). Then (1)(1) If If F(M(b))F(M(b))=O,(M(b)) =Op(M(b)) and and P Pfl nM(b) M(b) E EQ*(M(b), P(M(b), p)p) for for each each b bE E B#, B', thenthen Z(P) < n Op(M(b)) bEB# Solvable signalizersignalizerfunctors functors 235

(2)(2) WeWe havehave

QnOp(M(b)))c V(p, n). (bEB# Proof.Proof AssumeAssume thethe hypotheses ofof (1)(1) andand letlet Q(b) == O,(M(b))Op(M(b)) andand N(b) N(b) =_ O(NG(Q(b))).0(NG(Q(b))).By44.10.1, By 44.10.1, M(b)isaHalln-subgroupofN(b),soasM(b) is a Hall 7r-subgroup of N(b), so as P(b) P(b) = = PnP n M(b) EE Sylp(M(b))Syl,(M(b)) by the hypotheses of (I),(1), P(b)P(b) E Syl,(N(b)).Sylp(N(b)). Thus Q(b) <5 P(b)P(b) < ( P, P, so so Z(P) Z(P)

1#XEQ(nOp(M(b)) bEB# and N EE QxQx(G).By (G). By 44.13, 44.13,X X <( Op" Opz((CN(b), ((CN (b), X)) fo for each bb E B#, so soso byby 31.20.3,31.20.3, X <( O Opn pn(N), (N), establishing (2). (2).

(44.16)(44.16) Let M E P(G,Q*(G, n), 7r), a aE c,r(F(M)), n(F(M)), andand X (< F(M) with.7r(X)with n(X) = =a a andand Z(O.(F(M)))Z(O,(F(M))) <( X. X. Then Then for for each each N N E E Qx(G,.7r):Qx(G, n): (1)(1) [OP(X),[OP(X), Op(N)] O,(N)] =1= 1 for for eacheach p EE a. (2)(2) IfIfn(F(N)) 7r(F(N)) c E. a a thence thena == 7r(F(N)) n(F(N)) and and [Op(X),[O,(X), OP(F(N))] ==1 1 foreach foreach pEa.p E a. (3)(3) IfIf Tr(F(N))n(F(N)) cEa a and and laIla1 >> 1 1 then then XX <(F(N). F(N).

Proof. ForFor pp EE a,a, letlet XP Xp = = Op(X) Op(X) and and ZpZ, == Z(O Z(O,(M)). p(M)). ByBy 44.10.2,44.10.2, XqX, I 11 butbut Op(X)O,(X) $¢ F(N). Pick p # qq EE a andand let H == CN(Op(F(N))) CN(OP(F(N))) and K = HZq.HZ,. By (2),(2), XX ) = CH(F(H))C,y(F(H))=Z(F(H)). = Z(F(H)). AsAs [Zq,[Z,,OP(F(H))] OP(F(H))] == 1, wewe havehave ZqZ,

(44.17) Let 7rn c.7r(0),E n(0), HH E E Q* Q*(G, (G, 7r),n), X, L EE Q(G, ir),n), and and CF(H) CF(~)(X) (X) (< X <5 F(H) n n L.L. ThenThen (1) [Op(X),[OP(X), Op(L)] O,(L)] = 1 1 for for eacheach p E.7r(F(H)).E n(F(H)). (2) rr(F(L)n(F(L) n H) E97r(F(H)) n(F(H)). 236 SignalizerfunctorsSignalizer functors

(3) If7r(F(L))Cn(F(H))thenIfn(F(L)) 2 n(F(H))thenn(F(L)) 7r(F(L))=n(F(H))andeither =n(F(H))andeither In(F(H))I == 1 oror XX 5< F(L). (4) SupposeSuppose 11 # V EE &(Op(L)Q(Op(L) nfl O,(H))Op(H)) with with NH(V) NH(V) E E Q*(NG(V), P(NG(V), 7r)n) and L E Q*(G, n). 7r). Then Then n(F(H)) 7r(F(H))=7r(F(L)), =n(F(L)), and either n(F(H))7r(F(H)) = {p}(p} or XX_( < F(L),F(L), L LE E HIHI, , andandL=H L = H if ifn=n(0). 7r = 7r(9).

Proof. PartPart (1) (1) follows follows fromfrom 44.16.1,44.16.1, andand (3) follows from 44.16.244.16.2 and44.16.3.and 44.16.3. As

[Op(L) nfl H,H, OP(X)IOP(X)] ==1 1 byby (1)(1) and CF(H)(X)_( > 1 andand YY _( < F(H'). F(H`). By (3),(3), XX (< F(L).F(L). As OP(Y)=OP(F(L))OP(Y) =OP(F(L))

(44.18) Let 7rn 2C n(0),7r(9), U E V(p,V(p, n),7r), and H,H, LL EE Q*Pu(G, (G, n)70 withwith NL(U)NL(U) EE Q*(NG(U),P(NG(U), n).yr) Then (1) If F(L)F(L)=OP(L) = Op(L) then F(H)F(H)=OP(H). = Op(H). (2) IfIf qq E En, Jr, V V E E V(q,V(q, Jr),n), NHNH(V) (V) EE P(NG(V),Q*(NG(V),n), jr), and and [U,[U, Vlv] 5< U fln VV then either (i) p p == q q andand F(H)F(H) and and F(L) F(L) are are p-groups, p-groups, oror (ii) L E H'HI and ifif n 7r=7r(O)then = n(0) then H H=L. = L.

Proof. As UU EE V(p, n),7r), U _(

(44.19) IfIf B isis aa hyperplanehyperplane ofof AA andand HH E E M(b,M(b, 7r)n) for for each each b b E E B#B# thenthen H E E 7-l IFt(n). (7r ).

Proof. LetLet p EEn, 7r, QQ EE Sylp(H), Sylp(H),and and Q Q 5 P P E E Q*(G, P(G, p).p). As As HH E E M(b,M(b, 7r),n), we have CH(b) EE p(C~(b),Q*(CG(b),n), 7r), SO so CQ(b) CQ(b)=CP(b).= Cp(b). Thus Thus PP = QQ byby 44.8.1.44.8.1.

(44.20) LetLet 7rn Cc 7rn (0).(0). ThenThen eithereither (1) R(7r)IFt(n) #0 0, oror (2) There There exists exists aprime aprime p p suchsuch thatthat F*(M)F*(M) == O Op(M) p(M) for each MME E M (a,(a, 7r)n) and each a E E A#.A'.

Proof. LetLet aa E E A#.A'. By 44.14.1 there is U E U*(a, n),7r), and and byby 44.14.2,44.14.2, UU isis a p-group for some prime p. ByBy minimality of U andand Exercise 8.9, CA(U) == B is a hyperplane of A. Thus U <5 M(b)M(b) EE M(b,M(b, 7r)n) forfor each each b b E E B#.B#. By By 44.14.4, U E V(p,V (p, n),so7r), so UU 5< Op(M(b)).ByO p(M(b)). By 44.12thereis44.12 there is U(b) U(b) EU*(~,E U*(b, n)7r)with with U(b) U(b) 5 < F(M(b)), andand by 44.14.3, [U,[U, U(b)] U(b)] ==1. 1. Let H EE Q*(G,P(G, 7r)n) with NH(U) E Q(NG(U),Q*(NG(U), n).7r).If If F(H) F(H) is a p-group then by 44.18.1, so isis F*(M)F*(M) for for each each M M E E Q*Q*,(G, (G, 7r). n). In particular F(M(b)) =_ Op(M(b)).O,(M(b)). Further, for Cc EE A' A#and and MM EE M(c, M(c, n), 7r), from from thethe proof of 44.1444.14 there is U(c)U(c) EE U(a, n)7r) with with U(c) 5< M. By symmetrysymmetry there is aa hyperplanehyperplane B(c) of A centralizing U(c),U(c), and and asas BB nn B(c) # $ 1, we have U(c)U(c) 5< ~(b)'M(b)` forfor some i E I.I. As F(M(b))F(M(b)) isis a a p-group,p-group, soso isis F(H(c))F(H(c)) forfor H(c) H(c) EE Q*(G, P(G, 7r)n) with NH(c)NHCc)(U(c)) (U (0)E EQ*(NG(U(C)), Q*(NG(U(c)),n) 7r) by by 44.18.2. 44.18.2. Hence Hence F*(M) F*(M)=Op(M),= O,(M), as U(c) 5

(44.21) ThereThere is is a a prime prime p p suchsuch that that forfor eacheach a E E A#A# and eacheach M E MM(a, (a, 7rn (9)),(O)), F(M) == O O,(M). p(M).

Proof. IfIf not,not, byby 44.2044.20 therethere isis HH E7-l(7r(9)). E IFt(n(0)). But then for p EE7r(0) n(0) and P EE Sylp(H),Syl,(H), PP EE Q*(G,P(G, p),p), soso for for aa E E A#,A#, Cp(a) E Sylp(O(a)),Sylp(9(a)), and hence 0(a)9(a) =_ (Cp(a):(Cp(a): pp EE 7En(0)) (0))5 < H, H, contrarycontrary to 44.9.2.

In the remainderremainder of this section pickpick p as as inin 44.2144.21 andand letlet 7rn = 7r n(0) (9) -- (p). IfIf HH E E 7-1(7r),IFt(n),pick pick q q E E7r(F(H)), n(F(H)), Q E Q*(G,P(G, q), and BB aa hyperplanehyperplane ofof AA with Zz = Z(Q) Z(Q) nn Oq(H) Oq(H) n CH(B) CH(B) #0 1. As H E IFt(n),R(rr), H E M(b,M(b, ir)n) for each bEB#.b E B'. In this casecase let let M(b) M(b)=Hfor= H for bbEB#. E B'. If X(n)7-l(7r)= =0 0 then by 44.20 there is a prime q such that F*(M) == Oq O,(M) (M) forfor each M E M(a, 7r)n) andand eacheach a a E E A#.A'. Pick QQ EE Q*(G,Q*(G, q), B B a a hyperplane hyperplane of of AA with Z = Cz(Q)(B) CZ(~)(B) #01, 1, and and for forb b E E B#,B', M(b) E M(b, 7r)n) with QQ nn M(b) M(b) EE Q*P (M(b), q). By 44.15.1, Z 5

SoinSo in anyany case,case, Z (< OqO,(M(b))foreach (M(b)) for each b E B'.B#. HenceHence byby 44.15.2,44.15.2, Z E V(q,V(q, 7r).n). For a E A#,A#, pick pick MaMa E E M(a, M(a, n(0))7r(9))and and let let PP EE Q*(G,Q*(G, p). p). AsAs II (< Ma and I isis transitivetransitive on Q*(G, p), PP nn Ma Ma EE Q*(Ma,Q*(Ma, p). Thus Z(P)

Next, (2,(Z, I) <( 0(b)8(b) <5 Mb Mb forfor each b E B#, soso (2,(Z, I)I) actsacts on P0.PO. Let U == (ZI(z'}, ), and Wthe)/V the set set of of W W E E Q(G) Q(G) withwith I5I < NGNc(W), (W), W = Op Op(W)U (W) U and Po 5< Op(W).Op (W).

(44.22) Po E V(p,V(p, n(8)), 7r(0)),and and F(M) F(M) = =Op(M) Op(M) forfor each M E Pp0Q* (G).

Proof. FirstFirst Po Po EE V(p,V(p, n-(9))n(8)) byby 44.15.2.44.15.2. LetLet LL EE Q*(G)Q*(G) withwith 0(NG(PO))6(NG(Po)) (< L. As Po EE V(P, V(P,n(8)), 7r(9)),Po Po 5

(44.23) Let M, L E Q>o(G)Q* (G) andand Op(M)Op(M) <( R R EE Q(M, p) with with B(NG(C))~(NG(C)) <( L for each 11 # C char R.R. ThenThen M = L.

Proof. LetLet XX = ROp,F(M). ROp,F(M). AsAs QQ =Op(M) = Op(M) <( R, R, R R E E Sylp(X), Sylp(X), andand byby 44.22,44.22, Q == F(M), so so QQ = = F(X). F(X). Thus Thus by by Thompson Thompson Factorization,Factorization, 32.6,32.6, we have X == Nx(J(R))Cx(Z(R)), soso XX <5 LL byby hypothesis.hypothesis. As Q ==Op(X), Op(X), QQ = = Op 0, (L n M). M). By By 44.10.3,44.10.3, 0(NG(C))O(NG(C)) 5< M forfor eacheach 11 # C char Q. In particularparticular No,(~)(Q)NoP(L)(Q)5

(44.24) Let LetWEW,S=Op(W),Y=OP(W),LEQ*(G)with W E W, S =Op(W),Y = OP(W),L E Q*(G)withB(N~(Y)) 0(NG (Y))

Proof. RecallRecall ZZ E E V(q, V(q, 7r),n), so so Z Z< (Oqn 0," (M) (M) == O{ O1p,ql(M) p,q} (M)by by definitiondefinition of V(q, n).n). By construction ZZ 5< Z(Q)Z(Q) for some Q E Q*(G,Q*(G, q),q), andand asas I5I < M, Q n M M E E SylqSyl,(M). (M). Thus as ZZ <( O{p,q} OIp,,)(M), (M),Z Z 5 O OP,,(M) p,q (M) by 31.10.31.10. HenceHence U = (ZI)(2')

Nx(Y) 5< L. LetLet TT ==Op(L). O p(L). Then NT(R) 5< M byby 44.10.3, so [NT(R)*,[NT(R)*, Nx*Nx,(Y*)](Y*)] 5< NT(R)* nn Nx*Nx,(Y*)(Y*) = 1.

Therefore by the A x B Lemma, NT(R)* 5< CM.(X*)CM*(X*) 5< X*, SOso NT(R)NT(R) 5< R and hence TT <5 R.R. FinallyFinally byby 44.10.3,44.10.3, 0(NGB(Nc(C)) (C)) I< M forfor eacheach 11 # CCchar char R,R, so M=LM = L by by 44.23. 44.23.

If Mb = Mb' Mb, for all b, b' E B',B#, thenthen thethe argumentargument establishing 44.21 supplies a contradiction. Therefore Therefore as as Mb Mb EE Q*Q*&,y1(G) UI(G) for for each each b b E E B#, B#, IQP0U,(G)l ] Q>ouI(G)I > 1. 1. Pick W W E WW with S=Op(W)S = O,(W) maximalmaximal subjectsubject toto IQ*1(G)IIQ*,,(G)l> > 1.1. SetSet YY = OP(W) and pick M EE Q*(G) with 0(NGB(NG(S)) (S)) 5

(44.25) For each N EE Q*Q&,(G) I(G) -- {M}, {MJ, SS is is the the unique unique maximal maximal WI-invariant WZ-invariant member of Q(N, p).

Proof. SupposeSuppose S S < < So So E E Q(N, p)p) with with WI WZi < NG NG(So). (SO). ThenThen S < Nso(S), so without loss, S S <9 So. So. ThereforeTherefore SoSo <5 0(NG(S))B(NG(S)) < 5 M. M. Let Let WO Wo == WSo. WSo. ThenThen WOWo E E W W and and M, M, NN EE PWOIQw0I(G),(G), contradictingcontradicting thethe maximal choice of S.

(44.26) (1)(1) Q*(NG(y)) QZ;(NG(Y))(G) (G)contains contains aa uniqueunique member L. (2) Qe(NG(s))(G)QZ;(NG(S))(G) = {M}. (3) Q*PwI(G)= (G) =I {M,M, L}.LJ.

Proof. LetLet L L E E QB(NG(y))(G) Q&NG(Y))(G)and and N N E E PwI(G) Q** (G) - - {MI.{M}. ByBy 44.25,44.25, 02(N)02(N) <5 S,S, ! so by 44.24.2,44.24.2, N N == L. Thus (3)(3) holds holds and and as as I Pw1(Q*i(G)I (G)) > > 1, 1, M M # L, L, so so (1) (1) andand (2) hold.

We are now in a position to obtainobtain aa contradictioncontradiction and and hencehence establishestablish thethe Solvable 2-Signalizer Functor Theorem. By 44.26.2,44.26.2, B(NG(C))B(NG(C)) 5 < M forfor eacheach 1 ## CC charchar S,S, while while byby 44.25, OpOP(L) (L) 5::S S. S. Hence Hence M M = = LL byby 44.23, 44.23, contradicting contradicting 44.26.3 and the choice ofof W withwith I IQ*Pw1(G)I (G) I >> I.1. This completes the proof of the Solvable 2-Signalizer Functor Theorem.

Remarks. GorensteinGorenstein introduced introduced the the concept concept of of thethe signalizersignalizer functor and ini- tiated the studystudy ofof thesethese objectsobjects [Gor[Gor 1].11. HeHe waswas motivatedmotivated by earlier work of Thompson.?hompson. Goldschmidt [Gol 1,1, GolGo1 2]21 simplifiedsimplified somesome ofof Gorenstein'sGorenstein's definitions and proved the SolvableSolvable 2-Signalizer Functor Theorem and the Solvable Signalizer Functor Theorem for r oddodd whenwhen Mrm,(A) (A) >> 3. Glauberman [Gl 1]11 established the general theorem for solvablesolvable functors. Bender gave a new short proof ofof thethe SolvableSolvable 2-Signalizer2-Signalizer Functor Theorem [Be 21;2]; the proof given here is based on his proof, althoughalthough ourour proof isis longerlonger andand moremore 240 SignalizerSignalizerfunctors functors complicated thanthan Bender'sBender's forfor reasons I'll explainexplain inin aa moment.moment. Finally, Finally, McBride [Mc] proved thethe SignalizerSignalizer FunctorFunctor TheoremTheorem subject toto thethe hypothesishypothesis that the composition factorsfactors of 0(a)9(a) are known simple groups. The Classifica- tion Theorem allows this hypothesis to be removed. Our proof is longer than Bender's because because Bender uses thethe ZJ-Theorem,ZJ-Theorem, whereas we use Thompson Factorization, since thethe ZJ-Theorem isis not proved in this text.text. ThereThere areare atat leastleast twotwo advantagesadvantages toto thisthis approach: approach: First,First, thethe increaseincrease in thethe lengthlength ofof thethe proofproof causedcaused byby notnot appealingappealing toto the the ZJ-Theorem ZJ-Theorem isis probably probably less than the length of the proof of the ZJ-Theorem, although of course the ZJ- Theorem is of interest in its own right. Second, the change gives some insightinsight into how Thompson Factorization isis used in the the literature,literature, andand ThompsonThompson Factorization is used often, while the ZJ-Theorem is not.not. In the first edition of the book, we gave a proof ofof thethe SolvableSolvable SignalizerSignalizer Functor Theorem for all primes which was similar to the proof given herehere butbut still more complicated. In this edition we have opted forfor lessless generality in thethe hope that aa simpler proof willwill betterbetter exposeexpose thethe underlyingunderlying concepts.concepts.

Exercises forfor chapter 15 1. Let GG = A7, A7, A == ((1,((1,2)(3,4), 2)(3, 4), (3,4)(5,6)),(3, 4)(5, 6)), and and 0(a) 9(a) = = Os(CG(a))03(CG(a)) forfor aa Ec A#.A'. ProveProve 09 is a solvable A-signalizer functor on G which is not complete. 2. Prove Prove lemmalemma 44.7. (Hint:(Hint: Assume the lemma is false and choose P, Q Q E E Q(G,Q*(G, p)p) not conjugate underunder O(CG(A))9(CG(A))with with P P nfl Q maximal subject to these constraints. ProveProve PP nfl Q # 1 byby observingobserving Cp(a)Cp(a) # 1 # CQ(a) for some a E A',A#, and and considering considering 0(a). 9(a). Then Then consider consider NG(P NG(P fl nQ)/(P Q)/(P flfl Q) and use 44.3.) 3. Let Let AA bebe aa noncyclicnoncyclic abelian r-subgroup of a finite groupgroup GG and 09 an A- signalizer functor on on G.G. For For B B < 5 A A let let WB WE == (9 (0(b): (b): bb E E B#)B') andand letlet WW = = WA. WA. Prove (1) IfIf BB is is a a noncyclic noncyclic subgroup subgroup of of AA then then WW = = WB. WE. (2) IfIf 6(a8)0(as) = = 6(a)9 0(a)g for for each each g gE EG G and and a aE EA#, A', thenthen F2,A(G) r2,~(G) <5 NG(W), NG(W), where I'2,A(G)r2,~(G) == (NG(B):(NG (B):B B 5 < A,A, m(B)m(B) 2> 2).2). (3) AssumeAssume thethe hypothesishypothesis ofof (2)(2) andand letlet VIG(A,MG(A, r')r') denote the set of A- A- invariant rt-subgroupsr'-subgroups of G.G. ProveProve VIG(A,MG(A, rt) r') 5c NG(NG(W).W). IfIf 09 is complete, G = F2,A(G) r2,A(G) and O,t(G)Or'(G) = 1,1, prove 8(a)9(a) ==1 1 for for all all a a c- E A#.A'. 4. LetLet A A bebe anan elementary abelian abelian r-subgroupr -subgroup ofof a a finite finite group G with mm(A) (A) >_> 3.3. Prove (1) IfIf aa EE A#A' andand CGCG(a) (a) isis solvable thenthen CG(a)CG (a)is is balanced balanced for for thethe prime primer. r. (See section 313 1 for thethe definitiondefinition of balance.) (2) IfIf CG(a)CG(a) is is balancedbalanced for the prime r forfor eacheach a cE A#A# then 90 isis anan A-A- signalizer functor on G, where @(a)9(a) ==Or,(CG O,!(CG(a)). (a)). Solvable signalizersignalizerfunctors functors 24241 1

(3) Assume CG(a) is solvable forfor eacheach aa E EA' A# andand GG = = F2,A(G). I'~,A(G). Prove Or1(CG(a))Or'(CG(a)) <5 Or'(G)Or!(G) forfor each each a a c- E A#, and if O,!(G)Or'(G) = =1 1 then then F*(Cc(a))F*(CG(a)) = Or(CG(a)). Or(Cc(a)). 5. LetLet r r be be a a prime, prime, G G a a finite finite group,group, EraErs =E A A <5 G, G, andand 7r n aa set set ofof primesprimes with rrV7r.ForEr2=B en.For E,z Z B

(nO,r(CG(b)) bEB#

Assume U~,(~)(B)ac0(,)(B)

for EP2EP2 =N= B 5< AA and aa EE A'.A#. Prove yBy~ is anan A-signalizerA-signalizer functorfunctor andand a(B) <5 a(D) a(D) forfor each each pair pair B, B, D D of of hyperplanes hyperplanes of of A.)A.) 16

Finite simplesimple groups

To my mind the theorem classifying the finitefinite simple groups is thethe mostmost im-important result in finite group theory. As I indicated in the preface, the Classifi- cation Theorem is the foundation for aa powerful theory of finitefinite groups which proceeds by reducing suitable group theoretical questions to questionsquestions aboutabout representations ofof simple groups. The final chapter of this book isis devoteddevoted primarily toto aa discussiondiscussion of of thethe Classification Classification TheoremTheorem and and thethe finitefinite simple simple groups themselves. Sections 45 and 46 introduce two classes of techniques useful in the study of simple groups. Section 45 investigates consequences of the fact that each pair ofof involutionsinvolutions generates a dihedraldihedral group. The two principalprincipal results of thethe sectionsection areare thethe ThompsonThompson OrderOrder Formula Formula andand thethe Brauer-FowlerBrauer-Fowler The-The- orem. The Thompson OrderOrder FormulaFormula suppliessupplies aa formulaformula forfor thethe orderorder ofof a finite group with at leastleast twotwo conjugacyconjugacy classesclasses of involutionsinvolutions in termsterms ofof thethe fusionfusion of those involutions in the centralizers of involutions. TheThe Brauer-Brauer- Fowler Theorem shows that there areare atat mostmost aa finitefinite numbernumber ofof finitefinite simplesimple groups possessing an involution whose centralizer is isomorphic to any given group. SectionSection 46 considersconsiders the commutingcommuting graph graph onon thethe setset ofof elementaryelementary abelianabelian p-subgroupsp-subgroups of of p-rank atat leastleast kk in a group G. The determination of the groups forfor whichwhich this graphgraph isis disconnected disconnected for small k plays a crucialcrucial rolerole inin thethe Classification Theorem. Section 47 contains a brief description of the finite simple groups. The groups of Lie type areare described asas groupsgroups withwith aa splitsplit BN-pairBN-pair and generated by root groups satisfyingsatisfying a weakweak versionversion ofof thethe ChevalleyChevalley commutatorcommutator relations.relations. TheThe lastlast portion of section 47 explores consequences of these axioms. In particular thethe existenceexistence of LeviLevi complementscomplements is derived,derived, it isis shownshown thatthat thethe maximalmaximal parabolics are the maximal overgroups of SylowSylowp-groups p-groups of groups in charac- teristicp,teristic~, andand the the Borel-Tits TheoremTheoremis is (essentially) (essentially) provedproved for for finite finite groups of Lie type. This last result sayssays thatthat in aa finite group of Lie type and characteristic p, eacheach p-local isis contained inin aa maximalmaximal parabolic. Finally section 48 consists of a sketchy outline of thethe ClassificationClassification Theorem. This discussion provides aa nice illustration of of many of the techniques developed in earlierearlier chapters.chapters. Involutions ininJinite finite groups 243

45 InvolutionsInvolutions in finite groups Section 45 seeks to exploit the following property of involutions established in Exercise 10.1:10.1:

(45.1) LetLet x x andand yy bebe distinctdistinct involutions involutions of a group G. Then (x,(x, y) y) is is aa dihedral group of order 21xyl.21xy 1.

Throughout section 45, G will be assumed to be a finite group. To beginbegin let'slet's look more closely at 45.1:

(45.2) Let x and y be distinctdistinct involutions involutions in in G, G, n n = = Ixyl,IxyI, and and D D = = (x,(x, y).y). ThenThen (1) EachEach elementelement in D - (xy) (xy) is is an an involution. involution. (2) IfIf nn is is odd odd then then D D is is transitive transitive on on its its involutions, involutions, soso inin particularparticular x isis conjugate toto yy in D. (3) IfIf n n is is even even then then eacheach involutioninvolution in D is conjugate to exactly oneone ofof x,x, y,y, or z, where z is the unique involutioninvolution inin (xy).(xy). Further Further z z E E Z(D).Z(D). (4) IfIf nn is is even even and and zz isis thethe involutioninvolution inin (xy)(xy) then xz isis conjugateconjugate toto xx inin DD if and onlyonly ifif n n =- 0 0 modmod 4.

Proof. LetLet uu == xy xy andand UU == (u). (u). Then Then uX uX == u-1 u-' soso VX vX == v-1v-' forfor each v E U. Then (vx)~(vx)2 == VvvX VX= vv-'= vv-1 = =1, 1, so so (1) (1) holds. holds. Further Further for for w Ew U,E U,(VX)~ (vx)'== vx vxw =_ VW-'wXxvw-1wXx = vw-2x,soxD VW-~X,soxD = ={v2x:v (v2x: E u U}.E U]. In Inparticularifnparticular if n is is odd oddthenxD then XD = D - U U and and U U contains contains no no involutions, involutions, so so (2) (2) holds.holds. So take n eveneven and let z be the involution in U. Then

D - U {V2x: V E U} U {V2ux: V E U} with ux = y, y, so so D D - -U U= xD= xD U yDU y andand (3) (3) holds. holds. Finally Finally zx zx E E XD x preciselyprecisely when z is a square in U; that is whenwhen nn = 00 mod 4.

(45.3) LetLet GG bebe ofof even order with Z(G)Z(G) = 1, 1, let let m be the number of involutionsinvolutions in G, and n = Gm. IGllm. Then Then G possessesG possesses a proper a proper subgroup subgroup of of index index at at most most 2n2.

