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Uni International 300 N. Zeeb Road Ann Arbor, Ml 48106 8426500

Woldar, Andrew J.

ON THE MAXIMAL SUBGROUPS OF LYONS’ GROUP AND EVIDENCE FOR THE EXISTENCE OF A 111-DIMENSIONAL FAITHFUL LYS-MODULE OVER A FIELD OF CHARACTERISTIC 5

The Ohio State University Ph.D. 1984

University Microfilms International300 N. Zeeb Road, Ann Arbor, Ml 48106 ON THE MAXIMAL SUBGROUPS OF LYONS' GROUP AND

EVIDENCE FOR THE EXISTENCE OF A

111-DIMENSIONAL FAITHFUL LyS-MODULE OVER A

FIELD OF CHARACTERISTIC 5

DISSERTATION

Presented in Partial Fulfillment of the Requirements for

the Degree Doctor of Philosophy in the Graduate

School of The Ohio State University

By

Andrew J. Woldar, B.S., M.S.

The Ohio State University

1984

Reading Committee: Approved by

Ronald Solomon

Koichiro Harada

Sia K. Wong Department of Mathematics Dedicated to My Mother and the Memory of My Father

George Morris Woldar

(1910-1964) ACKNOWLEDGEMENTS

I wish to express my gratitude to Richard Lyons and Peter Landrock for their helpful and insightful sug­ gestions, and particularly to my adviser, Ronald Solomon, for his guidance throughout the work. VITA

December 12, 1947 Born - New York, New York

1976 B.S., The City College of New York, C.U.N.Y., New York, New York

1976-1984 Teaching Associate, Department of Mathematics, The Ohio State University, Columbus, Ohio

1978. M.S., The Ohio State University, Columbus, Ohio

1976,1982,1983 Recipient of Summer Fellow­ ship, The Ohio State University, Columbus, Ohio

1983 Recipient of Graduate Student Alumni Research Award

FIELDS OF STUDY

Major Field: Mathematics

Studies in Group Theory. Professor Ronald Solomon

Studies in Representation Theory. Professor Ronald Solomon TABLE OF CONTENTS

Page ACKNOWLEDGEMENTS...... iii VITA ...... iv

LIST OF TABLES ...... vi

LIST OF FIGURES...... vii

LIST OF SYMBOLS...... viii

INTRODUCTION ...... 1

CHAPTER

I. THE MAXIMAL LOCAL SUBGROUPS OF LYONS’ GROUP. . 6

1. 2-Local Analysis ...... 18 2. 3-Local Analysis ...... 35 3. 5-Local Analysis ...... 52 4. Local Analysis for Remaining Primes. . . 84 5. A Complete Set of Maximal Local Subgroups for LyS...... 8 6

II. ON THE MAXIMAL SUBGROUPS OF LYONS’ GROUP . . . 89

1. Some Non-Local Analysis...... 91 2. Conclusions...... 127

III. EVIDENCE FOR A 111-DIMENSIONAL IRREDUCIBLE LyS-MODULE OVER A FIELD OF CHARACTERISTIC 5 . 146

1. 5-Modular Analysis of M e ...... 152 2. The 5-Decomposition Matrices for 2n (5 ^ n ^ 11) and A x x ...... 172 3. The Character Tables and 5-Decomposition Matrices for N = E 3S *Mii and S = 3 2 + lt-(SL(2,5)*Z8) ...... 210 4. Evidence ...... 237 5. Minimal Degree for a Faithful Character of L y S ...... 263

LIST OF REFERENCES ...... 270 v LIST OF TABLES

Table Page

1. Centralizers of Prime Order Elements .... 14

2. The Character Induction-Restriction Table for (Me, 1^ ( 3 ) ) ...... 154

3. The Character Induction-Restriction Table for (Mc,M22) ...... 1 5 5

4. The Character Induction-Restriction Table for (.3,Me) ...... 1 6 9

5. 5-Cores of Partitions of Size n ( 5 ^ n ^ l l ) ...... 1 7 4

6 . The Character Table for N = E 35»M1 1 ...... 214

7. The Character Table for S =3 2+lf • (SL(2 , 5)*Za ). 218 LIST OF FIGURES

Figure Page

1. 2-Local Structure of Ly S ...... 15

2. 3-Local Structure of LyS ...... 16

3. 5-Local Structure of Ly S ...... 17

vii LIST OF SYMBOLS

A is a subgroup of B

the elements of A not contained in B

split extension of A by B

non-split extension of A by B

direct product of A and B

central product of A and B

the subgroup generated by the elements Xj,f ..Xjj of some group K

the group generated by the subgroups A lf..,An of some group K

the element y-1xy of K where x , y K

the element x_1xy of K where x , y K

the subgroup <[x,y]> of K where x ranges over A and y ranges over B

(We call [A,A] the commutator subgroup or , derived group of A.)

the index of B in A

the normalizer of H in K

the centralizer of H in K

the centralizer of x in K

(When K = LyS , we write N_,(H) , Cr (H) , and Cr(x) for N„(H) , C„(H) , and C„(x) respectively.)

viii Z(K) the center of K

Op(K) the largest normal p-subgroup of K

0(K) the largest normal odd-order subgroup of K

^(K) the subgroup of the p-group K generated by its elements of order p

®(K) the Prattini subgroup of K K* the non-identity elements of K

|K| the order of K (We call x a p-element if | | = pn for some prime p and positive integer n.)

|K|p the order of a Sylow p-subgroup of K ir(K) the set of prime divisors of | K | (We call K a p -group if p is a prime not contained in tt(K). Also we call x a p -ele­ ment - or p-regular - if p is not contained in ir(). )

x ~ K y x is K-conjugate to y

(When K = LyS we omit the subscript "K" .)

XK the set of all K-conjugates of x

SP Sylow p-subgroup Frob(n,m) Frobenius group with kernel of order n and complement of order m

(q = pn) Elementary abelian p-group of order q

symmetric group of degree n

alternating group of degree n

the two-fold cover of A X1 (in the sense of Schur) 2 { i ^ , . . , ijj} symmetric group on the letters {ij,..,in}

A{i^ f ••>in) alternating group on the letters { i j i n}

Ln (q) PSL(n,q), the group of projective special n x n matrices with entries in GF(q)

Un (q) PSU(n,q), the group of projective special n x n unitary matrices with entries in GF(q2)

SU±(2,5) group of 2 x 2 unitary matrices with entries in GF(25) and having determinant ±1

Me the sporadic simple group of McLaughlin of order 2 7 »36 *53 *7*11

Me the triple cover of Me (in the sense of Schur)

Janko's 2nd sporadic simple group, of order 2 7.3 3 .5 2 . 7

^11 ’^12 » 2 the three smallest of the five sporadic simple groups of Mathieu

GF(q) (q = pn ). the Galois field of q elements

«a+b special p-group of order p a+b with center of order pa

(When a = l we call pa+k extraspecial.)

(X»4»)k the inner product (defined by the standard orthogonality relations) applied to the characters x anc* f of K

X+H the restriction of x to H where x is a character of an over-group of H

X+M the character x induced to M where x is a character of a subgroup of M

% the principal character of K Irr(K) the set of ordinary irreducible characters of K IBrp (K) the set of irreducible p-Brauer charac­ ters of K

ik (0) the inertia group in K of the character 0

X < * X is a constituent of ij; deg(x) the degree of the character x

xi INTRODUCTION

With little doubt, one of the most significant ad­ vances in modern algebra is the classification of finite simple groups. Believed to be at a state of completion, the classification represents the culmination of a century- long endeavor, riddled with great periods of inactivity and skepticism. We elaborate briefly on it, so as to establish a context for the work which follows.

The classification reveals the existence of seventeen infinite families of finite simple groups: sixteen families of "finite simple groups of Lie type" and the alternating groups. The groups of Lie type possess a unified classifi­ cation unsurpassed in beauty and elegance. The first person to realize the possibility for such a classification was

Chevalley. Defining certain groups as being generated by formal exponentials of nilpotent elements in a simple Lie algebra, Chevalley was able to give a unified description of nine infinite families. His work was later extended, through the collective efforts of Steinberg and Ree, to in­ clude seven new families (the so-called twisted relatives of the nine aforementioned families) giving a total of

1 sixteen in all. These then are the families of groups of

Lie type.

For obvious reasons, it would be desirable to obtain a unified classification for all finite simple groups. The alternating groups pose no serious threat here. The prob­ lem, however, stems from the existence of twenty-six finite simple groups, none of which lie in any of the aforementioned seventeen infinite families. These are the so-called spo­ radic groups and, as their name suggests, they appear to defy a unified classification. Perhaps the work of Tits offers the most promise in overcoming this obstacle. Tits' main contribution was to define a family of geometrical ob­ jects (Buildings) which could be completely classified

(under the additional hypotheses of spherical and rank ^ 3) and which admit the groups of Lie type as flag-transitive automorphism groups. Attempts to extend the work of Tits to the sporadics have been pursued by Beukenhout, Ronan, Kantor, and others. (Independent approaches are currently being studied by Harada and Lepowsky ) . Thus the current deluge of attention devoted to the study of sporadic groups is well justified. Generally, it is because of their pathological nature that sporadic groups are so delightfully intriguing and the focus of so much theoretical investigation.

Another area of considerable interest concerns pro­ viding a computer-free construction of each of the sporadic 3 groups. In 1979, Griess successfully constructed (by hand!) the largest sporadic group - the so-called "Monster" -

F l • As F i involves no fewer than twenty of the spo- radics, a computer-free construction is thereby obtained for each of these. Lyons’ group LyS remains one of the few sporadic groups for which no such construction is known. It is my hope that this work will help to provide some insight into this problem. If a computer-free con­ struction of LyS is ever to be achieved, it appears that the (alleged) 111-dimensional module analyzed in Chapter III may be the critical object in the construction. Perhaps a

LyS-invariant geometry can be defined on this module with sufficient structure so as to ensure that the automorphism group of said geometry be LyS itself. Still, the geometry must be intrinsic to the module if it is to provide an inde­ pendent construction of LyS .

Some time ago, following the lead of my adviser, Pro­ fessor Solomon, I searched for a low-dimensional faithful

LyS-module over any field. It soon became apparent that 5 was the natural characteristic for the underlying field, and by considering the 5-modular characters of an appropriate 3- local, a lower bound of 1 1 0 was readily achieved for the de­ gree of such a module. Not long after, in a conversation with David Benson, I learned that the group theorists at

Cambridge University had constructed what they believed to be a set of generating matrices for LyS inside GL(111,5). 4

Although a verification of this seemed to be (at that time)

beyond the capabilities of their computer, they still were

able to furnish a convincing probablistic argument that the

matrices they had constructed did the trick. It was at

this point that I became interested in performing a thorough

investigation of the related module (Chapter III). My

hope was that I could perhaps gain new insights from the

modular character theoretic point of view, a new slant on

accruing evidence for the existence of such'a module. In

this pursuit, the results of Chapters I and II (which treat

the maximal subgroup problem for LyS) were put to good use.

More specifically, Chapter I treats the determination and

classification of a set of representatives for the maximal

local subgroups of LyS (those local subgroups of LyS which

are maximal with respect to the property of being local).

A complete determination and classification is achieved. In

Chapter II, an attempt is made to classify the maximal non­

local subgroups of LyS (normalizers of non-solvable charac­

teristically simple subgroups of LyS ) . Here I meet with

only partial success, but still I am able to eliminate

several candidates for maximality mainly by local arguments.

In addition, a complete classification of a different type

is achieved. Specifically, I am able to describe all the maximal subgroups of LyS containing a Sylow p-normalizer

for every p e'n'(LyS). (This mirrors the theory of parabolic subgroups in finite BN-pairs of characteristic p , where the maximal parabolics are precisely the maximal subgroups containing a Sylow p-normalizer.) It is my hope that the results of Chapters I and II may someday contribute to a complete resolution of the maximal subgroup problem for

Lyons 1 group.

Finally, I wish to remark that the notation and ter­ minology used in the text are. chiefly standard. Nonethe­ less, a list of symbols is provided for the sake of com­ pleteness. Books and articles are referred to by the use of brackets; e.g. [n] refers to the n-th entry in the List of References. CHAPTER I

THE MAXIMAL LOCAL SUBGROUPS OF LYONS’ GROUP

Our main objective in this chapter will be to clas­ sify all maximal local subgroups of LyS, Lyons' sporadic simple group. A p-local subgroup of a group G is by defi­ nition the normalizer Nq(P) of a non-trivial p-group P of

G. We call a subgroup of G local if it is a p-local sub­ group of G for some p €ir(G). Finally, H is a maximal local subgroup of G if H is maximal (under inclusion) among all local subgroups of G . Now in classifying all maximal local subgroups of a group G , we first observe that it is only necessary to consider elementary abelian p- groups in the role of P above. Indeed, given an arbitrary p-group P f 1 in G, we consider its subgroup fijCZCP)).

From its definition, the latter group is easily seen to be an abelian group of exponent p , i.e. an elementary abelian p-group of G . Furthermore, nj(Z(P)) is characteristic

in P whence NQ (P) is contained in Nq(J21 (Z(P))) . Thus in our search for maximal local subgroups, we need only consid­ er groups of the form ^(ZCP)) rather than the more gener­

al P .

6 7

In sections 1, 2, and 3 we treat the analysis of

2-, 3-, and 5-local subgroups respectively; in section 4 all remaining local analysis is performed. Section 5 consolidates the results of the previous four sections and provides us with the desired classification (theorem

1.47) which we briefly summarize below.

(Main Result) LyS has (up to conjugacy) ten

maximal local subgroups, the isomorphism types of

which are given as follows:

A two 2-local subgroups, viz. Aj x , E 23\GL( 2 ,3 )

three 3-local subgroups, viz. Z 3\Aut(Mc),

E 3 s* (Mn x Z 2) , 3 2 + l* • (SL( 2 ,5)*Z8 )*Z2 l + i* two 5-local subgroups, viz. 5 •SL(2,9)*ZIi ,

E53\SL(3,5)

one 31-local subgroup, viz. Frob(31,6)

one 37-local subgroup, viz. Frob(37,18)

one 67-local subgroup, viz. Frob(67,22)

We further remark that every maximal p-local sub­ group of LyS for p € {3,5,31,37,67} is in fact maxi­ mal local, while, in contrast, there exist certain maxi­ mal 2-locals in LyS which fail to be maximal local (in­ asmuch as they are properly contained in 3-locals). 8

We conclude our introductory discussion with a list of facts which can readily be found in [13] and which we shall assume throughout the chapter. This is followed by

Table 1, which gives representatives for certain LyS- classes (those relevant to the maximal subgroup problem) and the isomorphism types of their corresponding central- izers. Finally, in figures 1, 2, and 3, we supply flow­ charts which render a synopsis of the p-local structure of

LyS for p = 2,3,5. These flowcharts are designed to pro­ vide the reader with a concise overview of the material presented in the first three sections of this chapter. 9

1.1. Let x be a non-central 2-element of A1]L.

The order of x is then determined from the

cycle structure of its image x in A 1X as

follows:

cycle structure of x order of x

(ab)(cd)(ef)(gh) 2

(ab)(cd) 4

(abcd)(efgh) 4

(abed)(ef)(g h )(ij) 8

(abcd)(ef) 8

(abedefgh)(ij) 8

A 1.2. Let x be a non-central element of Ax and

A A let Z(A1X) = . Then x is A1}-conjugate

to x z .

1.3. Let S 2 denote a Sylow 2-subgroup of LyS. Then

Z(S2 ) & Z2 . 10

1.4. Aut(Mc) has a unique class of subgroups isomorphic

to E24 which is the union of two Mc-classes of

such subgroups. Aut(Mc) contains no subgroup iso­

morphic to E 2 5 .

1.5. A Sylow 3-subgroup S 3 of LyS can be chosen to

be the group generated by elements u^, u 2, u 3>X1-X2-

Wi,w2 of LyS, each of order 3, subject to defin­

ing relations [w^Xj] = [w2 ,x2] = u2, [w1 ,x2] =

[x1 ,w2] = u1n2, [wj ,u3] - x p [w2 ,u3] = x|, all

other commutators of generators being trivial. We

further define two subgroups of S3, namely

V = and S = .

1.6. Let y be any element of S 3 not contained in V.

Then Cv(y) = .

1.7. There exist precisely 22 elements in S 3 of LyS-

type 3!- Moreover, they are all contained in V.

1.8. Every element of order 3 in S 3 lies in 0 3(Cq(u2 ))

or V. 11

1.9. Z(S3 ) = .

1 .1 0 . Fix an involution z of LyS and identify c q (z )

A with Ajj. Then any element of order 3 in CG(z)

whose image modulo is a 3-cycle in A 1]L is of

LyS-type 3j. A particular representative is given

by v x where ?j = (123). The class of 3 2-

elements of LyS has representative v 2 where

v 2 = (123)(456).

1 .1 1 . Uj and u 2 can be chosen in 1.5 so as to satis­

fy u 2 = v 2 and = .

1 .12. Suppose z , u 2 are as in 1.10 and 1.11. Then

it is possible to choose a complement W to

0 3 (Cg(u2)) in cq(u2) such that z€W. Assume

this has been done. Then z acts fixed-point-

freely on the Frattini quotient of 0 3(CG (u2 )), as

do all elements of order 5 in W.

1.13. A Sylow 5-subgroup S 5 of LyS can be chosen to be the group generated by elements i x,g2,g3,gh,

g 5 ,f2 of LyS, each of order 5, subject to defin­

ing relations [g2 ,g5] = [g3 ,g5] = [g^,gg] = fj. 12

[g±,±z] = for i=3,4,5, all other commutators

of generators being trivial.

1.14. Let R = . Then R = 5 1+lf, R has

exponent 5, and R = R/$(R) is a uniserial -

module.

1.15. cG(fi>g2) = i>g 2 'g 3 »g 4 >f2 > and NG () is a split extension of CG(flfg2) by GL(2,5).

1.16. f 1 ,f2 can be chosen in CG (Z ) ( z as in 1 . 1 0 )

so as to satisfy ^ = (12345), f2=(12345)(6789 10),

where bar denotes modulo . When this is done,

fx is of LyS-type 5 ; f 2 is of type 52<

1.17. Let f x be chosen as in 1.16. Then NG()

is a split extension of R by a group Q ^ C q Cz ),

Q x = , where modulo , Q is the alternat­

ing group on the letters {6,7,8,9,10,11} and

u = (2354)(6798) under the identification of

Cq(z) with A X1 in 1 .1 0 .

1.18. Every 5x-pure elementary abelian subgroup of

Sg which contains fx and has order 25 is NG(R)-conjugate to . 13

1.19. Let v, v* be Sx-elements which together gener­ ate a Sylow 3-subgroup of Q ( Q as in 1.17).

Then Cr(v) S 5 1 + 2 has center and exponent

5. Moreover, all elements of CR(v)'^ are of

LyS-type 52, and for f 6 CR(v)\, CG(f) :i's

a split extension of xcR (v*) by Z 3 .

1.20. Let P = xCR (v*). Then P is the unique

Sylow 5-subgroup of CG(f), P is isomorphic to

Z 5 x51+2, Z(P) = , and [P,P] = .

1.21. Let S31, S37, S g 7 denote Sylow 31-, 37-, and

67-subgroups of LyS respectively. Then

INq (S 31) • ® 31 I = ® 3 7 I = and |NG (S67) : S67| =22. Moreover, each of

these Sylow subgroups is self-centralizing in

LyS. 14

TABLE 1. Centralizers of Prime Order Elements

representative class centralizer

z 2 A 11

U 1 31 Me

u 2 3z 3 2 +^-SU±(2,5)

*1 51 5! + ^ »SL(2,9)

*2 52 (Z5 x5 1 + 2 )-E3 s 7 Z ?xSL(2,3)

e H i e”1 1 1 2 Z11XE3

a 31 Z31 b 37 Z 3 7 c 67 Z6 7 15

E an elementary abelian 2-group of LyS z,t,r,x as in lemma 1 . 6

Yes

No

May assume

Nr (E) is contained Yes in some 3-local

subgroup of LyS

No

N q (E) S E 2 3\GL(2,3)

FIGURE 1. 2-Local Structure of LyS 16

E an elementary abelian 3-group of LyS

Yes E — 3 -pure Nq (E) S Z 3\Aut(Mc)

No

May assume

Yes Yes

No No

Nr (E) < Nr (V) Ng(E) < Nq(V) or Nq()

FIGURE 2. 3-Local Structure of LyS 17

E an elementary abelian 5-group of LyS

Nr (E) is conjugate to a No 5,-pure subgroup of Nr() or

Yes

No

5 1 +lf • SL(2 , 9 ) • Z

Yes

Nr (E) is conjugate No to a subgroup of

Yes

E ~

Nq (E) & E 5 3\SL(3,5)

FIGURE 3. 5-Local Structure of LyS 18

1. 2-Local Analysis

The purpose of this section is to show that every elementary abelian 2-group E of LyS satisfies one of the following three conditions:

(i) NG ( E ) S A llf

(ii) Nq (E) & E 2 3 \GL(3,2),

(iii) Nq (E) is contained in some 3-local subgroup of LyS.

As Cg (z) = A 11 (see Table 1), it will often be desirable to identify these two groups, as well as their respective quotients CG(z)=CG(z)/ and Alx. This shall be done freely throughout the section, and the reader should anticipate no difficulty in recognizing when these identifications are being made.

Lemma 1.1. LyS has a unique class of fours-groups.

Proof. We first show that C e Cg (z ) is transitive on its non-central involutions. Thus let a and b be involu­ tions of C with a^z, b^z. By 1.1, a and b are each products of four disjoint transpositions in A xl.

As A n is 9-transitive on its letters, there clearly exists an element x € C such that ax = b whence ax and b lie in the same (non-trivial) coset of in 19

C. By 1.2, ax ~c b, so that a b as claimed.

Now given two arbitrary fours-groups A and B of

LyS, we may first assume z€AnB (as LyS has a unique class of involutions), say A = , B = . Thus by the above, Ac = = . = B for some c in

C, and the lemma is proved.

General hypotheses for lemmas 1.2- 1.6, E shall repre­ sent an elementary abelian 2-group of LyS of order at least 4. t and x will be fixed elements of- C^(z) of respective orders 2 and 3 and having respective images t = (12)(34)(56)(78) and x = (9 10 11) in A j j .

By virtue of lemma 1.1, we shall also assume that E contains the fours-group .

Lemma 1.2. E < C Q (E) < N G ().

Proof. Clearly E < Cq(E) ^ CG(t) where C denotes the group Cq(z). It therefore suffices to show CG(t) is contained in Nq(). Let y€CG(t). Then y centralizes t and consequently permutes each of the subsets {1 ,2 ,...,8 } (the orbit of t on {1 ,2 ,..,1 1 } ) and {9,10,11} (the fixed point set of t on {1,2,..11}).

Thus for any i in {1,2,...,8 } we clearly have that i^ -1 is in {1 ,2 ,..,8 } as well, and as x fixes 20

{1 ,2 ,..,8 } pointwise, we get iy xy = iy y = iT = i, i.e. xy fixes {1,2,..,8 } pointwise. Thus xy = x or x - 1 whence xy must be equal to one of x,x- 1 ,xz, or x- 1 z. But x has order 3 while each of xz,x_1z has order 6 . Thus xy = x or x " 1 , and y € N Q () as desired.

Lemma 1.3. Suppose E ^C^(x). Then N^(E) is con­ tained in some 3-local subgroup of LyS.

Proof. Clearly x€CG(E) whence < C q (E) by the preceding lemma. Since Cq (E) < C^(t), we have

|Cq (E) | 3 ^ |Cc (t) | 3 ^ |Cg-(t) | 3 . Under our identification of C with A1]L, (^(t) is identified with the group

C. ((12)(34)(56)(78)). But the latter group is contained A i l in C ((12)(34)(56)(78)) 3 (Ejg'Z,*) x Z 3 which has 3- part 9 . Thus |C^(t')|3 ^ 9, and it follows that the

3-part of C(j(E) must be at most 9 .

Now if is characteristic in CG (E), then

< x > o N g (E) and the result follows at once with NG(E)<

Ng(). We therefore assume there exists a € Aut(CG (-E)) with a?^ . As is normal in CG(E), we have

ao C G(E)a = Cg(E) whence 0 < C G (E) as well.

But by order, CT must be a Sylow 3-subgroup of 21

C^(E). As a normal Sylow subgroup is trivially charac­ teristic, we conclude that ° is characteristic in Cq (E), s o normal in NQ(E). NG(E) < NQ( a) now follows and the lemma is proved.

Lemma 1.4. Nq () = Aut(Mc) where the double bar denotes modulo .

Proof. As CQ(x) = Me (1.10; Table 1), it suffices to show N q () = A u t ( C G (x)). Let $: NQ ()— *-Aut(CG (x)) denote the action of NG() on CG(x) via conjugation.

We show first that Ker $ is trivial.

Let a € Ker $. Then [CG(x),] = 1 whence

[CG(x),] < . We therefore have [CG(x),,CG(x)]

< [,CG(x)] = 1 and [,CG(x),CQ(x)] < [ ,CG(x)] =

1 , and it follows from the three-subgroup lemma that

[CG (x),CG (x),] = 1. But as CG (x) is a non-split extension of Z 3 by the simple group Me (Table 1), it is immediate that [CG(x),CG(x)] = CG(x) whence

1 = [CQ(x),CG(x),] = [CQ(x),]. In particular,

[x,a] = 1 so that a is in fact in CG (x). But then

[CG(x),] = 1 implies a€Z(CG(x)) = . Thus we conclude a = 1 and Ker $ = 1 as claimed.

To complete the proof it obviously suffices to show |NQ()| = |Aut(CQ(x))|. As it is well known 22

that |Aut(Mc):Mc| = 2, we obtain at once

2 = | Aut(CG (x)):CG (x) | = |Aut(CG (x) ) :NQ ( ) | x

|Ng ():Cq (x )| .

Now consider the element r of Cq(z) which maps onto r = (13)(24)(78)(9 10) in AX1. Clearly xr = x7 =

(9 10 11)(13)(24)(78)(9 10) = ( 1 0 9 xl) = Thus xr = x - 1 ( xr = x-1z is impossible as x - 1 has order

3 and x_1z has order 6 ). We therefore have that r is an element of NG () not contained in CG (x) , whence CG (x) ^ N G (). But then from above we obtain

|Ng (< x > ):CG(x)| = 2 and |Aut(CG(x)):NG()| = 1.

Thus |Aut(CG(x))| = |NQ(< x > ) | as desired, and the proof is complete.

Lemma 1.5. Suppose E ^ CQ(x). Then |E| = 8 .

Proof. We first show x and t commute. Clearly x = (9 10 11) and t = (12)(34)(56)(78) commute whence x^ = x or xz. As x has order 3 and xz has order

6 , x^ = x is assured. Thus < E ncG(x), and as

E £ CG(x) we immediately obtain |E| Ss 8 .

Suppose, by way of contradiction, that |E| ^ 1 6 and choose A < E, A S Elg. By lemma 1.2, E < NG().

Thus, letting double bars denote images modulo , we obtain A ^ E < NG() = Aut(Me), (the isomorphism 23

having been established in lemma 1.4). But we now see from 1.4 that E must have order 16 and so be con­ tained in Cq(x). This gives E < Cq(x), an obvious contradiction. Thus |E| = 8 as asserted and the lemma is proved.

Lemma 1.6. Suppose E ^ Cq (x ). Then E~ where r = (13)(24)(78)(9 10) in A .

Proof. The proof is accomplished in four steps.

Step 1. Let u,w,a,b be elements of Cq (z ) having respective images u = (13)(24)(57)(68),

W = (15)(26)(37)(48), a = (13)(24)(58)(67), and b = (18)(27)(35)(46). Then T = is a subgroup of CQ(z,t) of order 2 6 .

Proof of step 1. From direct calculation, we obtain

ut = (14)(23)(58)(67), wf = (16)(25)(38)(47)

at = (14)(23)(57)(68), bt = (17)(28)(36)(45).

Therefore by 1.1, the elements ut,wt,at,bt are all involutions, as are u,w,a,b, and t. Thus [g,t] = g“ 1t ~ 1gt = gtgt = (gt) 2 = 1 for g€{u,w,a,b), whence

T < C£,(z,t). It remains to show that T has the pre­ scribed order.

Clearly is normal (in fact central) in T, so we may consider T/, which we shall denote by 24

double bars: T = . By direct calculation, we obtain uw=(17)(28)(35)(46), ua=(56)(78), ub=(1526)(3847), wa=(1827)(3645), wb=(1324)(5867), ab=(15)(26)(38)(47).

Thus (uw)2 =(ua)2 =(ab)2=l and (ub)2 =(wa)2 =(wb)2 = t . This clearly implies that each of uw, ua, ub, wa, wb, ab are involutions of T , whence (as u, w, a, b are involutions) we have [g,h] = (gh) 2 = 1 for all g,h € {u,w,a,b}.

Therefore T is abelian of exponent 2. We show T has order 16, whence |T| = 6 4 follows immediately.

First we observe that = x = E1+. Indeed we otherwise have u = w whence u = w or wF, a contra­ diction. Similarly, as a ^ E , a £ T5F , we also have

= x=Elt. Thus to complete the proof of step 1 we need only show n = T. But n

^ T implies that at least one of u, w, uw occurs in the following listing: a, b, ab, at=(14)(23)(57)(68), bt=(17)(28)(36)(45), abt=(16)(25)(37)(48). This not be­ ing the case, the proof of step 1 is complete.

Step 2. r normalizes T. Moreover, T* is a Sylow

2 -subgroup of C£,(z,t).

Proof of step 2. It is easily verified that conjugation by r fixes ~t and interchanges u with a and b with w . This proves r normalizes T = T/ whence r normalizes T . Suppose r € T . Then r = 25

(13)(24)(78)(9 10) is an element of T = .

But it is immediate that each of "t, u, w, a, b fixes the set {9,10,11) pointwise whence of course T does as well. As r moves 9, we obtain a contradiction.

Thus S = T* has order 27. As Ft=(14)(23)(56)(9 10), we conclude from 1.1 that rt is an involution. There­ fore [r,t] = r _ 1t_1rt = rtrt = (rt) 2 = 1, i.e. r € C G (z,t).

We need only show S € Syl 2 (CG (z,t)). But if S • is not

Sylow in CG (z,t) , then a Sylow 2-subgroup of CG (z,t) has order 2 8 and so is in fact Sylow in LyS . This of course implies the center of a Sylow 2-subgroup of LyS is of order at least 4- ||, contradicting 1.3 .

The proof of step 2 is now complete.

Step 3. All involutions of S--T are conjugate in S .

Proof of step 3. We first show |C=(r)| = 4 and

|Cg(r)| = 8 . As (ua)r = urar =au= (56)(78) = ua and

(wb)r = wrbr = bw = (1423)(5768) = wbt , we see at once that (ua)r = ua and (wb)r = wb whence ua,wb€C^(r) .

We show these elements are distinct. Indeed ua = wb implies ua = wb or wbt. Each of the elements ua, wb, wbt has been previously calculated, and it is easy to see that ua f wb , ua ^ wbt . Thus ua and wb are dis­ tinct involutions in C^(r) whence |C^(r)|St4. We now calculate: 26

(13)(24 )( 57)(68) "r u= (13 ) ( 24 ) (78) (9 10) =(1 3 )(2 4 )(56)(9 10)

rw=(13)(24)(78)(9 I0 )(l5)(26)(37)(l*8) =(57)(68)(34)(9 10)

I now claim r, ru , rw are distinct elements of S .

Indeed r = r11 implies r = ru or rut (neither of which holds as rut = (14)(23)(78)(9 10) ), r = rw im­ plies r = rw or rwt (neither of which holds as rwt =

(12)(58)(67)(9 10) ), and ru = rw implies ru = rw or rwt (neither of which holds from the calculations above).

Thus r, ru , rw are three distinct elements, each of which is T-conjugate to r . Let J2 denote the set of

T-conjugates of r . As |ft| must divide the order of the 2-group T , we see that |S2|^4. Thus by the orbit- stabilizer theorem, we get |C^(r)| = |T|/|ft|^16/4 = 4.

As the reverse inclusion has already been established, we have |C^(r)| = 4. Finally as [S :T| = |S:T| = 2, we see that [Cg(r):C^(r)| ^ 2. But r € C^(r)^C^(r) whence

C^(r) has order 8 as claimed.

We are now ready to verify the claim of step 3. As

|S| = 32 and |C^(r)| = 8 , the orbit-stabilizer theorem establishes that r has four distinct S-conjugates. As r <£ T and T ^ S , all of these conjugates are involutions in S^T . But we shall now show S'-T contains pre­ cisely 4 involutions, whence the result follows. Let m, n be distinct involutions in S"-T . Then we can 27

express in and n in the form m = ** = ^2r

( t’1 ,t’2€ T ) as all elements of S--T lie in the unique non-trivial coset Tr of T in S . Now [^,r]

= t ^ 1r “ 1t^r = t^rt^r = (tTr ) 2 = T ( i=l,2 ). (Recall that T has exponent 2 whence t7 1 = t ^ .) Thus t ^ , 1 2 are distinct elements of C^(r). As C^(r) has order 4,

S --T must contain precisely 4 involutions and step 3 is proved.

Step 4. E ~ .