Proof. LetLet I I be be the the set set of of involutionsinvolutions ofof G,G, RR thethe setset ofof elements of G inverted by a member of I, and (xi :: 0 0 5< i <5 k) k) a a set set of of representatives for the conjugacy classes of GinG in R. R. Pick Pick xo xo = = 1 1and and let let mi mi == IxG I,I, letlet Bi Bi bebe thethe setset of pairs (u, v)v) with u, v EE II and and uv uv ==xi, xi, and andlet let bi bi = = I Bi IBil. I. Observe firstfirst that that if if u, u, v v E E I I thenthen eithereither u u = = vv andand uvuv == 1 or (u,(u, v) is dihedral. InIn eithereither casecase u u invertsinverts uv, uv, so so uv uv EE R.R. ThusThus countingcounting I I xx I inin twotwo 244 Finite simple groups different ways, we obtain:obtain:

k (a) m2=II x71 = kmibi. i-0 Moreover there is an involution ti inverting xi and ifif (u,(u, v) E BiBi then u inverts xi and v = uxi. uxi. Hence Hence the the map map (u, (u, v) v) -r H u isu anis aninj injection ection of of Bi Bi into into ti tiCG CG(X~) (Xi) soSO bi

(b) m2 5 0 thenthen Ek by hypothesis xi V6 Z(G), soso mimi == I IGG :: CG(xi)I CG(xi)l L> s.S. HenceHence IGI(GI > 1 ~f=~o miml 3> 1 + ks,ks, whichwhich II recordrecord as:as: (c) k < (IGI - 1)/s. Combining (b) and (c) gives:gives:

(4(d) m2 5 < (lGI(IGI (IGI(IGI - -1)/s)+m. l)/s) + m. Then, asas nn == IGIIGllm, /m, itit follows follows from from (d) (d) that that s 5s 2, s <5 2n2.2n2.

(45.4) Let G be aa finitefinite simplesimple group of even order, tt an involution inin G, and n = I lCG(t)l.CG(t)I. ThenThen IG1IGI (< (2n2)!. (2n2)!.

Proof. By 45.3, G possesses possesses a properproper subgroup H,gfH 9f indexindex atat mostmost 2no,24, where no = I I G l/rnI /m andand rnin is the number of involutions inin G.G. Then Then m m > q I ItGtG 1I == (GIG : CG(t)l,CG(t)1,SO so no no i < IGJIJG:IGI/IG : CG(t)JCG(t)1 = = n. Represent G as a permutationpermutation group onon thethe cosetcoset spacespace G/H.G/H. Then k == IIG/HI G/HI 5< 2n2, and as G is simple the representation is is faithful. SoSo GG isis isomor-isomorphic to a subgroup of the symmetric group Sk, andand hencehence JGIIGI (< k! <5 (2n2)!. (2n2)!.

As an immediate corollary to 45.4 we have the following theorem of Brauer and Fowler:

(45.5) (Brauer-Fowler [BF])[BF]) LetLet H bebe aa finitefinite group.group. ThenThen therethere existsexists atat most a finite number of finite simple groups G with an involution t suchsuch thatthat CG(t) = H.

Recall that the Odd Order Theorem of Feit andand ThompsonThompson sayssays thatthat nonabeliannonabelian finite simple groups are ofof eveneven orderorder andand hencehence possesspossess involutions. involutions. TheThe Brauer-Fowler Theorem suggestssuggests it it isis possible to classify finite simple groups Connected groups 245

in terms of thethe centralizerscentralizers ofof involutions.involutions. ThisThis approachapproach willwill bebe discusseddiscussed inin more detail in sectionsection 48. If G hashas moremore thanthan oneone classclass ofof involutionsinvolutions it is possible to make a much more precise statementstatement than 45.5:45.5:

(45.6) (Thompson(Thompson OrderOrder Formula) AssumeAssume G has kk 2> 22 conjugacyconjugacy classes of involutions xxf, 1 <5 ii <5 k, k, and and define define ni ni to to be be the the number number of of orderedordered pairspairs (u,v)with(u, v) with u Ex1E xf, ,v u Ex2,andxiE xf, and xi c E (uv). (uv). ThenThen

k IGI = ICG(xi)IICG(x2)I ni/ICG(xi)I i=1

Proof.ProoJ LetLet II be be the the set set ofof involutionsinvolutions inin G and QQ == xGxf x x2x:. .Then J X G JJ X G J IG : CG(xi)I IG : CG(x2)I (*) ICI = 1 2 = For (u, v) E Q,a, uu $ vG, soso by 45.2 there is a unique involution z(u, v) in (uv). Hence QQ isis partitionedpartitioned byby thethe subsetssubsets QZ Q, consisting of those pairs (u,(u, v)v)

with zz = z(u,z(u, v).v). ThusThus IQI1Q1 = C,,,>-+ZEI IQz1.lQzl. Further Further 1Q,1IQ,I == IQx,1Q,1 Ifor for z z Ec xc,xP, soso Y IQzl = IxGIIQxiI = IG:CG(xi)Ini ZExG Therefore 101 k (**) = IG:CG(xi)Ini. i=1 ' Of course the lemmalemma followsfollows fromfrom (*) (*) andand (**).(**).

Notice that the integer ni can be calculated if xxy g nfl CG(xi),CG (xi ),j j= = 1,1,2, 2, is knownknown for each i. HenceHence thethe order of GG isis determineddetermined by thethe fusionfusion of xl andand x2x2 inin CG(xi), 1 1 5< ii <5 k. k. In In particular particular the the order order of of G G can can be be determined determined fromfrom thethe fusion of involutions in local subgroups.

46 ConnectedConnected groups In this section G isis aa finitefinite group and p isis aa prime.prime. If Q is a collectioncollection of subgroupssubgroups ofof G permutedpermuted by G via conjugation,conjugation, define g(Q)5(Q) toto be be the the graph graph on on Q Q obtained obtained by by joining joining A A to to B B if if [A, [A, B] B] == 1.1. Evidently G is represented asas aa group of automorphisms of g(Q)5(Q) viavia conjugation.conjugation.

(46.1) Let A be a G-invariant collectioncollection ofof subgroupssubgroups of of G G and and H H 5< G. Then the following areare equivalent:equivalent:

(1) HH controls controls fusionfusion in H fln A and NG(X) _(< H forfor eacheach X E H fln A.A. (2) HH flrl HgHg fln A A isis empty forfor g g EE G G -- H.H. 246 Finite simple groups

(3) TheThe members members of of HH flil AA fixfix a unique point in thethe permutationpermutation representationrepresentation of G on G/HGIH by by right right multiplication. multiplication. (4) HH flil A A isis the the unionunion ofof aa setset Fr ofof connectedconnected components of of LB(A), 2(0), andand rnrfl Fgrgisemptyforg is emptyforg E cG G-H. - H.

Proof. AssumeAssume (1) and let g cE GG withwith HH flil HgHg fl il A A nonempty. nonempty. ThenThen there is X E H fln A with Xg 5< HH so, so, asas HH controls controls fusionfusion in H flil A,A, Xghxgh == X forfor some hh E H. Then gh EE NG(X)NG(X) <5 H,H, so so g g EE H.H. Thus Thus (1) (1) implies implies (2).(2). Assume (2) andand considerconsider the the representation representation of of G G on on GI G/H.H. Then XX E H flil A fixes HgHg if and only ifif X <5 Hg,Hg, whichwhich holds precisely whenwhen gg Ec H byby (2).(2). Thus (2) implies (3). Assume (3).(3). Then, Then, for for A A E E H H ilfl A, A, {H){H} = = Fix(A), so NG(A)NG(A) 5< H, as H is the stabilizer inin GG of thethe cosetcoset H. InIn particular ifif B E A is incidentincident to A in LB(A)1(0) then then B B < 5 C(A) C(A) < 5H, H, so so B BE EH Hfl ilA.A. Thus Thus H H fl il AA is is the the union union ofof somesome set rF of connectedconnected components components of of g(A). 2(0). Further if A E 6'8 E Fr andand 8g6'g E Fr then Ag (< H,H, so so {H}{H) == Fix(Ag) Fix(A8) = = {Hg} {Hg) andand hence hence gg EE H.H. So So (3) (3) implies implies (4). (4). Finally assume (4).(4). If If X X E E H H ilfl A A andand g g EE GG withwith Xg X g<5 H, then XX E 6'9 E I'r andXg EE 6"8' EE r,I',so8'SOB' == 8g 6'8 EE I'nFg.Hence,rnrg. Hence, by(4),gby (4), g EE H.So(4)implies(1).H. So (4)implies (1).

If k isis aa positivepositive integerinteger writewrite Sk&l(G) (G) for thethe setset ofof allall elementaryelementary abelianabelian p- subgroups of G of p-rank at least k. GG isis saidsaid toto bebe k-connectedk-connected for thethe prime p if 0(61k(G))g(&j!(G)) is connected.

(46.2) Let G be aap-group. p-group. Then (1) GG isis 1-connected1-connected forfor the prime p. (2) IfIf m(G)m(G) >> 2 2 then then therethere is Ep2Epz E= U 2.2. (3) IfIf pp = = 2 2and and G Ghas has a normala normal E8-subgroup E8-subgroup then then GG is is 2-connected 2-connected for the prime 2. (4) IfIf pp = = 3 3and and m(G) m(G) > >3, 3, then then G G is is 2-connected 2-connected for for the the prime prime 3.3.

Proof. Part (1) follows fromfrom the the fact fact that that Z(G) Z(G) #,- 1. Assume m(G) m(G) >> 2;2; by Exercise 8.48.4 therethere is is Ep2Ep2 E = U9U< G. G/CG(U) ( SL2(p)SLz(p) and SL2(p) is of p-rank 1. So if A E 63(G)&;(G) thenthen m(CA(U)) 2> 2,2, soso m(CA(U)U) > 2.2. HenceHence m(CG(U)) > 2.2. Also, if uU ,-# EE E E S' &,P(G) (G) isis in in the the same same connected connected component as U, then there is E # D EE °z8; (G) (G) adjacent toto EE andand m(CG m(CG(E))(E)) 2 > m m(DE) (DE) 2> 3. Thus to prove (2) it remains to show that each A EE 613&f(G) (G) isis in in the the connected connected component of U.U. ButBut A, A, CA(U), CA(U), CA(U)U,CA(U)U, UU is is a apath path in in _T(6'2P(G)), LB(&;(G)), soSO thethe proof of (2) is complete. Assume p = 2 2 andand E8E8 - E Va V9 G. G.Let Let E EE 6122(G);E &$(G); toto proveprove (3)(3) we must exhibit apathfroma path from EE toto V.V. ForFore e cE E,E, m m(Cv(e)) (CV (e)) > 22 by Exercise 9.8, and wewe maymay assume there isis ee Ec E -- V, V, so so E, E, (e, (e, CV Cv(E)), (E)), Cv Cv(e), (e), V V is is a apathinpath in 2(d0z g(&;(~)). (G)). Connected groups 247

Similarly under the hypothesis of (4) there is EslE81 = VV 9a G G byby ExerciseExercise 8.11, 8.1 1, after which we can use Exercise 9.8 and thethe argument ofof thethe lastlast paragraphparagraph toto establish (4).

Given a p-group P actingacting asas aa groupgroup ofof automorphismsautomorphisms of G,G, andand givengiven aa positive integer k, define I'P,k(G)~P,L(G) == (NG(X):(NG(X): X X 5< P, m(X) > k)k) I'Pr:,,(~) k(G) = (NG(X):(NG(X): X X 5< P, m(X) > k,k, m(XCP(X))m(XCp(X)) > k).k). The following observation is anan easy consequenceconsequence of of ExerciseExercise 3.2.1: 3.2.1:

(46.3) Let H <( G, G, P P E ESylp(H), Sylp(H), and and k ka apositive positive integer. integer. Then Then eithereither ofof thethe following implies thatthat PP EE Sylp(G):

(1) m(P) >> kandFP,k(G)kandrp,k(G) <5 H (2) m(P)m(P) >> k k and and I'Pr$.k(~) k(G) 5< H.H.

(46.4) LetLet H <5 G,G, PP EE Sylp(H),Sylp(H), andand m(P)m(P) >> k. k. ThenThen thethe following following are are equivalent:

(1) FP,k(G)~P,L(G) 5< H. (2) HH controls controls fusion fusion in in olk&i(H) (H) and NG(X) (< HH forfor eacheach XX EE 6k&:(H). (H). (3) mmp(H p(H ilfl Hg) Hg) << kforeachgk for each g E E G- G - H. (4) Each Each member member of of &'k &f(G) (G) fixes a unique point in the permutation represen-representation of G byby rightright multiplication on GIH.G/H.

Proof.Prooj PartsParts (2),(2), (3),(3), andand (4)(4) are are equivalentequivalent byby 46.1,46.1, exceptexcept inin (4),(4), °k&:(G) (G) should be &:(H).gk(H). ThisThis weakerweaker versionversion of (4)(4) evidentlyevidently implies (1) and (1) and 46.3 show P EE Sylp(G),Sylp(G), from from which which thethe strongstrong versionversion of (4)(4) follows. It remains to show (1) implies (2).(2). Assume rP,L(G)FP,k(G) 5< H andand let XX EE °k&;(H). (H). ItIt sufficessuffices to show thatthat ifif gg E G with Xg 5< H,H, then then gg EE H.H. By By Sylow's Sylow's TheoremTheorem wewe may assume (X, Xg)Xg) 5< P. ByBy 46.3,46.3, PP EE Sylp(G). Sylp(G). By By Alperin's Alperin's Fusion Fusion Theorem Theorem therethere existsexists Pi EE Sylp(G), 11 5 k, SOso gig; EE NG(PNG(P fl f Pi)Pi) 5< FP,k(G)rP,L(G) 5< H. Hence gg == gl ...... g g, E EH, H, so so that that (2) (2) holds. holds.

If P EE Sy1p(G)Sylp(G) then then FP,k(G)rP,L(G) isis called the k-generatedk-generatedp-core p-core ofof G.G. ByBy SylowSylow the k-generatedp-corek-generatedp-core ofof G is determined up toto conjugacyconjugacy inin G.G. TheThe hypothesishypothesis of 46.4 are equivalent toto thethe assertionassertion thatthat HH contains the k-generatedp-corek-generated p-core ofof G, and hence, if H is a proper subgroup of G, thatp-core isis aa properproper subgroupsubgroup of G. By 46.1, if the k-generatedp-core isis proper then G is k-disconnected for 248 Finite simple groups the prime p, although thethe converse need not be true. However inin thethe followingfollowing special case the converse isis valid:valid:

(46.5) Let P EE Sylp(G)Sylp(G) and and assumeassume PP isis k-connected. k-connected. ThenThen (1) ek&;(rp,k(G)) (1, p,k(G)) isis aa connectedconnected component component of of _q((ffk LB(&:(G)) (G)) and FP,k(G)rp,k(G) is is thethe stabilizer in G of thatthat connectedconnected component.component. (2) GG isis k-disconnectedk-disconnected for the prime pp if and only if G hashas aa properproper k- generated p-core.

Proof. EvidentlyEvidently (1) (1) impliesimplies (2).(2). Let H == i,P,k(G). ~P,~(G). ByBy 46.1 46.1 and and 46.446.4 ek&:(H) (H) is the union of connected componentscomponents of of _T((ffk LB(&;(G)), (G)), while, while, asas PP isis k-connected, &[(P)(9k (P) is is contained contained in in some some component component A A and and of of course course H H <_< NG(A).NG(A). Hence &:(H)ek (H) CE AA byby Sylow'sSylow's Theorem. So A = ,gk &:(HI (H) andand HH = NG(A) NG(A) byby 46.1.4. That is (1) holds.

If H is a proper subgroup of G satisfying any of the equivalent conditionsconditions ofof 46.4 with k = 1, 1, then then wewe saysay HH is is a a strongly strongly p-embedded p-embedded subgroup of G. As a direct consequence of 46.5 and 46.2.1 we have:

(46.6) GG possesses aa strongly p-embeddedp-embedded subgroupsubgroup if if and and onlyonly ifif G isis 1-disconnected for the prime p.

(46.7) Let mp(G) >> 2,2, P P E ESy1P(G), Sylp(G), and and let let (92(G)° &;(G)' denotedenote thethe setset ofof sub-subgroups XX E &;(G)o' (G) withwith mp(XCG(X))9(XCG(X)) >> 2. Then (1) e2&:(G)' (G)° isis the the set set of of points points ofof e2&:(G) (G) which are not isolated in LB(&:(G)).(o'2 (G)). (2) Sz&;(ri,,(G))' (I'P 2(G))° is a connectedconnected component ofof LB(&;(G))0(ff2 (G)) andand ri,2(~)rP 2(G) is the stabilizer in G of this connected component. (3) LB(&;(G)')1((ffz (G)°) is isconnected connected if ifand and only only if if G G = = I';,,(G).F , 2(G).

Proof. Part (1) isis trivial,trivial, asas isis (3)(3) givengiven (2).(2). SoSo itit remainsremains toto establishestablish (2).(2). By 46.2.2, &;(P)'e (P)° isis contained contained in in a a connected connected componentcomponent AA ofof ?(ff2LB(&;(G)). (G)). Thus HH = rPr;,,(G) 2(G) <5 NG(A).NG(A). LetLet rF == &:(H)'.ez(H)°. ASAs &;(P)'ff2(P)° Ec AA andand H actsacts on A, I'r CE A A by by Sylow'sSylow's Theorem. Theorem. IfIf AA 0# 17 r there is xx E F,r, Y E A - 17 r with X and Y adjacent inin LB(A).1(A). WithoutWithout loss X E &;(P)',ff2(P)°, so Y 5< CG(X) (< H. Hence, as mp(CH(Y))mp(CH(Y)) 2> mp(XY) >2 3,3, YY EE F,r, aa contradiction.contradiction. So A == F. r. Thus, Thus, to to complete complete the the proof proof ofof (2),(2), itit remainsremains toto show show that if X, Xg E rI' then g E H. SupposeSuppose X, XgXg EE r. Then Then NG(X) 5< HH >2 NG(X9) NG(Xg) so there isis Ep3EP3 = AA i:< CG(X)CG(X)

Lemma 46.7 says that if mp(G)m p(G) > > 2 and P EE Sylp(G),Sylp(G), then, then, neglectingneglecting thethe isolated pointspoints of LB(&:(G)),1((g2 (G)), G G is is 2-connected 2-connected for for the the prime prime pp preciselyprecisely when G == ~O,,,(G).F 2(G) These observations are very usefuluseful inin conjunctionconjunction withwith thethe Signalizer Signalizer FunctorFunctor Theorem, as the next two lemmas andand ExerciseExercise 16.116.1 indicate.indicate.

(46.8) LetLet 17r bebe thethe setset ofof elementselements aa of G of order p withwith Mp(CGmp(CG(a)) (a)) > 22 andand let 89 be aa mapmap fromfrom r17 into into the the set set of of p'-subgroups p'-subgroups of of G G such such that, that, for for all all a, a, b b E E rF with [a, b] = 11 andand allall g EE G,G, 9(ag) B(ag) == 9(a)9 8(a)g and and 9(a)B(a) fl fl CG(b)CG(b) <( 9(b). 8(b). LetLet P EE Sylp(G),Sylp(G), assume G = F I';,,(G), , 2(G), Op-(G)Op!(G) = 1,1, and and either

(1) 9(a)8(a) isis solvable solvable for for each each a a E E 17, r, or (2) the the SignalizerSignalizer Functor Functor TheoremTheorem holds on G.

Then 8(a)9(a) = 11 for each aa E r.F.

Proof. For A A E E (P3d':(G), (G), 8 9 isis an A-signalizerA-signalizer functor byby hypothesis.hypothesis. For For B EE (ffz&;(G)' (G)° define define WEWB= = (8(b):(9(b): bb E B').B#). Then there existsexists AA EE 6'3 &:(G) (G) with B <5 AA and, and, by by Exercise Exercise 15.3, 15.3, WB WE == WA WA andand FA,2(G)rA,,(G) < ( NG(WA). NG(WA). In particular if B andand DD are are distinct distinct membersmembers of S2&:(G) (G) adjacent inin LB(&;(G))1(ez (G)) then then BD EE 6'3&f(G) (G) so WB == WBDWBD == WDWD.Thus, Thus, as F0°I';,,(G) 2(G) == G,G,46.7 46.7 says WB = WW is independentindependent ofof thethe choicechoice of of B B EE &;(G)'.(9z (G)°. Thus Thus G G = = r!,,(~) rP 2(G) (< NG(W) as NG(B) <( FA,2(G) rA,,(G) < ( NG(W). NG(W). But,But, byby (1)(1) andand thethe SolvableSolvable SignalizerSignalizer Func- tor Theorem oror byby (2),(2), WW isis aa p'-group,p'-group, so W (< Op,(G).Opf(G). As Opf(G)Op,(G) == 11 by hypothesis, and as 8(a)9(a) < ( W W for for each each a a E E 17, r, thethe lemmalemma is established.

(46.9) LetLet 17r bebe thethe setset ofof elementselements aa of G of order p withwith mp(CGmp(CG(a)) (a)) > 2, letlet P EE Sylp(G),Sylp(G), andand assume I';,,(G)F 2(G) == G, G, Op,(G)Op!(G) == 1, 1, CG(a) CG(a) isis balanced balanced forfor the prime p forfor eacheach aa E E 17, r, and either

(1) Op'(CG OPt(CG(a)) (a)) isis solvable forfor eacheach aa E r,F, or (2) the the SignalizerSignalizer FunctorFunctor TheoremTheorem holds on G.

Then Opt(CG(a))Op'(CG(a)) == 11 for each aa E 1.r .

Proof. ForFor aa E E F r let let 9(a) 8(a) == Opr(CG(a)). Op~(CG(a)). ByBy Exercise Exercise 15.4.2,15.4.2,8(a) 9(a) fl CG(b) (< 9(b)8(b) for each a, b b E E 17r withwith [a, b] = 1. 1. Hence Hence 46.846.8 completescompletes the proof.

47 TheThe finitefinite simple groups Section 47 describesdescribes a a listlist .7CYC of finite simple groups. The Classification The- orem says that any finite simplesimple groupgroup isis isomorphicisomorphic to to a a member member of of YC. X. The proof of the ClassificationClassification Theorem is far beyond the scope of this book,book, butbut there is aa brief outlineoutline ofof thatthat proofproof inin thethe finalfinal sectionsection ofof thisthis chapter.chapter. 250 Finite simple groups

.XC can be described as the following collection of groups: (1) TheThe groupsgroups ofof primeprime order.order. (2) TheThe alternatingalternating groups A,An ofof degreedegree nn >> 5. (3) The The finitefinite simplesimple groupsgroups ofof Lie type listedlisted inin Table 16.1. (4) TheThe 2626 sporadicsporadic simplesimple groupsgroups listedlisted inin TableTable 16.3.16.3. The groups of prime order areare thethe abelianabelian simplesimple groupsgroups (cf.(cf. 8.4.1).8.4.1). TheyThey are the simplest of the simple groups. By 15.16,15.16, the alternating groups A,,An, nn >2 5,5, are are simple. simple. TheThe permutationpermutation representation of of AnA, of degree n is an excellent tool for investigating the group, and, as in 15.3 15.3 and Exercise 5.3, to determine its conjugacy classes. Lemma 33.15 describes the covering group and Schur multiplier of A,.An. ExerciseExercise 16.2 says Aut(A,) = S,,,S,, unlessunless nn == 6.6. The groups of Lie type are the analogues of the semisimple Lie groups. This class of groups is extremely interesting; for one thing by some measure most finite simple groups are of Lie type. We've alreadyalready encounteredencountered examplesexamples ofof these groups; thethe classicalclassical groupsgroups areare ofof LieLie type.type. The groups of Lie type can be described in termstenns of various representations: as groups of automorphisms of certain Lie algebrasalgebras oror fixedfixed points ofof suitablesuitable automorphisms of such groups; as fixed points of suitable endomorphisms of semisimple algebraicalgebraic groups;groups; asas groups of automorphisms ofof buildings with finite WeylWeyl groups;groups; asas groupsgroups withwith aa BN-pairBN-pair andand aa finitefinite Weyl Weyl group. group. I'll I'll taketake a modified version of this last point of view here to avoid complications. A finite group of Lie type satisfies the following conditions:

(Ll) GG possesses possesses a a TitsTits systemsystem (G,(G, B,B, N,N, S)S) withwith aa finitefinite Weyl groupgroup WW = N/(B fl fl N). N). The The Lie Lie rankrank of of G G is is the the integer integer I JSJ.S!. (L2) HH == B B fl n N N possesses possesses a a normal normal complement complement U in B. (L3) There isis aa rootroot systemsystem Z E forfor W and aa simple systemsystem n r for EZ withwith S = (sa:(s,: a E n)jr) wherewhere sas, is thethe reflectionreflection throughthrough a. a. LetLet E+ bebe thethe positive system ofof n.rr.There There exists exists an an injection injection a a F-+H UaU, ofof EZ intointo thethe set of subgroups ofof GG suchsuch thatthat for for each each a a EE ZE andand ww EE W, HH <5 NG(Ua)NG(U,) and (Ua)'(U,), == U. U,,. Each Each member member of ofU Ucan can be be written written uniquely uniquely asas aa productproduct

l-IaEC+u,,with u,ua EE Ua,U,, inin some fixed ordering ofof ZE+ + respecting height. (LA)(L4) TheThe root groups (U,:(Ua: aa E E+)Z +) satisfy satisfy [Ua, UU] C (Z(Ua) fl Z(UU)) f Uia+lp,a,/R E E+, where the product is over all rootsroots iai a ++ jBj,8 with i and j positivepositive integers.integers. We will be most interested in groups which alsoalso satisfy:satisfy: (L5) Let Let w0 wo bebe thethe unique member of WW ofof maximal lengthlength in WW (cf.(cf. Exercise 10.3). ThenThen G G == (U, U'O),UWo), B nfl B'OBWO = H, H, and and WW isis irreducible.irreducible. TheThe$nite finite simple groups 25125 1

Some observations about these axioms are inin order.order. DefineDefine a TitsTits systemsystem T == (G, (G, B,B, N, N, S) S) to to bebe saturated if ow,,flwB'BW = = H. H. By By Exercise Exercise 14.9.5, 14.9.5, TT isis saturated ifif andand onlyonly if if BB nfl BWOB"'0 = = H.H. So thethe conditioncondition U u nfl UWOU"'0 == H inin L5 can be replaced by the hypothesis that TT isis saturated.saturated. Next, from ExerciseExercise 10.3,10.3, Zf E+wo wo == E-.Z-. Hence, Hence, by by L3,L3, 47.1.1,47.1.1, andand 30.7,30.7, UaU, = U U fl fl U"'0,° U wOSa for a EE r.n Moreover. Moreover the the proof proof ofof 47.1 depends onon LLl-L4 1-LA butbut not L5. By 30.9 eacheach yy E EZ cancan bebe written y == aw aw for for some some a a E ETr, n, Ww EE W.W. So, by L3, U, == UaN, U,, _= (Ua)"' (U,)W = = (U (U fl n UwOS°)'. UwOS~)W. Hence L1-LAL1-L4 determine the root groups U,,Uy, y EE EZ uniquely.uniquely. Conversely if T isis aa saturated saturated Tits Tits systemsystem satisfyingsatisfying L2L2 thenthen byby ExerciseExercise 14.10 wewe can can define define U, Uy = = (U (U n n UWOSa)W, U'0'°)', wherewhere y y == aw, a EE r,n, andand w w EE W.W. Then, by ExerciseExercise 14.1014.10 parts (7) and (9), L3L3 isis satisfied.satisfied. Thus L3 can be dispensed with ifif we assume LlL1 and and L2L2 with T saturated.saturated. EquivalentlyEquivalently L1,L1, L2,L2, and L5 imply L3.L3. The properties L1 and L2 essentiallyessentially characterize the finite groups of Lie type (cf.(cf. [FS],[FS], [HKS],[HKS], andand [Ti]).[Ti]). There areare somesome ratherrather trivialtrivial examplesexamples of groups of Lie rank 11 satisfyingsatisfying L1-L5Ll-L5 whichwhich areare not not ofof LieLie type. type. The The finite finite groups of Lie typetype satisfyingsatisfying L5 andand possessingpossessing a trivialtrivial centercenter areare listedlisted inin Table 16.1. Column 1 lists G. The parameters qq and n are an arbitraryarbitrary prime power and positive integer, respectively, unlessunless somesome restrictionrestriction isis listed explicitly. Column 2 lists the order of a certain central extension G ofof G.G. UsuallyUsually G isis simple,simple, Ge isis thethe universaluniversal covering group of G, and d(G) isis the order of the Schur multiplier ofof G.G. InIn anyany eventevent ]GIIGI == I61/d(G).~el/d(~). Finally,Finally, columncolumn 4 lists the Dynkin diagram of the root system EZ ofof G.G. TableTable 16.216.2 explains explains thethe notation An,A,, B,B,,, etc. etc. The The nodes nodes of of the the Dynkin Dynkin diagram diagram are are indexed indexed byby thethe setset nr ofof simplesimple roots. roots. DistinctDistinct nodesnodes aa and and ,B B areare joinedjoined byby anan edgeedge ofof weightweight 2(a, ,8)(,8, B)(B, a)/(a, a)(,8,a)(B, ,B)B) = map rn,~ -- 2. 2. It It turns turns out out Isasp Is,sSI I == man, rn4, so, so, if if wewe ne-ne- glect the arrows in the diagram, the Dynkin diagram of ZE isis just thethe CoxeterCoxeter diagram of the Weyl group W of EZ oror G,G, asas defined in section 29. By 30.9, each member of EZ isis conjugateconjugate to to aa member of rn under under W, W, and and it it turns turns out out that that there there are one or twotwo orbitsorbits of WW onon EZ dependingdepending on on whether whether thethe DynkinDynkin diagramdiagram has no multiplemultiple bondsbonds oror hashas multiplemultiple bonds, bonds, respectively.respectively. Indeed ifif aa andand ,B B are joined byby anan edgeedge of weight 11 then a isis conjugate conjugate to to ,B B under (sa,(s,, so).sS). If the diagram has multiple bonds, roots in differentdifferent orbits are of different lengths, and hence are called long or short roots. The arrowarrow in the diagram points to the shortshort root in thethe pair joined. Let q be a power of the prime p.p. p isis calledcalled the characteristic ofof G.G. RecallRecall that G cancan bebe defineddefined inin termsterms of one of severalseveral representations. Consider forfor the moment the representationrepresentation of G asas aa groupgroup ofof automorphismsautomorphisms of aa LieLie algebra L. LL isis aa Lie Lie algebra algebra over over aa field field ofof characteristiccharacteristic pp obtainedobtained fromfrom a simplesimple Lie algebra L' overover thethe complexcomplex numbers. L' hashas anan associatedassociated rootroot 252 Finite simple groups