Proof of step 4. Recall that < E by assumption.

As E ^ C(j(x) we have |E| = 8 by lemma 1.5. But

E.^. Nq () by lemma 1.2, so that we can write E =

with y € N q ( )-^ Cq (x ) . If y€T = T/ , then y€T = T/ whence y fixes {9,10,11} point- wise. But then certainly y centralizes x = (9 10 11) whence y centralizes x ( xy = xz cannot occur as x has order 3 while xz has order 6 ). This contra­ dicts E ^ CG(x). Thus ygT, and as y may be chosen in S by virtue of Sylow's theorem, we have yeS-^T .

As y is an involution we see from step 3 that y is conjugate to r by an element s of S . Thus ys is in and Es = s = = as claimed. This completes the proof of step 4 and so also that of lemma 1 .6 . 28

Lemma 1.7 Nq() = E 2 3 \GL(3,2)

Proof. The lemma is proved in four steps.

Step 1. is self-centralizing in LyS .

Proof of step 1. Recall that t = (12)(34)(56)(78) , r = (13)(24)(78)(9 10) , tr = (14)(23)(56)(9 10) where as usual the bar denotes modulo . Now consider an arbitrary element g of C^(t,r) where C denotes the group CQ(z) . As g centralizes each of t , r , tr , we see that g stabilizes (as a set) each of the orbits and fixed point sets corresponding to r , t , and tr respectively. As a consequence we have

(i) C^(t,r) < Z{l,2,3,4,5,6,7,8}xZ{9,10,ll}

(ii) CG (t,r) ^ Z{l,2,3,4,7,8,9,10}xZ{5,6,ll}

(iii) C^(t,r) < Z{l,2,3,4,5,6,9,10}xZ{7,8,ll}

Let i € {1,2,3,4} , g€C^(t,r) . By (i) ig £ {9 ,10,11} , by (ii) ig £{5,6,11}, and by (iii) ig £{7,8,11} . Thus ig must lie in {1,2,3,4} and g permutes the set

{1,2,3,4} . For j€{5,6}, jg € {5,6,11} (by (ii)) , while jg ^ 11 (by (i) or (iii)). Thus g permutes the set {5,6} . The fact that each of {7,8} , {9,10} , and {1 1 } are g-invariant (so C^-(t, r)-invariant) follows easily from arguments similar to those above.

Thus we conclude that CG(t,r) is a subgroup (under the

A usual identification of CG(z) with A j-j ) of 29

I{1,2,3,4}xZ{5,6 }xi{7,8 }x£{9,10} . Therefore C^(t,r) is clearly a (2,3)-group. Now suppose g € C^(T,r) has order

3. By the above inclusion, we see that g€Z{l,2,3,4} whence g fixes precisely one letter of {1 ,2 ,3,4} , say i. But as g and T commute, i^ = (i®)'*' = i®^ = i^® =

(i^)® , whence i^ is fixed by g as well. As i is the unique such letter, we obviously have i^ = i . But this is a contradiction as t clearly moves every letter of {1,2,3,4} . Thus C^(t,r) can have no element of order 3 and is therefore a 2-group. We conclude from this that CG(z,t,r) is a 2-group as well. Now let P be a Sylow 2-subgroup of CG(z,t) which contains

CG(z,t,r) . By Sylow's theorem, there exists h € C G(z,t) with CG(z,t,rh) = CG(z,t,r)h ^ Ph = S , where S is the

Sylow 2-subgroup of CG (z,t) described in lemma 1.6. We now show r £ T (with T as described in step 1 of lemma 1.6). We do this by way of contradiction. Suppose then that r*1 6 T whence P* € T . As T has been shown to fix {9,10,11} pointwise, this is certainly the case for rh. But h€C^(t) whence h fixes {9,10,11} set­ wise, and we have (9 h )rl1 = 9*1 . But as hrh = hh_1rh = rh , we also have (9^)r = 9ril = 10 . Putting these two equations together yields the contradiction 9*1 = 1 0 *1.

Thus r*1 £ T as claimed. But now as seen in step 4 of 30

h. s lemma 1.6, there exists s€S such that =

. Thus CG(z,t,r) = CG(z,t,rk)s < Ss = S and it suffices to show is self-centralizing in S .

We do this presently.

We first recall that the elements ua,wb€T have images ua = (56)(78) and wb = (1324)(5867) in .

Thus t1ia = t , r113- = r , t™*3 = t , and r ^ = tr , whence each of ua, wb normalizes . It follows that ua and wb are distinct elements of Ng().

Clearly wb£ Cg(z,t,r) from above. We easily calculate uar = (13)(24)(56)(9 10) whence by 1.1 , (uar) 2 = 1 .

Also 1.1 implies ua has order 4 and as (ua) 2 = 1, we see that (ua) 2 = z . Thus rua = (ua)-1rua =

(ua)3rua = zuarua = zuaruarr = z(uar)2r = zr . This proves ua 0 Cg(z,t,r). We have produced two elements ua, wb in Ng() '-Cg(z,t,r) and as wb(ua ) -1 =

(1324)(5768) is not in Cg(z,t,r) , we see that ua and wb lie in distinct non-trivial cosets of Cg(z,t,r) in Ng() As S is a 2-group, this proves

|Ng():Cg(z,t,r)|^ 4 . But |S:Ng()| =

|s| . As ~ in S if and only if r ~h in S = S/ , we see from step 3 of lemma

1.6 that |®| = |cclg=(r)| = 4. Thus | S: Cg (z, t, r) |

= |S:Ng()|(Ng():Cg(z,t,r)| ^16. As |S| = 31

27, we immediately obtain |Cg(z,t,r)|^ 8 .* This of course proves = Cg(z,t,r) = CG(z,t,r) as claimed.

Step 2. |Nc()| ^ 2 6 *3.

Proof of step 2. Consider the elements k, h of Cq(z) with images k = (12)(34), h = (13)(24) in A11.

Clearly k^ = kr = k^1 - k , = hr = h , and therefore

= E 16 . This of course gives |C^(t,r)|^16.

Now consider the natural embedding N^()/C^(t,r)

*■ Aut() & Z3. We show the quotient contains elements of order 2 and 3. Let g = (123)(597)(6 10 8 ) and d = (1324)(57)(68)(9 10). Then tg = tr , V 1 = t , rg = t , and r^ = tr . Thus g, d are elements of

NG() "»-CG(t,r) of the appropriate orders. We con­ clude that |NG ()| = |N^-( ) :CG (t\r) | |CG (t,r) | St

6*16 = 253 and it follows that Nc() has order at least 263 as asserted.

Step 3. |NG() |^2 6 -3-7 .

Proof of step 3. We first show that the seven involutions in are pairwise conjugate in NG() . So let y # z be an arbitrary involution in . As

LyS has a unique class of involutions, there exists an element g in LyS such that y® = z . Thus z is in

® and g ^ ^(z) * ^s was shown earlier 32

(lemma 1 .1 ), CQ(z) is transitive on its non-central involutions. Thus there exists h € CG(z) such that zgh _ £ ^ (Note z^ f z since = z.) Observe that

^k contains the fours-group . (Indeed t = z®*1 and z = y^*1. ) Therefore, if we can show

®^^CQ(x) , we will have ®^ — in

Cg(z,t) by step 4 of lemma 1.6. But £k< C^(x) implies x € CG (z, t, r )Sk whence x^®*1) € C G(z,t,r) = .

, a contradiction. Thus there exists 5, 6 Cg(z,t) such that ^*1& = . But then ghJl € N^( ) and ygll£ = = z whence y~z in NG() as claimed. As certainly contains all Nq()- conjugates of z , we see that the NG()-class con­ taining z has precise order 7, and hence 7 is a divi­ sor of |Nq()| . Step 3 now follows from this and step 2 .

Step 4. Nr() is an extension of by f ^ GL(3,2) .

Proof of step 4. We have the standard embedding

NG()/CG (z,t,r) -- *■ Aut() = GL(3,2) . Thus the order of the quotient is bounded above by 2 3 »3*7 , the order of GL(3,2) . We now easily see that

|Nq()| ^ 2 3 *3*7 |CG (z,t,r)| = 2 6 *3*7 . Combined with step 3 this gives |Nq()| = 2 6 *3*7 and the 33

embedding is surjective. Thus NG() is an exten­ sion of CG(z,t,r) by GL(3,2) and as CG(z,t,r) =

by step 1 , the lemma follows.

Lemma 1.8. NG () = E 2 3 \GL(3,2) , (i.e. the exten­ sion in lemma 1.7 is non-split).

Proof. Suppose, by way of contradiction, that NG() splits over . Then every Sylow 2-subgroup of

NG () splits over ; in particular we can write P = *D where P 6 Syl 2 (NG ()) is cho­ sen so as to contain the element a which maps onto a =

(1342)(59)(6 10)(78) in Alx. (Note that tTa = r and ra = tf whence a normalizes and as a consequence a € NG().) Since D = P/)/

= GL(3,2) , D is a dihedral group of order 8 . (Indeed

D„O is the isomorphism type of a Sylow 2-subgroup of

GL(3,2).) We now express a as a = be where b 6 . and c € D . Thus c = ba and c is in the coset

a of in P . We claim that all elements of this coset have order 8 . Indeed a = za =

(1342)(59)(6 10)(78), ta = zta = (23)(5 10 6 9), ra = zra = (14)(596 10), and tra=ztra=(1243)(5 10)(69)(78), so the claim follows from 1.1 . But c1* = 1 as D has exponent 4. This gives the desired contradiction and we 34

conclude that the extension is non-split as claimed.

Theorem 1.9. Let E be an arbitrary elementary abelian

2-group in LyS . Then one of the following holds:

(i) N g (E) s A n

(ii) Ng (E) 5 E 2 3\GL(3,2)

(iii) Nq (E) is contained in some 3-local

subgroup of LyS .

Moreover, (i) occurs precisely when E = E 2 , (ii) occurs when E is LyS-conjugate to = Eg , and (iii) occurs in all remaining cases.

Proof. As LyS has a unique class of involutions, we may assume z €E . Thus if E = E2, then E = and

Nq(E) = Cq (z ) s A , i.e. (i) obtains. Suppose now that

|E|^4. As LyS has a unique class of fours-groups, we may now assume ^E whence E ^N^() ( t, x as described immediately preceding lemma 1.2 ). Now if E is contained in CQ(x) , then NQ(E) is contained in some

3-local subgroup of LyS ( lemma 1.3 ), i.e. (iii) ob­ tains. If on the other hand E ^ cq(x) » t*1® 11 by lemma 1.6 we have E~ . The isomorphism in (ii) now follows from lemmas 1.7 and 1.8 for this particular E .

The theorem is thereby proved. 35

2. 3-Local Analysis

In this section we shall show that the set

{Nq(),Nq(),Ng(V)} constitutes a complete set of maximal 3-local subgroups of LyS , where ux , u 2 ,

V are desribed in 1.5 . That is to say if H is an arbitrary 3-local subgroup of LyS , then H is LyS- conjugate to a subgroup of one of N(^(), Nq ( ),

Nq (V) . Furthermore, we shall establish the following isomorphisms:

Ng ( ) s Z 3\Aut(Mc)

ng (u 2> ) s 32 + tt • (SL(2,5)*Z 0 ) • Z£

Ng (V) = E 3 5 .(M1 1 xZ2 )

We begin with a series of lemmas.

Lemma 1.10. Let u 3 , S be as described in 1.5.

Then (au^ ) 3 = [a,[a,U3 ]]. where a€S and n = ±1 .

Proof. Let a be an arbitrary element of S and let n€{-l,l} . Then [a,[a,u“]] = a" 1 [a,u“ ] “ ^[a.u^] = a" 1u 3na " 1u^aaa" 1UgI1au^ = a ~ 1u 3na “ 1u 3 au“nau^ = a" 1a" 1u?au:I1uZnau^ , where the last equality follows from 3 3 o o the fact that a normalizes the abelian group V whence a'^“a and u"n commute. But a- 1a -1 = a (as S has O <3 exponent 3) and u^u ' 11 = u^ (as u^ has order 3) .

Thus from above [a,[a,u^]] = a“1a~^^au^u^aug = 36

au^au^au^ = (au^ ) 3 as claimed.

Lemma 1.11. With notation as in 1.5 , let v€SHV,

w€#, ne{-l,l} • Then (vwu“ ) 3 ^ 1. In partic­

ular, every element of S 3 not contained in either S or

V is of order 9 .

Proof. As vw € S , we may apply lemma 1.10 to obtain

(vwu^ ) 3 = [vw,[vw,u°]] . Thus if (vwu“ ) 3 = 1 , then

clearly vw and [vw,u“] commute. But [vw,u“] € [S3 ,V]

< V as V

[vw,u^] € Cy(vw) = ( 1.6 ). As v,Uj 6 V commute we have [vw,u°] = w " 1 v“ 1ujnvwu^ = w~ 1u 3 nwu° = [w,u3] .

Thus (vwu“ ) 3 = 1 implies [w,u^] € < u 1 ,u2> . We show

this cannot occur by direct computation of [w,u^] for

all elements w of # and n€{-l,l} . Indeed

from the defining relations for S 3 it is straightforward

to calculate the following: [w1 ,u3] = x x, [wj1 ,u3] =

u ^ x " 1 , [w2 ,u3] = x 2 , [w^.Uj] = u 2 1 x 2 1 , [wi1w 2 ,u3]

= ^ ^ 2 X ^ X 2 , [WjW^.Ug] =u^ 1 x 1 x“ 1 , [WjWg ,u3 ]=u1u 2 x 1 x 2 ,

[w^wj1,^] = UjU^x'^x ^ 1 . As w normalizes V we see

that w - 1u 3w and u 1^1 commute. Therefore we obtain

[WjU"1] = w _ 1u 3w u ‘ 1 = u “ 1w _ 1u 3w = [u3 ,w] = [w.Ug]"1 , and

the values for [WjU"1] follow easily from the above: 37

[w^U^1] = xj1, [w^.uj1] = U 2 X^ , [W2 .U3 1] = x" 1 ,

[w"1, ^ 1] = U 2 X 2 , [w‘ 1W 2 ,U3 1] = UjU^XjXg 1 ,

[WjW^1, ^ 1] = U 2 x ^ 1x 2 , [w1w 2 ,Ug1] = U j 1u 2 1x 1x 2 ,

[w^w^.uj1] =uj 1u 2 x 1x 2 .

In any case, [w,Ug] € is contradicted whence (vwu^ ) 3 ^ 1 as desired.

Now let y be an arbitrary element of S 3 v.(SuV) .

As S 3 * S US u 3 U Su" 1 and y£S , we can express y as y = su^ with s€S . But s 0 V. (Indeed s€V implies y€V, a contradiction.) Thus s lies in some non-trivial coset of S n V in S . As S = U { ( S n V ) w : w€}, we can express s as s = vw for v€SnV and w€#.

We therefore have y = vwu“ whence y 3 f 1 by the first part of the lemma. As S 3 has exponent 9 , y has order

9 as claimed.

Lemma 1.12. Let E be an arbitrary elementary abelian

3-group in S 3 . Then either E < V or E < S .

Proof. Suppose E ^ V, E ^ S and choose x€E'— V, y € E ^ S . By lemma 1.11, x, y, xy (each of order 3) must be contained in SuV . As x^V we have x€S, and as y £ S we have y € V . Now xy is contained in either S or V . But xy€ S implies y = x- 1 (xy) € S , a contra­ diction, while xy€V implies x = (xy)y- 1 e V , again a 38

contradiction. Thus E S , V cannot simultaneously occur, and the lemma is proved.

Lemma 1.13. Let E be a 31-pure elementary abelian

3-group in LyS , i.e. every non-identity element of E is of LyS-type 3j . Then E = E 3 and Nq(E) is iso­ morphic to E 3\Aut(Mc) .

Proof. We may certainly assume ux 6 E whence E ^CqCUj)

(■ Uj as described in 1.5 ). Moreover, as S 3 is a

Sylow 3-subgroup of C^Cuj) , we may further assume that

E ^ S 3 . Thus to prove the lemma, it suffices to show Sg contains no 31-pure subgroup isomorphic to Eg .

By 1.7 , S 3 contains precisely 22 3 1-elements,

A all of which lie in V . By the structure of A1;l , and by conjugating inside S 3 , we are able to produce an ex­ plicit list of these elements: Uj, UjU^1, Uguj1, UgU^Xj,

UjXj, u 2 u " 1x 2 , u 3x 2 , u - 1u 3 1x 1 x 2 , u 1u 2U 3 1x][1x 2 , u 1u 2 1u 3x 1x 2 ,

UjUj^u^XjX^ 1 , and their respective inverses. It is now apparent from our list that given any 3^element y of

S 3 '^, yux must be of type 32. Thus E = E 3 as claimed. By lemma 1.4, NQ()/ = Aut(Mc) where x in Cq(z) has image x = (9 10 11) in A X1 . We see from 1.10 that x is a 31-element of LyS whence

Nq(E) is an extension of E = E 3 by Aut(Mc) . As 39

Cq(E) = Me is a non-split extension of E by Me , we conclude that Nq(E) 3 E 3\Aut(Mc) , i.e. the extension is non-split.

Lemma 1.14. N^() < Nq() .

Proof. From the defining relations for S 3 ( 1.5 ) it is easily seen that = Z(S) , where we recall

S = . From 1.8 it follows that

S = 0 3(Cq(u2)) whence S (so also Z(S) ) is a charac­ teristic subgroup of Cq(u2) . Thus we have that Z(S) is normal in Nq() and the result follows.

General hypotheses for lemmas 1.15-1.23. E shall denote an arbitrary elementary abelian subgroup of S 3 which contains u 2 and has order at least 9 .

Lemma 1.15. Assume n l is contained in E and further­ more that E is a subgroup of S . Then N^(E) is a subgroup of Nq() .

Proof. Consulting the list of the 22 elements of type

3 X in S 3 (lemma 1.13), we see that (ux,uj1 .UjU^1 ,uj*u2} is the complete set of 31-elements contained in E .

Thus for n€NG(E) , un and (u1u 21)n are again 3J- elements of E and so are contained in this set. It 40

therefore follows that n = n = whence n € NG() as desired.

Lemma 1.16. NQ () & 3 2 + tf • (SL(2,5)*Z8 )-Z2 .

Proof. We accomplish the proof of the lemma in seven steps.

Step 1. S) .

Proof of step 1. We first show S = 0 3 (CqCUj,u2)) .

From 1.9 , we see that Z(S3) = whence

S 3 < CQ (u1 ,u2 ) . As C (,( u 1 ,u2) < CQ(u2) and S) . As is normal in CG (u2) it follows that CG(u2) < NG() and Cg(u2), a contradiction. Thus S = 0 3 (CG (ux,u2 )) as desired. We now simply observe that S is characteristic in CG(ux,u2) whence normal in NG () as asserted.

Step 2. |NG()/S| = 2 6 *3*5 .

Proof of step 2. Let X = NG(), a, b € CQ(z) with a = (23)(78) , b = (56)(78). Then clearly a , b € X and the members of {u2 ,uxu 2 >uYlu21} are -conjugate. 41

As CgCUg.) < X , we have |X| = |X:Cx(u2)||CG(u2 ) | =

|u^||Cq(u2)| = 2 6 *37 *5, and the proof of step 2 follows.

Step 3. Let W be a complement to S in Cq (u 2) chosen so as to satisfy z€W < C^z) • Then Z(Wx) = where

denotes the commutator subgroup [W,W] of W .

Proof of step 3. That W can be chosen as suggested above is noted in 1.12 . Now W = Cq (u 2)/S = SU±(2,5) whence S [SU±(2,5),SU±(2,5)] = SL(2,5) . As z € Z(W) and Z(SL(2,5)) & Z 2 , the proof of step 3 is now complete.

Step 4. Wx centralizes .

Proof of step 4. Clearly Wx normalizes U 2 > * denote by 0:W -- ^■Aut() the action via conjugation of W on . As W/Ker© is isomorphic to a sub­ group of Aut() = GL(2,3) , W/Ker© is solvable.

Thus there exists a positive integer n such that

(W/Ker©)n = 1 whence Wn ^ Ker© . (Here (W/Ker©)n and

Wn denote respectively the n-th commutators of the

groups W/Ker© and W .) But as Wx =SL(2,5) is per­

fect, we see at once that Wn = W ^ “ 1 = Wj . Thus W x is

contained in Ker© and step 4 is proved.

Step 5. Let bar denote modulo . Then under the

identification of C&(z) with Axx prescribed in the 42

previous section, Wx = A{7,8,9,10,11> , the full alter­ nating group on the letters {7,8,9,10,11} .

Proof of step 5. From step 4, W, < C.(<(123),(456)>) = A i i <(123)>x<(456)>xA{7,8,9,10,11} • (We have used here the fact that = <(123)>x<(456)> , since =

(by 1.11) .) Now Wj = W x/ = W 1/Z(W1) =

SL(2,5)/Z(SL(2,5)) = PSL(2,5) S A 5 . In particular Wj is simple. As WxnA{7,8 ,9,10,11} is normal in , we must have either WxnA{7,8 ,9,1 0 ,1 1 } = 1 or .

But Wj/WjH A{7,8,9,10,11} S WjA{7,8 ,9,10,11}/A{7,8 ,9,10,11}

< C.(<(123),(456)>)/A{7,8,9,10,ll}= <(123)>x<(456)> S E9, fli l whence W xn A{7,8 ,9,10,11} = . Therefore W x is con­ tained in A{7,8,9,10,11} and the result follows as Wj is isomorphic to A 5 .

Step 6 . Let L = where a€CG(z) has image a = (1425)(36) in A X1 . Then L s SL(2,5)*Z8 .

Proof of step 6 . It follows from step 5 that a central­ izes Wj ; thus L = W x and

W, normalizes = ZD (the isomorphism following from 1 8

1.1 ), we clearly have c^ 1 C {a,a“ 1 ,a 3 ,a“ 3}. Therefore

|W1 :C^(o)| = |a^1 |^4. As SL(2,5) has no subgroup of in­ dex 2 or 4 , we conclude that Wx = C^(a) whence L is a central product of with . As a *1 = 1. we have z = o'* € n Wj , which shows the product is not 43

direct. The proof of step 6 is now complete.

Step 7. Let K = where y eCj,(z) has image y = (14)(25)(36)(10 11) in A 12 • Then K is isomor­ phic to (SL(2,5)*Zg)*Z2 and NG () = S»K .

Proof of step 7. We note first that each of y, ~a nor­ malizes = <(123)>x<(456)> whence y,a € N^().

As L is clearly a subgroup of Nq(), we have that

SK is contained in NG (). Now |SK||SnK| = |s||K| so we have |SK|3 | S n K | 3 « |S|3 |K| 3 = 3 7 . As | SK | 3 = 3 7 and SnK is a 3-group , we have SOK = 1. Therefore

2 6 • 3• 5 = | SK/S | = |K/SnK| = |Kj > |L| = 2 s -3-5 whence

SK/S = Ng()/S follows from step 2. This proves

Nq () = S*K , and as |K| = 2 6 *3*5 , it easily fol­ lows that K = L* = (SL( 2,5)*Zg) *Z2 . As S = 3 2 + lf , the lemma follows at once.

Lemma 1.17. Let S denote the Frattini quotient of S

C i.e. S = S/), and let L be as described in step 6 of lemma 1.16. Then v^ = S# for v € .

Proof. As v^ is certainly contained in , it suffices to show |vL | = 80 . Observe ( 1.12 ) that z and all elements of order 5 in W act fixed point freely on S .

As L = W 1 * (so that and commute element-wise) we see that all elements of L not in Wx must have even 44

order. Thus all elements of order 5 in L are in fact-

in Wx (so in W ) and act fixed point freely on S .

This of course proves 5 does not divide the order of

C^(v) . We now show 4 also fails to divide this order.

Let T be a Sylow 2-subgroup of CL(v) . Then TDWX

is a 2-group of Wj = SL(2,5) whence either TnWx = 1 or z CTOWj ( SL(2,5) has a unique involution ) . But as z acts fixed point freely on S we have z ^ C^(v) whence TnWj s 1 . We thereby obtain the following:

T = T/T nWj =• TW^Wj < L/Wj = / = Z^ , i.e. T is isomorphic to a subgroup of Z^ . Let t € T . As T is contained in W 1 we can express t as t = wa 1 for some w€Wj and some integer i . As t 4 = 1 and [w,a]

= 1, we have w 1* = a-1** whence w u € Wj n = . Thus w 8 = 1 and w is a 2-element of Wj = SL(2,5) . But

SL(2,5) has Sylow 2-subgroups isomorphic to Q 8 (so no elements of order 8 ) and it follows that = w 4 =

1 . As a has order 8 , this implies i is even whence a2*- € . We therefore obtain t 2 = (wa^ ) 2 = w 2 a2* € W .

But as was shown earlier, T nW x = 1 whence t 2 = 1 .

As t was arbitrarily chosen in T , which from above embeds in Zh , we have proved |T|^ 2 , and 4 is not a divisor of |CL(v)| . But then by virtue of the equa­ tion |vL | = |L:Cl(v)| , we see at once that 2^*5 = 80 must divide |vL| whence v^1 = asclaimed. 45

Lemma 1.18. Suppose E < S and u ^ E . Then there exists t €L such that E^ < VnS .

Proof. Choose x€E'^. . By assumption x€S whence x is a non-identity element of the Frattini quotient S .

(Note as ux^E , En$(S) = .) By lemma 1.17 there exists t € L such that x* = Xj , with x x as described in 1.5 . Therefore x^ lies in the coset x 1$(S) of

$(S) = in S whence x^ is an element of E ^ n V .

As t normalizes S , our assumption E

E^ < S . Now suppose there exists an element y of E^ not in V . As y€S— V we immediately obtain Cy(y) =

from 1.6 . As Etn v ^ Cy-(y) we see from above that x ^ e CyCy) = , a contradiction. This proves

E^ < V and, as we have already shown E^< S, the lemma follows.

Lemma 1.19. With E , t as in lemma 1.18, V n S is normal in Cq (E1:) .

Proof. As VnS is abelian, it follows from lemma 1.18 that VnS is a subgroup of C^E^) . Recalling our general assumption that u2 € E , we have Cq (E) < C G (u2 )

< NG () . (The latter inclusion follows as is normal in Cq(u2) .) Thus as t € Ng() , we see that CG (Et ) is a subgroup of NG () and conse­ 46

quently CqCE1') normalizes each of and S .

Thus u°, u 2 € < u 1 ,u2>< VnS and x^, x° € S for all c in CqCE^) . As V n S = , the lemma will follow once we show x^, x 2 € V . Now let x be the element of E for which x^ = Xj (lemma 1.18). As x^ is in E* , we certainly have xtc = xt for all c in

Cg(E ) . But x = x^ implies xx = x u with ue whence x^ = xtcuc = x^uc € V . ( € V was shown in the previous lemma. ) Now if x2 g V , then Cy(x|) = from 1.6 . In particular x^e whence xx is contained in c_1= < u x,u2> , a contradiction.

Therefore x 2 eV and the lemma follows.

Lemma 1.20. V is the unique subgroup of S 3 which is isomorphic to E g5 . Therefore V<>K if and only if

V is characteristic in K for any subgroup K of LyS containing V .

Proof. Let W < S 3 be isomorphic to E g5 and distinct from V . Choosing w € W ^ V , we have ( by 1.6 )

Cy(W) = . In particular, W O V < .

But |W/WHV| = |WV/V| ^ IS 3/VI = 9 whence

[W | = |w/w nv I |W n v | ^ 81 , a contradiction. This proves the first statement of the lemma. The second statement follows easily from the first along with Sylow's theorem: 47

Indeed let 9 6 Aut(K) and let T x, T 2 be Sylow 3-subgroups of K which contain V , V 9 respectively. Then there exists k€K such that = T 2 whence .V^, V 0 are two subgroups of T isomorphic to E 35 • We conclude at once that Vk = Ve whence V<»K implies V is characteristic in K. The converse statement is obvious.

Lemma 1.21. Let E and t be as in lemmas 1.18, 1.19.

Then at least one of the two groups V , V n S is characteristic in Cq (E^) .

Proof. Suppose V n S is not characteristic in CJ^E'*') .

Then we can choose 9 € Aut(C^CE^)) such that (VnS )0 f

V n S . As V n S is normal in Cq(E^) by lemma 1.19, it follows that (V nS)9

V ^ T . But then T 0 is Sylow in (^(E^) , so there exists an element c of CG(E^) with T = T0C .

We therefore have V , V0C , subgroups of T , each isomorphic to E 35 , whence by lemma 1.20 V = V0C .

But as ( V n S )9 is normal in CG(Et) , we clearly have

(VnS)0c - (VnS )9 whence (VnS)e < V9c = V . Thus

( V n S ) ( V n S )0 < V ; in fact (VnS)(VnS)e = V as 48

0 VnS , (V nS) are distinct subgroups of V isomorphic to Egi* . This proves V is normal in Cq CE*) whence characteristic in CG (Et ) by lemma 1.20. The proof is now complete.

Lemma 1.22. With notation as in the previous lemma,

NqCE^) is contained in either NG(V) or Nq() .

Proof. By lemma 1.21, V or V n S is characteristic in

CG(Et) so normal in Nq CE^) . Thus either N(,(Et) < n q (v ) or Nq CE^) < NQ(V nS) . To prove the lemma, it therefore suffices to show NQ(VnS) is a subgroup of NG() .

We do this presently.

Checking our earlier-derived list of 3x-elements in S 3 (see proof of lemma 1.13), it is immediate that the Sj^-elements of V n S are precisely u x,u^ 1 ,u xu 2 1 , and u ^ u 2 . The elements of Ng(VnS) must clearly permute these four elements among themselves. We thereby obtain n = n = for all n in

NG(VnS) . Thus n € NG'( ) and the lemma follows.

Lemma 1.23. Suppose E ^ S . Then NG (E) < N q (V) .

Proof. First we recall our general assumption that u2€ E whence CQ(E) < C G(u2) < • Therefore u^,u|,x^,x2€ S for all c€Cg(E) . We now show u 3 € S 3 from which 49

V° < S 3 will follow. Choose y€E ^ S . Then

S 3 = S u S y u S y -1 whence u 3 can he expressed in the form u 3 =syn for some s€S and n€{-l,l} . Clearly yc = y for all c € Cq (E) . Therefore, as sc e S , u° = sc (yn )c

= scyn 6 S 3 as desired. It is now immediate from lemma

1.20 that Vc = V , i.e. V is normalized by C^(E) . As

E ^ S by assumption, lemma 1.12 implies E < V , whence

V is a subgroup of CG(E) . We therefore have V o Cg (E) .

The lemma now follows easily from lemma 1.20.

Theorem 1.24. Let E be an arbitrary elementary abelian

3-group of LyS . Then NQ(E) is conjugate to a subgroup of (at least) one of the following 3-local subgroups of

LyS : Ng () , NG () , nq(V) . Furthermore, if

H and K are chosen to be distinct members of Tl =

{NG ( ),NQ (),NQ(V)} , then H is not LyS-conju- gate to a subgroup of K . (Thus n is a complete set of representatives of the maximal 3-local subgroups of

LyS.) Finally, the following isomorphisms obtain:

(i) Ng () = Z 3\ Aut(Me )

(ii) Nq () == 3 2+ 1W S L ( 2 , 5 ) * Z 8 )*Z2

(iii) Nq (V) s e 3s•(m11xz2 ) . 50

Proof. The isomorphisms given in (i), (ii), and (iii) fol­ low respectively from lemmas 1.13, 1.16, and [13j?p.545-6 ].

Let E be as in the theorem statement. If E is 3 j-pure then E = E 3 and NG (E) — NG () by lemma 1.13. Thus we assume E contains 32-elements . By conjugating if necessary, we may assume u2€ E and E < S 3 . Now E =

implies NG(E) < N Q() (lemma 1.14). Other­ wise E has order at least 9 and we divide the balance of the proof into two cases.

Case one: E ^ S . If ux € E , then lemma 1.15 implies NQ(E) < NQ () . If u1 $ E , then either

NG (E) :£ N g () or NQ (E) < NG (V > by lemma 1.22.

Case two: E ^ S . Here NQ (E) < n q (v ) follows from lemma 1.23.

It remains to show there exist no containment rela­ tions among conjugates of distinct members of n

Lagrange's theorem disposes of all but two possibilities, namely NG(

) conjugate to a subgroup of NQ(), and Nq(V) conjugate to a subgroup of NQ( ) . We rule these out simultaneously by the following argument.

Suppose A ^ Ng() for some g € LyS where

A € {Nq ( ) ,Ng(V)}. Then clearly S 3 < NQ () and as |Nq(< u | > ) :CG (u®)| = 2, we have S 3 < CG (u®) .

But this implies u^ € Z(S3) = . As |(u^)^G(V)| 51

= 22 (since ^q(V) controls fusion of its 3x-elements

[13,p.546]) and |(u^)^G()| = 4 (proof of lemma

1.16, step 2), we obtain the contradiction 4 ^ |(u*j>)^|

^ |(u®)^G()| = 2. The proof of the theorem is now complete. 52

3. 5-Local Analysis

In this section we show that {NqCcE^) , NG ()}

is a complete set of representatives of the maximal

5-local subgroups of LyS . ( flsg 2 ,g3 are described in

1.13 .) This is to say if E denotes an arbitrary 5-

group of LyS , then NQ(E) is conjugate to a subgroup

of either NG() or NG() , and neither of

N(,() , Nq ( S 3 >) is conjugate to a subgroup of

the other.