Table 16.1 Finite groups of Lie type with L5 and trivial center

G IGld(G)IGId(G) = IGII~I d(G) ZE

Jr An(q) qn(n+l)/2 rl(gi+l - 1) (n+1,q-1) An i=1 q"2 Bn(q), q odd fl(g2, - 1) (2, q - 1) Bn i=I

Cn(q) q"2 fl(g2i- 1) (2, q - 1) Cn i=1 n-1 Dn(q), n > 2 qn(n-l)(qn - 1) (g2i - 1) (4, qn - 1) Dn i=1 E6(q) g3fi(g12 - 1)(q9 - 1)(q8 - 1) (3, q - 1) E6 (q6 - 1)(q5 - 1)(q2 - 1) E7(q) g63(g18 - 1)(g14 - 1) (2, q - 1) E7 (812 - 1)(g10 - 1) (q8 - 1)(q6 - 1) (q2-1) g120(g3° - 1)(g24 - 1) E8(q) 1 E8 (q20 - 1)(g18 - 1) (814 - 1)(g12 - 1) (q8 - 1)(q2 - 1) F4(q) g24(g12 - 1)(q8 - 1) (q6 - 1)(q2 - 1) 1 F4 G2(q) q6(q6- 1)(q2 - 1) 1 G2 n 2An(q) qn(n+1)/2 rl(gi+l -(-1)i+l) (n + 1, q + 1) C[n+l/2] i=1 2B2(q), q = 22,n+1 q2(q2 + 1)(q- 1) 1 Al n-1 2Dn(q), n > qn(n-l)(q" + 1) rl (q2i - 1) (4, q" + 1) Cn-1 i=1 3D4(q) g12(g8 + q4 + 1)(q6 - 1) (q2 - 1) 1 C2 2E6(q) g36(g12 - 1)(q9 + 1)(q8 - 1) (3, q + 1) F4 (q6 - 1)(q5 + 1)(q2 - 1) 2F4(q), q = 22m+I gl2(g6 + 1)(q4 - 1) 1 dihedral 16 (q3 + 1)(q - 1) 2G2(q), q = 32m+1 q3(q3 + 1)(q- 1) 1

system E'C' withwith DynkinDynkin diagramdiagram V. C'. If If G G is is of of type type A, A, B, B, C, C, D, D, F, E oror G then G is called an ordinaryordinary Chevalley group group and and C E == V.C'. On On thethe otherother handhand the groups of type "X'X areare called called twisted twisted Chevalley Chevalley groups. groups. mX "X(q) (q) is the fixed points on anan ordinary Chevalley groupgroup ofof type X(qe) of a suitablesuitable automorphismautomorphism of order m, and inin thisthis casecase XE isis notnot equalequal to to XI.V. It turns outout alsoalso thatthat U is a Sylow p-subgroup of G, where p isis thethe characteristic of G (cf. 47.3). TheThejnite finite simple simple groups groups 253

TableTable 16.216.2 Some Some Dynkin Dynkin diagrams diagrams

1 2 n-I n A. o--- ... - B 1 2 n-2 n- In C 1 2 n-2 n-1 n

n-1 1 2 -<

1 2 3 5 n-I n En 0-0

1 2 3 4 F4 o ate-c

1 2 G2 o

1 122 DihedralDihedral 1616 ^- 66 ^

TheThe classical classical groups groups areare groupsgroups ofof LieLie type; type; namely namely An(q) An(q) = = Ln+1(q), Ln+l(q), Bn(q)Bn(q) = P22n+1(q),PQ2n+l(q), for qq odd,odd, Cn(q)Cn(q) = PSP2n(q), PSp2n(q), Dn(q)Dn(q) == PQ2n(q), f'filn(q), 2An(q)2~n(q) == Un+1(q), Un+l(q), andand 2Dn(q) Dn(q) = = PS22n(q). P!2,(q). AlsoAlso the the groups groups 2B2(q) B2(q) were were first first discovereddiscovered as as permutation permutation groupsgroups by by M. M. Suzuki Suzuki and and hence hence are are called called Suzuki Suzuki groupsgroups and and sometimes sometimes denoted denoted by by Sz(q).Sz(q). ThereThere are are some some isomorphisms isomorphisms among among the the groupsgroups of of LieLie type,type, andand ofof groupsgroups ofof LieLie typetype withwith alternatingalternating groups. groups. AlsoAlso somesome centerlesscenterless groupsgroups ofof LieLie typetype areare not simple. Here's thethe listlist ofof such exceptions; we've alreadyalready seenseen a number ofof them:them:

A1(q) = B1(q) - C1(q) 2A1(q) - L2(q), B2(q) - C2(q) PSP4(q), D2(q) - PS24 (q) = L2(q) x L2(q), 2D2(q) = PS24 -(q) --- L2(q2), 2A3(q). D3(q) - A3(q), 2D3(q) InIn particularparticular recall recall L2(2) L2(2) and and L2(3)L2(3) areare notnot simple,simple, andand ofof coursecourse neitherneither isis L2(q)L2(q) x L2(q).L2(q). AlsoAlso C2(2) %- S6, S6, andand 2A2(2) 2~2(2) -% PSU3(2) Psu3(2) are are not not simple. simple. IfIf 254 Finite simple groups

G = G2(2) G2(2) or or 2F4(2)'~~(2) then IGIG : G(I))G(')I = 22 withwith G(I)G(') simple. (G2(2))(1)(~~(2))(') E= U3(3). (2~4(2))(1)(2F4(2))(1)is is called called the the TitsTits group. group. '~'(2)2B2(2) is is aa FrobeniusFrobenius group of order 20,20, and hence solvable. 12G2(3) I2G2(3) : :2G2(3)(1)1 2G2(3)(I)I= = 33 withwith '~~(3)(')2G2(3)(1 ="= L2(8). All other centerless groupsgroups ofof Lie Lie type type are are simple. simple. L2(4) L2(4) E = L2(5) L2(5) E - A5,As, L2(9) E= A6,Ag, L4(2) E- A8, A8, U4(2) U4(2) =E PSp4(3), PSp4(3), and and L3(2)L3(2) ==" L2(7). L2(7). This This exhausts exhausts thethe iso-isomorphisms among groups of Lie type and betweenbetween groups of Lie typetype andand alternating groups. Later in this section we will explore further consequences of the properties Ll-L5,El-L5, usingusing thethe theory theory ofof BN-pairsBN-pairs developeddeveloped in chapter 14. But first let's take a look at the sporadic simplesimple groups.groups. The sporadic simple groups and their orders are listed in Table 16.3. At present there is no nice nice class class ofof representationsrepresentations availableavailable to describe the sporadic groups inin a uniform uniform manner,manner, although some recent work on thethe

Table 16.3 The sporadic simple groups

Notation Name Order

M11MII Mathieu 2424.32.5.11 .32.5.11 M12Mil 26.33.5.11 MZZM22 2727.32.5.7.11 .32.5.7.11 M7.3M23 27.32.5.7.11.2327 .32.S.7.11.23 M24 210.33.5.7.11.23210.33.S.7.11.23 J1 Janko 23.3.5.7.11.19 J2 =HJ Hall-Janko f.27 33.33.52.7 .52 .7 J3 =HJM Higman-Janko-McKay 2.35.5.17.19z7 . 35 .S . 17 . 19 J4 Janko 2212". .3333 .5.7. .5 .7 .l13. 113 23.29.31.37.43 .23-29-31-37-43 HS Higman-Sims 29.32.53.7.1129.32.~3.7.11 Mc McLaughlin 27.36.S3.7.11 Sz Suzuki 213.37.52.7.11.13213 .37 .s2.7.11.13 Ly = LyS Lyons-Sims 2828.37.S6.7.11.31.37.67 . 37 .56 .7-11-31-37-67 He = HHM Held-Higman-McKay 21021° .33 33 .52.S2 .73. 73 . , 17 17 Ru Rudvalis 214.33.53.7.13.29214 . 33 . S3 .7 . 13 .29 O'N = O'NS O'Nan-Sims 29.34.5.73.11.19.31 Co3Cog = .3.3 Conway 210.37.53.7.11.2321° .37 .S3 .7. 11 .23 CotCoz = .2.2 218218.36.S3.7. 36 .53 .7-11-2311 .23 Col = .1 .l 2212" .39.S4.72.11.13.23 .39 .54 .72 . 11 .13.23 M(22) = F22 Fz2 Fischer 217.39.52.7.11.13217.39-~2~7.11.13 M(23) == F23 FZ3 218.313.52.7.11.13.218 .313 .s2.7. 11. 13. 17.2317 +23 M(24)' = F24 FZ4 221.316.52.73.11.13.17.23.292" . 316 .S2 .73.11. 13. 17.23 .29 F3 == E Thompson 215.310.53.72.13.19.31215 . 310.53 . 72 . 13. 19.31 F5 == D Harada-Norton 214.36.56.7.11.19214. 36 .s6.7.11.19 F2 == B Baby monster 241 .313 .313.56.72.11.13.17 .s6 .72 . 11 . 13 . 17 .19.23.31.47.19 .23-31-47 F1 ==M M Monster 246246. . 320320 .59 59 .76..76 . 112. 133133. . 1717 .19.23.29.31-19-23-29-31-41-47-59-71 .41 .47.59.71 ThejniteThe finite simplesimple groups 255 representations of the sporadic groups on geometries may eventually lead to such a description.description. InsteadInstead thethe sporadicsporadic groupsgroups werewere discovered discovered inin various various ways. The Mathieu groups all have multiply transitive permutation representations; specifically M,M isis 3, 3,4, 4, or or 5-transitive 5-transitive ofof degreedegree n.n. J2,Jz, Mc,Mc, HS,HS, Sz,Sz, Ru, M(22), M(23),M(23), andand M(24)M(24) allall havehave rank-3 permutation representations. Hence each is aa groupgroup ofof automorphismsautomorphisms of aa stronglystrongly regularregular graphgraph viavia thethe con-construction of section 16.16. The Conway groups act on the Leech lattice, a certain discrete integer submodule of 24-dimensional Euclidean Euclidean space.space. The The other other spo-sporadic groups were discovereddiscovered throughthrough thethe studystudy ofof centralizerscentralizers of of involutionsinvolutions (cf. section 48). In the remainder of this section assume G satisfiessatisfies the conditions L1-L5. We'll see thatthat thesethese conditionsconditions together with the theory in chapterchapter 1414 havehave aa number of interesting consequences. A subset A0 of EC isis closedclosed if, forfor eacheach a, a, ,!?,B E E A 0 withwith a a ++ ,!?P E E,C, wewe have a ++ P ,!? EE A.A. A A subsetsubset of r ofof 0A is is an an ideal ideal ofof 0A if, if, whenever whenever a EE F,r, ,B ,!? EE A,0, anda+pand a +,B E E,C,thena+,!? then a +,B E r.DefineUAr. Define Uo == (Ua:(U,:a a E A).0).

(47.1) Let 0A be be aa closedclosed subset of C+E+ andand r anan ideal ideal ofof A.A. ThenThen (1) UoUA = = 11aEAUa, naeAU,, withwith thethe productproduct inin anyany order,order, andand eacheach elementelement inin UoUA can be written uniquely as a productproduct lIaEAu,,11aEoua, u,ua EE A0 (for(for anyany fixedfixed ordering of A).0). (2) Urdur Uo.UA (3) HH

Proof. HH acts acts on on UaU, byby L3,L3, so (3) holds. PropertyProperty L4 L4 says says Ur Ur4 l] UA.U. Recall Recall the definitiondefinition of of the the height height function function h hfrom from section section 30, 30, and and let let a aE E A 0 with h(a) minimal. ObserveObserve A' 0' = 0A - -{a} {a} isis an an ideal ideal of of A. A. Thus Thus Uo-4UAl: Uo,UA, so so UAUo == Ua U,UAf, UA,, andand then,then, by by induction induction on on the the order order of ofA, 0, U U = = r1,,,nSEAUg UP with respect to some ordering of A. NowNow L3L3 andand ExerciseExercise 16.316.3 completecomplete thethe proof.

(47.2) LetLet w W E EW,n W, nE EN N with with Hn Hn = =w, w, and and A 0= =E+ C+ ilfl C-w.E-w. ThenThen A0 isis a closed subset of EC andand eacheach elementelement ofof BwBBwB can be written uniquely in the form bnu,bnu, b b E E B, B, u u E E UA. U.

Proof. Observe C+E+ andand E-C- areare closed closed subsets subsets ofof E,C, the the imageimage ofof aa closedclosed subset under an element of W is closed, and the intersectionintersection of closed subsets isis closed. closed. So So A 0 is is closed. closed. Similarly Similarly r =r =C+ E+ - - A0 = E+ C+ fl n E+w C+w is is closed. closed.

By 47.1,47.1, UU = = UrUA Ur UA so SO BwB BwB = =I: BwHUrUA BwHUr UA = B((u~)w-')wB((Ur)' ')wUoUA ==: BwUo BwUA as (u~)w-'(Ur)'"-' = UrW-lUrw-[ 5 U. < U.Suppose Suppose bnu bnu = anv,= any, a, a,b Eb EB, B, u, u, v vE E UA. U. Then Then 25625 6 Finite simple groups

a-'ba-1b =(vu-')"-I= (vu-1)n-' E B n (UA)w-'(U~)W-'5 < BB nn UWoU'Oas as AW-' Aw-1 & C C-E- == E+wo, C+wo, by by Exercise 10.3.10.3. But But by by L5, L5, B B il flBWO B'O= = H,H, so BB ilfl UWO Uw° = = 1. HenceHence a-'b a-1b = 1,1, sosoa=bandu=v. a = b and u = v.

If G isis aa finitefinite group of Lie type in characteristic pp satisfying L5L5 thenthen UU E Syl,(B).Sylp(B). Thus the next result shows UU E Syl,(G).Sylp(G).

(47.3) If G is finite and U E Sylp(B)Syl,(B) then U EE Sylp(G).Syl,(G).

Proof. For w E W letlet A(w)A(w) == E+C+ flil E-w.C-w. Then,Then, by by 47.2, IGI = IBI E WA(w)l WEW AsAsU U E~Syl,(B),~B~,=~U~and~U~(~)~OrnodpifU~(~~# Syl p(B), I B I p= I U I and I UA(w) l- 0 mod p if UA(w) # 1.l.ButU~(,)= But UA(w)= 11 if and only ifif A(w) = 0,0, and, and, by by 30.12, 30.12, this this happenshappens preciselyprecisely when w = 1. 1. Hence [GIIGI =- IBlmod(pIUI), IBlmod(plUI), soso UU EE Sylp(G).Syl,(G).

Recall fromfrom section section 43 43 that, that, for for J J& C n, rr,Sj SJ = = (sj:(s1: j j E J) and and further thatthat WjWj == (Si)(Sj) and and PiPj = = (B, (B, Wj) Wj) are are parabolics parabolics of of W W and and G,G, respectively. respectively. Let CEJ bebe

the setset ofof roots roots spanned spanned by by J Jand and let let $rJ *J = = C+E+ - E C J. j. ObserveObserve *J$rJ is is anan ideal of C+E+ and letlet VjVi = U*,.U$, .Similarly Similarly EiCf isis a a closed closed subsetsubset of CE and set UJUJ = UE+.U,;. Finally setset Lj LJ == (WJ,(Wj, UJ).

(47.4) (1) PiPj = = NG(VJ) NG(Vj). (2) LJL isis a a complement complement to Vj inin Pj. (3) (Li,(L j, HUj, HUj, N, N, Si) Sj) is ais Titsa Tits system system satisfying satisfying L1-L4 L1-L4 withwith respect respect toto the root systemsystem CjEJ and the simple systemsystem J.J.

Proof. By By 30.20, 30.20, EJ C isis a aroot root system system with withsimplesystem simple system JJ and and Weyl Weyl group WJ.Wj.

As$rJisanidealofC+,AsYf Jis an ideal of E+, VjGUandUVi :!l U and U = = UJVJby47.1.ThenPJ UJ Vi by 47.1. Then PJ = = (U, (U, Wj)WJ) =_ (Lj,(LJ, Vj).VJ). Finally, Finally, for for sj sj E ESj Si and and a aE E$rJ, 1frj, asj asj = =a a- - 2(a, 2(a, j)j/(j, j)j/(j, j)j) E E 1/rj, 1+9j, because aa has a positive projectionprojection on on n 7r - - J, and and hencehence asasj j does does too.too. ThusThus U: U."U: = U.,,U,, <5 Vj, VJ, so so Wj WJ <5 NG(Vi). NG(Vj). Therefore Therefore PjPj = (U,(U, Wj) Wj) <_( NN(Vi).NG(Vj). So, byby 43.7,43.7, NG(Vj) NG(VJ) = = PK PK for for some some K K& C n 7r with with J J& C K. K. If If k kE E K K - - J thenthen k EE 1/rj$rJ andand kskksk = = -k,-k, soSO U_kUPk = Ukk U: (< Vj.V. ButBut then U_kU-k 5< Uw0UWOn fl Uu = 1. 1. So K = JJ and and hence hence (1)(1) holds.holds. LetLetA A = HUJ,ClaimHUJ,ClaimLJ LJ = =A WjAWJA.ItsufficestoshowwUJwl A. It suffices to show w Ujw' C_& AWjAAWJA == Y for all w, w1w' E Wj.Wj. ThenThen byby inductioninduction on the lengthlength of w',wl, itit sufficessuffices toto show wUJsjw Ujs &j CY Yfor for each each w WE EWj Wj and and j jE E J.J. Let A == CjEJ - {j}. {j). ByBy 30.7, AsjAs jc C Cf,E , soso Uo U: == UAsj UA,, C& Uj. UJ. Thus Thus as as Uj UJ = =U UjUA,j UA, itit sufficessuffices to show wUJSIwUJs.. & C Y.Y. Similarly,Similarly, if if l(wsj) l(wsj) > > l(w),1(w), then then by by 30.10, 30.10, jw-' jw-1 > 0,0, soso The jinitefinite simplesimple groupsgroups 257

-I Uj'-' == Uj", U,,-I 5< UjUJ andand hence wU;s; w Ujsj = = U;-'WS;Uj'-'wsj cG Y. Y. SoSo assumeassumel(wsj) l(wsj) << UwJ l(w). Then byby inductioninduction onon l(w),l(w), wsws,U;s, j Uj sj Cc Y.Y. NowNow ififAW;A AWjA is a group then U:UJ' cG AWjA, AW; A, soso

wUjsj = wsjUjj c ws1AWWA = wsjA U wsjAsjA C Y.

Thus it sufficessuffices to to showshow AWj AW; A is a group. By Exercise 10.3, sys" = = si si for for some some i iE E 7t, n, andand wo wo isis an involution. LetLet XX == PTO.Pi". I'llI'll show show X X n il PjPi = =AWjA, AW; A, which which will will show show AWjA AW; A is is aa groupgroup andand hence establish the claim. By 43.7,43.7, PiPj = B B U U BsBs; j BB = VVjT, j T, wherewhere TT = AWjA W,A. A. Sim- ilarlyilarlyX X = = BWOUBWOsjBWO BwOUBWOs;BWO = = BWO BWOUSjBW0SiBW'0,and,UsjBWoSj BWo, and, ass; assjE X, E sjXX,sjX = = X, X,so so X == s s;BWOUBWoS~BWO.Letj B'OU BWOSJ BWO. Leti-2 =Q =C--{-j).ThenUWO E - - {- j }. Then UWO = = U_;UQU_ j Uu soUwOSjso Uw°S;= = UjUQUj Uc2 by by 30.7,30.7, andand hence BWOS;BWoSjBWO B'O = = U; Uj BWO. BwO.Next Next B B il nsj s BWO j Bw0= = (Bwo (Bwo iln sjwoB)wos jwoB)wo and,and, as as wo wo # 0 sjwo,s jwo, BwoBnBsjwoB Bw0B n Bs jwoB = = @by0 by 43.2. SoSo BnsiBWOB nsj B'O == @.Hence0. Hence BnXBnX = BnB'Os;BIO BnBWOsjBWO == BnU;B"'0BnUjBWO = Uj(BnBw0)U;(BnBWO) = = UjH == A. A. Assj,ACX,TCX,soXnPj=TVjnX=T(VjnX)c_T(BnX)=ASS^, A c X, T c X, SOX~P; = TV, nx = T(V; n x) c T(B nx)= TA == T. ThusThus thethe claimclaim isis atat lastlast established.established. It's now easy to see thatthat (L(LJ, j, A, N, Sj)Sj) is is a a Tits Tits system.system. TheThe only one of the BN-pair axiomsaxioms whichwhich isis notnot evident evident is is BN3. BN3. But, But, for for s sE E Sj,Sj, w E Wj,

SAWsAw GC (BwB(BWB u BswB)BSWB) n AWjAWjA A = AwA AwA UU AswA with the equality following from 47.2 and 43.2. Similarly it is clear that L2-L4 inherit to L j.j. So (3)(3) holds. WesawVJgPJ=(LJ,VJ),soPJ=LJVJ.By43.2and47.1,VJnLJ=We saw Vj< Pj = (L j, Vj), so Pj = L jVj. By 43.2 and 47. 1, Vj n L j= V.Vj nil AA == 1, 1, establishing establishing (2). (2).

The subgroup LLj is is calledcalled a Levi factor factor ofof thethe parabolicparabolic Pj, Pj, and Vj is the unipotent radicalradical ofof Pj.Pj.

(47.5) (1) HUjHUj n il u,'-Up = =1, 1,where where xo xo is is the the element element of of WjWj of of maximalmaximal length. (2) CH(UJ)CH(UJ) = kerHU,(LJ) ~~~Hu~(LJ>. (3) If If GG is is finite, finite, Z(G) == 1, 1, and and U U E E Syl Sylp(B) p(B) for for somesome prime p, then Vj = Op(Pj) = F*(Pj) F*(Pj) for for each each JJ Cc Tr. n.

Proof. LetLet AA == HUj andand L == L L j. j. ByBy ExerciseExercise 10.3,10.3, (Cf)xo (Ei )xo = = CEJ J, soso UpUJ° iln B <5 U"'°UWo nil B B == 1. 1. Thus Thus (1) (1) holds. holds. LetLet D == kerA(L). kerA(L). Then DD <5 AAnAxO=Hby(1).So[D,Uj] > 1 1then, then, by by induction induction on on 1, I, EE <5 kerHU; kerHu, (Li) (Li) for eacheach ii EETr. n. 25825 8 Finite simple groups

Thus EE <( C(Uf) C(U,") for for eacheach x eE Li,Li, so so L, Li = = (H, (H, UiL') u:') < 5 NG(E). Nc(E). Therefore Therefore G = (Li,(Li, B: B: i eE n)n) < (NG(E), NG(E), so so E E < (D. D. So So take take l 1= =1, 1,and and consider consider the the action of G on BGB' by conjugation. By BN3, G = B B U U BwB,BwB, so B isis transitivetransitive on BGB' -- {B). {B). ByBy 47.2, BwB = BwU, BwU, soso BGB' - {B) {B) = = BW'U. B~'. Hence, as E <( Bw Bw and U (< C(E), E m).> m). Observe Observe 52,E a, is an idealideal in in C+, E+, so so V,,, V = Vim VQm < B. Indeed by L4, U,,U, V,+lVm+1 9 < B B for for each each aa EE Qm.M,. SoSo alsoalso UaVm+14U, V,+l

$jVri Gc C(E)(y). Hence,Hence, by Exercise 10.4, y centralizes (C),(E), a a contradiction.contradiction. Therefore Sj fl il Y is empty. However,However, by by 43.7, 43.7, Y Y = = PKPK for for some some K K G C J so as SiflY=1,K=0andY=B.SjnY=l,K=IZIandY=B. So X 5< B.B. Next, as LLj is a complement to to Vj Vj in in Pj, Pj, XVj XVj = = Vj(XVj ilflL Lj) J) and, asas XVjXV j5 < B,B, XVj XVV il Ll Lj 5 < kerA(Lker,l(Lj)j) == CH(UJ). CH(Uj). AssumeAssume the the hypothesis hypothesis of (3). ThenThen HH is a p'-group andand HoHo = XVj XVj flil LjL inducesinduces inner inner automorphisms automorphisms on the p-groupp-group VJ,Vi, so so [Ho, [Ho, Vj] Vj] = = 1.1. Thus Thus Ho Ho = = CH(UJVj)CH(UJVJ) = = CH(U)CH(U) =_ kerB(G). But,But, as as G G = = (u'),(UG), kerB(G)kerB(G) = = Z(G)Z(G) = 1.1. I've shown CG(Vj)CG(VJ) (< ViVj underunder thethe hypothesishypothesis of (3). Thus to complete thethe proof of (3) itit remainsremains onlyonly to to observe observe that, that, by by (2), (2), Vj Vj = = O,(Pj).Oy(Pi).

(47.6) UG = UwEw(Uw)U

Proof.ProoJ ByBy 43.7,43.7, G G = = nWEwBwU, nWEwBwU, soso the the remark remark holds.holds.

(47.7) IfIf UU (< X = (UG(u' ilfl X) (< GG then H <( NI(X) NG(X) andand HX = PjPj for for somesome J cG .r. n. In In particular particular (P,-{r}: (P,+): ii EE .r) n) is is the the set set of of maximal maximal subgroups subgroups of GG containing U.

Proof.ProoJ By 47.6, XX = (Ud: (ud: dd EE D)D) for for some some D D C G W. W. So,So, as as H H normalizes normalizes Uw for each ww E W, HH normalizes X.X. SoSo HXHX 5< G and, as B = HU HU <5 HX, HX, HX == Pj for for some J cG n, n, by by 43.7.1. 43.7.1. Let M be aa maximalmaximal subgroup of of GG containingcontaining U U and and X X == (UG(u' flil M). M). Then XX9 < M, so by maximality of M either M = NG NG(X) (X) or X9X < G. In the latterlatter case GG = (U,(U, Uw0) UwO) == X,X, aa contradiction. So So HX HX ( < NG(X) NI(X) = MM andand hencehence M = PjPj for for some some JJ c G 7r. n. By By maximality maximality of M, J == 7r n -- {i) {i) for for some some ii EE 7r.n.

If GG isis aa finitefinite groupgroup ofof LieLie typetype inin characteristiccharacteristic p then, then, asas aa consequenceconsequence of a theorem ofof Bore1Borel andand TitsTits [BT],[BT], eacheach p-localp-local subgroupsubgroup ofof GG isis containedcontained in a maximal parabolic. This result can be derived using L1-L5L1-L5 togethertogether with TheTheJinite finite simple groups 259

threethree extra properties of finitefinite groups of Lie typetype listedlisted asas hypotheseshypotheses inin thethe next lemma:

(47.8) AssumeAssume thethe followingfollowing hold:hold:

(a) GG is is a a finite finite andand UU EE Sylp(B).Sylp(B). (b)(b) Let Let I'r be be the the set set of of nontrivialnontrivial p-subgroupsp-subgroups RR ofof GG withwith RR == Op(NG(R)). Op(NG(R)). Assume forfor eacheach R R EE rF thatthat RR == P flil Q Q forfor somesome P, Q Q EE Sylp(NG(R)).SylP(NG(R)). (c) UU=(Ui:iE7r). = (Ui: i E n).