We begin with the following lemma which determines

all possible orders for a 5l-pure elementary abelian sub­

group of LyS .

Lemma 1.25. Let E be a 5 l-pure elementary abelian

subgroup of LyS . Then |e| ^ 5 3 . Furthermore |E| =

5 2 implies NG (E) is LyS-conjugate to a subgroup of

NG ( ) .

Proof. By our assumption that E is Sj-pure , we may

certainly assume f x € E whence E ^(^(fj) . As S 5

is Sylow in CgCfj^) , we may further assume E < Sg

(where S 5 is the Sylow 5-subgroup of LyS described in

1.13). Without loss of generality, we may certainly

assume that |e| St 52 . By 1.18 , every 5 1 -pure E 52

in Sc which contains f is itself contained in R = ^ 1 53

g 3 > gi+> g5> • Thus for any x € E we have x€ ^ R , i.e. E ^ R . But now by 1.18 (and the fact that g 2 is a Sj-element) we may assume < E . Thus

E < CR (g2) = , a non-abelian group of order

51*, which proves |e| ^ 5 3 . Now suppose |E| = 5 2 whence E = . Then from 1.15 we see that cG (f 1 ,g2 ) = ! ,S2 > S 3 * Stj, f 2> » and direct computation using the defining relations for Sg given in 1.13 shows [CG(fj,g2 ),CQ(fx,g2)] = . But then

is a characteristic subgroup of cq(e) =

CG (f 1 ,g2) whence normal in NG(E) , and NG(E) is con­ tained in NG () as claimed. This concludes the proof of the lemma.

Lemma 1.26. There exists at most one class of 5x-pure

E g 3 1s in LyS .

Proof. We consider all elementary abelian 5-groups of

R of order 5 3 which contain . Clearly each such group E is contained in cR(g2) = ’ which has center <^i»S2> and exponent 5 , whence E may be expressed as E = with x€nf .

It is easy to show that there are precisely six such groups, and as [f x,f 2] = [g2 ,f2] = 1 ’ tliey are permuted under the action of conjugation by . But g^z•f = 54

g 4g 3 whence f 2 fails to normalize gif> • Thus it follows that | | = | :N( ) |

=■ 5, and the set of six E 5 3 's in R which contain

decomposes into two -orbits having lengths

5 and 1 and representatives and respectively. (Note from 1.13 that [f2 ,g3] = g 2 whence

is -invariant.) We show below that not both and are Sj-pure . As every

51-pure elementary abelian subgroup of LyS may be as­ sumed to contain and be contained in CR(g2) =

i > S2 > S3 ,glt> (see proof of lemma 1.25), we see that if a

5 1-pure E 53 in fact exists, it must be conjugate to one of or . Therefore, we will have shown that there exists at most one such class.

Suppose, by way of contradiction, that both the groups and g2’g3 > are 5 i“Pure • It then follows that every non-identity element of CR(g2) ~

x ,g2 >g3 ,gtt> is of LyS-type 5X . From Table 1 ,

CG(fj) = R*Q with Q = SL(2,9) , and by 1.19 , if we choose v of type 3j in Q , then C^(v) is extra­ special of order 5 3 and exponent 5 . Now

lcR(v):CR(v)n C R(g2)| = |CR(v)CR(g2 ):CR(g2)| ^|R:CR(g2)|

= 5, whence |CR(v)nCR(g2)| ^ 5 2 . By the above, all elements of (CR(v)nCR(g2))# are of LyS-type 5j. But 55

from 1.19 all elements of CR(v) — are o:f LyS- type 5 2 • As these two sets obviously have non-empty intersection, a contradiction is reached. We may now conclude that at most one of the groups ,

is Sj-pure and the lemma follows from our discussion of the previous paragraph.

Lemma 1.27. NQ (R) = NQ () = 5 1 + 4 *SL(2,9)‘Z^ .

Proof. As = Z(R) , we surely have N^(R) contained in N<-,() . The reverse inclusion follows at once from the fact that R o N^() . More explicitly, N(^()

- R*QX where Qj is generated by Q = SL(2,9) and an element u of Cq(z) mapping onto (2354)(6798) modulo

( 1.17 ) . (Note that as in section 1 we are identi­ fying Cq(z) with A X1 .) But from 1.17 we see that

Q ^(^(fjjZ) was originally chosen so as to satisfy Q =

A{6 ,7,8 ,9,10,11} . This immediately shows Qu = Q whence

Qu = Q . Moreover Q n = 1 , as otherwise we would have u z 6 Q ( u has order 4 by 1.1 ), whence u =

(25)(34)(69)(78) would be an element of Q , a contradic­ tion. Thus = Q* = SL(2,9)*Zlf and the lemma follows as R = 5 1+t* . 56

Remark. The following ten lemmas will be committed to proving is a 5 1-pure E s3 , a far from trivial task. In the discussion which follows, we lay much of the groundwork which will be critical to our success.

As R = is extraspecial, commuta­ tion in R induces a non-degenerate skew-symmetric bi­ linear form A on the Frattini quotient R , regarded as a 4-dimensional vector space over GF(5) .

A: R x R ---► GF(5) is defined by A(u,v) = i ( u,veR ) where i is the unique element of GF(5) which satisfies fi = [u,v] . Every element n of NQ(R) clearly induces a linear action 0(n) on R . As n normalizes , we have f^ = fj for some integer l , 0 ^ a ^ 4. Thus for any two elements u,v€R , we clearly have A(u,v) = i where f* = [u,v] and A(0(n)u,0(n)v) = k where f^ =

[un ,vn ] . But f* = [u,v]n = (fj-)n = (f? ) 1 = (f* ) 1 = f^ 1 whence A(0(n)u,0 (n)v) = £A(u,v) . This proves 0(n) is an element of the generalized symplectic group GSp(R) , which consists of all linear operators on R which pre­ serve A up to scalar multiplication. As a consequence, we obtain a homomorphism 0 of Nq(R) into GSp(R) .

Now with an appropriate choice of basis for R , we can represent A by a skew-diagonal matrix. Q =

{gg ,g3 ,g^gs) does the trick nicely. With respect to 57

0 , A is represented by

* « l

y = “ 1 i

- 1

» *

Along with this choice of basis, there corresponds a canonical isomorphism of GSp(R) onto GSp(4,5) , where

GSp(4,5) is defined to be the group of all matrices of GL(4,5) satisfying the property a^ya = aly for some a in GF(5) . Identifying GSp(R) with GSp(4,5), we thereby obtain an embedding of Ng(R)/Ker0 into

GSp(4,5) . We claim Ker0 = R . Clearly R is contained in CNg (R )(R) = Ker0 . It is easy to see that any 5; - element which centralizes R must in fact centralize R .

Indeed if x € (R) has order m with (5,m) = 1 , then for any u € R we have ux = u whence ux = uf a for III m i some integer i , 0 ^ i ^ 4 . But then u = ux = uf whence 5 divides mi . As (5,m) = 1 we have that 5 divides i whence i = 0 . Thus ux = u for all u €R and x 6 Cg(R) as claimed. Now Cq(R) < C^Cfj.gg) =

( 1-15 ) , i.e. Cq (R) is a 5-group.

Therefore, from the above, ^Nq ( R ) ^ ^ •’'•s a 5-group as well. As R is maximal abelian in S 5 , we have estab­ lished the reverse inclusion. Thus R = Ker0 and 58

Nq (R)/R can be viewed as a subgroup of GSp(4,5) .

Notation conventions. By virtue of the preceding remark, we adopt the following conventions for use in the balance of this section.

A (i) N shall denote the quotient N/R where

N = N q (R) .

(ii) X shall denote the group GSp(4,5) . Thus

X = {a€GL(4,5): a^ya = aly for some a€GF(5)}.

Arbitrary elements of X shall be denoted

by lower case Greek letters: a, 8 , S, ...

(iii) We shall regard N as a subgroup of X .

Under this identification, n will be the

matrix representing the action of n on R

with respect to the basis Q =

(iv) GL(4,5) shall be abbreviated by simply GL .

Moreover, for any subgroup Y of GL , we

shall denote by D(Y) and U(Y) the groups

of diagonal and upper triangular matrices of

Y , respectively.

(v) Elements of D(GL) shall be expressed in the

form diag(a,b,c,d) rather than the more 59

cumbersome

Lemma 1.28. Nx () ^ U(X) .

Proof. From the defining relations for S 5 ( 1.13 ) , it is easy to see that f 2 is given by

1 1 111

1

t 4

Let a be an arbitrary element of Nx() • As

R is a uniserial -module ( 1.14 ) , we see that

V x = , V 2 = , V 3 = represent the unique proper -invariant subspaces of R . As the action of f2 on R coincides with that of f2 on

R , we see that V1, V 2 , V 3 are unique -invariant subspaces of R of dimension 1, 2, 3 respectively.

Now f2a = af2J for some integer j . Therefore for every i , 1 ^ i ^ 3 , we have f 2a(Vi ) = af2J (V^) = 60

0 (V-j_) , whence by uniqueness, a(V^) = . It is now immediate that a € U(X) .

Lemma 1.29. Let 6 be the matrix of GL given by

1 1

2 3 1

4 4

3

Then 6 is an element of order 4 in Nx() .

Proof. It is routine to verify that 5 has order 4 and also that 6^y6 = 3Iy whence 6 € X . From the easy calculation

3 4 1

2 2 3 5f: V 4 2

3

we see that 6 normalizes <^2> as asser‘t;e

Lemma 1.30, <5e = diag(l, 2 ,4 , 3) for some e € U(X) , with 6 as in lemma 1.29.

Proof. It is immediate that D(X) is a Sylow 2-subgroup of U(X) . (Indeed |U(X):D(X)| ^ |U(GL):D(GL)| = 56, 61

while D(X) is a 2-group as D(X) < D(GL) = 2^xZ^xZ^xZ^ . )

Thus, as 6 is a 2-element of U(X) (lemmas 1.28,1.29), there clearly exists an element e in U(X) with S£ in

D(X) . With Vlf V 2 , V 3 as in the proof of lemma 1.28, we therefore have 6e(vi) = Vi for all i . Let 6e ! denote the restriction of <5e to ( 1 ^ i ^ 3 ) .

Then 5e |V! = (e" 1 | Vx ) (

6G is 2 . Continuing in this manner gives the desired result. 62

Lemma 1.31. The following conditions are satisfied:

(i) Z(X) is a Sylow 2-subgroup of Cx(f2)

(ii) Z(X)<6 > is a Sylow 2-subgroup of Nx()

(iii) |N-( ) ( 2 = |Nx ( ) [ 2 = 2** .

Proof. Certainly Z(X) is contained in Cx(f2) .

A Let now o be an arbitrary 2-element of Cx(f2). By lemma 1.28, a is upper triangular, say • * a b c d

e f g a = h i f

j * ✓S A vith a,b, . . . , j €GF(5) . As af 2 = f2cr we calculate: • a a+b b+c b+c+d a b+e c+f d+g

e e+f e+f+g e f+h g+i+j

h h+i h i+j

j 3 « . 4 whence a = e = h = j easily follows.

But as D(X) is Sylow in U(X) , there exists an element x of U(X) such that oT is diagonal. As a and aT have the same eigenvalues, we therefore have aT = al whence a = al and a € Z(GL). As (al)ty(al) = a 2Iy, a € X as well so u€XnZ(GL))5Z(X), proving (i). 63

It is trivial to check that 62 ^Z(X) ( 6 as de­ scribed in lemma 1.29). Thus Z(X)n< 6> = 1 and

Z(X)<6> = Z(X) x & x Zk . As |Nx():Cx (f2)| = 4, it follows that |P:Z(X)| 4 for P € Syl2 (Nx ()) , whence (ii) follows. (Recall that 6 €Nx() by lemma 1.29.)

We now prove (iii). It has already been established that |Nx()|2 = 2 4 . Suppose 3 divides the order of

N^(). Then as jN^():C^(f2)| = 4 , we easily see that 3 divides |Cjj(f2)| as well, whence N contains an element of order 15 . But then as N = SL(2,9)*Zlt , every such element of N would by necessity be contained in a subgroup of N isomorphic to SL(2,9) . We show presently that SL(2,9) contains no elements of order 15 whence 3 cannot divide |Njj()| . Indeed let A and

B be elements of SL(2,9) of respective orders 3 and

5 with AB = BA , and let V denote the 2-dimensional vector space on which SL(2,9) acts naturally. As x 3-l=(x-l)3(mod 3), we see that A must have minimal polynomial (x-1)2 and Cy(A) must be a 1-dimensional subspace of V . But AB = BA clearly implies Cy(A) is

B-invariant. We now consider the cyclotomic polynomial

$(x) = x 4+ x 3+x 2+x+l . It is not hard to show that $(x) is irreducible over GF(3), whence (by elementary field theory) $(x) possesses no linear factors over GF(9) . 64

For this reason, the minimal polynomial for B must divide (x) and therefore be an irreducible quadratic over GF(9) . Thus B must act irreducibly on V , contradicting the existence of Cy(A) above. As dis­ cussed, we conclude that 3 does not divide |N^()| whence 32 divides the index n = |N:N^()| . But also n divides |N: Z(X) | = |N|/22 *5 = 2lf*32 , so that n = 9,18,36,72, or 144. As n=l(mod5) by Sylow's theorem, we conclude that n = 36 and it easily follows that |N^()| = 24 *5 . This clearly proves (iii), and the proof of the lemma is complete.

Lemma 1.32. N/Z(X) is isomorphic to Eg .

Proof. Let P be a Sylow 2-subgroup of N^() .

Then by lemma 1.31 (iii), P is Sylow in N^() as well. Thus, as Z(X) is a normal 2-subgroup of

Nx() , we have Z(X) < P. In particular, Z(X) is contained in N , whence Z(X)

gives N/<-I> = N/ = Q / = Q 1/ where Q x is isomorphic to SL(2,9)«Zlf (see lemma 1.27). As z in­ verts R , it also follows that CD(z) = whence it 1

CN (Z) = Crq(z) = CR(z)CQ (z) = *Q 1 . (Recall that

N = RQj with Q x < Cq (z ) .) We therefore obtain

Q x/ = CN (z)/ & CN (z)/ = Nc ()/ where C = Cq (z ) and bar denotes modulo * (We have used here the fact that Nq (R) = NQ() which was proved in lemma 1.27.) Combining the above results yields

N/<-I> = Nc()/ .

We now analyze the latter quotient, shifting our perspective to modulo . (We shall denote images modulo by double bars.) This gives the following.

Nc()/ = Ng()/ S Na (<(12345)>)/<(12345)>.

Now clearly (<(12345)> * A{6,7,8,9,10,11})•<(2354)(67)> normalizes <(12345)> and has order 2 6 *32 *52 . As

lAii:Na (<(12345)>)| = |cclA(<(12345)>)| = ( V ) - 3 ! , it follows that |N.(<(12345)>)| = 2 6 »32 *52 , whence the A i i above group is in fact the full normalizer of <(12345)> in A.. . Thus N .(<(12345)>)/<(12345)> is canonically 11 Ai i isomorphic to A{6,7,8,9,10,11}•<(2354)(67)> . 66

In what follows we shall abbreviate by E and A the symmetric group and alternating group on the set

{6,7,8,9,10,11} , respectively. We also let h denote the element (2354)(67) of A X1 , and we set Y = A* .

We now define Y — >-£ by the rule $(y) = a (67)^ where y € Y is uniquely expressible in the form y = ah1, a€A. We claim $ is a group epimorphism with kernel

. Indeed for ah 1 , xh~ € Y we have $(ahi)$(xhJ)

= a(67)-*-x(67) j while ^(ahixhj) = $(ahixh~ihi+j) = ahiTh~i(67)i+j = CT(67)iT(67)-i (67)i+J = cr(67)ix(67) J .

Thus $ is a group homomorphism. As Z=A*<(67)> , $ is clearly surjective with a(67)i = $(ahi) . Finally

Ker$ = {ah-”- € Y:a(67)i = $(ah-*-) = 1} = {ah* € Y.:a = 1, i even}

=

as claimed. We conclude that Y/

= Z = Eg .

A Consolidating our results, we have N/<-I> =

Nc()/ = Y with Y/

= Z.g . Now as

Z(Y)

/

< Z(Y/

) 3 Z(Z) = 1 , we have Z(Y) contained in

. As h 2 = (25) (34) , the reverse inclusion is obvious from the definition of Y , so we have equality: Z(Y) =

. Thus Y/Z(Y) = Zg . We now observe that Z(X)/<-I> < N/<-I>nZ(X/<-I>) < Z(N/<-I>)

= Z(Y) . As Z(X)/<-I> = Z 2 = Z(Y) , we have from above

Z(X)/<-I> = Z(N/<-I>) and we may now conclude that 67

N/Z(X) = (N/<-I>)/(Z(X)/<-I>) = (N/<-I>)/Z(N/<-I>)

= Y/Z(Y) = Z 6 , the desired result. The proof is now

complete.

Lemma 1.33. Let P be a Sylow 2-subgroup of N which

contains S , a Sylow 2-subgroup of N^j() . Then

S is normal in P .

Proof. From lemma 1.31 we see that a Sylow 2-

A A subgroup of N^() is Sylow in N^() as well,

so is conjugate to Z(X)x<6> = Z^xZ^ . Thus S/Z(X) is isomorphic to Zu. As N/Z(X) is isomorphic to Ze

(lemma 1.31), we see that P/Z(X) is isomorphic to

D a x z2 (the isomorphism type of a Sylow 2-subgroup of

Ze ). As every Zh in D 8 x Z 2 is clearly normal, we have S/Z(X) <3 P/Z(X) and the result follows. 68

Lemma 1.34. There exists a Sylow 2-subgroup P of N and an element t of U'(X) such that PT^ N^T( ).

Proof. Choose and fix S and P as in lemma 1.33. As

S is Sylow in N^.() (lemma 1.31 (iii)) there exists an element p of NY() such that Sp = .

( € Syl2 (Nx()) by lemma 1.31 (ii).) Thus

SPe = e = with e as in lemma 1.30.

As P < Nj^(S) , we have PT < N^T(ST) = N^T( ) for t = pe. Moreover t € U(X) , since p € U(X) by lemma

1.28 and e€U(X) by lemma 1.30. This concludes the proof of the lemma.

Lemma 1.35 Let a and 3 be the elements of GL given by • s 0 1 1

1 0 0 1 a = 3 = 0 1 f -1 0

1 0 1 ✓ With notation as in the previous lemma, PT K where

K is defined to be the subgroup of GL generated by

D(X), a , and 3 .

Proof. We accomplish the proof in several steps.

Step 1. D(X)

Proof of step 1. It is trivial to verify the following:

a_ 1diag(a,b,c,d)a = diag(b,a,d,c) ,

3- 1diag(a,b,c,d)3 = diag(a,c,b,d) ,

a 2 = I , 32 = diag(l

(a3)tf = -I

0 0 1 0

1 0 0 0

ot3 = ,(aS)2 = 0 0 0 1 -1

0 -1 0 0 -1

Thus we see that each of a , 3 normalizes D(X) whence

D(X) o K , and denoting images modulo D(X) by the symbol , we have a 2 = $ 2 = (a$)^ = 1 whence a , $ satisfy the defining relations for Dg . As (a$ ) 2 ? a ,

= has order at least 8 , so is isomorphic to Dg . Step 1 is thereby proved.

Step 2. CGL(

Proof of step 2. Let e^ (1 ^ i ^4) denote the i-th basis vector of Q and let A^ denote the eigenvalue for 6e with corresponding eigenspace If ? is an arbitrary element of Cr,T(5e) , then we have 6es(e^)

= ?6e(ei) = cAi(e±) = A^Ce^ , i.e. ^(e ± ) € , the eigenspace corresponding to A^ . Thus £ € D(GL) as asserted. 70

Step 3. (i) Cg l (D(GL)) = D(GL) = CQ L (D(X))

(ii) CX(D(GL)) = D(X) = CX (D(X)) .

Proof of step 3. As D(GL) is abelian we clearly have

D(GL) < Cql(D(GL)) < Cq l (D(X)) < CGL( 6 e) . By step 2

C g ^ ( 6£) < D(GL) , so each of the above inequalities is

in actuality an equality. This proves (i). (ii) is

proved upon taking intersections with X .

Step 4. Ng l (D(GL))/D(GL) == .

Proof of step 4. It is well known that NGL(D(GL)) is

precisely the subgroup M = D(GL)«P of GL where M

and P here represent the subgroups of all monomial and

permutation matrices of GL respectively. (A matrix is

monomial provided each of its rows and columns contains

precisely one non-zero entry; a permutation matrix is a

monomial matrix with all non-zero entries equal to 1 .)

As each permutation matrix induces a rearrangement of

basis vectors and visa versa , we see that P is natural­

ly isomorphic to E^ . The result follows.

Step 5. NX(D(X))/D(X) is isomorphic to a proper sub­

group of E^ .

Proof of step 5. As D(X) = CX

we see that NX(D(X))/D(X) is the automizer of D(X) in

X , which can be viewed as a subgroup of the automizer of 71

of D(X) in GL : NQL(D(X))/CQL(D(X)) . Thus it suf­ fices to show Nq l (D(X))/CgL(D(X)) satisfies the asser­ tion of step 5. But C^(D(X)) = D(GL) by step 3 (i) , whence NQL(D(X))/CQL(D(X)) = NQL(D(X))/D(GL) . We show

Nq ^(D(X)) is proper in NGL(D(GL)) whence the result will follow from step 4. Suppose then, by way of con­ tradiction, that Nq ^(D(X)) = Ng l (D(GL)) . Consider the matrix t given by

0 0 1

1 0 0

T 0 1 0

As t is a permutation matrix in GL , we certainly have t€Ngl(D(GL)) whence by assumption t€NqL(D(X)) . But then as Se = diag(l,2,4,3) is in D(X) , we must have

T_16eT € D(X) as well. We readily calculate t -16£t = diag(2,4,1,3). But now setting p = t _16£t gives

PtYP -1

-1

Thus p^yp cannot be expressed as a scalar multiple of

Y , a contradiction since p is in X . Step 5 now 72

follows as discussed.

Step 6 . K = NX(D(X)) .

Proof of step 6 . Prom step 1 it follows that K is con­ tained in Nqjj(D(X)) . It is routine to verify

at-ya = -y B^y B = Y whencea,B€X ; thus we obtain K < N X(D(X)) . But then

K/D(X) is a subgroup of NX(D(X))/D(X) with K/D(X) isomorphic to DQ by step 1, and NX(D(X))/D(X) iso­ morphic to a proper subgroup of by step 5. This clearly implies K/D(X) = NX(D(X))/D(X) and the result follows upon taking preimages.

Step 7. PT < K with P T as in lemma 1.34.

Proof of step 7. We first observe that D(X) ^ Cx(Z(X),6 e )

— CX (6 E > = X n C G L () and it follows that

NX () is a subgroup of NX(D(X)) = K . As

PT ^ N^x() by lemma 1.34 and Nx ^ X , the assertion follows. This completes the proof of lemma 1.35. 73

Lemma 1.36. With notation as in the previous lemma, a e P tD(X)/D(X) .

Proof. We first claim that PTnD(X) = .

Recall that x was originally chosen so as to satisfy

^ PT with PT € Syl 2 (NT) . Now Nx can be

expressed as S*T where S = SL(2,9) and T = Z^ .

Setting H = PxnD(X) , we see at once that H n S is an abelian subgroup of S of exponent 4 (as D(X) < D(GL) and D(GL) = Z 4 x zh x Zh x ). As Sylow 2-subgroups of

SL(2,9) are generalized quaternion, it follows that

|HnS| ^ 4 , whence |H| = |H:Hr»S||HnS| ^4|H:HnS|^

4 |ST:S| =16 . As is a subgroup of H of order 16, we get H = as claimed.

It now follows that PTD(X)/D(X) S P x/PxnD(X) =

Px/ has order 4 . By lemma 1.35, PxD(X)/D(X)

is a subgroup of K/D(X) = = D 8 . We now see from

the defining relations given in step 1 of lemma 1.35 that

PxD(X)/D(X) is equal to one of <<*&> , , or

<3,(ag)2> , the three subgroups of of order 4.

We show is the correct choice by the process

of elimination.

Suppose first that g € PxD(X)/D(X) . Then Px must contain an element of the form 74

0 b a = c 0

But then as PT normalizes we must clearly

have 6 ea € . An easy calculation reveals

6 ea = a _1diag(l,2,4,3)o = diag(l,4,2,3)

whence there exists a€GF(5) and b, 0 ^ b ^3, such

that diag( 1 ,4 , 2 ,3 ) = aldiag(lk,2k,4 k, 3k) ^ Bu ^. -then

u i.. al = 1 implies a = 1 , while 3 =3(mod 5) implies

b = 1 . This of course leads to the contradiction

diag(l,4,2,3) = diag(1,2,4,3) . We conclude that $

cannot lie in PTD(X)/D(X) .

Next suppose ag 6 PTD(X)/D(X) . Then PT must

contain an element of the form

0 0 a 0

b 0 0 0

P = 0 0 0 c

0 d 0 0

One easily computes 5ep = diag(2,3,1,4) which, as above,

must be an element of since p normalizes

. Thus as before, there exists a€GF(5) and b, 0^b^3, such that diag(2,3,1,4) = aldiag(lb , 2b ,4b , 3^) . 75

al^ = 2 implies a = 2 whence a2l3=3(mod 5) implies b = 2 . We thereby obtain the contradiction diag(2,3,l,4) = diag(2,3,2,3) and we conclude that o$ * P TD(X)/D(X) .

As we have shown neither of $ , a$ is contained in PTD(X)/D(X) , we see at once that this qoutient must equal . The lemma now follows.

Lemma 1.37. is 5^-pure , so represents the unique class of Sj-pure elementary abelian sub­ groups of LyS .

Proof. As R is extraspecial ( 1.14 ) , every non­ trivial coset of in R is completely fused in

R . It therefore suffices to show all non-identity

_ _ A elements of are fused in N where, as usual, bar denotes images modulo and circumflex de­ notes images modulo R . (Recall that g 2 is a 5i- element of LyS ; see for example 1.18 .)

We consider the six subgroups of order 5 in

. Clearly normalizes (®-g- see the defining relations for S 5 (1.13)), so permutes the six subgroups mentioned above. As normalizes , but not > it is immediate that the set of six E 5 's in decomposes into 76

two -orbits having respective sizes 1 and 5 and representatives and • It therefore suffices

/ v to show these two orbits are fused under the action of N

We accomplish this presently.

From lemma 1.36, we see that PT must contain an element of the form

0 a

b 0

0 c

d 0 and as PT :£ NT it follows that t£t _1 e N . Now as t is in U(X) (lemma 1.34), is both t- and

^-invariant, so t£t" 1-invariant. Therefore t^t -1 normalizes • Suppose is t?t" 1-invariant.

Then T?x“ 1( ) ® whence Ct' 1() = T _ 1() .

But x € U(X) implies T~ 1() = . This yields

^( ) = , an obvious contradiction to the shape of £ . We conclude that t£t -1 fails to normalize

. We have produced an element t£t _1 € N (so with preimage in LyS ) which normalizes but moves

. Thus the six subgroups of order 5 in are indeed fused under the action of LyS , and there­ fore all elements of # are conjugate. As earlier discussed, this proves all elements of i^f are 77

of LyS-type 5X , i.e. is 5x-pure . The

claim that represents the unique class of

5 -pure E s3's in LyS now follows from lemma 1.26.

The proof of the lemma is now complete.

Lemma 1.38. Let H = N G (E) where E = •

Then E = CQ(E) and H/E SSL(3,5) .

Proof. We first observe that E < CG(E) < C G(f1 ,g2 ) <

CR (g2) , where the last inclusion follows as in the proof of lemma 1.25. As E is maximal abelian in cr(S2 ) = S2 >S3 ,Sn> , we conclude that E = CQ(E) as claimed. Set B = NG () • Then, as seen in the proof of lemma 1.25, B < H . Now B is precisely the stabilizer in H of , one of the 31 E 52's

in E permuted under the action of H . Thus we have

|H:B| = |H | -£ 31 . As H/E = NG(E)/CG(E) is iso­ morphic to a subgroup of Aut(E) = GL(3,5) , we see that

|H| = |H:E||EI is a divisor of |GL(3,5)||E| = 273*5631.

But from 1.15 , |B| = 2 5 »3«56 whence [H:B| divides

22*31 . We show presently that |H:B| = 31 . Recall

from the previous lemma that the six E5's in

are fused into one -orbit . Thus the six

E 5 2 1s in E which contain <^!> are fused under the

action of L , where L is the full inverse image in 78

LyS of . Moreover, as each of f£ and t?t_1 normalizes , we see that L < H = Nq(E) .

We therefore have |^| ^ \^\ ^ 6 , and it follows from above that |H:B| = |H | = 31 . But this of course implies |H| = 25*3*56*31 whence |H/E|

= 25-3*53*31 . Thus H/E is isomorphic to a subgroup of index 4 in GL(3,5) . As SL(3,5) is the unique such subgroup, the result follows.

Lemma 1.39. With notation as in lemma 1.38, H is isomorphic to E 53\SL( 3 ,5 ) .

Proof. By the previous lemma, H is an extension of

E = E _3 by SL(3,5) . We suppose by way of contradic- tion that the extension splits, and as such we identify

SL(3,5) with a subgroup of H . It is easily verified that Y as given below is an elementary abelian 5-group in SL(3,5) of order 25 .

a Y 1 b : a,b€GF(5)

1

Moreover, CE(Y) is clearly a 2-dimensional subspace of

E whence C.-,(Y)Y = E _4 , a result of the extension iii 5 splitting. But now from lemma 1.25 it is apparent that

Ce(Y)Y is not Sj-pure . Let f 6 CE(Y)Y be a 5 2-ele- 79

ment . Then clearly Cg(Y)Y < C^(i) . As a Sylow 5- subgroup of C^(f) has isomorphism type Z 5 x 5 1 + 2 (by

Table 1) we obtain the desired contradiction and the extension is non-split as claimed.

Lemma 1.40. Let E be an arbitrary elementary abelian

5-group of LyS which is not 5l-pure . Then |Cq(E)|s is equal to either 5 3 or .

Proof. As E is not 5x-pure , we may assume E con­ tains the 52-element f described in 1.19 . Thus

Cq(E ) < C Q (f), and |Cg (E )|5 ^ | CQ ( f) | s = 5 4 follows.

As E ^ CG (f) j we easily see that E is contained in

P , the unique Sylow 5-subgroup of CG(f) ( 1.20 ).

Thus Z(P)E < c q (e) • Now E — z(p) > then certainly

P < CG (E) whence |CG (E ) |5 = 51* . If E^Z(P) , then

Z(P)E itself has order at least 5 3 whence J CG(E) |5

^ 5 3 . The lemma is thereby proved.

Lemma 1.41. With f as in 1.19 , let E be an ele­ mentary abelian 5-group of LyS containing f , and suppose |CG(E) | 5 = 51*. Then N G (E) < NG () .

Proof. As |Cq(E)I5 = 5 4 we see that P must be the unique Sylow 5-subgroup of CG(E) whence P is charac­ teristic in Cg(E) . But [P,P] , being a characteristic 80

subgroup of P , must be characteristic in C^(E) as

well. Thus [P,P] ^ Nq (E) and the result follows as

[P,P] = ( 1.20 ) .

Lemma 1.42. Let f , E be as in lemma 1.41 and

suppose |Cq(E)|5 = 5 3 . Then either Nq(E) < Ng()

or Nq(E) is conjugate to a subgroup of N^( ) .

Proof. The method of proof will be to show Nq(E) is

conjugate to a subgroup of NQ() under the

assumption Ng(E)^ NG() . We first treat the case

where E has order 5 3 . In this case we plainly have

Z(P) < E , for otherwise Z(P)E = P , a contradiction

since P is non-abelian. Thus f,fj€ E ( 1.20 ).

Now as R is extraspecial with center , it is well known that the elements of any non-trivial coset

g of in R are fused in R . We shall use

this fact repeatedly in what follows. Our first applica­

tion of it yields that all elements of not in

are of LyS-type 5 2 . We now choose n € NG (E) ,

n 0 NG() . As f“ is clearly a 5^element of E

not contained in <^i> (so not contained in by

the above) we have E = . We claim that all

elements of E not contained in are of LyS-

type 5 2 . Indeed let x be such an element. Then we 81

* can express x in the form x = f^(f^)^fk with i,j,k

integers. As all elements of (f^)3fk are fused,

we have x ~ (f“ )Jfk . But certainly (f°)^fk is con­

jugate to f^(fn 1)k (under n-1) . Again using the

fusion argument, this time on the coset (fn 1)k ,

we see that f^(fn )k ~ (fn )k . (Note that the argu-

n — 1 — I ment applies here as f € En = E < P < R . ) Finally

(fn 1 )k — fn 1~ f . Consolidating our results, we have

x ~ f , i.e. x is a 52-element as asserted. Thus all

elements of E not contained in are indeed

52-elements . Applying the fusion argument once more,

this time to non-trivial cosets of in ,

we easily see that is Sj-pure . As NG(E)

clearly permutes the 5x-pure E^'s in E , and as

is the unique such subgroup, we conclude that

Nq (E) ^ NG () . The result now follows from the

fact that NG() is conjugate to a subgroup of

NQ () by lemma 1.25.