Then the followingfollowing hold:hold:

(1) (NG(R):(NG(R): RR EE F)r) is is the the set set of of properproper parabolicparabolic subgroups of G. (2) IfIf QQ is is a a nontrivial nontrivial p-subgroup p-subgroup of of G G then then there there exists exists a a proper proper parabolic parabolic M of GG withwith NG(Q)NG(Q) < 5 M M and and Q Q :S 5 Op Op(M). (M). (3) F*(X)F*(X) = = Op(X) O,(X) forfor eacheach p-local X of G, ifif Z(G)Z(G) = 1.1.

Proof. LetLet R R E E F r and and M M = = NG(R). NG(R). ByBy hypothesis hypothesis therethere are P, Q Q EE Sylp(M)Sylp(M) with R = P P fl n Q. Q. By By 47.3, 47.3, U U E E Sylp(G), Sylp(G), so so withoutwithout loss P <5 U. U. Similarly, Similarly, by 47.6, Q <5 U"Uwu forfor some some w w E EW, W, U u E E U, U, and and replacing replacing R R by by R"-`R"-' wewe maymay assume QQ 5< U'.UW. ObserveObserve RR = UU fln U'.UW. ForFor ifif not, R < UU fln U"'UW nfl M <5 P flil Q Q = = R, R, a acontradiction. contradiction. Let A = E+ Z+ fl n E+wZ+w andand S2'2 = = Z-E- flfl E+w. Z+w . Then Then UwU' = = UoUQ UAUn with with UoUA 5< R,R, so R = Uo(U UA(U fl n UQ)U,) Ec Uo(UUA(U fl n U"'°)UWo) == Uo.UA. That isis RR == U.UA. I'll I'll show show UTUlu' <5 M M for for each each i iE E n nand and some some vi v, E EU. U. Then, Then, by by (c) (c) and and Exercise Exercise 8.12, 8.12, UU <5 M. M. Also Also HH < 5 N(Uo) N(UA) < 5M, M, so so M M is isa parabolica parabolic by by 43.7, 43.7, and and (1) (I) holds. holds. Let i EE 7r;n; it remains to show U,Ui <5 M.M. If If notnot U; U, -$ R, so i & A. Hence -i-i EE (E+)w(Z+)w byby ExerciseExercise 10.5, soso U-,U_i 5< U'.Uw. By By 47.5, 47.5, U U fl il US,US' = = U*< U+<1 (U,(U, US! US'), ), where ,kI) = E+ Z+ - - {i). {i). So So U_i U-, < 5USi USa < 5NG(U NG(U fl il USi)US,) and and ofof course course U_1 U-, <5 U"'Uw <5 NG(Uw),NG(Uw), so so U-,U_i acts acts on on U U n fl Uw U"' n flUS' Usi = =U+ U* n fl UA Uo = = UA U. = R,R, as A Gc ,.I). That That is is U_i U-, << M. M. Observe nextnext that, that, as as P P E E SylP(M), Sylp(M), P P = = U ilfl M. IfIf PP <5 U,,U+ thenthen even PP = U,,U+ flil M so,so, asas U_iU-, actsacts onon U,,,U+, U_;U-, actsacts on P. ButBut then,then, asas PP EE Sylp(M), Sylp(M), U_iU-, <5 P,P, contradicting contradicting U nfl UwOUlo = 1.1. So P -$ U*.U+. ConsiderConsider thethe parabolic Pi.P, . By 43.7, Bs;BSl is a maximal subgroup ofof P,,Pi, while, as P -$ U*,U+, MMn il PiP, -$ BS1.BSi. Hence,Hence, as BSlBsi == HU-,HU_iU* U+ 5< (M(M nil Pi)U*, P,)U+, itit followsfollows thatthat PIPi == (M nil Pi)U,,. P,)U+. Therefore s,vsiv E MM forfor somesome vv Ee U, so U:Ui" = = (U-,)SIU(U-i)'1° 5< PP. ThisThis completescompletes the the proof proof ofof (1).(1). Arguing by by induction on on the order of Q, there exists R eE Fr with with QQ < 5 R R and and NG(Q) NG(Q) < 5 NG(R), NG(R), so so (1) (1) implies implies (2). (2). Finally (2), 47.5.3,47.5.3, andand 31.1631.16 implyimply (3). (3). 260 Finite simple groups

48 AnAn outlineoutline ofof the ClassificationClassification Theorem Let K. 9 Cbe be the the list list of of finite finite simple simple groups groups appearing appearing in in section section 47.47. SectionSection 48 provides a brief outline of the Classification Theorem, which asserts:

Classification Theorem. Theorem. Every Every finite finite simple simple group group is is isomorphic isomorphic to to a a membermember of K.X.

The usual procedure for classifying a collection ofof objects isis to associate to each member of the collection some family of invariants, proveprove thatthat eacheach objectobject is uniquely determined by itsits invariants, and determine which setssets of invariantsinvariants actually correspond to objects. The invariantsinvariants used to classify thethe finitefinite simplesimple groups are certain local subgroups of the group, usually the normalizers of suitable subgroups of prime order, particularly centralizers of involutions.involutions. A rationale for this approach can be obtained from the Odd Order Theorem of Feit and Thompson and the Brauer-Fowler Theorem (Theorem 45.5).

Odd Order Order Theorem. Theorem. (Feit-Thompson (Feit-Thompson [FT])[FT]) Groups of oddodd orderorder are are solvable.

The Odd Order Theorem says that every nonabelians simple group G is of even order and hence possesses an involution t.t. The Brauer-Fowler Theorem says there is only a finitefinite numbernumber ofof finitefinite simplesimple groupsgroups GoGo possessingpossessing anan involutioninvolution to with CGo(tO)CG0(to)Z = CG(t). In practice, with aa smallsmall numbernumber ofof exceptions,exceptions, GG is the unique simple group with such a centralizer.centralizer. Even in the exceptional cases at most three simplesimple groupsgroups possess the samesame centralizercentralizer (e.g. L5(2),L5(2), MZ4,M24, and and HeHe allall possesspossess an involution withwith centralizer L~(~)/D;).L3(2)/Dg). Exercise 16.6 illustrates how the isomorphism type of a simple group can be recovered from the isomorphism type of the centralizers of involutions.involutions. So centralizers of involutions provide a set of invariants upon which to base the ClassificationClassification-Theorem. Theorem. For For variousvarious reasons reasons it it turns turns out out toto bebe better to enlarge this set of invariants to include suitable normalizers ofof subgroups of odd prime order. To be more precise,precise, define aa standardstandard subgroup of a groupgroup GG for thethe primeprime pp to bebe aa subgroupsubgroup HH of G such thatthat HH = CG(x) CG(x) forfor somesome element x of order p, H has a unique component L, and CH(L) has has cycliccyclic SylowSylow p-groups. p-groups. StandardStandard subgroups provideprovide thethe principal set of invariants for the classification of the finite simple groups. However certaincertain small groupsgroups eithereither possesspossess nono standardstandard subgroupsubgroup oror cannotcannot be effectively characterized in terms of this invariant. Such groups are characterized by other methods. HenceHence we we havehave ourour firstfirst partition of the simple groupsgroups for purposes of thethe classification:classification: the partition intointo generic groups andand smallsmall An outlineoutline ofof thethe ClassificationClassiJication Theorem Theorem 26126 1 groups. I'llI'll definedefine the the appropriateappropriate measure measure of of sizesize in in a a moment; moment; but but beforebefore thatthat anotheranother partition. When possible, we'd like to characterize a simple group G in terms of a standardstandard subgroupsubgroup forfor thethe primeprime 2.2. ButBut oftenoften GG possesses possesses no no such such subgroup. subgroup. G is said to be of characteristic p-type p-type if if F*(H) F*(H) == Op(H)O,(H) forfor each p-local subgroupsubgroup H H ofof G.G. For For example, example, if if GG is is of of LieLie typetype andand characteristiccharacteristic p, p, we we sawsaw inin 47.8 that GG isis of characteristic p-type. InIn particularparticular if if GG isis ofof characteristic 2- 2- typetype thenthen itit possesses no no standardstandard subgroup subgroup for for the the prime prime 2.2. OurOur second partition of the simple groupsgroups isis thethe partition intointo groupsgroups ofof eveneven andand odd characteristic, wherewhere byby definitiondefinition GG isis ofof eveneven characteristiccharacteristic if if GG is is of of characteristiccharacteristic 2-type 2-type andand GG isis ofof oddodd characteristiccharacteristic otherwise. otherwise. DefineDefine the 2-local p-rank of G to be m2,m2,,(G) p(G) == max{mmax{mp(H): p(H): His a 2-local of G},G), andand definedefine e(G) == max{m2,p(G): max{m2,,(G): pp odd}.odd). In aa groupgroup ofof LieLie typetype andand characteristiccharacteristic 2, e(G) isis aa goodgood approximationapproximation of thethe LieLie rank.rank. A group G of even characteristic is small ifif e(G)e(G) <5 22 andand genericgeneric otherwise.otherwise. A groupgroup of oddodd characteristiccharacteristic is small if GG isis 2-disconnectedz-disconnected forfor thethe prime 22 andand generic otherwise.otherwise. Thus we have a four part partition of the finitefinite simple groups forfor purposespurposes ofof thethe classification.classification. The proof of the Classification Theorem proceeds by induction on the order of thethe simple simple groupgroup toto bebe classified. classified. Thus Thus wewe considerconsider aa minimal minimal countercounter example G toto thethe ClassificationClassification Theorem; that is G is aa finitefinite simplesimple group of minimalminimal order subject toto GG @¢ X.X. DefineDefine aa finitefinite groupgroup H H to to be be a a.7C- X- group if everyevery simplesimple section of H is is inin .7CX (a section of H is is aa factor factor groupgroup A/B,A/ B, where where B B <5 A A <5 H). H). Observe Observe that that every every proper proper subgroup subgroup of of ourour minimalminimal counterexamplecounterexample is is aa XX-group. -group. This This property property will will be be usedused repeatedly.repeatedly. IfIf GG isis aa finitefinite simple group with m2(G) 5< 2,2, aa moment's thoughtthought shows G toto bebe 2-disconnectedz-disconnected for the prime 2. If m2(G) > 2 2 then, then, by by 46.7,46.7, neglectingneglecting isolated points in the graphgraph -T(ez(G)),g(8?j(~)), G G isis 2-disconnected2-disconnected forfor thethe primeprime 22 precisely when G $# I' r:,2(G), , 2(G), wherewhere P P E ESy12(G). Sy12(G). As As aa mattermatter ofof fact, if G is simple and m2(G) > 2,2, itit cancan bebe shownshown that G is 2-disconnectedZdisconnected for the prime 2 if and only if GG hashas aa properproper 2-generated2-generated 2-core.2-core. ThusThus toto classifyclassify thethe smallsmall simple groups ofof odd characteristic itit sufficessuffices to prove the following two results:

Theorem 48.1. 48.1. IfIf GG is is a a nonabelian nonabelian finite finite simple group with m2(G) 5< 2 then either: (1) aa SylowSylow 2-group 2-group ofof GG is is dihedral, dihedral, semidihedral, semidihedral, or or Z2» Z2,, wr wr Z2,22, andand GG "= L2(q),L2(4)7 L3L3(4), (q), U3U3(4), (q), q odd, oror M11,M11, or (2) GG 2 U3U3(4). (4). 262 Finite simple groups

Theorem 48.2.48.2. LetLet GG bebe aa nonabelian nonabelian finitefinite simple group withwith rnz(G)m2(G) > 22 and assume G has a proper 2-generated 2-core. ThenThen eithereither GG isis a group of Lie type of characteristic 2 and Lie rank 1 1 (i.e.(i.e. L2(2"), U3(2"), oror Sz(2")) or GZG= J1.J1.

In brief, Theorem 48.1 is proved by using local theory to restrict the subgroup structure of our minimal counterexample G. G. AtAt thisthis point there is either enough information available to concludeconclude G E 3%.C oror G possesses many TI-sets. TI-sets can be exploited usingusing character theory (along thethe lineslines ofof 35.2235.22 oror Frobenius'Frobenius' Theorem) to derive a contradiction oror toto restrictrestrict thethe structurestructure ofof GG furtherfurther and show GG EE 3%. X. It's interestinginteresting to note that character theory plays an important rolerole inin the proof of Theorem 48.1, but is rarelyrarely usedused inin thethe remainderremainder ofof thethe classification.classification. In essence character theory is used to deal with small groups such as thethe groups of Lie rank 1 and some groups of Lie rank 2. The generic groups of higher dimension can be identified using geometric or quasigeometric techniques. The local theory used in the proof of Theorem 48.1 is of two sorts. Transfer and fusion techniques are used to pin down the structure of a SylowSylow 2-group of G andand thethe fusionfusion ofof 2-elements;2-elements; seesee forfor exampleexample Exercise 13.2.13.2. One such tool, used sparingly in the proof of Theorem 48.1 but frequently in laterlater stagesstages of the classification, isis Glauberman'sGlauberman's Z*-Theorem:

Glauberman Z*-Theorem. Z*-Theorem. [G13]. [GI 31. Let Let G Gbe be a afinite finite group group and and tt anan involutioninvolution in G such such that t t isis weaklyweakly closedclosed inin CG(t).CG(t). Then t* E Z(G*), where G* = _ G/Oz4G).G/OT(G).

Recall a subset S of G is weakly closed inin a subgroup HH of G (with respect to G) if SGS' nfl H == {S}. {S). TheThe proofproof ofof thethe Z*Z*-Theorem -Theorem usesuses modular character theory and is beyond the scope of this book. The second kind of local theory used in the proof of Theorem 48.1 involves an analysis of subgroups ofof G of odd order using signalizer functorfunctor theorytheory oror some variant of that theory. Notice the Odd Order Theorem is one step in the proof of Theorem 48.1, since groups of odd order are of 2-rank 0. When Feit and Thompson proved the Odd Order Theorem signalizer functor theory did not exist; instead they generated their own techniques, which in time evolvedevolved into signalizer functorfunctor theory.theory. I'llI'll illustrate the signalizer functor approach inin the generic case a littlelittle later.later. The techniques used to establish Theorem 48.2 are ratherrather different.different. TheThe proofproof depends heavily onon the fact that each pair of involutionsinvolutions generate a dihedral group. This observation makesmakes possiblepossible aa numbernumber ofof combinatorialcombinatorial andand groupgroup theoretic arguments of the flavorflavor of sectionssections 45 andand 46.46. ExercisesExercises 16.516.5 andand An outline of the ClassificationClass$cation TheoremTheorem 263

16.6 illustrate some of thesethese arguments.arguments. IndeedIndeed ExerciseExercise 16.616.6 establishesestablishes aa very special case of Theorem 48.2. It would be nice to have the analogueanalogue of Theorem 48.2 for odd primes, but as nothing in particular can be said about groups generated by by aa pair of elements of of odd prime order, a different approach is required. Let's turnturn nextnext toto thethe genericgeneric groupsgroups ofof oddodd characteristic.characteristic. ObserveObserve that by Exercise 16.116.1 a generic group G ofof oddodd characteristiccharacteristic possesses possesses anan involutioninvolution t such that OZ',E(CG(t))O2',E(Cc(t)) #:O2'(Cc(t)). OZ'(CG(t)). I encourageI encourage you you to to retrace retrace the the stepssteps in the proof ofof thisthis exercise;exercise; thethe proofproof providesprovides aa goodgood illustrationillustration of of signalizersignalizer functor theory.theory. Similar arguments reappear inin thethe proofproof ofof TheoremTheorem 48.148.1 andand in the analysisanalysis ofof thethe genericgeneric groups groups of of eveneven characteristic.characteristic. This brings us toto thethe followingfollowing fundamentalfundamental property property ofof finitefinite groups:groups:

Bp-Property. Let Let pp be be aa prime, prime, GG aa finitefinite group with Opf(G)Op,(G) = = 1,1, and x an element of order p inin G.G. ThenThen Op',E(CG(x))Opf,~(CG (x)) =Op'(CG(x))E(CG(x)).=Opt (CG(x))E(CG(X)).

The verification of the Bp-Property is one of the most critical and difficult steps in thethe classification.classification. Only the B2-PropertyBz-Property isis established directly;directly; forfor odd p, the Bp-Property follows follows onlyonly asas aa corollarycorollary toto thethe ClassificationClassification by by inspectioninspection of the groups in X. (Notice(Notice that that by by Exercise Exercise 16.4, 16.4, to to verify verify thethe Bp-PropertyBp-Property it suffices to consider the case where F*(G) is is simple.)simple.) Observe next that the B2-PropertyBz-Property follows from the following result:

Unbalanced GroupGroup Theorem.Theorem. LetLet GG be be a a finite finite groupgroup withwith F*(G)F*(G) simplesimple which is unbalanced for the primeprime 2.2. ThenThen F*(G) is a group of Lie type and odd characteristic, A2n+i, AZ*+~, L3(4),L3(4), oror He.He.

For to verify the B2-PropertyBz-Property it suffices toto assumeassume F*(G) is simple by Exercise 16.5. If Ozl(CG(t))O2'(Cc(t)) = = 1 1for for each each involution involution tt inin G, G, then then the the B2-Property B2-Property isis trivially satisfied, soso we may assume G isis unbalanced.unbalanced. Hence,Hence, byby thethe Unbal-Unbal- anced GroupGroup Theorem,Theorem, F*(G)F*(G) E YC.X. But, But, by by inspectioninspection of of thethe locallocal structure of Aut(L) for L E X, ifif F*(G) F*(G) E E X 3% then then G Gsatisfies satisfies the the B2-Property. Bz-Property. supposeSuppose for the moment that the UnbalancedUnbalanced GroupGroup Theorem, andand hencehence also the B2-conjecture,Bz-conjecture, is established. The B2-conjectureBz-conjecture makes possible ma- nipulations which prove:

Component Theorem. (Aschbacher(Aschbacher [As 11)1]) Let G bebe aa finitefinite groupgroup withwith F*(G) simple simple satisfyingsatisfying the the B2-PropertyB2-Property and possessing an involution t such that 02',02',E(CG(t)) E (CG (t)) # OZ'(CG(t)).O2' (CG (t)).The The G G possesses possesses aa standardstandard subgroupsubgroup for the prime 2.2. 264 Finite simple groups

Actually the definition of aa standard subgroupsubgroup has has toto bebe relaxedrelaxed aa littlelittle toto makemake the Component TheoremTheorem correctcorrect asas stated above, but the spirit isis accurate.accurate. Notice that at thisthis stagestage wewe havehave associatedassociated toto eacheach genericgeneric groupgroup ofof oddodd characteristic the desired set of invariants: its collection of standard subgroupssubgroups for the prime 2.2. ItIt remainsremains toto characterizecharacterize simple simple groupsgroups viavia thesethese invariants. invariants. Thus we must consider:

Standard form form problem problem for for (L, (L, r): r): Determine Determine all all finite finite groups groups G G possessingpossessing a standard subgroupsubgroup HH forfor thethe primeprime r r withwith E(H) E(H) Z = L.

If G is our minimal counterexample andand HH a standard subgroupsubgroup ofof G,G, thenthen HH is a K-group,.X-group, so E(H)/Z(E(H)) E E X. YC. AsAs we we know know the the universal universal covering covering group of eacheach membermember of of 3%, X, we knowknow E(H),E(H), and toto proveprove G G Ee YCX it it remains remains to treat the standardstandard formform problemproblem forfor eacheach perfectperfect centralcentral extensionextension ofof eacheach member ofof x. X. How does one retrieve the structure of G from that of the standard subgroup H? Let L = E(H). E(H). Then Then H/CG(H) H/CG(H) < 5Aut(L), Aut(L), so so we we have have good good controlcontrol over H/CG(H),HI CG(H), while as CG(H) has cyclic Sylow p-groups we have good controlcontrol of CG(H). Then analysis of fusion of p-elements ofof HH givesgives usus aa conjugate Hg Hg of H suchsuch that the intersection H fl Hg is large. Results similar to TheoremTheorem 48.2 allow us toto concludeconclude GG = (H,(H, Hg). Hg). This This information information cancan bebe usedused toto definedefine a representation of G on a subgroup geometry along the lines of the construction in section 3, or to obtain a presentation ofof G.G. ForFor exampleexample wewe mightmight representrepresent G on a building or show it possesses a BN-pair. IfIf so,so, thethe machinerymachinery inin chapterchapter 14 becomesbecomes availableavailable to to identify identify G G asas aa membermember of of 3%. X. It remains to prove thethe UnbalancedUnbalanced Group Theorem. By thethe Odd Order The- orem O2O2,(CG(t)) (CG(t)) is solvable for each involution t in G, soso by 46.9 and Exercise 13.3, ifif G is not balanced for the prime 2, also CG(t)*CG(~)* = CG(t)/O2'(CG(t))CG(t)/o2l(C~(t)) is not balanced for the prime 2 for some involution tt in G. Hence by 31.1931.19 there isis a componentcomponent L*L* ofof C(t)*C(t)* andand X*X* 5< N(L*)N(L*) suchsuch thatthat Autp(L*) is not balanced for thethe primeprime 2.2. ByBy inductioninduction onon thethe orderorder of of G, G, L*/Z(L*)L*/Z(L*) is one of the groups listed in the conclusionconclusion of thethe UnbalancedUnbalanced Group Theorem. This piece of information is critical; together with some deep local theory it can bebe used to produce a standard subgroup for the prime 2, reducing us to aa previous case. The groups of odd characteristic have been treated; let us turn nextnext to thethe groups of even characteristic.characteristic. If G is a generic group of even characteristiccharacteristic we seek to produce a standard subgroup for some oddodd primeprime p.p. More precisely p is a prime in the setset o-(G)a(G) where

or(G)a(G) = {p:{p: p odd,odd, m2,p(G)rn~,,(G) >Z 31.31. An outline of the ClassificationClass$cation TheoremTheorem 265

The actual definition of n(G)a(G) is is a a littlelittle more more complicated,complicated, but but againagain thethe defini-definition above is in the right spirit. Using signalizer functor theory and other local group theoretic techniquestechniques oneone shows:shows:

Theorem 48.3. 48.3. (Trichotomy (Trichotomy Theorem Theorem [As [As 2], 21, [GL]) [GL]) let let GG be a minimalminimal counter-counterexample to the Classification Theorem and assume G isis generic of even characteristic. Then oneone ofof thethe followingfollowing holds:holds:

(1) G G possessespossesses aa standardstandard subgroupsubgroup forfor some some p p Ee n(G).a(G). (2) ThereThere exists exists an involutioninvolution tt in GG suchsuch that that F*(CG(t))F*(Cc(t)) isis aa 2-group2-group ofof symplectic type.type. (3) GG is is in in the the uniquenessuniqueness case.case.

G is in thethe uniquenessuniqueness case if,if, for each p eE a(G),n(G), G G possesses possesses a a strongly strongly p-embedded maximalmaximal 2-local2-local subgroup.subgroup. RecallRecall that a p-group isis ofof symplecticsymplectic type ifif itit possesses possesses nono noncyclicnoncyclic characteristiccharacteristic abelianabelian subgroups,subgroups, andand thatthat groups of symplectic type are described completely in chapter 8. I've already discussed discussed brieflybriefly how one deals with standard subgroups. In casecase 2 of Theorem 48.3,48.3, thethe structurestructure of of F*(C(t)) F*(C(t)) = QQ isis determineddetermined from chapter 8, as is Aut(Q)Aut(Q) (cf.(cf. ExerciseExercise 8.5). 8.5). With With thisthis informationinformation andand somesome workwork oneone can recover C(t) andand thenthen proceed as though C(t) werewere aa standardstandard subgroup. The arguments used to deal with the uniqueness case and the small groups of even characteristic are quite similar.similar. They involve factoring 2-locals as the product of normalizersnormalizers of certain subgroupssubgroups of aa SylowSylow 2-group2-group ofof thethe local.local. The Thompson Factorization, discussed in sectionsection 32,32, is the prototype ofof such factorizations. The proof of the Thompson Normal p-Complement Theorem (39.5) and ofof thethe SolvableSolvable Signalizer Signalizer FunctorFunctor TheoremTheorem givegive somesome indicationindication of how such factorizations can bebe used.used.

Remarks. SeeSee Gorenstein's Gorenstein's seriesseries of of booksbooks [Gor[Gor 2,2, Gor 3]31 for a more detailed outline of the proof of the Classification Theorem and a more complete discussion of the sporadic simple groups. In particular [Gor[Gor 2, Gor 313] contain explicit references to thethe articlesarticles which,which, takentaken together,together, supplysupply aa proofproof ofof TheoremsTheorems 48.1 and 48.2 and the Unbalanced Group Theorem. Carter [Ca][Ca] andand Steinberg [St][St] are good places to learn about groupsgroups ofof Lie type. Exercises for chapter 1616 1. LetLet GG bebe aa finitefinite group withwith PP EE Sy12(G)Sy12(G) and and G G == 02(G).o~(G). Assume G == TrOp,2(~), , 2(G), m;?(G)m2(G) > 2,2, andand 02',E(CG(t))021,~(C~(t)) == O2(CG(t)) O2(CG(t)) for for eacheach involutioninvolution t inin G.G. Assume Assume O2,(G)Or(G) = 1. 1. Prove Prove GG is is of of characteristic characteristic 2-type;2-type; that is F*(H) = = 02(H) 02(H) forfor each each 2-local 2-local subgroup subgroup HH ofof G.G. 266 Finite simple groups

2. LetLet GG == A,, A, bebe the the alternating alternating group onon nn L> 55 lettersletters and A = Aut(G). Aut(G). Prove: (1) AA=S,, = S, ifif n n#6. # 6. (2) AIGA/G=E4ifn=6. E E4 ifn = 6. 3. LetLet (Gi:(Gi: 11

gi91 ...g,,... gn, withwithgi gi E Gi, andand (2) GmGm+1GrnGrn+'...... GnG,s < GG forfor each each rn, m, 1 1 5 < rn m 5

@(a,0(a,b,c,d):yb, c, d): y Hi-+ (ay (ay+b)/(cy+d) + b)/(cy + d) a,b,c,da, b, c, d cE F,adF, ad - - be bc ## 00 An outline of the ClassificationClassiJication Theorem Theorem 267

as in Exercise 4.10. Let H*={4(a,b,1,1):aEF#,bEF).H* = {#(a, b, 1,l):a E F', b E F). Prove: (1) TT is is elementary elementary abelian. abelian. (2) HH isis strongly strongly 2-embedded 2-embedded in G. (3) DDisinvertedbyu,~H~=q(q-1),andDisacomplementtoTinH. is inverted by u, I H I = q (q - 1), and D is a complement to T in H. (4) NG(D)NG(D) = = D(u) D(u) and and {H, {H, Hu} Hu} is is the the fixed fixed pointpoint set of D on X. (5) GG is is 3-transitive 3-transitive on on X X withwith onlyonly the the identity identity fixingfixing 33 oror moremore points. TT is regularregular onon XX -- {H}.{HI. (6) ThereThere isis anan isomorphism n:n: HH -++ H*H* and and aa bijectionbijection a: X -++ Y Y suchsuch that Ha == oo, oo, (Hu)a (Hu)a == 0, 0, andand (xa)h7r(xa)hn = (xh)a(xh)a forfor allall xx EE X andand h E H. (7) ThereThere exists vv Ec uD withwith (aa-')v (act-1)v = = a-la-'a-la-1 forfor allall a EE F#.F'. (8) nn extends extends to to isomorphism isomorphism of G and G* = L2L2(q) (q) with with vnvn = 4(0, #(0,1,1,0). 1, 1, 0). (Hints: Use Exercise 16.516.5 in the proof of (3) and (4). In (4) show NGNG(D) (D) =_ E(u) where where E == 02'(NG(D)) 02t(NG(D)) andand E(u)/D E(u)/D is is regular regular on on the the fixed fixed point point setset A0 ofof DD onon X.X. LetLet r be be the the set set ofof triples (i, x, y) such thatthat ii Ec uEuE and (x, y) is a cycle of i on A.A. CountCount r inin two two waysways toto getget IlAl 0 I == 2.2. UseUse (4)(4) andand Phillip Hall's Theorem, 18.5, toto getget TT regularregular onon XX -- {H}.{HI. Then complete (5) using 15.11.15.11. For (6) use (5) and the observation that,that, as D is regular on T',T#, D == EndGF(2)D(T)# ~nd~~(2)D(T)' and EndGF(2)D(T)= = F.)

Appendix

Solutions to selected exercises Chapter 3, 3, ExerciseExercise 5.5. First,First, asas aa isis of of orderorder nn andand (a)(a) isis transitive transitive onon AA = {Gi{Gi: :1 1 ( < i i( < n} n} of of order order n, n, (a) (a) isis regular onon A,0, soso renumbering renumbering ifif necessarynecessary we may take Gia=Gi+1,Gia = Gi+1, with with the the subscripts subscripts readread modulomodulo n. ThusThus a`-1:a'-': G1 Gi isis anan isomorphism.isomorphism. Next, by definition of "central product" in 11.1,11.1,

n G=fi1Yi:YiEGij with yi yjy, = yy, jyi yi for for i i# $ j,j, soso flini yiyi isis independent independent of of thethe order order of of thethe factors. factors. Now as ai-1:a'-': G1G1 + Gi isis anan isomorphism,isomorphism,

n (*) G =lFIxjai_1:XiEG1I. i=1 Define 7r: n: G1 G1 + G by xnx7r = = nifli xac-1.xui-'. ThenThen

x7ra =(flxat_1)a = FI XU'= X7r, i i as thethe productproduct isis independentindependent of the orderorder ofof itsits factorsfactors andand indicesindices areare readread modulo n. ThusThus G17rG 1n (< CG(a).CG(a). Similarly

fxai-1yai-1 = T7xai-1 f yai-1=x7ry7r, (xy)7r = fl(xy)a1-1=i i i i so 7rn isis aa homomorphism.homomorphism. Further, Further, if if x xE E ker(n) ker(7r) then then 1 =1= ni fli xai--', xai-1, so

1 x =(flxa') E G1 r1 G2G3 ... Gn < Z(G1), i>1 so ker(7r)ker(n) (< Z(G1). Claim CG(a)CG(a) == G17rZ, GlnZ, where where Z Z = = CZ(G)(a). Cz(G)(ff). WeWe just sawsaw G17rGin (< CG(a), soSO G17rZGlnZ I< CG(a).CG(ff). Suppose g E CG(ff).CG(a). BY By (*), (*), gg == flini xiai-1xiai-' for for some some xi xi E E G1. GI. Then x1u-1 = flxia`-1 =g =gai+1 = fxiai+'=x-iu.i i i 270 Appendix for eacheach j, j, wherewhere uj uj = = ni+-j fi0 -i xiai+jxia`+j EE G2G3 . . . Gn, soSO xlxI = u u-1 E G1 fl G2G3 Gn < Z(G1) and hencehence xj xj=xlzj = xlzj forfor some some zjEzj E Z(G1). Z(G1). ThereforeTherefore gg =xlnz, =xiJrz, wherewhere z z= = finj zjaj-1zjaj-l E Z(G).E Z(G). As As g, g,x17r xln E E CG(a), Cc(a), also also z z =g(xiJr)-1 =g(xln)-l E ECZ(G)(a), Cz(c)(a), com-completing thethe proofproof that that CG((X) CG(a) = G1nCZ(G)(01). GlnC~(~)(a). Let KK = Glrr Gln and and Z Z = = CZ(G)(a). Cz(G)(a). AsAs 7r:n: G1 -+ + KK isis a a surjective surjective homomor- homomor- phism and G1 is perfect,perfect, K is perfect and as ZZ isis abelianabelian Z(1)z(') = 1.1. Thus

CG((X)(1) = (KZ)(1) = K(1)Z(1) = K.

Thus itit remainsremains to to show show that that NAut(G)(G1) NAut(G)(Gl) f1n CA~~(G)(K)4 CAut(G)(G1).CA~~(G)(G~). LetLet L = G1GI and and MM == CG(L). CG(L). AsAS G2 G2.. ... Gn. G,, < 5M, M, G G= LM.= LM. Let Let 0 EB NAnt(G)(L)E NAnqc)(L) f1n CAut(G)(K).Then Then p ,Bacts acts on on M. M. For For x xEL, E L,y y=x-' =x-1 .x7r xn EEM, M, soso asas ,Bp acts on M, y-1y-l . yfiyp EEM. M. Now

y-1yN=x-17r x x-1,B x7rf=(x .x-1P)x" EL as ,BB centralizescentralizes xn xir E E K K andand LL 9< G.G. Thus (x . x-1,8)x'7x-'~)~~ == y-ly-1 . y,Byp EL EL fln M == Z(G), so x . x-'Bx-1,B EE Z(L), Z(L), so [,B,[B,L, L, L]L] ==1. 1. HenceHence [,B,[p, L] = 1 by 8.9,8.9, so indeed ,B p E CAut(G)(G1)CAut(~)(G1).

Chapter 4,4, ExerciseExercise 7. 7. (1) (1) Let Let f (x,f (x, y)y) _ =1:j Ci,j 'j ai,1xiai,jxiyj y3 EE VV = f [x,[x, y] andand g E G. By definition

.f (g7r) = E ai,i (xg)` (yg)j , i,i soasxgso as xg = = ax+byandygax + by and yg == cxex+dyforsomea, + dy for some a, b,c,db, c, d E F,F,wehave we have ff(gn)~ (g7r)E V. It is easyeasy toto checkcheck thatthat girgn preserves preserves additionaddition andand scalarscalar multiplication,multiplication, so g7rgn E EndF(V).EndF(V). For h E G,

.f (gh)ir _ .f (x(gh), y(gh)) = .f ((xg)h, (yg)h) _ .f (xg, yg)(h7r) _ (.f (g7r))(h7r), so (gh)ir(gh)n = girhir,gn hn, and and hencehence 7r:n: G -++ GL(V)GL(V) is is a a representation.representation. . . (2) Observe VnV, hashas basisbasis B,Bn = = {xlyn-"{xi yn-i::0 0 5 < i i _(< n).n}. Further, Further, (xiyn-')(pr) (xi yn-i)(gir) =_ (ax + by)iby)' (ex(cx ++ dy),-'dy)n-i with (k) (ax + by)k= ()a1xi(by)k_JE Vk, j i so (xiyn-i)(gir)(xiyn-')(gn) EE Vn.V,. Thus G acts on V, of dimension n + 1. 1. (3) For i E Z,Z, letlet 7i be thethe residue of i modulo modulo p; thatthat is is 00 <5 i < p andand iiimodp. - i mod p. Assume Assumep

UU={f = { f EVV:aij E V,: ai,j =0 =O for for all all ii suchsuch that ii >>r}. r]. Appendix 27127 1

yn_P yn-P+l Then 0 # U # Vn asas xpyn-pxP E U but xP-1XP-l yn-p+' 6 U. Claim U is G-invariant, so that G isis notnot irreducibleirreducible on on V,,.Vn. It It sufficessuffices to showshow (x'(xiyn-')gnyn-i)g7r E U for all i with 1i == s <5 r. To To dodo so,so, we we willwill show show that that all all -monomials monomials in in x1 xjgn g7r and and yjyjgn g7r are of the formform xtxtyj-'yi_t with with i t5 < 7. J. Hence as n - i i = = r r -- s, s, all all monomials monomials inin (x'(,i yn-iyn-' )g7r)gn =xignyn-'gn= x i g7r yn-i g7r are are of ofthe the form form xtyn-' Xt yn-t with with t 5s t < +s + (r (r - - s) = r. r. Let J=j = v,v, so so thatthat jj ==up up ++ v v and and xig7r xjgn = = (xng7r)Px"g7r. (xUgn)Pxugn.As As (f (f + h)Ph)P = fPf P ++ hP hP forfor all all f, f, g g E E V, V, allall monomials monomials in in (xng7r)P (xugn)P areare of of thethe formform xPtyP(j-t)soxptyp(j-'), so we can assume jj = v. Then i xjgn =_ (xgn)j(xgn)' = (ax ++ by)jby)' = = ck1 (L)(i)()k(by)J_k, (ax)*(by)j-*, k=O=O k) establishing the claim.claim. Finally assume nn < p;p; it it remains remains to to show show GG is is irreducible irreducible on on Vn.Vn. By hypothesis, X = {x, {x, y}y] is a basis for U. IdentifyIdentify g EE G with its matrix MX(g)Mx(g) and define

a), (1a 10 ga = ha= / (0 Then S={ga:aEF} = F=T={ha:aEF}. Let MiMi=(yjxn-j:0< = (yjxn-j:Oi j

Lemma. TT actsacts on on M,Mi forfor allall ii andand [Mi[Mi/Mi-2, /Mi_2, TI = M,-1 Mi-l/Mi-2 /Mi_2 for ii >> 1.

Proof.Prooj: xhaxh, = x x andand yhayh, = ax + y, y, soso n-i [y'xn-j,ha] _ (y' x'-' )ha - j-1 (I)yai_kXn_k ajyj 1xn-j+1 modMj_2 k=O' andifa0Ojthen aj00.

Corollary. IfIf ii >>O 0 andand w E MiM, - M_1 Mi-1 thenthen [w, [w, T]TI = =Mi-1. Mi -1.

Prooj:Proof. The proofproof isis byby inductioninduction on on i.i. If Ifi =i 0=O then then w Ew (xE (xn)n) <5 Cv C"(T), (T), so [w[w,, TIT] = 0 0 = M-1. M-l. AssumeAssume thethe result holds at ii --1. 1. By By thethe Lemma,Lemma, [w,[w, h] h] EE MiplMi-1 - Mi-2 Mi-2 forfor some some h hE E T, T, so so by by the the induction induction hypothesis, hypothesis, Mi_2Mi-2 == [w, h, T]TI 5< [w, T],TI, soSO [w,[w, TIT] = Mi-iMi_1 asas dim(Mi-l/Mi-2)dim(Mi-1/Mi-2)= =1. 1. 272 Appendix

Now SL(U) == H = (T,(T, S) andand we we willwill showshow HH isis irreducible irreducible on on Vn.Vn. Let Let 0 #W W be be an an FH-submodule FH-submodule of of Vn. Vn. ByBy the the Corollary Corollary and and symmetrysymmetry betweenbetween T andand S,S, yn yn EE [W, S] or WW = (yn),(yn), soso inin either either casecase ynyn EE W. Hence by the Corollary, Mn-1Mn-1 == [W, TI,T], so VnVn == (yn(yn,, MnP1)Mn-1) < W. (4) Suppose Supposeg g Ec ker(nn).Thenxnker(nn).Then xn == xng7r xngn == (xg7r)n,soxg7r (xgn)",so xgn = Xx hx forforsome some nth root of unity A.h. SimilarlySimilarly ygn yg r == µy.py. ThenThen xyn-'xyn-1 = =(x~-')~n (xyn-1)g7r= = hpn-'),µn-1 xYn-l,xyn 1, soSO h = pl-" = It.p.

Chapter 4,4, ExerciseExercise 10.10. (1)(1) and and (2):(2): First First observeobserve that if I isis thethe identity identity matrix inin G then zq5(I)z'(I) = = z z forfor all all z z c E 17, r, so 0(I)@(I) =1= 1 is is the the identity identity ofof thethe monoid S ofof allall functionsfunctions fromfrom rr into F.r. Next, if A == (at,1)(ai,j) andand BB == (b;,1)(b;,j) areare inin G G then then for for z zE c F F withwith al,2z+a2,2al,2z+a2,2 # 0: Ca1,1z + a2,11 z(P(A)(P(B) = (P(B) (a 1,2Z + a2,2 /J) (a1,1z + a2,1)bi,I + (al,2Z + a2,2)b2,1 (al,iz + a2,1)b1,2 + (al,2Z + a2,2)b2,2 -(a1,ib1,1 +a1,2b2,1)z +a2,ib1,1 +a2,2b2,1 (ai,1bi,2 + a1,2b2,2)z + a2,1b1,2 + a2,2b2,2

If a1,2zal,2z ++ a2,2 = = 00 thenthen zq5(A)zo(A) = oo, m, so so zO(A)O(B)= oo0(B)=bi,i/b1,2

On the otherotherhand hand asas ai,2zal.2z + a2,2 a2,2 = 0 0 and det(A) # 0, a1,2 # 0, soSO z Z= = -a2,2/a1,2 -a2,2/ai,2 and a2,1b1,i ai,1a2,2b1,1/a1,2 bi,i(ai,2a2,i a1,1a2,2)=b1,1/b1,2. z.P(AB) - - - a2,ib1,2 - ai,ia2,2b1,2/a1,2 b1,2(a1,2a2,1 - a1,ia2,2)

Finally, ooo(A)O(B)mq5(A)q5(B) _= (a1,1/a1,2)0(B) (a131/al,2)q5(B) andand if if a1,2al,2 # 0 then

b1,1a1,1/a1,2 + b2,1_b1,1a1,1 + b2,1a1,2 (a1,1/a1,2)(P(B) = oo(P(AB). b1,2a1,1/a1,2 + b2,2 b12a1,1 + b2,2a1,2=

On the other hand if a1,2al-2 == 0 then al,la1,1 # :0 0 and and a1, al, i1 /al,2/ai,2 = oo, m, so so bi,iai,i +b2,1a1,2 oo(P(A)(P(B) _ oo(P(B) = b1,1/b1,2 - = oo-P(AB). b1 za1 1 + b2,2a1,2 Thus we have verified thatthat q5:0: GG -++ SS is is a a monoid monoid homomorphism.homomorphism. HenceHence as G is a groupgroup andand @(I) 0(I) ==1, 1, G* G* _= O(G) q5(G) is is aa subgroup of SS andand q5:0: GG +-+ G*G* is a surjective group homomorphism. AsAs G*G* is is a a subgroup subgroup of of S,S, each each O(A) @(A) EE G*G* is invertible inin S and hence (1)(1) holds. Further to complete thethe proofproof of (2) it remains to show ker(q5)ker(o) isis thethe groupgroup of scalarscalar matrices.matrices. So let A E ker(o).ker(q5). AppendixAppendix 273273

ThusThus zo(A)zq5(A) = = z z for for all all z z E E F. r. In In particular particular oo m = = ooo(A) mq5(A) = = a1,1 a1,1/a1,2r /ai,2, soSO a1,2 al,2 = O0 #a1,1. al,l. Also 00=O = Oq5(A) (A)=a2,1/a2,2, = a2,1/a2,2, soso a2,1 a2.1 =O= 0 #a2,2. a2,2. Finally Finally a1,1 +a2,1 1 1/a2,2, a1,2 + a2,2 soso a al.1 1,1 == a2,~a2,2 and hence AA = = a al,l 1,11 I is scalar.scalar. (3)(3) ByBy constructionconstruction a: Q C2 -++ F r is is a a bijection. bijection. LetLet HH bebe the the stabilizer stabilizer in in G G of of Fx1 Fxl E E Q. C2. By By 13.5, 13.5, G G is is 2-transitive Ztransitive on on S2, C2, soso GGisprimitiveonC2 is primitive on Q byby 15.14.15.14.Hence Hence H isismaximalinG maximal in G byby5.19,soG 5.19, so G = (H, (H, t), t), wherewhere tt cE G G interchanges interchanges x1 xl andand x2. Thus toto showshow (wg)a(t)g)a = (wa)$(g) forfor eacheach g EE G,c, it suffices to show (wg)a(t)g)a = (coa)0(g)(wa)@(g) for each g cE H andand gg == t. t. EachEach hh EE HH isis the the product product ofof aa scalar scalar matrix matrix and and aa matrix matrix

_(a0 g aEF#, bEF, b 1 soso itit sufficessuffices to show (wg)a(t)g)a == (wa)o(g). (wa)@(g). ButBut 0(g): q5(g): z zv--> H azaz + + b, b, so so g g fixesfixes Fx1Fxl andand 0(g)q5(g) fixes fixes oo m = = (Fx1)a. (Fx1)a. Further, Further, for for co w = = F(Xx1 F(hxl + + x2), x2), ,X h E F, (wg)a = F((a.1. + b)x1 + x2)a = a), + b = )fi(g) _ (wa)o(g)

Next,Next, tt interchangesinterchanges x1 xl andand x2, soSO t has cycles (Fx1,(Fxl, Fx2) andand (F(hxl(F(ax1 +x2),+ x2), F(a.-1x1~(h-lxl + x2)), x.~)), a.h EE F', F#, onon C2. Q. Then Then q5(t): 0(t): z z H H l/z,1/z, so @(t)0(t) hashas cyclescycles (oo,(m, 0)0) andand (X,(h, h-')A 1) on r.F. Hence as (Fx1)a == m,oo, (Fx2)a == 0, and F(.lx1F(hxl + x2)a x2)a =A, thethe proofproof isis complete.complete.

ChapterChapter 5, 5, Exercise Exercise 6. 6. (1) (1) Induct Induct on on in. m. By By Jordan's Jordan's Theorem, Theorem, 15.17, 15.17, G G is is 2-transitive2-transitive on X, so so thethe resultresult holdsholds whenwhen inm = 1.1. Again by Jordan, GxG, is primitiveprimitive onon XX - {x} {x) = X' for Xx EE Y. Y. So So as as Y' Y= = Y -- {x} {x) is of order min - 1 1 withwith Gx,y'G,,yf == GGy primitiveprimitive on on X' X' - - Y'=Y' = X X -- Y, Y, Gx G, isis in-transitive m-transitive onon X' byby inductioninduction onon in.m. Thus G isis (m(m + 1)-transitive 1)-transitive on on XX byby 15.12.1.15.12.1. (2)(2) LetLet tt bebe a a transposition transposition oror cyclecycle ofof lengthlength 33 inin GG andand setset Y Y ==Fix(t). Fix(t). ThenThen t E GyG and (t)(t) is is transitive on X - Y. Y. ThusThus GyG isis primitiveprimitive on X - Y Y as as IXIX - YI Y I isis prime. prime. ThereforeTherefore GG isis (n(n -- 2)-transitive 2)-transitive on on X X by by (1),(I), so so G G contains contains thethe alternatingalternating group group by by 15.12.4.15.12.4. (3)(3) Let {x}{x) == M(a)M(a) nrl M(b), y =xu,=xa, zz =xb,=xb, and and A A = = {x, {x, y, y, z}. z). Now Now Fix(a) rln Fix(b) = Fix((a, b)) C5 Fix([a,Fix([a, b]). b]).

IfIf Vv EE M(a) M(a) -- {x, {x, y} y) then v, va-'va-1 E M(a) - {x} {x) C 5 Fix(b), Fix(b), so so v[a, b] = va-lb-lab = va-lab = vb = v.

Similarly if v cE M(b) -- {x, {x, z} z) then then v,v, vb-1 vb-' EE Fix(a), Fix(a), soso v[a, b] = va-lb-lab = vb-lab = vb-1b = v. 274 Appendix

Thus XX - A A C G Fix([a, Pix([a, b]). b]). Next ~ext xb-1xb-' EE Mov(b) - {x} {x} C G Fix(a),Fix(a), soso Y[a, b] =Ya-lb-lab =xb-lab =xb-lb =x. Similarly xu-'xa-1 E Fix(b), so x [a, b] =xa-lb-lab =xa~lab =xb = z.

Finally z =xb=xb E Mov(b)Mov(b) -- {x}{x} Gc Fix(a)Fix(a) andand yy =xa=xa EE Fix(b),Fix(b), so z[a, b] = za-lb-lab = zb-lab =xab = yb = y. Thus (y, x, z)z) isis aa cyclecycle ofof lengthlength 33 inin [a,[a, b], b], completingcompleting the the proofproof ofof (3).(3). (4) AssumeAssume G does not contain the alternating group on X and IIY Y I I> > n/2; n/2; we must derive a contradiction. AsAs II Y I I> > n/2,n /2, rF = X X -- Y Y has has order order less less than than I YI Y I, I, so byby minimalityminimality ofof IYIY1, I, Gr Gr # # 1.1. Pick Pick a a E E G#,, G,, y E M(a),M(a), and set A == Y - {y}. {y}. Then IlAl A I< < IYI,IYI, so so by by minimality minimality of of IYI, IYI, Ga Go # #1. 1. pick Pick b bE EG#,. G. AsAs GyGy ==1, 1, y E M(b), while M(a) fl M(b) c (X - F) n (X - A) = X - (I' U A) = {y}, so M(a) flfl M(b) M(b) = = (y). {y}. Thus Thus [a, [a, b] b] is is a a 3-cycle 3-cycle by by (3),(3), so so (2) (2) supplies supplies aa con-contradiction, establishing (4). (5)PickY(5) Pick Y asin(4)andletHas in (4) and let H =Sym(X)y.ThenIHI=rn!,wherern== Sym(X)y. Then I H Im!, where m = IIX-YI X -Y I andbyand by (4),(4), rnm > > [(n[(n ++ 1)/2]. Now H fln GG=Gy = Gy == 1, 1, soso

ISym(X)1 > IHGI = IHI IGI = m! I GI, establishing (5).(5).

Chapter 6,6, ExerciseExercise 2.2. (1)(1) LetLet GG bebe thethe semidirectsemidirect product of H byby A.A. WeWe must show there exists anan A-invariant Hall Hall n-subgroup n-subgroup of of H.H. Let A be the set of Hall n-subgroups of H. ByBy Hall'sHall's Theorem,Theorem, 18.5,18.5, there is K EE A and H isis transitivetransitive on A, soso byby aa FrattiniFrattini argument,argument, GG = HNG HNG(K). (K). By By thethe Schur-Schur- Zassenhaus Theorem 18.1, there exists a complementcomplement BB to NH(K) in NG(K)NG(K) and BgB9 == A for somesome gg EE G.G. HenceHence JJ = K9 Kg isis anan A-invariant Hall n-subgroupn-subgroup of H. (2) We must show CH(A) is transitive onon thethe set Fix(A) of fixed point of A on A. ButBut byby Schur-Zaussenhaus,Schur-Zaussenhaus, NH(K) is transitive onon thethe setset A'Ac fl NG(K) of complements to NH(K)NH(K) in in NG NG(K), (K), so SO by 5.21,5.21, NH(A)NH(A) is is transitivetransitive on Fix(A). (3) We must show eacheach A-invariant n-subgroup n-subgroup XX ofof H isis containedcontained inin aa member of Fix(A). TheThe proof proof isis by by inductioninduction on on I I G 1,I, soso assumeassume G is a minimal counter example. LetLet MM be a minimal normal subgroupsubgroup ofof GG containedcontained inin HH and G* == G/M.G/M. ByBy minimalityminimality of G there there exists exists anan A-invariant A-invariant HallHall n-n- subgroup Y* ofof H* containingcontaining X*. Now IYI,IYI, == I IHI,, HI,r, so Y contains a HallHall n-subgroup of H, andand ifif HH :h # Y Y thenthen XX isis contained contained inin anan A-invariantA-invariant Hall Appendix 275 n-subgroup7r-subgroup ofof YY by by minimality minimality of of G.G. ThusThus HH == Y, so HIM is is a a 7r-group n-group andand hence H = JM, so H* = J*. By 9.4, M isis aa p-group forfor somesome primeprime p.p. If If Pp EE Tr n thenthen HH isis a a 7r-group, n-group, so H is an A-invariant Hall n-subgroup7r-subgroup of of H H containingcontaining X.X. ThusThus pp 0$ 7r,n, so XX is aa Hall n-subgroup7r-subgroup of of MX. MX. Let Let Z Z= = JJ fl XM.XM. As As H* H* = = J*, Z* == X*, X*, so so Z Z andand XX are A-invariant Hall 7r-subgroupsn-subgroups of XM, andand hencehence by (2) there is g E CH(A) with ZgZ9 = X. Thus JgJg is is anan A-invariantA-invariant HallHall 7r-subgroupn-subgroup ofof H containing X. (4) Part (4) does not depend on p andand hence remains valid.

Chapter 6,6, ExerciseExercise 3.3. AsAs VV isis thethe permutationpermutation modulemodule for for G G = = Alt(I)Alt(I) onon II and F = GF(2), GF(2), wewe cancan identifyidentify VV withwith thethe powerpower set of I, and and forfor u,u, v v E E V,V, u + vv isis thethe symmetricsymmetric difference of u and v; that is, u+v=uUv-(u fl v). Define the weight ofof vv toto bebe thethe orderorder I Iv 1I of of thethe subsetsubset vv ofof I. (1) Let 0:A0 # W be an FG-submoduleFG-submodule of VV andand Ww E W#W# ofof weight m.m. By 15.12.3, G is (n - 2)-transitive 2)-transitive on I, so so asas n >> 2, 2, GG isis transitive on m-subsets of I andand hence onon vectors of weight m. Thus for each v of weight m, v E ww G G GC W.W. Assume w # I, I, the the generator generator of of Z. Z. Then Then there there is is v v of of weight weight m rn suchsuch that that ww flfl vv is of order rnm - 1, 1, so so w w + + v v is is of of weight weight 2. 2. So So as as w w + + V v E E W,W, wewe maymay taketake rnm == 2. But as w is ofof weightweight 2, (wG)(wG) = U. U. Namely,Namely, if 0 # uu EE U,U, pickpick ii EE u, and observeobserve u u = = xiEi +O j,,{i, jEU{i, j}.j). Thus WW = UU oror V.V. (2) and (3):(3): FirstFirst observe that G == 02(~).02(G). IfIf n >> 4 4 this this follows follows because G is simple byby 15.16. IfIf nn == 3 or 5 it isis easyeasy toto checkcheck GG == o~(G)02(G) directly.directly. Observe nextnext thatthat if if n n is is odd odd then then I 10 $ U, U, soso VV = = UU @® ZZ andand hence hence 0 U E = U is of dimensiondimension n n - - 1. On thethe otherother hand, hand, if ifn nis is even even then then 0 <0 > 3. 3. LetLet HH == G1G 1 and and J J = = II - { {I}, 1), so that H =Alt(J) = Z A,i_1. A,-1. ForFor j jE E J, J, let let vj v, = = {1, {I, j}. j}. Then Then {vj: {vj: jj E E J} J} isis a a basis basis forfor U with vjhv, h == Vjh vjh for h E H, soSO U isis thethe permutation module for H of degree n - 1. 1. In In particular,particular, by ExerciseExercise 4.6.1, 4.6.1, v v = = C,,,EjEJ vivj generatesgenerates Cu(H),CU(H), andand by induction onon n,n, (v)(3) = CU(H).Cu(H). Notice if n is odd then v == J soso (J) (I) == CC, Cu(H), (H), while if nn is even thenthen v == I, so so CUCu(H) (H) == 0. 276 Appendix

Assume nn isis odd,odd, so so that that U U Z = 0U as anan FG-module. FG-module. Suppose Suppose w W E ECw(H) Cw(H) - - 0U and let MM= = (wG). (wG). By By ExerciseExercise 4.6,4.6, MM isis aa homomorphichomomorphic image ofof V as an FG-module. As CM(G)( < Cw(G)Cw(G) = = 0, 0, 0 0 # 0 [M,[M, GIG] _( < [W, [W, GI G] = = 0, U, so so aa GG is irreducible on on 0, U, 0 U=[M, = [M, G]GI = = Cw((H, K)) = Cw(H) n CH(K),CH(K), so dim(Cw(H nfl K)) <5 22 andand d <( 1. 1. Therefore dim(W) = d + dim(U) < n - 1= dim(V ), so again W == V, completing the proof.