It remains to prove the result for E of order at

most 5 2 . We first observe that for such E , E^Z(P).

Indeed E :£ Z(P) implies P < ^G(E) which contradicts

|Cg (E)|5 = 5 3 , our main assumption of the lemma. Thus

Z(P)E is an elementary abelian group of order 5 3 to

which the former case applies. It therefore suffices to 82

show N q (E) is a subgroup of NG(Z(P)E) . Now as

Z(P) < ^g (E) < CG(f) with Z(P)

It therefore follows that Z(P)E o NG(E) whence NG(E) is contained in NG (Z(P)E) as desired. The proof of the.lemma is now complete.

Theorem 1.43.• Let E be an arbitrary elementary abelian

5-group of LyS . Then NG(E) is conjugate to a subgroup of either NQ() or NQ() . Moreover, neither of NG() , NQ() is conjugate to a subgroup of the other. (Thus {NQ(),NQ()} is a complete set of representatives of the maximal 5-local subgroups of LyS.) Finally, the following two isomorphisms obtain: NG() s 5 1 + lt *SL(2,9)-Z^

(ii) NG ()£E53\SL(3,5)

Proof. (i) follows from lemma 1.27; (ii) is proved in lemma 1.38. Suppose now that E is 5j-pure . If

E = E 5 we certainly have NQ(E) ~ N&(); E = E 52 im­ plies Ng(E) is conjugate to a subgroup of NQ( ) by lemma 1.25; and E = E 5 3 implies NG(E) is conjugate 83

to Nq( ) by lemma 1.37. Finally if E is not

5j-pure , we may assume E contains an element f (as

described in 1.19), and the result follows from lemmas

1.40, 1.41, and 1.42.

It remains only to show that Ng() is not

conjugate to a subgroup of N^( ) and converse­

ly. Both of these, however, are immediate consequences of Lagrange's theorem. The proof of the theorem is now

complete. 84

4. Local Analysis for Remaining Primes

In this very brief section we analyze the p-local

structure of LyS for p a prime dividing |LyS| ,

p ^ 2,3,5 . Each such prime divides the order of LyS

to only the first power, so for each p there exists

a unique local subgroup NG(Sp) , where Sp = Zp is a

Sylow p-subgroup of LyS . We show NG(S7) is conjugate

to a subgroup of CQ(z) ( z as in section 1 ), NG(S1X)

is conjugate to a subgroup of N(,() ( u x as in sec­ tion 2) , and NG(S31) , NG(S37) , and NG(S67) are all

Frobenius groups.

Lemma 1.44 . NG'(S7) is conjugate to a subgroup of

CG (z) .

Proof. From Table 1 we have CG(S7) = S 7 xL, where L

is isomorphic to SL(2,3) . As L is a 7'-group and

all elements of CG(S?) outside L have order divisible by 7 , we see that L is characteristic in CG(S7) .

But then also Z(L) = Z 2 is characteristic in CG (S7) ,

so normal in Nq(S7) . The result follows.

Lemma 1.45. Nq C S ^ ) is conjugate to a subgroup of

CG (U l ) . 85

Proof. Prom Table 1 we have Cq(S;l i) = Sn xK with

K = S 3 . As K is an 11;-group and all elements of

Cq (S11) not in K have order a multiple of 11 , we

see that K (whence also [K,K] ) is characteristic

in CG(SX1) , so normal in NG(SX1) . Clearly [K,K]

= where u is an element of order 3 . But

S1]L < C(j(u) whence by Table 1 , u must be a 3X-

element of LyS . (Indeed 11 does not divide the

order of CG (u2 ) .) Thus NG (SXX) < NQ () ~ NQ ()

and the lemma is proved.

Lemma 1.46. NQ (S3x) S Frob(31,6) , NQ (S37) 3 Frob(37,18),

and Ng (S67) = Frob(67,22), where Frob(n,m) denotes a

Frobenius group with kernel of order n and complement

of order m .

Proof. Let p e {31,37,67} . Clearly each NG(Sp) is

an extension over Sp ; as p is relatively prime to

the order of NG(Sp)/Sp each of these extensions splits.

By Table 1 each Sp is self-centralizing in LyS. There­

fore, denoting by Hp a complement to Sp in NG (Sp ) , we see that each element of Hp acts fixed point freely

on Sp ( p = 31,37,67 ). Thus for each p , NG(Sp) is

a Frobenius group with kernel Sp ; the orders of the

corresponding complements are given by 1 . 2 1 . 5. A Complete Set of Maximal Local Subgroups for LyS

The following theorem is the culmination of all work done in earlier sections.

Theorem 1.47. With M 1 ,M2 ,...,MX as described below,

{Mj,M2 ,...,MX0} constitutes a complete set of represen­

tatives of the maximal local subgroups of LyS . This is

to say every local subgroup of LyS is conjugate to a

subgroup of some M^_ and {M 1 ,M2 ,...,M^q} is minimal with respect to this property.

local subgroup P isomorphism type order A M 1=N(,( < z > ) 2 2 8 *34 *52 *7*11 A 11

2 E 2 3\ G L (3,2) 2 6 •3•7 m 2=ng<r>)

2 8 -37 •5 3 *7*11 M3= NG ( ) 3 Z 3\ Aut ( Me )

M 4=Ng (V) 3 E 35 *(Mi! x Z2) 2 5 *37 *5*11

M 5=Nq () 3 3 2 + 4 •(SL(2,5)*Z8)• Z 2 2 6 •3 7 • 5

5 5 1+ 4 -SL(2,9)-Zh 2 6 *32 *56 M 6=NG (:)

M 7=NQ () 5 E 5 3\SL(3,5) 2 5 •3•56 •31

m 8=ng (s3 i) 31 Frob(31,6) 2.3*31

37 Frob(37,18) 2•3 2 •37 ^9-^G ^ ®3 7 ^ 67 Frob(67,22) 2-11-67 M 10“^G ^®67^ 87

Proof. By theorems 1.9, 1.24, 1.43 and lemmas 1.44,

1.45, 1.46, every local subgroup of LyS is conjugate to a subgroup of some . To show {M1 ,M2 ,...,M} is minimal with respect to this property, we must show

is conjugate to a subgroup of Mj if and only if i = j . By virtue of theorems 1.9, 1.24, 1.43 and

Lagrange's theorem no such inclusions exist, except for possibly the following:

(i) is conjugate to a subgroup of M 3

(ii) M 2 is conjugate to a subgroup of Mx

(iii) M 2 is conjugate to a subgroup of

(iv) M 0 is conjugate to a subgroup of M„. o 7

None of these inclusions actually occur, as we now show.

Suppose (i) obtains. Then Nq () < N q () for

A some g in LyS . As N^() = is perfect, we have

Nq () = [Nq ( ),N q ()] < [NQ(),Ng ()] =

Cq(uS) . This however is a contradiction, as |NG()j2

= 2 8 while |CQ(uf) |2 = 2 7 . Thus is not conjugate to a subgroup of M 3 .

Suppose (ii) holds, i.e. NG() < NG() for some g in LyS . Then < NQ() = CG(zg) whence zg € CG() = . As the quotient

Ng ()/ is isomorphic to GL(3,2), it is clear that NG() is transitive on the seven 88

involutions in . Thus there exists an element

n of NG() such that z®n = z . But then we

have NG () = NG ()n < N^,() = N^,() =

Cq (z ). This is an obvious contradiction as z is not

central in NG() . Thus (ii) does not occur.

Suppose (iii) obtains, i.e. NG () )

for some g in LyS . We fix h e NG() of order

7. As x 7-l = (x-1)(x3+x+l)(x3+x 2+l) with each of

x 3+x+l, x 3+x2+l irreducible over GF(2). , we see that h must act trivially or irreducibly on . Since

is self-centralizing in LyS , h must act ir­

reducibly. But then [,] = . (Indeed

[,] is an -invariant subspace of

not equal to 1 .) We therefore have =

[,] ^ [NG(),NG()] = CQ(u|) whence

uf € CG() = , a contradiction. Thus M 2

is not conjugate to a subgroup of M 3 .

Finally suppose the inclusion in (iv) occurs. Then

there exists an element g of LyS such that NG(S31)

is contained in . Setting M = we clearly have

NG^S 3i5 = NM (S3 i) ' Therefore lM:NM(s31)l = 2^.56, and as a consequence of Sylow's theorem we must have that

2lt*56 is congruent to 1 mod 31, an obvious contradic­

tion. We conclude that (iv) cannot obtain and the theorem

is thereby proved. CHAPTER II

ON THE MAXIMAL SUBGROUPS OF LYONS' GROUP

Any attempt to classify the maximal subgroups of a given finite group G has a natural division into two parts, namely the determination and classification of all maximal local subgroups of G , and that of all maximal non-local subgroups of G . In general, if M is maximal

in G , we choose K to be a minimal (non-identity) normal

subgroup of M . As M itself is normal in M , the exis­

tence of such a subgroup K of M is assured. Now K is clearly characteristically simple, as any characteristic

subgroup H of K would indeed have to be normal in M whence equal to 1 or K (by the minimality of K ). But

then from the classification of characteristically simple

groups, we would have that K is isomorphic to a direct product of n copies of some simple group A , i.e.

K = AxAx...xA ( n times ). If A is solvable, then K

is in fact elementary abelian, and the classification of

M is accomplished via local analysis on G . (For G =

LyS, this was the content of Chapter I.) In the present

89 90 chapter we perform some non-local analysis on LyS , namely we concern ourselves with the case where A above is simple non-solvable, whence M = N q (K) is a (maximal) non-local subgroup of LyS . Although a complete classification of the maximal non-local subgroups of LyS is far from achieved, we list some results which have a direct bearing on the maximal subgroup problem. Among these is a classi­ fication of all maximal subgroups containing a fixed Sylow normalizer for each relevant prime. 91

1. Some Non-Local Analysis

Our first result shows that every characteristically

simple non-solvable subgroup of LyS is in fact simple.

Proposition 2.1. There is no subgroup of LyS isomorphic

to A x b with A , B simple non-solvable .• As a conse­ quence, every non-solvable characteristically simple sub­

group of LyS is simple and every maximal non-local sub­

group of LyS is the normalizer of some simple group.

Proof. Let K = Ax B be a subgroup of LyS with A , B sim­

ple non-solvable. Then for any prime p dividing the order

of B , there clearly exists a subgroup A x C of A x B

with C s Zp . Thus A x c . is isomorphic to a subgroup of

Cq (x ) for some element x of LyS having order p . In

particular C^(x) is non-solvable. But a glance at Table

1 of the previous chapter shows that Cq(x) is non-solv­ able only for elements of LyS-type 2, 3j, 32, and 5J .

Thus it (A) (the set of prime divisors of [ A | ) is a sub­ set of {2,3,5} , and by Burnside's classic paqb-theorem we see that the two sets are in fact equal: tt(A) = {2,3,5}.

Moreover, by a symmetric argument, tt(B) = {2,3,5} as well.

As 5 divides j B| we see from the above discussion that

A must be isomorphic to a subgroup of (^(fi) which by 92

Table 1 is isomorphic to 5 1+1* »SL(2 ,9) . By using the

fact that A is simple non-solvable, it follows at once that A must be isomorphic to a subgroup of PSL(2,9)s

A 6 whence A = A 5 or A = Ag . In any case A (so also

CG(fi) ) contains a fours-group. As (^(fj) has general­

ized quaternion Sylow 2-subgroups, this is an obvious con­ tradiction. This proves the first statement of the propo­ sition; the second statement is an immediate consequence of the special case A = B.

Lemma 2.2. LyS contains copies of each of the following simple groups: L 2 (5), L 2 (7), L 2 (8 ), L 2 (9), L 2 (ll), L 3 (5),

U 3 (3), G 2 (5), M 1l .

Proof. It is well known that LyS contains a group iso­ morphic to G 2 (5); indeed it is this group from which Sims constructed LyS [17] . From [13,p . 565] it is obvious that G 2 (5) contains copies of L 2 (7) and U 3 (3) .

(More explicitly, Z 3 *PGL(2,7) and U 3(3) are isomorphism types for two of the double-point stabilizers in the action of LyS on the cosets of H in LyS , H = G 2 (5) .) That

L 3 (5) can be realized as a subgroup of G 2 (5) is noted in

[ 8 ] . 93

We have already seen that LyS contains copies of

x (indeed a copy exists in the maximal 3-local Nq(V) of Chapter 1, Section 2). As Mxx contains subgroups isomorphic to A 5( = L 2 (5)), A g ( S L 2 (9)), and L 2 (ll), we have only to verify that L 2 (8 ) occurs in LyS .

But surely L 2 (8 ) can be embedded in A 2 2 (in fact in A g) via its action on PG(1,8), the projective line of order 9.

As L 2 (8 ) has trivial multiplier, we see at once that

A L 2 (8 ) lifts to Z 2 x L 2 (8 ) in A 11 under this embedding..

Thus L 2 (8 )'s indeed occur in LyS and the lemma is proved.

Lemma 2.3. Each of PSp(4,3) and L 3 (4) occurs in

LyS.

Proof. By [3] , each of PSp(4,3) and Lg(4) can be realized in Me . (Note that L 3 (4) is isomorphic to a one-point stabilizer under the natural action of M 22 on its 22 letters.) As PSp(4,3) has multiplier Z 2 , we see that its full inverse image in $c is isomorphic to

Z 3 x PSp(4,3), i.e. PSp(4 ,3) occurs in lie , so also in

LyS . 94

It is far less obvious that L 3 (4) occurs in LyS .

As L 3 (4) occurs in Me , and as L 3 (4) has multiplier

Z4 x z12, we see that the full inverse image of a copy of

L 3 (4) in Me has two distinct possibilities in m'c :

Z 3 x L 3 (4) (corresponding to L 3 (4) splitting over Z3), and SL(3,4) (corresponding to the unique non-split ex­ tension of L 3 (4) over Z 3).

We next observe from [3,p.62] that Me contains a copy of L 3 (4)*Z2 where the complement Z 2 may be chosen so as to contain the field automorphism of L 3 C4 ) • We identify the subgroup L* of Me with L 3 (4)*Z2 so that for all A in L , Aa = A , where A is defined by

[cTfJ] for A = [oti-j 3 ( 1 ^ i,j ^ 3, a^j € GF(4) ) and dT[J denotes the image of a^j under the unique field automorphism of GF(4) having fixed field GF(2). Now

A A suppose the full inverse image L of L in Me is iso­ morphic to SL(3,4) . We let a denote a fixed preimage of a in Me so that L* pulls back to L* . We claim that a inverts Z(L) . Indeed for any T € L we

A ^ A A have Tc = T mod Z(L) whence T° = a(T)IT for some a(T) in GF(4)# . As a acts as an (outer) automorphism on L, we have

a (ST) 1ST = (ST)& = S<*T<* = a(S)ISa(T)IT = a(S)a(T)IST 95

whence a(ST) = a(S)a(T) for all S, T in L .

Thus a induces a homomorphic action from L into the multiplicative group of the field GF(4), i.e. from

SL(3,4) into Z 3 . By the structure of SL(3,4) , a

A must be trivial, i.e. a(T) = 1 for all T e L . Thus

A ^ A A Ta = T for all T in L ; in particular a must invert

Z(L) as a = a " 1 for all a€GF(4)#. But Z(L) = Z(Mc) whence a must centralize Z(L) . This obvious contra­ diction yields that L splits over Z(L) . Thus L is

A isomorphic to L 3 (4) * Z 3 as discussed, and Me (so LyS ) contains a copy of L 3 (4).

Lemma 2.4. L 2 (31) and L 2(125) do not occur in LyS .

Proof. It is well known that L2(31) and L2(125) con­ tain elements of order 16 and 63 , respectively. As LyS fails to contain elements of either order, the lemma follows.

Lemma 2.5. Let A < LyS be isomorphic to A 5 and let b and c be elements of A having respective orders 3 and 5 . Then b is of LyS-type 3 2 and c is of type 96

Proof. A well known fact from the theory of ordinary

characters enables us to determine the number of distinct

ways an element z of LyS can be expressed as a product

z = xy where x and y belong to prescribed conjugacy

classes. More explicitly, if G is any finite group and

Kj, K 2 , K 3 any three conjugacy classes of G , we denote

by #(K 1 ,K2 ,K3) the cardinality of the set

{(x,y):x e K L,y € K 2 ,xy=z} where z is a fixed (but arbitrary) representative of K 3 .