Chapter 7, 7, Exercise Exercise 5. 5. Let Let u u = = ax2 ++ bxy bxy ++ cy2 cy2 and and vv = rx2 rx2 + sxy SXJ + ty2ty2 be inin W. Then Q(x+y)=(b+s)2-4(a+r)(c+t) = (b2 (b2 - 4ac) 4ac) + (s2 (s2 - 4rt) 4rt) ++ 2bs 2bs - 4(at 4(at ++ rc) rc) = TOQ(u> ++ Q(v) Q ++ f f(u, (~3 v).v). Evidently f isis bilinear bilinear and and symmetricsymmetric and and Q(,ku)Q(hu) = A2Q(u) h2~(u) for h Et F. To prove (W, Q) isis nondegeneratenondegenerate itit sufficessuffices toto observeobserve thatthat XX == [x2, {x2, xy,xy, y2}y2} is a basis for W with x2,x2, y2y2 singular,singular, Q(q)Q(xy) = = 1,1, ff (x2,(x2, y2)y2)= = -4,-4, andand xyxy is orthogonal to x2 and y2. (2) Let Ca bl g= E dJG. c To proveprove (2),(2), it it sufficessuffices to to check check Q(wga) Q(wga) = =,k(ga) h(ga)Q(w) Q(w) and and f f (wga, (wga, zga) zga) =_ h(ga)A(ga) ff (w,(w, z)z) forfor w, w, z z t E X, X, where where h(ga) k(ga) = = det(,g)'det(g)2 =ad = ad -- bc. For example Q(x2ga)~(x~ga) == Q((ax Q((ax + + by)2) by)2) = = Q(a2x2 ~(a'x' + + 2abxy 2abxy ++ b2y2) b2y2) = 4a2b2 - 4a2b2 = 0 = X(ga)Q(x2), Q(xy(ga)) = Q((axQ((ax + + by)(cx by)(cx + dy)) dy)) = Q(acx2~(ac2 ++ (ad + bc)xy ++ bdy2)bdy) = (ad(ad + bc)2 hc)' -- 4abcd 4abcd = (ad - bc)2 bc12 == gga)h(ga)Q(xy), Q(xy), Appendix 277

and

f (x2ga, y2ga) = f ((ax + by)2, (cx + dy)2)

= f(ax2 + 2abxy + b2y2,c2x2 + 2cdxy + d2y2)

= 8abcd - 4(a2d2 + c2b2) _ -4(ad - bc)2 = )(ga)f (x2, y2).

(3) Observe thatthat uu isis singularsingular iffif b2 = 4ac, so the set S of singular points inin W is So U {Fx2},{FX~}, wherewhere

So=So = {~(e~x~{F(e2x2 + 2exy 2exy + y2): y2): e~e E F}.F}. The subgroup T={(i ?)F}

of G isis transitivetransitive onon So,So,so so G G is is 2-transitive 2-transitive onS on Sandhence and hence A(W) A (W) = = GaAF2,,,2. Ga OF,Fy2. Now if hh EE OFx2AFX2,,,2, Fy2, then then as as Fxy Fxy = = (x2, (x2,y2)l, y2)', Mx(h)MX(h) is is diagonal diagonal withwith x2hx2h == px2,µx2, y2h ==.lµ-1 Ap-l y2, and xyhxyh = = Axy, 5xy, for for some some µp E E F#,F', wherewhere kh =_) h(h). (h). OnOn the otherother handhand gaga E E AFx2,Fy2, A,,2,,,2, where

(1 0- g )EG, o

andx2ga =x2,=x2, y2gay2ga =hp-2y2,andxyga==µ-2y2, andxyga = Ap-'xy,sohsoh =ga.pI E Ga.S. Therefore (3)(3) holds. (4) If U isis aa nondefinitenondefinite 3-dimensional orthogonalorthogonal spacespace over over F F thenthen U == HID, where where H H is isa ahyperbolic hyperbolic line line and and D D = = Fd Fd is is definite. definite. Multiplying Multiplying thethe quadratic form QuQU onon UU by a suitable scalar, wewe maymay assume assume QU(d)Qu(d) = 1.1. Thus (U, Qu)Qv) r- (W, cw, Q).Q). (5) Part (4) implies (5),(3, since if F isis finitefinite oror algebraicallyalgebraically closed,closed, thenthen nono 3-dimensional orthogonal space space over F is is definitedefinite by lemmas 21.3 and 20.10.1. (6) Let A = O(W, A(W, Q).Q). By By ExerciseExercise 4.7.4, 4.7.4,

ker(a)ker(a)={aI:aEFand = {ata E F and a2 a2=1}=(-I).= 1) = (-I).

Thus G(')CYG(1)aZ - L2(F),L2(F), soso as A0 = SGa SGa andand S (< Z(A),Z(O), A(')0M = G(')aGO)a Z= L2(F).L 2(F). As 0/0(1)A/A(') isis abelian, abelian, rg rg E E rO(1) rA(') for each r E R and g E A,0, soso rrgrfl E A(').0M. ButBut by 43.12.1, L2(F) isis simplesimple unless I FI F I= = 3, 3, so so except except possiblypossibly inin that case, 0(1) A(') = (rrh: (r9: hh EE 0(1)).A(')). Finally,Finally, if II FFl 1 == 3 then 0) r=(-1 ER 0 1 278 AppendixAppendix

with center Fxy,Fny, and r invertsinverts g_ (11) EG 1 1

ofof orderorder 3, so again GO)aG(')m is generated by conjugates ofof r+rrg == g-lg-1 of order 3.

Chapter 8, 8, Exercise Exercise 11. 11. Let Let V V be be a a normal normal elementary elementary abelianabelian subgroupsubgroup of G of maximal rank, H = CG(V),CG(V), and G* == GIH.G/H. By By 23.16,23.16, VV == Q1(H). C21(H). AsAs m(G)m(G) >> 2,m(V) 2,m(V) > > 2 2by23.17,by 23.17, so so we wemay may assumeassumem(V) m (V) ==3. 3.Thus Thus G* <( GL(V GL(V), ), soso m(G*) <5 m(GL3(p)) = 2. 2. LetEp.LetE,. EAsGwithnA > 3.AsV=Q1(H),AnH=AnVisofrank3.A~V=C2~(H),AnH=AnVisofrankat at mostmost 2.2. AlsoAlso m(A*)m(A*) <( m(G*)m(G*) = 2, 2, soso 4 < m(A) = m(A*) + m(A n H) < 2 + 2 = 4

andand hencehence all all inequalities inequalities areare equalities,equalities, so m(A)=4m(A) =4 andand m(A*) m(A*) = m(AnH)=2.m(A n H) = 2. LetLet BB == A An n V. V. ThenThen A A centralizes centralizes the the hyperplanehyperplane B B of of VV and and m(A*)m(A*) = = 2, 2, so so A*A* is thethe fullfull groupgroup ofof transvections transvections withwith axisaxis B.B. ThereforeTherefore [A,[A, V]VI = B B == Cv Cv (a(a) ) foreachforeacha a E E A-B.A- B. Suppose Suppose D D = ZEp4 Ep4 with with D D

ChapterChapter 9, 9, Exercise Exercise 6. 6. (1) (1) Fix Fix vi vi c EVZ, &, ii == 1, 1,2, 2, and and define define

fV,,U2: V1 x V2 -* F,

(u1, u2) H f1(v1, u1)f2(v2, u2)

AsAs f1fl andand f2 f2 areare bilinear, bilinear, so so is is f,,,,,,2, f,, ,, , so so byby thethe universaluniversal property ofof thethe tensortensor product,product, f,,,,, extendsextends toto aa linearlinear map map f f,2: ,,,,: VV -* -+ FF with with f ,,,,11I,,,2(u1 ,(ul @I (9 u2) = f,,,,,,2(u1,fv, ,vz (u 1, u2). Similarly V1xV2-*v1 x vz -+ Vv*,

(v1, v2) H fVhV2 AppendixAppendix 279279 isis bilinear, bilinear, so so there there is is f fE EHom(V, Hom(V, V*) V*) withwith f f(vl(v1 (9 @ v2) v2) == fv,,,,. ThereforeTherefore f f:: V V x x V V -* -+ F definedF defined by by f (x,f (x, y) y) = =f (x)(y) f(~)(~) is isbilinear bilinear and and f (V1 ® v2, u1 (9 u2) = 1(V1 (9 v2)(ui (9 u2)

U U 2 ) -

AsAAs s the the fundamental fundamental tensors tensors generate generate V V = = V1 Vl 0@ V2, Vz, ff isis unique unique subject subject to to this this property.property. Finally Finally f (ul ® u2, v1 (9 v2) = f (u1, vi)f (u2, V2)= (-f (VI, ul))(-f (v2, u2)) = f(v1, ul)f(v2,u2)= f(V1 0 v2, u10 u2), soso as as the the fundamental fundamental tensors tensors generate generate V1 Vl ® @ V2,V2, f f isis symmetric. symmetric. (2)(2) FirstFirst f f(v1 (vl ® @ v2, v2, V1 VI @(9 v2)v2) = = fl(v1,fl (vi, vl)f2(v2,vl)f2(v2, v2)v2) = = 0 as fi isis symplectic. symplectic. ButBut ifif char(F) char(F) : #2, 2, recall recall from from Chapter Chapter 7 7 that that the the unique unique quadratic quadratic form form QQ associatedassociated to f satisfiessatisfies Q(x)Q(x) == f f (x, (x, x)/2.x)/2. Thus Thus Q(v10 Q(vl @ v2) v2) == f f(v1(9 (vl @ v2, v2, v1® vl @ V2)/2v2)/2 = 0, 0, asas desired. desired. On On the the other other hand: hand:

Lemma.Lemma. If If char(F) char(F) = = 2, 2, X X is is a abasis basis for for an an F-space F-space U, U, f f is is a asymplectic symplectic form form onon U,U, and and ax a, EE FF for for each each x x E E X, X, then then there there is is a a unique unique quadraticquadratic form form QQ on on UU associated associated to to ff such such that that Q(x) Q(x) ==a, ax for for each each x x E E X. X.

LetLet XiXi == {vi, {vi, ui)ui) be be aa basis basis forfor V1;Vi; then X == X1 @®X2 X2 is a basis for V,V, soso byby thethe lemmalemma therethere isis aa uniqueunique quadratic form Q on VV associatedassociated to f withwith QQ(xl (x 1 @0 x2) = 0 0 forfor each each x, xi E E Xi. Xi. Now Now for for w w c V1,E Vl, ww = =avla v 1 ++ bu bul 1 andand Q(w (9 v2) = Q(a(v1 (9 v2) + b(u1 (9 v2)) 2 == a2Q(vi a Q(vl (9 @ v2) ~2) ++b2Q(ul b2e(ul (9 @ v2) vz) + + abf abf (v1 (vl ® @ v2, vz,u1 ul (9 @ V2) v2) =O.= 0. SimilarlySimilarly Q(w Q(w (9 @ u2) uz) = = 0, 0, so so for for z z = = cv2 cv2 + + due duz E E V2, V2, Q(w (9 z) = Q(c(w (9 v2) + d(w (9 u2)) == c2 c2 Q(w Q(W (9 @ v2) v2) + + d2 d2e(w Q(w (9@ u2) u2) + + cdf cdf (w (w ® @ v2, vz, w w (9 @ u2) u2) = = 0 0 completingcompleting the the proof proof ofof (2).(2). (3)(3) Pick Pick XiXi toto be be a a hyperbolic hyperbolic basis basis for for Vi. V,. ThenThen x1 xl ®@ x2x2 isis orthogonalorthogonal toto x1xl ®@ Y2y2 andand yly1 @®x2 xz for all xi,xi, yi Ec Xi, while while

ff(v1 (V1 ®@v2,ul@~2)=1= v2, u1 (9 u2) =1= f (V1f(v1 ® @Uz,U1 u2, u1 ®@v2), v2), soso XX isis aa hyperbolichyperbolic basis basis forfor V.V. (4)(4) LetLet gg = = (g1, (gi, 92)g2) E E A.A. ThenThen f(V1 (9 v2)gn, (ul (9 u2)gn) = f(v1g1 ® v2g2, u1g1 (9 u2g2) = fi(v1g1, u1g1)f2(v2g2, u2g2) = A(g1)fl(vl,ul)?(g2)f2(v2,u2) A(g1)X(g2)f (vl ® V2, U 1 (9 u2), 280280 AppendixAppendix

soso asas thethe fundamental tensors tensors generate generate V, V,gn gn E EA(V, 0(V, f) f) with h(gn)) (grr) =_ .l(gi))l(g2).h(gl)h(gz). SimilarlySimilarly

Q((vl (9 v2)(g7)) = Q(v1g1 (9 v2g2) = 0 = Q(v1 (9 v2)

asas Q(u1Q(u1 (9@ uz)u2)=0 = 0 forfor allall uiui EVi,E Vi, soSO byby thethe lemma,lemma, gngn alsoalso preservespreserves Q.Q. OfOf coursecourse gngn Ec O(V, Q)Q) iffif 1=,1(g)1 = h(g) ififf ,l($2) h(gz) =,l(gi)-1.= A(~~)-'. AlsoAlso gg EE ker(rr)ker(n) iffif x1xl (&X2=(XI@ xz =(XI (&x2)g7r@ xz)gn =xlgl=xlg1 O@ X292 xzgz forfor allall xixi EEX,, Xi, so so asas XX isis aa basis basis for for V, V, itit followsfollows that g cE ker(rr)ker(n) if iff gi gi ==pi µi I Ifor for Ui ui E E F# F' withwith A2 p2 == µ p; i 1 . (5)(5) The The map map

Ta:T,: VlV1 @ ® vz V2 + - V,v, (u,(u, va) H vv ®ua@ ua isis bilinear,bilinear, so there exists aa uniqueunique tt == t,t,, Ec End(V)End(V) withwith t,(uta(u @(9 va)va) = = v @® ua forfor allall u,u, v v c E V1. Vl. Evidently t2t2 = 1. 1. Further,Further, asas aa is is an an isometry, isometry, f ((vi (9 va)t, (ul (9 ua)t) = f (v ® via, u (9 ula)= fl(v, u) f2(vla, ula)

== fl fl(V1, (VI, u1)fz(va,ui)f2(va, ua)= f(vl f (vl (9 @ va, va, u1 U1 ® @ ua),

sotso tis is anisometry. an isometry. Next, Next, we we may may choose choose notation notation so so that that via vl a = =v2 vz and and u u11 a == u2, UZ, soSO

(v1 (9 v2)t = (v1 (9 vla)t = (v1 (9 via) = v1 ® v2.

SimilarlySimilarly t t fixes fixes u u 1®u2 1 @ uz andand interchangesinterchanges v1®u vl @uz 2 andand u1ul ®v2,@ vz, so t is aareflection reflection oror transvection.transvection. LetLet gg=(gl, = (g1, 1)1)~ E A,,01, so that gn EE O1n.Aln. Then (u (9 va)gt = (v (9 ua)gt = (vgi ® ua)t = u ® vgla = u ® (vagia*), soso as Ala* = 02,A2, gtg' E Azn.O2n. Thus (Aln)'(Din)t = LenAzn and and similarly (Gin)'(Gin)` = Gen.Gzn. (6)(6) and and (7):(7): ForFor FvFv a apoint point in in V1 Vl let let Fv Fv ® @ V2 V2 == {v {v ®@ u u : :u uE E V2} V2) andand setset G1L1= = {Fv{Fv@ ® V2V2 : : vv EE V141.v:).For For Fu Fu apointa point in V2 define define VlV1®@ FuFu and G2Lz similarly. ObserveObserve GL = G1 L1 U LZG2 isis aa setset ofof totallytotally singularsingular lines in V, as Q(vlQ(vl (9 @ v2)v2) = 0 0 for allall vivi cE Vi. Vi. Claim the singular vectorsvectors areare thethe fundmentalfundmental tensors tensors v1vl ®@ v2vz and GL is the setset of all totallytotally singularsingular lines.lines. For if x EE V# is totally singular thenthen 00:0 # x'x1 fln Fv®Fv@ V2,V2, soSO x E (v@u)'(v(gu)1 forfor some some u u E E V20. V$ Now Now as as V V is is 6dimensional 4-dimensional hyperbolichyperbolic space and yy == v @® uu isis totallytotally singular,singular, we we have have y' y1== Fy Fy I I H,H, where H is a hyperbolichyperbolic line, and if H1H1 andand H2Hz areare thethe two two totallytotally singularsingular points in H,H, then Fy + H1 H1 andand FyFy ++ H2 Hz areare thethe twotwo totallytotally singularsingular lineslines throughthrough Fy and containcontain all totally singular pointspoints in in y'.y1. ThusThus xx Ec FyFy ++ Hi for some i, and and as FvFv ®@ V2 VZ and and V1Vl ®@ Fu areare totallytotally singularsingular lines through y, wewe maymay taketake x EE Fv ®@ V2.Vz. This showsshows each totally singularsingular point is aa fundamentalfundamental tensor,tensor, and each totally singular line l1 is one of the two totally singular lines Fv ® @ V2 Vz Appendix 28281 1

and V1Vl @ ® Fu Fu through aa pointpoint F(v F(v 80 u)u) onon 1,I, so L isis thethe setset ofof allall totallytotally singular lines. Observe G1 L1 and G2Lz are thethe twotwo classesclasses of maximalmaximal totally singular subspaces of VV described in 22.13, soso byby 22.13,22.13, the the subgroup subgroup r F of of X E = = A(V,0(V, Q) acting on G1L1 andand L2G2 is is of of index index two two in in X, E, with with X E == r(t).F(t). Next by Exercise 7.1.1,7.1.1, GiGl == SL(&).SL(V, ). Further Further if ifXi X, = _ (vi, (v ul)ui) isis ourour hyper-hyperbolic basis, µp E E F#F' andand gµ,, g,,i EE GL(V;) GL(&) withwith v;VigLL,i gµ,; = v;Vi andand uiUigCL,i gµ,, =pi,= µu;, then gµ,?g,,i EE AiAi withwith h(g,,i))(gµ,i) = µ,II.9 soSO Ai =SL(V,)(gµj: µ E F#) =GL(VV).

Now 01Aln n <5 ro, ro, thethe stabilizer stabilizer in in F r of of V1®Fv2 Vl @ Fvz and and V1® Vl @ Fu2. Fuz. For For x cE VlVl ®@ Fv2,Fvz, define fx cE (V1 @® Fu2)* by fx(y)= = f (x, y). Then the map x Hi-* fx is a define f, (Vl Fuz)* by f,(y) f (x, y). Then the map x f, is a To-ro- isomorphismisomorphism ofof V1Vl @® FvzFv2 with (Vl(V1 @(9 Fuz)*, Fu2)*, so so as as Cr(Vl Cr(Vi @(9 Fvz)Fv2) nn Cr(ViCr(Vl ®@ Fu2)Fuz) = 1, 1, ro ro isis faithful faithful on on V1® Vl @ Fv2,Fvz, and and thus, as 01Aln n acts acts faithfully faithfully as as GL(V1(9GL(Vl @ Fv2)Fvz) on V1Vl @® Fvz,Fv2, it follows that roF0 == Aln.Otn. Further Further 0271A2n isis 2-transitiveZtransitive on thethe points of A2Az andand hencehence alsoalso onon G2,CZ, soSO Fr == F0L r0A2n 27i = O7i.An. This completes the proof of (6).(6). Finally c2Q == c2(V,Q(V, Q) is thethe derivedderived groupgroup of of O(V, O(V, Q),Q), andand O(V, O(V, Q) =_ (O(V, Q) n r)(t) with with ro =O(V, Q) n F = ((gl, g2)7r:,1(g1)=A(g2)-'} by (4) and (6). Thus Foro == (G1G2)2 (G1G2)n Tom, . Tn, where where T T = = (gµ7r: {g,n: µ p E EF#} F') andand gN, g, = (gµ,1,(g,,l, 9,,-1,2)g,-~,~).Furtherg,m FurthergN,22r=hlh2n =h1h22r E(G~Gz)~,E (G1G2)n, wherevlhlwhere v1h1 =~l.-~vl,u~hl = µ-1v1, u1h1 = µupul, 1, v2h2 == ~VZ,µv2, and u2h2UZ~Z = µ-1p-luz, u2, andand g`µg; == g,-I,g.-I, so so [Tn, [Tn, tlt] i< (G1(G1Gz)n G2)jr and 1 hence Q = Q(V,Q(V, Q)Q) <5 (G1G2)n. (GIGz)n. Conversely, Conversely, ifif IIF1 FI > 33 thenthen G;TrGin 2= SL2(F)SLz(F) is perfectperfect byby 43.12.1,43.12.1, soso Q Q = = (G (GIGz)n. 1 G2)jr.If IIF1 F I= = 33 itit can be checked directly that (G1Gz)n(G1G2) r 5

Chapter 9, 9, Exercise Exercise 9. 9. By By Exercise Exercise 9.1, 9.1, a =a =f 0 f E @ HOMFGE Hom~~(v, W, V0*)v'*) and as f ## 0, 0, a:0a # 0. 0. Thus Thus as as G G is is irreducible irreducible on on V,V, aa is is an an isomorphism isomorphism byby Schur's LemmaLemma 12.4.12.4. LetLet orCJ = 0* O* regardedregarded asas anan automorphismautomorphism of of GL,,GL,(F), (F), the group of invertible n by n matricesmatrices over F, and and regardregard GG as as a a subgroup subgroup of of GL,, GL,(F). (F). AsAs VV -2 VQ, V", g°g" = = gB* gB* for for all all g gE E G G and and some some B B E E GL,, GL,(F) (F) fromfrom the the discussiondiscussion in section 13.13. ThusThus

gal gB*(BQ)* = (gB*)° = g°(Bo )* = = g(BB°)* so VV 2 VQ2.v"'. However, However, 6' 0 commutescommutes with with thethe transpose inverseinverse mapmap * onon GL,,(F),GL,(F), so V°zv"' ==((V*)*)Bz ((v*)*)" 2 = V"V° as (V*)* 2- V.V. Similarly if V 2= VoV' thenthen V*V* -2 VB* v'* =2 V, V, so SO by by Exercise Exercise 9.1.4, 9.1.4, GG preservespreserves aa nondegeneratenondegenerate bilinear bilinear formform onon V.V. ThatThat isis (1)(1) holds holds inin this this case,case, so so we we may may V02, assumeassume V V 7 V0.v'. ThenThen as as V V 2 - v", mrn is even, since otherwise 0 6' is is a apower power of of B2.6''. 282 Appendix

Let K be the fixed fieldfield of of 0'. 02. Then Then (0') (02) = = Gal(F/K), Gal(F/K), so by 26.3, V == F ®KUBKU for some irreducibleirreducible KG-submodule U of V. As 0 is an automorphismautomorphism of K of order 2, Exercise 9.1.4 says G preserves a nondegenerate hermitian symmetricsymmetric form on U.

Chapter 9, 9, Exercise Exercise 10. 10. Let Let G1 G1 = = G2 G2 = = G. G. As As or a is is 1-dimensional, 1-dimensional, or a is irreducible and by hypothesis 7rn is irreducible, so by 27.15, aor €3 ® n n isis anan irreducibleirreducible representation ofof G1G1 xx G2.G2. Recall that if we we identifyidentify GG withwith thethe diagonaldiagonal subgroup {(g, g): g E G)G} of G1 x G2 via the isomorphism g i-+H (g,(g, g), then the tensor product representation or a €3® nn of G is thethe restriction of thethe tensortensor product representation aor €3 0 nr ofof G1G1 xx G2G2 to to thethe diagonaldiagonal subgroupsubgroup G, so it remains to show thethe diagonaldiagonal subgroupsubgroup GG isis irreducibleirreducible onon UU €3® V, wherewhere U and V are the representation modules for ora andand n, respectively.respectively. But if u E U'U# thenthen UU == CuCu and thethe mapmap (o:lo: au au €3 ® v v H H avav is an isomorphism of UU ®8 V with VV suchsuch thatthat x(g,x(g, g)(o g)(a ® €3 n)cp n)(o =X(g)(x(p(g7r))= h(g)(x(o(gn)) for g E GG and x E U €3® V,V, wherewhere u(ga) u(ga) == h(g)u) (g)u and h(g)X(g)E E c'. C. Namely xx == au €3®v, v,

x(g, g)(a (9 7r)lp = (auger, vgn)1P = (a) (g)u, vgn)co = a) (g)vgn and xcp(g7r)x(o(gn)=avgn. = avg7r. ThusThus (oco induces induces aa bijectionbijection (o:lp: W W H H W(oWlo ofof the CG-CG- subspaces ofof U €30 VV andand V,V, soso asas GG isis irreducibleirreducible on V,V, itit isis alsoalso irreducible on U ®€3 V.

Chapter 10,10, Exercise 6. (1) First, First, w w = =1 1 .w w with with l(1) = 0, sosow w <5 w. Thus 5< is reflexive. Second,Second, ifif uu (< w then w =xu= xu with 1(w)l(w) == l(x)1(x) +l(u),+l (u), so as l(x)1(x) 2 > 00 with equality iffiff xx == 1, we have l(w)l(w) 2> l(u)1(u) with with equalityequality iff iff x x = =1 1 and u = w. Then if also ww (< u,u, by symmetry, l(u) l(u)2 > l(w), l(w), soso u u = = w. w. That That is, is, (< isis anti-anti- symmetric. Finally,Finally, assume assume u u ( < w w ( < v. v. Then Then w w = =xu xu and vv == ywyw with 11(w) (w)= _ l(x)I(x) + 1(u) l(u) and l(v)1(v) = = l(y)1(y)+ + l(w),1(w),so so v v = = ywyw == yxuyxu with 1(v) =1(y) + 1(w) =1(y) + 1(x) + 1(u).

Further ll(yx) (yx)5 < l(y)1(y) + + l(x),1(x), while as v = yxu, yxu, l(yx) > 1(v) -1(u) = 1(y) + 1(w) - (1(w) - l(x)) =1(y) + 1(x), so l(yx) == l(y)1(y)+ + l(x)l(x) and hencehence uu (< v,v, provingproving (< isis transitive.transitive. (2) Let w bebe maximalmaximal with respect toto (< andand supposesuppose w # wo. Then by Exercise 10.3.2, l(w)1(w) << IP1,/PI, so byby 30.12, therethere isis aa E PP with aw E P. ThenThen by 30.10,1(rffw)30.10, l(raw) > 1(w), l(w), so raw >> w, w ,contradicting contradicting the the maximalitymaximality ofof w.w . (3) First, (b) and (c) are equivalent by 30.10. Next, ifif 2(rw)l(rw) 5< l(w), then by 30.10, l(w) l(w) == l(rw) + 1,1, so as l(r)1(r) == 1,1, we have l(w)1(w)= = l(rw)l(rw) + + l(r)1(r) and and w w = = rr rw,. rw, so so rw rw < ( w. w. That That is, is, (b) (b) impliesimplies (a). AppendixAppendix 283283

AssumeAssume rwrw 5

ChapterChapter 11, 11, Exercise Exercise 5.5. ForFor 11 <5 u <5 v <5 n, let U,,,U,,,,, = = U, U® @ ®U,+l Un+i €3.. ® . ®@ U.U,. ForFor k = 2j or or 2j2j - -1, 1, let let j j(k)(k) = = j jand and 1(k) l(k) == j j + + 2 2 or or j j+ + 1, 1, respectively. respectively. LetLet U2 U1U' = Ul,j-1,Ul,j-l, U2 =- Uj,l(k)-1, u.,,l(k)-l, andand U3u3 = Ul(k),n. U1(.),,. ThusThus V == U1U' ®€3 U2U' ®€3 U3u3 andand sk = I €3 ak €3 6. where 6. = y €3 . . . €3 y andand a. Qk(1+a®y)/,/2-= (1 €3 /3 + a! €3 y)/& oror (a(a! +,8)/,/2- + /3)/& forfor k keven even or or odd, odd, respectively. respectively. LetLet a!+B YP'YS asaa! Baa Yaa aa=-a ,, b-b=- c=c=- , d-d=- e=e=- . 1/Z &' 1/Z' &' &' CheckCheck that y/3y,8 = = -/3 -,8y y and ya!ya = -my.-ay. HenceHence yaya = = -ay. -ay. Similarly, Similarly, a,8 a/3 = = -- ,8a. pa. Check Check also also that a, ,8,/3, y, and aa are involutions,involutions, b2 b2 = = - - 112112=e = e2,2, ac ac + ca=a,ca=a,de+ed=e,c2+ de+ed=e, c2 + 1/21/2=c,andd= c, and d 22 + 1/21/2=d.= d. (1)As(1)As y2= 1, at =(y @. . . @ Y)2=Y2 @. +.@ Y2= 1. Similarly, ifif kisk is oddodd thenthen Qkat = a2 ==1, 1, while while if if k k is is even even then then 21®,82+a®(.8Y+Y.8)+a2®Y2=1, a;= (~@B;@Y)~-~@B~+.@(BY+YB)+~~@Y~1®,8+a®y - = 1, I ( 2 ) 2 asas ,82 == a!'a2 == y2 ==1 1 and and ,8y/3 y = -y,8.- y/3. Therefore Therefore Sk=(I ®ak(g k)2=I XQk NextNext let V1=U1,j_1,V' = Ul,j-l, v2V2=Uj,j+i,= Uj,j+l, andand V3=Uj+2,n.v3= Uj+2,,. IfIf k=2jk = 2j - 1 1 then then Sksk=I = I ®@(a@y)@6L9where6L=y@...@ (a (9 y) where k = y ® ® y andyandsk+l=I@ak+l@6k+l. Sk+1 = 1 ® ak+1 0 k+1 ThusThus SkSk+1sksk+l =I = 10@ (a ®y)ak+1@ y)ak+l 8 6Ltk+1 Withwith 6;6k+l=SklSk+1 y2 = y2 @ ®...... @ ®Y2 y2 = =I I asas Y2 == 1. T~USThus toto showshow (S.S.+~)~(SkSk+l)3 = -I= -Iwe we must must show show x3 x3 = = -I, -I, wherewhere x=(a®y)ak+l=(a®y)((l®,8+a®y)/,/2-)=a®b+c®1. Equivalently we must show x2x2 + + 11= = x. x. But But x2=a2®b2+(ac+ca)®b+c2®1=-1/2+a®b+c2®1, as a2 = 1, b2 = -1/2, and ac + ca= a, so as c2 + 1/2 = c, indeed x2+1=a®b+c2®1+(1®1)/2==a®b+c®1=x, completingcompleting thethe proofproof thatthat (S2j_1S2(s2j-ls2j)3 j)3= = -I. -I. 284 Appendix

Similarly let let k k = = 2j. 2j. ThenThen sk Sk = = I 10€3 ak ak €3 0 Ckk andand sk+l=Sk+1 =1I €3 ®(1 (1 €3® a) a) €3 0 Ck+l,k+1, so sksk+l=sksk+1=1® I C~J ak(1ak(l €3 0 a) 0 €3 k (k(k+l, k+1, and and it it remains remains to show y2y2 + + 11= = y, where (1®P + -Y)(1®a) y=(lark (l0 a)=

But y2=1®d2+a®(de+ed)+a2®e2=1®d2-1/2+a®e, as de + ed = e, a2 = 1, and e2 = - 1/2. Thus as d2 + 1/2 = d, y2+ 1=a ®e+ 1 ®d2+(1 ®1/2)=a ®e+ 10 d = y, completing the proof of (1).(1). (2) LetLet kk i1(k)i < j(i), i(i), or (b) kk =2j= 2j and i = 2(j ++ 1). 1). ForForif if jj == j(k)j (k) then2j then 2j - - 1 ik< k < < ii -- 1152j(i)- < 2j(i) - 1,1, so so j <5 j j(i) (i) - 1 1 and and hencehence ll(k)= (k) = j+ci j + e < jj(i)unlessc=2(sothatk=2j)and (i) unless e = 2 (so that k = 2j) and j(i)=j (i) = j+j + 1. 1.InthelatterIn the latter case, asas j(i)= j (i) = j j ++ 1 and kk =2j = 2j < ii -- 1, 1, we we have have ii ==2(j 2(j + + 1), I), establishing establishing the claim. In case (a)(a) let let Vl Vl == Ui,j-1, Ul,j-1, V2 V2 = Uj,1(k)-1,= U;,l(k)-1, V3 V3 == U1(k),j(i)-1, Ul(k),j(i)-1, V4 V4 = Uj(i),i(i)-1, Uj(i),l(i)-lr and V5Vg = = Ul(i),n. UI(i),n.Then Then V =V Vl= V1€3 . ®. . €3® V5 and and s, sr = =sr,1 s,l €3 ® . .0 . €3Sr,5, s,5, where where Sr sr induces sr,ts,, on V.V, .Further,Further, Si,,, s~,~ = Sk,1 sk, 1 = = II forfor uu <( 3, 3, sk,2 Sk.2 = ak, ak, si,4 s~,~ = ai, ai ,and and si,5 si,~ and ~k,~,Sk,,,,v v > > 2, are of the formform y ®€3.. . ®. €3y. y.In Inparticular particular Sk,wSi,w Sk,WSi,w == si,wsk,w Si,WSk,w forfor w # 4, whilesk,4si,4while sk,45i,4= -~~,~s~,4,sinceay= -si,45k,4, since =ay -ya, = -ya, By ,ay = = -yB,anday -y,a, and ay = = -ya.-yet. Therefore (sksi)'(SkSi)2 = = -I. J. IncaseIn case (b), let W1=Wl = Ul,j_1,Ul,j-1, W2W2 = Uj,j+2, U;, j+2, andand W3 W3 = Uj+3,n. Uj+3,n. Again V = W1® W1 €3 W2 ®W3 and Sr = sr,1 ®Sr,2 ®sr,3 with I, sr,3 of the form y ®... ®y, and W2@W3ands, =s,,1€3~,,~€3~~,3withs,,~Sr,1 = = I,s,,~oftheform y €3...€3y,and sk,2=(lOp®Y+a®Y(&Y)1-12, si,2=(l®1®P+1(&a0Y)/J. It~t remainsremains to showshow (siSk)2(s~s~)'= = -I,-I, soSO we we must must show show Sk,2Si,2 sk,zsi,z = = -si,z~k,z.-si,24,2. ThisThis follows because because pa Pa = =-up, -up, a ay y = = -ya, -ya, andand By,ay == -yB.-y,a. (3) AsAs (sksi)'(Sksi)2 = =-I -I for for i i >> k + 1,1, we have -I-I EE G(1).G('). Let Let G G = = GI(-I). G/(-I). Then G = (sk: (Sk: 1 i< k << 2n), 2n), and and by by (1) (1) and and (2), (2), Isis lSiSjI j I ==mi,j, mi, j, where mi,i = 1, 1, mi,i+1=mi+l,i=3,andmi9j=2forli-mi,i+1= mi+l,i = 3, and mi, j = 2 for Ji - jjl J > 1.l.Thusby30.19,G Thus by 30.19, G =2 Sen. Szn.Let Let n >> 2.2. As As GG - 2Stn Szn with with 2n 2n > > 4, 4, G(1) G(')= = G(')(1) 2= AenAh is a nonabelian simple group by 15.16. ThusThus asas -I-I EE G(1), G('), G(1)G(') isis quasisimplequasisimple or or G(')G(1) = = (-I) (-I) x x G(2) G(~) with G(2)G(') 2= A2n.Az,. In the latter casecase the the projectionprojection of of 5153 $1~3 onon G(~)G(2)is is an an involution, involution, which is impossible, asas (~1~3)'(5153)2= = -I. -I.

Chapter 11, 11, ExerciseExercise 10. 10. We We firstfirst observeobserve Z, isis subnormal subnormal in in GyG, andand hence also in Gy,ZG,,, foreachzfor each z E I',,ry, sincesince ZX Z,

Then Z,Zx I 0(Gy,,). 8(GY,,). ForFor asas ZxZ, isis subnormal in GyG, andand Gy,Z,G,,,, wewe have E(G,)E(G,,,)E(Gy)E(Gy,,)

Gy < NGy(0(Gy)) = NGy(0(Gx)) < Gx and by symmetry, GxG, (< Gy,G,, contradicting our hypothesis that GxG, # G,.Gy. Thus 8(G,)0(Gx) == 1, so Oq(Gx)Oq(G,) = E(Gx)E(G,) = 1.1. ButBut ZxZ, # 1, so F*(Gx)F*(G,) # 1 and hence Op(GX)Op(G,) # 1 for some prime p. However we have shown Op(Gx)Oq(Gx)= =1 1 unless ZxZ, or ZyZ, is a p-group,p-group, soso interchanginginterchanging the roles of x andand yy ifif necessary,necessary, we may assume ZxZ, is a p-group. Next Qx 9a Gx,y, Gx,y, so g(Qx)0(Qx) 5<0(Gx,y)=0(Qy). 8(GX,,)= 8(Qy). HenceHence

0(Qx) < n 0(Qy) < 0(Zx) =1, yErs as ZxZ, isis aa p-group. p-group. That That is is 0(Qx) 8(Q,) =1.= 1. But But Zy Z, < (Qx Q, <5 Gx,y G,,,

Chapter 11, 11, ExerciseExercise 11.11. LetLet Yy E Y,Y, F= .F_ {Gx,{G,, G,),Gy), andand rr = F(G, r(G, .T) F) bebe thethe coset geometry defined inin section 3. IfIf G,Gx == GyG, then Y == yGx = yGy yG, = {y}, (y), contradicting the hypotheses that YY is nontrivial. Thus GxG, # G.G,. As G is primitive on X,X, G,Gx isis maximal maximal in in G G by by 5.19. 5.19. Thus Thus if if 1 10 # HH ga GxG, thenthen by maximality ofof G,,G, NG(H)NG(H) == Gx G, oror G. G. But But inin thethe latter case, H

some primeprime p.p. Therefore we may assumeassume 2,Z, = 1. 1. ThusThus GxG, isis faithfulfaithful on

0Q= = U cr,. .

Y€~XyErx

Let mm = I IYI. Y I. Then Then mm == lIr,l= Fx I =IF, Ir,/ I forfor each z E r,,rx , so 101131i

Chapter 12,12, ExerciseExercise 5.5. (1) (1) The The possible possible cycle cycle structuresstructures for elementselements of G are: 1, (1, 2), (1, 2, 3), (1, 2, 3, 4), (1, 2, 3, 4, 5), (1, 2)(3, 4, 5), (1, 2)(3, 4) giving a setset SS = = {gi {gi: : 1 I< ii (< 7)71 of elements suchsuch thatthat IgiI gi I =I = i fori for i i << 7 andand lg71971= I = 22 by by 15.2.4.15.2.4. Further Further SS isis aa set set of of representatives representatives forfor thethe conjugacyconjugacy classesclasses of G by 15.3.2.15.3.2. (2) The representation of G on XX isis ofof degreedegree 55 andand 2-transitiveZtransitive by 15.12.2.15.12.2. Arguing as in Exercise 5.1, G has aa transitivetransitive representation of degree 6 on the coset spacespace G/H,GIH, where where H H = =NA(P) NA(P) and and P =P (g5)= (gg) E ESy15(A). Syl,(A). Namely,Namely, counting thethe number of 5-cycles, G has 2424 elementselements of orderorder 55 andand hencehence 6 SylowSylow 5-subgroups, so by Sylow'sSylow's Theorem,Theorem, IGI G : :HI H =I = 6. 6. Now Now thethe only possible cycle structurestructure forfor thethe elementelement ggg5 of of order order 5 5 on on the the 6-set 6-set GIH G/H is one cycle ofof lengthlength 55 and and one one fixed fixed point, point, so so P P is is transitive transitive on on GIH G/H - - {H}{H) and hence G is 2-transitive onon GIG/HH byby 15.12.1.15.12.1. Let 1/r@ bebe thethe permutationpermutation charactercharacter of G on X.X. ByBy ExerciseExercise 4.5.1,4.5.1, 1/r(g) @(g) is the number of fixed pointspoints ofof gg Ec G on X, soso

1/r(gi) = 5, 3, 2,1, 0, 0, 1 for ii ==1, 1, ...... , ,7,7, respectively. respectively. Similarly let rpip be be the the permutation permutation character character of of G G on on GI G/H. H. LetLet h h = = (1,2,4,3).(1, 2, 4, 3). By 15.3.1,g,h=(2,4,15.3.1,g5=(2,4, 1,3,5)=gi,soh~H 1,3,5)=g5,sohEHandH=P(h)andH=P(h) as as

I P(h)I = I P II h I =20= IGI/IG : HI =1 H1.

Thus the elements inin H areare ofof conjugateconjugate to gi,gl, g4, g4, g5,gg, and g7, soSO for all other gi, ip(gi)rp(gi) == IFixG/ff(gi)lIFixG/H(gi)I= = 0.0. By By Exercise Exercise 2.7, 2.7, if if Q Q E E Syl,(H) Sylq(H) then then IFix(Q)l IFix(Q)I = = I NG(Q) NG(Q): :NH(Q)l, NH(Q)I, so so as as HH == NG(P), ip(g5)rp(gs) == 1 while as (h)(h) EE Sy12(H) with NG((~)>NG((h)) E Sy12(G)Sy12(G) of of order order 8, 8, rp(g4) ip(g4) = = Irp(h)l= I c(h)I =2. 2. Thus Thus g4 g4 =(a, = (a, b, b, c, c, d)(e)(f) d)(e)(f) on G/H,GIH, so so g7g7 = g: 924 == (a, c)(b, d)d) and and hencehence ip(g7) rp(g7) = 2. (3) By ExerciseExercise 12.1,12.1, GG has has I GI G : :G(1) G(')I I linear linear characters.characters. LetLet A == Alt(X); by 15.16,15.16, A is simple,simple, and by 15.5,15.5, 1lG G : A1: A = I 2,= 2,so so A A = = G(') GM and and G G hashas two linear characters,characters, thethe principalprincipal charactercharacter X1 and the sign character X2,~2, where Appendix 287 by Exercise 12.1,12.1, ker(x2)ker(X2)= = A.A. ThatThat is is x2(a)X2(a)= =1 1 forfor aa E AA andand x2(g)X2(9)= _ - -1 1 (the(the primitive 2nd2nd rootroot of of 1)1) for for g g E c G G - - A. (4) By Exercise 12.1,12.1, X1~1 and X2~2 areare irreducible. By Exercise 12.6.3,12.6.3, X3~3 == 1/r@ - - X1~1 andand x4X4 = = rp p -- X1~1 are irreducibleirreducible characters. Then by Exercises 9.39.3 andand 9.10, Xi+2xi+2 = = x2xi,X2Xi, i i= = 3,4,3, 4, areare alsoalso irreducibleirreducible characters.characters. This gives 6 irreducible characters of G, exhibited in the first six rows of the table below. AsAs G has 7 conjugacy classes, G has 7 irreducible characters by 34.3.1. Thus it remains to determine X7.~7. LetLet nini == Xi(1);~~(1); by 35.5.3,

7 120=IGI= n? i=1 so n7 == 6, giving the first column of the table. Then we use thethe orthogonalityorthogonality relation 35.5.2 to calculate x7(gi)X7(g0 for for ii > 11 toto completecomplete the table.

Character Table Table of of S5Ss

91 g2 g3 g4 g5 g6 g7

X1 1 1 1 1 1 1 1 X2 1 -1 1 -1 1 -1 1 X3 4 2 1 0 -1-1 0 X4 5 -1-1 1 0 -1 1 X5 4 -2 1 0 -1 1 0 X6 5 1 -1-1 0 1 1 X7 6 0 0 0 1 0 -2

Chapter 13, Exercise 1.1. (1)(1) LetLet M M == CG(g). ThenThen 1 HxMlIHxMI = = (MIM : MHMHx I,1, where MHx isis thethe stabilizerstabilizer of of Hx Hx inin M.M. ThusThus MHxMHx = = M fln HxHX = = CH=(g), CHx(g), so (IHxMI,(IHxM 1, p) == 1 1if iff CHx(g) CHI@) containscontains a a Sylow Sylow p-subgroupp-subgroup of CG(g) iffif cH(gX-I)CH(gx-') contains a Sylow p-subgroupp-subgroup of cG(gX-I)CG(gx-') iffiff gX-Igx-' is extrernalextremal inin H.H. (2) Let Let A A = = HIK,H/K, a: H + A the natural map, and V the transfer of G into A via a. RecallRecall byby 37.237.2 thatthat VV isis aa groupgroup homomorphism,homomorphism, soso asas AA isis abelianabelian G(')Gf1> ( < ker(V)ker(V) and hencehence it suffices toto show gVgV # 1. Choose a set X of coset representativesrepresentatives forfor HH in G as inin 37.3.37.3. AsAs IgIIgI = = p the -1 length ni ofof thethe ith cyclecycle of g is 1 or p.p. If ni = p p then then 1 1= = gnh ga , ,soso gnixr gnlxl ' EE gn'gni K; that isis is gn'x` = (gn )a = (ga)n' = 1.

On the other hand nini = 11 if iff g gfixes fixes Hxi, Hxi, in in which which case case gxi gx;' ' E H. NowNow M-1 M = CG(g) permutespermutes thethe fixedfixed points points of of g g on on GIH G/H andand g("irn)-'g(x`m)-' = gx;l = gx1' for for 288 Appendix Appendix 289

each m E M, so if (HyjM: l < j < s} are the orbits of M on FixG/H(g) (5) Define (Hx,: l < i < t} then H = n BW. WEW s W a =r7 ((gy' )a)k, Then H is the pointwise stablilizer in G of E, so (a) and (b) are equivalent. =1 j1=11 By Exercise 10.6.2, w < wo for all w E W, so by (4), B fl BW0 < BW. Thus H = B fl BWO, so (b) and (c) are equivalent. where kJ = IHy,MJ is the length of the jth orbit. If p divides kj then(gyi')ki = (6) Let T = (G, B, N, S). Recall H = B fl N a N with W = N/H and for 1 = (ga)ki. On the other hand, if p does not divide ki, then by (1), gyi 'is w=nHEW,BW=Bn.Thus H 0, l(sw) > 1(w) by Exercise 10.6.3, Let X be the set of galleries in 8 from C to E of length n- 1. By 41.7 and so l(w-Is) = l(sw) > 1(w) = l(w-1). Also by Exercise 10.6.2, w-Is < wo, 42.3, d(C, Cw) = n and d(C, Cu) = n - 1, so d(C, E) = n- 1. Thus X C E and by Exercise 10.3.2, l (wos) < 1(wo). Therefore by Exercise 10.6.4, w-1 = by 42.3. As A acts on C and E, A acts on X. Further, Cu and Cw are the two (w-Is)s < wos. Then by Exercise 14.9.4, chambers in E through E, so as d(C, Cw) = n, Cu is the terminal member of each member of X. Thus A < Gcu = B" = BSW Ba=BfBWOS

(5) Define H=fl= n n BW. BW. W€WWEW Then H isis thethe pointwisepointwise stablilizer in G of E,C, so so (a) (a) and and (b)(b) areare equivalent. equivalent. By Exercise 10.6.2, ww 5< wo for allall ww E W, so by (4), B n BWBWO °_( < BW. B'. ThusThus H == B B n ilBWO,BWO, so (b) and (c) are equivalent. (6)Let(6) Let FT == (G, B, i?,N, S).RecallS). Recall H H = = BnNB n N r?a N with WW = = N/H andandfor for w=nH~W,B~=B~.ThusH_(B~forallw~W,so~_(~.w=nHEW,BW=B".Thus H

Chapter 14,14, ExerciseExercise 10.10. (1)(1) AsAs awuw >> 0, 0, l (sw)l(sw) > > I (w)l(w) by by ExerciseExercise 10.6.3, 10.6.3, so l(w-'s)l(w-1s) = l(sw)l(sw) >> l(w) l(w) = = l(w-1). l(w-'). AlsoAlso by by Exercise Exercise 10.6.2, 10.6.2, w-1sw-'s <5 wo,wo, and byby ExerciseExercise 10.3.2, l(wos) t(wos) 5< l(wo). ThereforeTherefore by by Exercise Exercise 10.6.4, 10.6.4, w-' w-1 = = (w-'s)s(w-1s)s

Ba=BnBWOS

(3) AsAs l(w)1(w) 5 < l(ws), l(ws), l(w-') l(w-1) 5 < l(sw-'), l(sw-1), so so by by 43.3.1, 43.3.1, SBW-' sBw-1 5 c Bsw-' Bsw-'B, B, and hence BB 5c sBsw-1Bws Bsw-' BW == BSBW. BSBW. Thus for bb E B, therethere existexist x x Ec BSB' and y EE BwBW with bb = xy.xy. ThenThen y == x-lb x-'b E EBw BW n i? BSBBSB C c B B by by Exercise Exercise 14.9.3, 14.9.3, soy EBnBWE BnBW andx andx=by-1 = by-' EEBnBs. B n BS. (4) Notice as as == -a-a < <0, 0, so so by by (2), (2), Ba B, ni? BSBS = = H. H. Also, Also, setting setting w = wos, wos, 1(w)Z(W) 5< l(wo)z(wo) == l(ws),~(ws), soSO by (3), B = (B(B nn BS)(BB~)(B n Bw)B~) = (B (B nn BS)Ba.B~)B,. (5)(5) If BaB, = H H then then byby (4),(4), BB == H(B H(B n i? Bs).BS). ThenThen as as s sis is an an involution, involution, BS == HS(BHS(B n BS)SBs)S == H(B nn BS) BS) = = B, B, contrary contrary to to BN4.BN4. (6)(6) If aw >> 0 0 then then BaB, <5 Bw-' BW-' byby (1). (1). On On the the otherother hand,hand, if aw << 0, 0, then then by (2), H = Ba B, nn Bw-',BW-', soSO BaB, 6 B'-'B~-' byby (5).(5). (7)(7) DefineDefine gyp: p: A + (D@ byby p:gyp: B," BaH i-H aw. aw. Let Let ,!? PE En r andand ww E W; we firstfirst show

(*)(* 1 BaB," == Bp Bp ififf ,!?=aw.= aw.

First, if B,"Ba = B Bg then then as as B Bg = B= n B Bw0'P, i? BWorB, BaB, = =Bw-' B"-' n i?Bw0'pw-' . Thus by (6), aw >> 0 0 < < awrewo. awrpwo. Therefore Therefore awroawrp << 0 0 by by ExerciseExercise 10.3, soso awaw = ,!? by 30.7. SimilarlySimilarly if ifaw aw = = ,!?P then awrpwoawrpwo >> 00 << caw, aw, soso by (6), B,Ba <5 BwBw-' ' n n Bw0'fw BWO~~W-' ' and hencehence B," Ba' 5 < BB i? n BWorBBw0'P = =Bp. B. AsAs ,!?w-' jw-1 = a,a, by by symmetry B;'B' ' < B,B,, so Bso = Bp BC = ,W B,". Thus Thus we we have have established established (*). (*). By (*), B,"Ba = B- B; ififf Ba B,"u-' U = Bp iffif awu-'awu-1 = P,!? ififf awaw == 3u. ,!?u. ThusThus pp is a well defined injection.injection. ByBy 30.9.3,30.9.3, p p isis aa surjection.surjection. As As B,"'p Ba "cp = = awu awu = (Ba(B,")pu, )cpu, pcp is is W-equivariant.W-equivariant. (8)(8) First notice

Bsw = (B n BwOw-'S)S = Bs n BwOw-'.

Next as aw << 0, 0, aww0 awwo > > 0 0by by Exercise Exercise 10.3. 10.3. Hence Hence by by (1),(I), Ba B, <5 Bwow-', BWow-I, so BaB, 0,O,l(wwo) l(wwo) <5 l(swwo)l(swwo)byExercise by Exercise 10.6.3,sol(wow-'s)10.6.3, so l(wow-'s) >2 l(w0w-').Thenl(wow-'). Thenby by Exercise ~xercise 14.9.3,14.9.3, BS BS n i? Bw0w-'BWoW-' <( B, B, so SO Bsw Biw = BS BS i?n Bwow-'BWoW-' -<< B ni? BwOwBwow-' = Bw.B,. Thus BaBswB,Biw 5 S; B,. Bw. On On the the other other hand hand by by (4), (4), B B = Ba(BB,(B nn Bs), BS), so SO as as Ba B, <5 Bwow_I, Bwow-I,

Bw = B n Bw0w_' = Ba(B n BwOw-' n Bs) = Ba(B n Bsw) C BaBsw

ThatThat is, B,BiwBaBsw = = B,.Bw. Finally Finally as as = = -a-a << 0, 0, so so by by (2),(2), BaB, ni? BSBS = H.H. There- There- forefore BaB, n BswBiW = BaB, n BSBS n Bw°w-'BWow-' == H. (9)(9) We applyapply (8) (8) with with w w = = wi-lwo, wi_1w0, a a= = ai, al, andand s s = = riri forfor eacheach 11 5< ii < n.n. ByBy Exercise 10.6.3,10.6.3,aiwi_lwo aiwi-lwo < 0sinceOsince l(riwi_1wo)l(riwi-lwo) = l(wiwo)l(wiwo) < l(wi_lwo).l(wi-lwo). ThusThus byby (8),(8),

(*) Bw;_iwo Bai Bwwo Appendix 291 and (**>(**) Bai fl B i,,. = H. Claim k B"'i_1 Ba $Wk (!) Wi-1W0 1wl-1 WkWO j =i for l1 < ( i i< < n nand and i i< (k k<

Br'wt_l Bw'-'w;_Iwo- B 'i_Ia; w,wO- -Ba;w;_i BW1 O so (!) holdsholds whenwhen kk == i. Assume forfor kk -- 1. 1. By By (*),(*),

BWk-1 BWk-1 BrkWk-1- B B Wk Wk_IWO- ak WkWO - akWk-1 WkWO' so by the induction assumption

k-1 k BWi-1 B BWk-1 B Wk wi-IwO- ajwj_I Wk_1WO - ajwj_,* BWkWO, i=i, 1=1 establishing (!). Notice that wn-lwown_lw0 = rn,rn, soso

BWn-1 = BW"-' = BW' W" 1WO rn a"' = Banwn-1

Substituting thisthis equalityequality into into (!) (!) with with k k = = nn -- 1,1, we we get

n-1 n (!!) BW'Wi_1 ' WO =H BajWj_1 BWW"-1' WO =H Bajwj_; j=i j=i

Further B,,0Bwo = B B fl i? BwOwOBWow0= = B, B, so so applying applying (!!) (!!) at at i i == 11 we conclude

n B = H Baj W j-1 j=1

Next by (**) and (!!),

H == ~wi-1 H" -= (Bai(B,, nfl B;~~~)~~-~BwiwO)w' 1 = Baiwi-1 B~~~~-~n BC:,~ I IBWiWO n Baiwi-1 II H Bajwj-i j=i+1

It remains to show +:1Jrr: i Hi H aiwi-1 aiwi_1 is is a a bijection bijection of of I I = {1,11, ...,. . . ,n} n} with (D+.@+. We sawsaw atat thethe beginningbeginning of of the the proof proof that that aiwi-lwo aiwi-1w0 < < 0,0, soso aiwi-1 aiwi_1 >> 0 and hence I1JrI+ Cc cp+,@+. ByBy ExerciseExercise 10.3,10.3, 1cI+J)@+I == n,n, so so it it remainsremains to to showshow iJr+ is an injection. But ifif aiwi-1aiwi_1 = akwk_1akwk-1 forfor somesome kk > i,i, thenthen ai =CYkWk_1Wi 11 =akrk_1...ri, 292 Appendix so 0 > airiair; = = akrk_1 akrk-1 . . ri+l,ri+1, contrary to Exercise 10.6.3, since l(rk ... ri+i) > l(rk-1 ... ri+i)

Chapter 15, 15, Exercise Exercise 4. 4. (1)(1) To To showshow HH == CG(a) CG(a) isis balanced balanced wewe mustmust showshow Or'Or! (CH(CH(X)) (X)) 5 < Or/Or'(H)(H) for each X of order r inin X.X. But as H isis solvable,solvable, thisthis isis immediateimmediate fromfrom 31.15.3 1.15. (2) We must show O0 is balanced. But for a, b b EE A#A' 0(b)nCG(a) = Or'(CG(b))nCG(a) = Or'(CCa(a)(X)) < Or'(CG(a)) = 0(a), where X = (a, (a, b),b), asas CG(a) CG(a) isis balanced balanced by by the the hypotheses hypotheses of of partpart (2).(2). (3) Let 80 be the signalizer functor defined in (2); O0 is a a signalizersignalizer functor functor by by (1) and (2). By thethe Solvable Signalizer Functor Theorem, O0 is complete;complete; thatthat is, there existsexists anan rl-subgroupr'-subgroup Y ofof GG suchsuch that that O(a) 0(a) == Cy(a)Cr(a) forfor eacheach a EE A#.A'. For B aa noncylic subgroup ofof AA definedefine Y(B)Y (B) = (0(b) (O(b) :: b b EE B#).B'). Now for g E G with a, ag EE A#,A', O(alg0(a)9 = Or,(CG(a))g0,!(C~(a))~ = Or'(CG(ag))Ort(Cc(ag)) = 0(ag),O(ag), so NG(B) <5 NG(Y(B)).NG(Y(B)). Further, by construction Y(B)Y(B) ( < Y,Y, while by 44.8.1, Y <_( Y(B),Y(B), soSO YY == Y(B). Y(B). Thus Thus NG(B) NG(B) < 5 NG(Y), NG(Y), soso I'2,A(G) rz,~(G) <5 NG(Y). NG(Y). But by hypothesis, GG = 1'2,A(G), r2,A(G), SOso Y <1< G,G, and and hencehence asas YY is anan r'-group,rl-group, Y _<< Orf(G). Or'(G).Therefore Therefore Or,(CG(a)) Or (CG(a))( < Y (< Or'(G).Orr(G). In particular if Or!Or (G) = 1 1 then then Or Or/ (CG(CG (a))(a)) == 1. 1. Hence, Hence, as as CG CG (a)(a) isis solvable,solvable, F*(CG(a)) = = F(CG(a))F(CG(a>> = Or(CG(a))Or(CG(a)).

Chapter 16, 16, Exercise Exercise 5. 5. (1) (1) By By definition definition ofof strongstrong embedding,embedding, H isis properproper of even order, soso therethere isis anan involutioninvolution i i EE H. Indeed,Indeed, 46.4.3 is one of the equivalent conditions for strong embedding, so (*) (HnH91isoddforgEG-H. Let j EE I;I; it it suffices suffices toto show j EE iG.iG. SupposeSuppose jG _cE H. Then for gg E G,G, jg EE HH n n Hg, Hg, so SO IH IH nn HgI Hgl is is even even and and hencehence gg EE HH by by (*). (*). As As thisthis holdsholds for each g EE G,G, G G = = H, H, contradicting contradicting H H proper. proper. This This contradiction contradiction shows shows there isis aa conjugateconjugate s s of of j j inin G - H, H, and and it it suffices suffices toto showshow s EE iG.iG. ButBut ifif not, then byby 45.2, lisllisI is even, so by 45.2.3 therethere isis anan involutioninvolution z z EE (is) centralizing i andand s. NowNow asas 46.4.1 isis oneone of the equivalent conditions defining strong embedding,embedding, CG(t) CG(t)5

(3) As uu 40 H, D D == H H n n H" Hu is isof of odd odd order order by by (*).(*). (4) Let Let jj E J. Then Then j EE DD andand j"jU = = j-1 j-' byby definition definition ofof J.J. Thus Thus (uj)2(~j)~ = ujuj == j"jjuj = = j-' j-'j j = 1,= so1,souj Uj E E I. I.As As j jE E D, D,uj Uj EE uDuDn n I,souJI, so uJ Ec uD n I.I. On On thethe otherother hand, supposesuppose d d E E DD with with ud ud EE I.I. ThenThen 11 == (~d)~(ud)2 == udud = dud, dud, so dud" = d-1d-' andand hencehence d d E E J; J; thatthat is, is, u uD D n n I I5 c uuJ. J. (5) Set Y = CG(j)(u).Cc(j)(u). AsAS J J C _C D D and and I DID1 I is odd by (3),(3), Ijl1 j Iis is odd, odd, soso as j 0# 1,1, 1I jl j I> > 2. Then as u inverts j,j, wewe havehave uu 04 CG Cc(j). (j ). On the other hand u E NG((j)),Nc((j)), soso YY <5 NG((j))NG((j)) andand hencehence CG(j)CG(j) Y. Suppose tt E CG(j) and letlet KK == HH nn Y. Y. ByBy construction, u u EE YY - - H, soso K isis properproper in Y and t EE YY soso KK hashas eveneven order.order. Further forfor yY EE YY -- K,K, y 04 H,H, so so IK IK n nKYI Kyl isis odd odd by by (*), (*), and and hence hence K K isis strongly strongly embedded embedded inin Y.Y. Thus applying (1) to Y, the involutions uu and t areare conjugateconjugate in Y.Y. This is impossible, as tt E CGCG(j) (j) <1< YY while u E YY - CG CG(j). (j ). (6) Let Let r,r, ss bebe distinct distinct involutions involutions in inuD. uD. Then Then rD rD = =uD uD and and 1 10# rsrs E D, SOso s= ErDnI =uDnI =J by (4). But if rCG(t)rC~(t) == SCG(t), sCG(~), alsoalso rsrs E E CG(t), CG(t), contrarycontrary toto (5).(5). That That is,is, rCc(t>rCG(t) #0 sCG(t).sCc(t). (7) and (8): Define nn = I I G : HHI I and C20 == {(i, {(i,x, x, y):y): i cE I Iand and (x, (x, y)y) is is a acycleini cycle in i on GIH).G/H}. As 46.4.4 is oneone ofof thethe equivalentequivalent conditions for strong embedding, i fixesfixes a uniquepointofG/H,soihas(n-1)/2cyclesoflength2.Thus~~~unique point of G/H, so i has (n -1)/2 cycles of length 2. Thus 101= =III III(n-1) (n -1) and 111III = =n1I nlI nnHI HI ==nm,so nm, so I0I1C21 =n(n= n(n -- 1)m. 1)m. On the other hand, ifif A = X X xx X X and and 8(x, 6(x, y)y) is is the the number number of of involutions involutions with cycle (x, y), then

101 = 8(x, y) (x,y) E A Now, upup toto conjugation conjugation in in G, G, we we have have x x= = H, H, and and if if 6(x, 8(x,y) y) # 0 00 thenthen y y = = Hu for somesome u u Ec I - H. H. Let Let 8 6 == 8(x, 6(x, y),y), u1,ul, ...... , , us us be the involutions with cycle (x, y), y), andand di d1 = = ului ulul for for 1 15 < i i 5< 6.8. ThenThen di dl E c D D 5< H, and ifif i i #0 jj thenthen by (6),(61, dl 1dj = utu1u1u3CG(t) = utu3CG(t) 0 CG(t), soSO didiCG(t) CG(t) #0 dJCG(t).djCG(t). ThusThus 82 S6 == m m = = I HIH : :CH(t)I, CH(t)l,so so H == DCH(t) DCH(t)andhence(2)implies(8). andhence (2) implies (8).

Chapter 16,16, ExerciseExercise 6.6. (1)(1) LetLet zz bebe anan involutioninvolution in Z(T). ByBy hypothesis,hypothesis, CG(Z)CG(z) is is anan elementaryelementary abelian abelian 2-group,2-group, so so as as T T <( CG(Z), CG(Z), it followsfollows that CG(Z)CG(z) == T. Thus (1) holds. Further as T is elementary abelian, each t EE ToT# is an involution inin thethe centercenter of of T,T, soso byby symmetrysymmetry between between t tand and z, z, T T == CG(t).CG(~). Therefore T isis aa TI-set in G.G. (2) As T is a TI-set inin G,G, NG(S)NG(S) 5 < H H for for each each 1 10 # S (< T, andand as T EE Sy12(G), thisthis is is one one of of the the equivalent equivalent conditions conditions listed listed in in 46.4 46.4 (with (with k k = = 1) for H toto bebe stronglystrongly embedded in G. (3) By (2), we may applyapply Exercise 16.5. In particular, byby part (3) of that exercise, D isis ofof oddodd order.order. ThusThus ICD(u)IICD(u)l is odd,odd, whilewhile asas uu isis anan involution,involution, CD(U)CD(u) isis aa 2-group,2-group, soso CD(u)CD(U) = = 1. HenceHence u inverts D. Adopt the notation of Exercise 16.5. ThenThen II nfl H == T T', #, soso m = I II I nfl HHI I = q - 1, 1, as as T T is is elementary elementary abelian abelian of of orderorder q. q. By By Exercise 16.5.2,16.5.2, H isis transitivetransitive onIflH,soq-1=IH:CH(t)I=IH:TI,andhenceon I n H, so q - 1 = IH: CH(t)l = IH: TI, and hence IHIIHI=ITI(q-1)= = ITI(q - 1) = q(q - 1). 1). As As uu invertsinverts D, J == uD uD is isof of order order IDI, IDI, whilewhile JJ isis the the setset ofof involutions with with cycle cycle (H, (H, Hu) Hu) onon X,X, soso byby ExerciseExercise 16.5.7,16.5.7, I JIIJI == m. Thus IDIID1 = IIJI J I =m= m =q-l.Therefore,asD= q - 1. Therefore, as D <5 HHwithTnD with T fl D = = 1 1asDisofodd as D isof odd order, we have ITDI = ITIIDI = q(q - 1) = IHI, so H == TD TD and and D D is is a a complement complement to T inin H. (4) First, asas T is noncyclic,noncyclic, qq > 2,2, soso II DlD I= = qq -- 11 >> 1. 1. Further, Further, asas uu invertsinverts D, DD isis abelian,abelian, andand TT isis abelian,abelian, while while HIT HIT S- D,D, as as D D is is a a complement complement toto TT inin HH byby (3).(3). Thus Thus HH isis solvable solvable andand DD isis aa HallHall 2'-subgroup2'-subgroup of H, soso DG nfl H == DH D~ by by Phillip Phillip Hall's Hall's Theorem Theorem 18.5. 18.5. LetLet M == NG(D) NG(D) and and u u E E S SE ESy12(M). Sy12(M). As As T T is is elementary elementary abelian, abelian, so so is is S,S, so if S 0# (u)(u) then then S S is is noncyclicnoncyclic and and hence D == (CD(s) (CD(s) : :S s E E S#) s') byby ExerciseExercise 8.1. This is impossible, as D 0# 1 1 while while CD(S) CD(s) = = 1, 1, since since CG(S) CG(S) isis aa 2-group2-group and II DlD I isis odd. ThusThus S = (u),(u), so so by 39.2, MM == (u)E,(u)E, wherewhere E == O(M). O(M). LetLet A = Fix(D) Fix(D) be the fixed pointspoints of of D D onon X.X. AsAs D~DG n fl HH = DH,D~, MM isis transitive transitive on AA byby 5.21. T~US Thusk k= = la1 JAI = = IM: IM:MfHI.ButM n HI.BU~ m fl H = NH(D) = NDT(D) = DNT(D) = D, as H == TD TD and and ND(T) ND(T) = = 1, 1, as as CD(t) CD(t) = = 1 1 for for each each tt EE To.T'. ThusThus MID is is regularregular onon A.A. Appendix 295

Let Fr be the set of triples (i, x, y) such that i EE uEuE and and (x,(x, y) y) is is a a cycle cycle ofof ii on A.. As [ElEl isis odd,odd, CE(U)CE(u) == 1,1,so SO UEuE=InMisoforderIEI,andasM/D= I n M is of order /El,and as MID is regular on A, each i EE uEuE is is regular regular onon A.A. Hence Hence therethere areare k members of Fr whose first entry is i, so IDIk2/2 II'I = I EIk = IDIIE : Dik = = (q -1)k2/2.

On the other hand, there are k(k -- 1) 1) choices choices forfor (x,(x, y),y), and and byby ExerciseExercise 16.5.7 16.5.7 there are mm == q -1- 1involutions involutions i i withwith cyclecycle (x, y), each inin iG,,,i Gx y= = iD i D c uE, so IFIIrl = k(k - 1)(q l)(q -- 1). 1). Thus Thus k k = = 2, 2, so so D D = = E Eand and (4) (4) is is established. established. (5) Let x = HH EE X,X, yy == xu, xu, and and z z E E XX - - {x} {x) = = X'. XI. By By Exercise Exercise 16.5.7,16.5.7, there is v E II withwith cyclecycle (x,(x, z),z), soso by symmetrysymmetry between uu and v, Fix(Do)Fix(Do) == (x, z}z) and Do = HH fln H° HU isis a a Hall Hall 2'-subgroup2'-subgroup of H. Therefore Therefore there is hh E H with DhD~ == Do DO byby Phillip Phillip Hall's Hall's TheoremTheorem 18.5, 18.5, as as observedobserved duringduring the proof of (4). Now by (4)

(x, yh}yh) = Fix(D)hFix(D)h == Fix(Dh) F~x(D~) == Fix(Do) Fix(Do) = = {x, {x, z}, z), so yh = z.z. Finally,Finally, asas HH = DT,DT, we we have hh = dtdt forfor somesome dd E D and t E T, so z = yh yh == ydt ydt = yt,= yt,and and therefore therefore T Tis istransitive transitive on on X. XI. But But by by 46.4,46.4, {x} == Fix(t) forfor eacheach tt EE TO,T', so T isis regularregular on IX'I.IX'I. As T is regular on X',XI, the action of D on X'XI isis equivalent to its action on T by conjugation byby 15.115.11. 1. ThusThus as as D D isis regular regular on on T', T#, D D is is regular regular on on X X - - (x, y},y), so G is 3-transitive on X andand onlyonly thethe identityidentity fixesfixes 3 or more points of X. (6) Let q == 2e, 2", and and regard regard TT as as an an e-dimensional e-dimensional vector vector spacespace over over GF(2)GF(2) and D as a subgroup of GL(T) as in 12.1. Let E == EndD EndD(T). (T ). As D is regular on T#,T', D isis irreducible, so by Schur's Lemma 12.4,12.4, E isis aa finitefinite division algebra over GF(2) of dimension at most e. Therefore byby 26.1, E is aa finitefinite field and f == I IE: E :GF(2)I GF(2)I <5 e. e. ButBut D#D' <( E#, E', soso

q-1=IDI:5 IEI=2f-1<2e-1=q-1, so E == F. F. Thus Thus T T can can be be regarded regarded asas thethe additiveadditive group ofof F,F, and D as the multiplicative groupgroup F',F#, andand thethe actionaction of of D D by by conjugation conjugation is is a: a: b b H H ab for a EE DD and and b b E ET. T. Thus Thus D Dand and T Tare are determined determined up up to to isomorphism isomorphism andand the representation ofof D on TT isis determineddetermined upup toto quasiequivalence,quasiequivalence, so the semidirect product H isis determineddetermined up to isomorphismisomorphism by 10.3.10.3. Hence there exists an isomorphism 7r:n: H 4 H* with DnD7r == D* =_ (O(a, (@(a, 0,O,0, 0, 1):1): aa E F#),F'}, and the map

,B: HID -+ H*/D*, Dt H D*(t7r) 296296 AppendixAppendix isis aa permutation isomorphism of thethe representationsrepresentations of of H H onon HIDHID and and H*H* onon H*/D*.H *ID*. Finally Finally thethe mapmap y: H*/D* - F, D*0(1, b, 0, 1) H b isis anan equivalence of thethe representationsrepresentations of H* on H*/D*H*/D* and and F,F, soso composing composing thesethese twotwo isomorphisms,isomorphisms, we get the isomorphism a ofof (6).(6). (7) As DD isis transitivetransitive on on X X - - {H,{H,Hu), Hu}, there there is is d dE E D D with with (la-')ud (la-1)ud == la-1.la-'. Let Let v v = = ud. ud. Note Note that that as as u u inverts inverts D D and and D D is is abelian, abelian, v v invertsinverts D. For a EE F#,F', letlet *(a)@(a) == O(a, $(a, 0, O,0, 0, 1). 1). Thus Thus @(a)-'*(a)-1 == '(a-1) *(a-') andand *(a)n-1@(a)n-' EE D,D, soso 1'(a)n-1@(a)nW' isis inverted inverted by by v.v. Therefore Therefore (aa-1)v = ((lf'(a))a-1)v = (la-1)*(a)n-1v = (la-1)v(r(a)n-1)v _ (la-1)(*(a)-1)n-1 = (lr(a-1))a-1 = a-1a-1 (8) As a:a: X +- Y Y is is an an isomorphism isomorphism ofof sets,sets, a*: Sym(X) Sym(X) +- Sym(Y) Sym(Y) is is anan isomorphismisomorphism of of groups, groups, where where a*: a*: g gH H a-lga. a-1ga. ThusThus a*:a*: L L + - L* L* isis an an iso-isomorphism, wherewhere LL = (H,(H, v),v), L* L* == (H*, (H*, v*),v*), and and v*v* == va*. va*. However, However, byby construction, v*v* fixesfixes 1 1 and and hashas cycles cycles (0, (0, oo), oo), (a, (a,a-'), a-1), for for a a EE FF - to, (0, 11. 1). That is, v*v* == (b(0,4(0, 1, 1,1,O) 0) E G*. FurtherFurther a* a* extends extends n n as as xahn xahn == (xh)a forfor all x EE XandhX and h EE H.ThusitremainstoshowthatGH.Thusitremains to show that G = = (H, (H, v)v) andand G* G* = = (H*, (H*, v*).v*). But as GG is 3-transitive3-transitive on X, GG isis primitiveprimitive on X by 15.14,15.14, so H isis maximalmaximal inin GG by 5.19, and hencehence indeedindeed GG == (H,(H, v).v). SimilarlySimilarly G* = (H*, (H*, v*).v*). References

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Symbol Page Symbol Page HHIG < G 2 (S) 2 1GIIGI 2 HLZGH=G 2 HgGH

p, n, n(G) 5 P', n' 5 Ep.EP. 5 mp(G) 5 0,O,(G), (G), 0-O"(G) (G) 5 Z(G) 5 QQn(G), ,(G),Qn(G) S2n(G) 5 AutcAutG (X) 5 6 ffaIR IR 6 C, Q,0,R LQ 6 G#G# 6 Aut(X) 7 r(G,UG, F) T) 8 Sym(X) 9 GL(V) 9 Inn(G) 11 Out(G) 11 Gy,GY, G(Y),G(Y), G1'GY 14 xG 13 Fix(S) 14 np, IGIplGlp 19 Sy1p(G)Sylp(G) 19 HGHaaG 23 H charchar G 25 [X, Y]Yl 26,26,45 45 [X, Y, ZI Z] 26 G(n)~(n) 27 Ln(G) 28 Zn(G)Z"(G) 28 S(A, G,G, n)r) 30 G*H 33 LwrG,Lwr G, Lwr, Lwr,G G 33 EndR(V)E~~R(V) 38 H~~R(U,HomR(U, V) 38 Soc(V) 38, 158 Fnxn 42 MxMx(g) (g) 42 Pdim 44 Tr, det 44 SL(V), PG(V) 44 PSL(V), PGL(V) 44 SLn(F), Ln(F) 44 PSLn(F), PGLn(F) 44 vV* * 47 B=BT 47 Mov(A)Mov( A) 54 CYC,Cyca 54 Alt(X) 55 U(G, V) 65 H'(G, V) V) 67 71 1~x7nn, lGlnIGI 1I 76 Rad(V) 76 O(V,OW, f A 1, O(V, OW, Q) Q> 78 300 List of symbols

Symbol Page Symbol Page 78 78 A(V, f,f), 13 A(V, Q) 78 J(X,J(X,f)f 78 A@AB 78 NFNF" 84 sgn(Q), sgn(V)sgn(V) 86,8786, 87 Spn(F),Sp,(F), Sp(V)SP(V) 88 SO(V),so(v), O(V), Q(V) 88,8988, 89 SPn(R),Spn(q), SUn(9) SUn(R) 89 OO(R),0:(9), SOO(R) so:(9) 89 cO(R)QE(9) 89 PSpn (9(q),1, PGUn (9(q) 89 po:(9),POO(R), POO(R)pQ:(9) 89 G+G + 97 (D(H)@(HI 105 Mod,.Mode. , DznD2 107 SD2,,,SD211, Q2"Q2n 107 P'+~,Pi+2n D8,-D:Qr Qs 110,111 L(V1,L(V1,...... ,, Vm;vm; V)V) 117 V@UV®U 118 UKuK,K@~U ,K®FU 119 K Jr7r 119 xu,7r a, vuVa 121 Grp(Y : W) 140 Kh),l(h),IR(h) l~(h) 143 W(C)W(E) 148 C+ 149 Comp(G) 157 E(G) 157 F(G) 158 O,(G)Oc,(G) 158 F*(G) 159 O,~,E(G>On',R(G) 159 -W(G) 162 J(G) 162 J"P(G, (G, V) 163 ac(B)aG(B) 175 char(G) 180 cl(G) 179 aG 188 &;(GI(Fk(G) 246 -q(X).@(XI 245 rP,k(G),r~,k(G), r:,k(G)rp,k(G) 247 &,P(G)Oek (G)o 248 e(G) 26261 1 mz.,(G)m2,p(G) 26126 1 IndexIndex

algebraicalgebraic integer 184 184 commutatorcommutator 2626 Alperin'sAlperin's FusionFusion Theorem 200200 complexcomplex 209 209 alternatingalternating group 5555 chamberchamber 209209 apartmentapartment 215215 chamberchamber graph graph 209209 automizerautomizer 55 connectedconnected 209209 automorphismautomorphism 7 7 reflectionreflection 212 212 thickthick 209209 Baer-SuzukiBaer-Suzuki TheoremTheorem 204204 thinthin 209 209 barbar conventionconvention 6 6 ComponentComponent Theorem 263263 basisbasis compositioncomposition factors 24 24 dualdual 4747 compositioncomposition seriesseries 2323 hyperbolichyperbolic 81 81 conjugateconjugate 33 orthogonalorthogonal 7979 CoprimeCoprime Action Theorem 7373 orthonormalorthonomal 7979 coveringcovering 168 168 BN-pairBN-pair 218218 CoxeterCoxeter complex 211 2 11 BorelBore1 subgroup subgroup 219219 diagramdiagram 141 141 Br-PropertyBp-Property 263263 groupgroup 142142 Brauer-FowlerBrauer-Fowler TheoremTheorem 244 244 matrixmatrix 141141 buildingbuilding 215215 systemsystem 142,142, irreducible irreducible 146146 WeylWeyl group 219219 criticalcritical subgroupsubgroup 108108 BurnsideBurnside Normal p-Complement cyclecycle 5454 TheoremTheorem 202 202 cyclecycle structurestructure 5454 Burnside pagb-Theorempaqb-~heorem 187187 dihedraldihedral group 141141 Cartan subgroupsubgroup 219 directdirect product 4 4 categorycategory 6 6 distancedistance 88 coproductcoproduct 77 dual space 4747 productproduct 77 Dynkin diagram 25125 1 Cauchy'sCauchy's TheoremTheorem 2020 centralcentral product 3232 edge 88 withwith identifiedidentified centers 3333 exactexact sequence 4747 centralizercentralizer 33 shortshort 47 chamberchamber 209209 splitsplit 4747 charactercharacter 4949 Exchange Condition 143143 degreedegree 179179 extension,extension, central 166166 generalizedgeneralized 180180 perfect 168168 inducedinduced 189189 universaluniversal 166 166 irreducibleirreducible 179179 extension problem 1010 character table 183 183 characteristic value 127 127 FG-homomorphism 3636 characteristic vector 127 127 FG-representation 3535 class function 179 179 field,field, perfect 9292 classicalclassical group 8888 FittingFitting subgroup 158158 ClassificationClassification Theorem 260260 generalizedgeneralized 159 159 CliffordClifford algebra 9595 fixedfixed point 14 14 CliffordClifford group 9696 flagflag 88 Clifford's Theorem 4141 foldingfolding 21121 1 cocyclecocycle 6464 opposite 212212 302 IndexIndex

form,form, bilinear bilinear 75 75 k-generatedk-generated p-corep-core 247247 equivalentequivalent 7878 metacyclicmetacyclic 203203 hermitianhermitian symmetricsymmetric 7676 nilpotentnilpotent 28 28 nondegeneratenondegenerate 7676 perfectperfect 27 27 orthogonalorthogonal 76 76 representationrepresentation byby conjugationconjugation 11 11 quadraticquadratic 7777 ringring 3636 radicalradical 76 76 sectionsection 261 261 sesquilinearsesquilinear 7575 solvablesolvable 27 27 similarsimilar 7878 skewskew hermitianhermitian 104104 HallHall 7rr-subgroup -subgroup 7171 skewskew symmetricsymmetric 7676 hyperbolichyperbolic pair pair 8080 symmetricsymmetric 7676 hyperbolichyperbolic planeplane 8080 symplecticsymplectic 7676 WittWitt index 7878 inductioninduction map 189189 FrattiniFrattini argument argument 2020 innerinner automorphism automorphism groupgroup 1111 FrattiniFrattini subgroup subgroup 105 105 integerinteger FrobeniusFrobenius complementcomplement 191 191 p-partp-part 1919 FrobeniusFrobenius kernelkernel 191191 nr-partr-part 7171 FrobeniusFrobenius NormalNormal p-Complementp-Complement involutioninvolution 55 TheoremTheorem 203 203 isometryisometry 7878 FrobeinusFrobeinus Reciprocity Theorem 190190 isomorphismisomorphism 77 Frobenius'Frobenius' TheoremTheorem 191 191 fusion,fusion, control 199199 Jordan'sJordan's TheoremTheorem 5858 Jordan-HolderJordan-Holder TheoremTheorem 24,24,37 37 G-isomorphismG-isomorphism 99 G-morphismG-morphism 99 K-group 261 gallerygallery 209209 Gaschatz'Gaschiitz' TheoremTheorem 313 1 LeviLevi factor 257 257 generalgeneral linearlinear group 99 LieLie rankrank 250250 generalgeneral unitaryunitary group 8989 linearlinear representation representation 99 geometrygeometry 8 8 absolutely irreducible 121 121 connectedconnected 88 dualdual 47 47 flagflag 88 envelopingenveloping algebra 42 flagflag transitive 88 indecomposableindecomposable 3737 morphismmorphism 88 inducedinduced 188 188 oriflammeoriflamme 9999 irreducibleirreducible 3737 polarpolar 9999 irreducibleirreducible constituents 3737 rankrank 88 principal 179179 residuallyresidually connected 88 regularregular 182 182 residueresidue 88 tensortensor product 119119 typetype 88 written over a field 124124 GlaubermanGlauberman Z*-Theorem 262262 linearlinear transformation transformation graphgraph 8 8 determinantdeterminant 4444 connected 88 m-linearm-linear 118118 connected component 88 nilpotent 128128 groupgroup algebra 3636 semisimplesemisimple 128 128 center 5 semisimplesemisimple part 131131 characteristic p-typep-type 261261 tracetrace 4444 characteristicallycharacteristically simple 2525 unipotentunipotent 128 128 classclass 2828 unipotent part 131131 complementcomplement 30 30 coveringcovering 168 168 Maschke'sMaschke's Theorem 4040 exponentexponent 55 maximalmaximal subgroupsubgroup 55 extension 30 modularmodular property of groups 66 free 138 module Frobenius 191 absolutelyabsolutely irreducible 121 121 LieLie type 250 250 complement 3737 k-connectedk-connected 246 246 composition factor 3737 Index 303

condensed 125125 transitive 1414 cyclic 38 38 transitive constituentsconstituents 16 extension 3737 Phillip Hall's Theorem 7171 homogeneous 40 n-group1r-group 5 homogeneous component 40 presentation 140140 indecomposable 37 projective geometry 4343 irreducible 3737 dimension 44 permutation 50 hyperplane 44 semisimple 38 38 lines 4444 simple 3737 points 44 socle 3838 projective generalgeneral linearlinear groupgroup 44 split extensionextension 3737 projective specialspecial linearlinear groupgroup 44 morphism 6 quaternion group 107 107 normal p-complementp-complement 202 normal seriesseries 22 Rainy Day LemmaLemma 2162 16 A-invariant 22 reflection 93 factors 2323 center 93 93 length 22 representation 9 normalizer 3 3 equivalence 9 faithful 99 Odd Order TheoremTheorem 260 quasiequivalence 9 orbit 1313 residue 8 8 orbital 5555 root group 250 diagonal 5555 root system 148 148 paired 5959 closed subsetsubset 255 self pairedpaired 59 idealideal 255255 ordinary ChevalleyChevalley groupgroup 252 ordering 149 149 orthogonal groupgroup 88 positive system 149 149 orthogonality relationsrelations 183183 simple system 149 149 outer automorphism group 1111 Weyl group 148148 root p-group 5 5 long 251 elementary abelianabelian 5 short 251 extraspecial 108 108 modular 107107 scalar matrixmatrix 44 special 108 108 scalar transformation 44 symplectic typetype 109109 Schreier ConjectureConjecture 160160 p-rank 5 Schur's Lemma 38 38 2-local 260 Schur multiplier 168168 parabolic subgroup 221221 Schur-Zassenhaus Theorem 70 partition semidihedral groupgroup 107107 G-invariant 18 18 semidirect product 30 30 nontrivial 18 18 signalizer functor 229229 path 88 complete 229229 length 8 8 solvable 229 229 permutation solvably completecomplete 229 even 55 similarity 78 odd 5555 simplex 209209 permutation group 5353 Sims Conjecture 176176 permutation representation 9 Solvable 2-Signalizer FunctorFunctor TheoremTheorem 229 by conjugation 15 15 space by right multiplication 15 15 hyperbolic 81 81 m-transitive 56 orthogonal 77 primitive 1818 symplectic 77 rank 55 unitary 77 regular 56 special CliffordClifford groupgroup 97 semiregular 56 special linearlinear groupgroup 44