Then the value #(K 1 >K 2 ,K3) can be obtained from the

character table of G by the following formula:

X(a)x(b)x(z)

(In the above, a, b, and z denote fixed representatives

of Kx, K2, K 3 respectively, and the sum ranges over all

ordinary irreducible characters of G .) In this manner

we are able to determine the values #(2,3 1 ,51) = 0,

#(2,3 j,52) = 0, #(2,32 ,5X) = 0, #(2,32 ,52 ) = 6875 with

2, 3i , 5.^ (i = 1,2) conjugacy classes of LyS .

Suppose now that A ^ LyS is isomorphic to A 5 .

Then there clearly exist elements a,b,c € A of respec­

tive orders 2, 3, and 5 such that ab = c . (Indeed,

a 2 = b 3 = (ab )5 = 1 are defining relations for A 5 .) 97

We see from above that this is only possible when b and

c are representatives for the LyS-classes 3 2 and 5 2 ,

respectively. As A 5 possesses a unique class of elements of order 3 , and also one of elements of order 5 , the

lemma follows at once.

Lemma 2.6. There is no copy of A^ in 32+I* *SU±(2,5) .

As a consequence, LyS has no subgroup of the form A x B where A = A 4 and B contains a 32-element .

Proof. Suppose, by way of contradiction, that A is a

subgroup of 32+t* *SU±(2,5) which is isomorphic to Ak .

Then clearly 0 3 (A) = 1 whence A can be embedded in the complement SU±(2,5) . But then (under this embedding)

Ei+ ( =[A,A] ) is surely isomorphic to a subgroup of SL(2,5)

( =[SU±(2,5),SU±(2,5)]). This is of course a contradic­

tion, as SL(2,5) has quaternion Sylow 2-subgroups

(and in fact a unique involution), so cannot contain a

fours-group. This proves the first statement of the lemma.

The second statement follows from the fact that the cen-

tralizer in LyS of a 3 2-element is isomorphic to a split,

extension of 3 2+4 by SU±(2,5) . 98

Lemma 2.7. LyS contains no copy of J2 .

Proof. Suppose J ^ LyS is isomorphic to J 2 . We then

observe from [ 8 ] that J must contain a subgroup

AxB where A = A^ and B = Ag . But lemma 2.5 implies

that all elements of order 3 in B are 32-elements of

LyS . This of course contradicts the result of the pre­ vious lemma and we conclude that LyS contains no sub­ group isomorphic to J 2 as claimed.

Lemma 2.8. A 7 's do not occur in LyS .

Proof. Suppose H ^ LyS is isomorphic to A 7 . Clearly

A 7 possesses the subgroup A{1,2,3,4}x <(567)> , where

A{1,2,3,4} denotes the complete alternating group on the letters {1,2,3,4} . Thus H contains a subgroup AxB with A = A{1,2,3,4} = A^ and B = <(567)> = Z 3 . But

(567) is certainly contained in a copy of A 5 in A ? ; thus B is contained in a copy of As in H , and it follows from lemma 2.5 that B is 32-pure . The exist­ ence of A x B is now contradicted by lemma 2 .6 , and we conclude that no copy of A 7 can occur in LyS . 99

Lemma 2.9. LyS contains no subgroup isomorphic to M22.

Proof. From [ 8 ] we see that M 22 possesses a unique class of elements of order 3 and a representative of this class has centralizer (in M 22 ) isomorphic to A 4 *Z3.

Let M be a subgroup of LyS isomorphic to M 22 . As

M 22 has E 9 Sylow 3-subgroups, we see that a Sylow 3-sub- group of M must be 32-pure . (Indeed LyS contains no

32-pure E9's by lemma 1.13.) Thus all elements of order

3 in M have LyS-type 3 2 . As M contains a subgroup

AxB with A = A 4 and B a 32-pure Z 3 , we arrive at a contradiction by virtue of lemma 2.6. The proof is now complete.

Lemma 2.10. Let M be any copy of M xl in LyS , and let denote the unique ordinary irreducible character of LyS of degree 2480 . Then (^4-M,1jj) ^ 1 .

Proof. That M11's in fact exist in LyS is shown in lemma 2.2 . We first determine the LyS-type of all conjugacy classes of M . There is no difficulty in de­ termining this for elements of order 2, 4, and 11 ;

LyS has a unique class of elements of order 2 and 4 , and it has two non-real classes of elements of order 11 .

As M has a unique class of elements of order 3 (some of which are obviously contained in a copy of A 5 in M ), 100

we conclude from lemma 2.6 that all elements of order 3

in M have LyS-type 3 2 . An element of order 6 in

M squares to a 32-element of LyS ; therefore elements of order 6 in M (which lie in a single M-class ) must have LyS-type 6 2 or 6 3 . We shall not attempt to determine which of these is the correct class; we shall

instead consider cases. By lemma 2.6 the unique M-class of elements of order 5 must consist of 52-elements , as there certainly exist elements of order 5 in a copy of A 5 in M . Finally we shall see that, for our pur­ poses, no determination is necessary for elements of order

8 in M .

Now consider ip as defined in the lemma statement.

From the information gathered above, we can easily compute the value for (ip +M, lj^) using the standard orthogonality relations of characters. (Note from the character table

for LyS [ 13 ] that ip assumes the value 0 on all elements of order 8 . This explains why it is not ne­

cessary to determine the LyS-type of any such element

in M . ) We obtain (ifHM,lM ) = 0 (if elements of order

6 in M have type 6 2 ) and (\JhM,1m ) = 1 (if ele­ ments of order 6 in M have type 6 3 ). In any case,

(i|)+M, ljj) ^ 1 , as claimed. 101

Lemma 2.11 There exists no copy of Ml2 in LyS .

Proof. The argument we use is entirely character theoret­ ic. Let N denote a subgroup of LyS isomorphic to M12.

We perform a similar analysis to that used in lemma 2. 10 to determine the LyS-type of elements of N . The only difficulties which arise involve elements of order 3, 6 ,

5, and 10 (as in lemma 2. 10 we can ignore elements of order 8 ). As elements of order 5 occur in a copy of

M1]L in N (and there exists a unique N-class of such elements), we see that all elements of order 5 in N are of type 5 2 . As any element of order 10 in N squares to a 52-element of LyS , we see that all such elements have LyS-type 102 . Now N has two classes of elements of order 3 : 3A (corresponding to central- izer size 54 ) and 3B . From [ 8 ] we see that

C^(u) = Z 3 x Ait ^or u € 3B . We therefore conclude from lemma 2.6 that u is of type 3: , i.e. 3B C 3 1 . As

LyS has no 3x-pure E9's (lemma 1.13), it must be the case that 3A C 3 2 • The LyS-type of elements of order

6 in N can now be determined by identifying the type of their respective squares (up to the ambiguity involving

6 2 and 6 3 discussed in the previous lemma). We are now able to compute (iJhN.In) where tJj € Irr(LyS), iJj(1)=2480.

We obtain (iJj+N,!^) = 3 (if elements of order 6 in N 102

which square into the class 3A are of type 6 2 ) and

(ip+N,ljy) = 4 (otherwise). In any case (^iN.ljy) ^ 3 .

But this clearly implies ( , 1M) ^ 3 for M a subgroup of N isomorphic to M x1 . As this contradicts lemma

2.10, we conclude that LyS cannot contain a subgroup N isomorphic to M 12 . The proof of the lemma is now complete.

Theorem 2.12. Let H be a non-solvable characteristi­ cally simple subgroup of LyS . Then H is isomorphic to precisely one of the following groups: L 2 (5), L 2 (7),

L 2 (8 ), L 2 (9), L 2 (11), L 3 (4), L 3 (5), U 3 (3), PSp(4,3), G 2 (5),

Mjx . Moreover, every such group occurs in LyS .

Proof. By Lagrange's theorem, proposition 2.1, and a theorem of Gorenstein and Harada [ 6 ], H is isomorphic to one of the following groups: L 2 (5)=A5, L2 (9)=A6, A?,

A8, A9, A1q, AX1, L 2 (7), L 2 (8 ), L 2 (ll), L 2 (31), L 2(125),

L 3 (4), L 3 (5), U 3 (3), U 3 (5), U 4 (3), PSp(4,3), G 2 (5), 1,

Mi 2 j M 22, J2 , Me. By lemmas 2.4, 2.7, 2.9, and 2.11, we are able to dispose of L 2(31), L 2 (125), J 2 , M22> and M 12 as possible isomorphism types for H . Lemma 2.8 eliminates

A 7 as a possible isomorphism type, and so also AQ, Ag,

Ajq, ah> U 3 (5), U^(3), and Me , each of which contains a copy of A7. This leaves precisely the groups listed in 103 the statement of the theorem. That each in fact occurs in LyS is the content of lemmas 2.2 and 2.3.

Remark. The balance of this chapter is devoted to show­ ing that normalizers of certain non-solvable simple groups in LyS can never be maximal.

Lemma 2.13. LyS contains a copy of A 5 whose central- izer is isomorphic to Z 2 .

Proof. Identifying CQ(z) with A X1 we consider the subgroup A of Cq(z) = CG(z)/ generated by a = (12)(34)(56)(78) and x = (139)(5 7 10). As ox =

(12349)(5678 10), and as a 2 = b 3 = (ab) 5 = 1 are defin­ ing relations for As , we see that A = = A 5 .

Now As has a unique class of involutions; thus every involution in A is a product of four transpositions and so by 1.1 has preimage in CQ (z) of order 2 .

Thus, denoting by A the full inverse image of A in

CQ(z), we see that a Sylow 2-subgroup of A splits over

. By a theorem of Gaschiitz, A splits over as well, whence A = A 5 * Z 2 . Let B be the subgroup of A isomorphic to A 5 . Then clearly z € Cq(B) . We claim in fact that Cq (B) = . First observe from lemma 2.5 that B contains an element c of type 52 . Thus

CG (B) < CQ (c) £= (51+2 x Z5 ).S3 . That CQ (B) is a 5'- 104

group is easily deduced from Table l. Indeed let g be

an element of order 5 in CG (B) (so that B < c q (s ) )•

For g of type this gives B *---►51+lf *SL(2,9) .

As 0 5 (B) = 1 we further have B *— >-SL(2,9) , a contra­

diction as SL(2,9) has generalized quaternion Sylow 2- subgroups. For g of type 5 2 we have B (5 1 + 2 *Z5) *Z3,

also a contradiction as, for example, the latter group is solvable. Thus Cq(b) is in fact embedded in S 3 , so by above C^(B) = or Cq(b) = , where v is of order 3 and is inverted by z . But it readily fol­ lows from lemma 2.6 that v must be of LyS-type 3 3 whence by [13,p.544], CG (z,v) = M x1 . In particular,

if Cq(B) £ then B must be contained in a copy M • of M X1 in CG(z) . Under the canonical homomorphism

CG(z) >CG(z) we therefore have B < M with B = A 5 ,

M = M n . As subgroups of A1: (again under the identi­ fication with CQ(z) ) , B and M each have a natural permutation action on {1,2,...,11} . But from the character table of M X1 it is immediate that only one such action is possible for M , that corresponding to the permutation character 1+ip , where ip is the unique self-dual irreducible character of M X1 of degree 1 0 .

We now readily conclude that any element g € M of order

3 must have precisely 1(g) + iKg) =1+1=2 fixed 105

points under this action. The desired contradiction now follows from the fact that x (defined earlier in the proof) is an element of order 3 in M having fixed point set precisely {2,4,6,8,11} . We therefore con­ clude that Cq(B) = and the lemma is proved.

Lemma 2. 14. Let f be a fixed 52-element of LyS .

Then there exist precisely 625 distinct copies of A 5 in LyS which contain f and have centralizer order divisible by 3 .

Proof. From lemma 2.6 we observe that any element of order 3 which centralizes a copy of A 5 in LyS must have LyS-type 3 2 . Fix x € CG (f) of type 3 2 . We calculate structure constants (as in lemma 2.5) for

CG(x)-classes to obtain #(2',3^,5^) = 125 where 2',

3^, and 5^ are respectively the unique classes of

CG(x) which correspond to the LyS-classes 2, 32, and

52. As a fixed element of order 5 in A 5 can be ex­ pressed as a product of an involution and element of or­ der 3 in precisely 5 ways (proof by calculation of the appropriate structure constant in A 5 ), we see that

CQ(x) contains precisely 25 copies of A 5 which con­ tain f . Now is a Sylow 3-subgroup of CG(f), so by Sylow's theorem we see that cg(f) contains pre­ 106

cisely 25 = |CQ(f):NCG(f^()| distinct copies of Z 3 .

This means f is contained in 25 distinct CG(y)'s where y is a 31-element of LyS . As we have just

shown that each such CG(y) contains 25 As's which

contain f, we obtain 625 = 25-25 such A5’s in all.

It remains only to show that the 625 A5's so obtained

are in fact distinct. We therefore let A denote a sub­

group of LyS isomorphic to A 5 with A ^ Cq (k ) n CG (y), x and y 3 1-elements. Thus x,y € Cq (A) . Recall from the proof of the previous lemma that C(j(A) is isomorphic to a subgroup of S 3 ; thus we have CG(A) = Z 3 or Z 3 .

In any case cq(a) has a unique 3-group whence =

and Cq(x) = Cq(y) . It now follows that the 625

As's in question are indeed distinct and the lemma is proved.

Lemma 2.15. Let A denote an arbitrary subgroup of LyS

isomorphic to A 5 and let f be a fixed 52-element of

LyS . Then the number of distinct conjugates of A which

contain f is given by 1500/t where t is the index

|Nq (A):A| .

Proof. Consider the set A = {(fx ,Ay): x,y€LyS, fx e Ay} .

We shall count A in two different ways. First let n

equal the number of distinct conjugates of A which 107

contain f . (Clearly this value is independent of the representative chosen for the class 5 2 .) As LyS con­ tains precisely |LyS:CG(f)| 52-elements in all, we have

[A| = n|LyS:CG(f)| . Next fix a conjugate A? of A .

Ay contains 24 elements of order 5 , all of which are conjugate to f by lemma 2.5 . As LyS contains exactly |LyS:NG(A)| conjugates of A , we have |A| =

24 |LyS:NG(A)| . Thus n|LyS:CG(f)| = |A| = 24|LyS:NQ(A)| so that n = 2 4 |CQ (f) |/ |NG (A)| = 24-54 -6/|NG (A)| =

1500/|NG(A):A| as claimed.

Proposition 2.16. Let A denote an arbitrary subgroup of LyS isomorphic to A 5 . Then Cq(A) f 1 and NG(A)

is not maximal in LyS .

Proof. First we recall from lemma 2.5 that #(2,3 2 ,52 ) =

6875 whence f occurs in precisely 1375 = 6875/5 copies of A 5 in LyS . Now suppose there exists a subgroup A of LyS isomorphic to A 5 with ^G(A) = 1 . Then NG(A)

is isomorphic to a subgroup of Aut(A) 3 £5 whence

|NG (A):A| ^ 2 . Thus by lemma 2.15, f is contained in

at least 750 conjugates of A in LyS, each of which

has trivial centralizer. By lemma 2. 14, f is contained

in 625 additional A 5 1s in LyS, each of which has

centralizer order divisible by 3 . As 750 + 625 = 1375, 108

we have thereby accounted for all A5 1s in LyS which contain f, and none of these A5 's have centralizer of order 2 . A contradiction is now immediately derived from lemma 2. 13. We conclude that Cq(A) 1 for all

A = A 5 in LyS . Now, as observed in the proof of lemma

2. 13, Cq(A) is isomorphic to a subgroup of E3. If 3 divides the order of C^(A), then C^(A) is isomorphic to either Z 3 or E3. In either case a Sylow 3-subgroup

of Cq(A) is characteristic in cq(A) whence normal in Nq(A). This gives NQ (A) < NG (), proving the non-maximality of NQ(A) whenever 3 divides |CG(A)| .

The only remaining case is CQ(A) = = Z 2 . As CG(A) is normal in NQ (A) we have NG (A) < NG (CG (A)) = NQ ()

= CG(z) and again NG(A) cannot be maximal in LyS. The proof of the proposition is now complete.

Lemma 2. 17. Given a subgroup A of A 6 isomorphic to

A 5, there exists a subgroup S of A 6 with S = and

A n S = At*.

Proof. First let A be a one point stabilizer in Ag, say the stabilizer of {6 >. Clearly A = A3. Now denote by B the double point stabilizer of the points {5,6} .

Clearly B < A and B = A4. As (12)(56) normalizes

B, we easily see that S = B*<(12)(56)> is isomorphic 109

to E^. Thus S satisfies the conclusion of the lemma.

Now if A is an arbitrary A 5 in A 6 , then there exists an outer automorphism a of A 6 such that Aa is in fact a one point stabilizer. Indeed it is well known that A 6 has two classes of A5's (corresponding to the two classes of 3-elements in Ag ) which are fused in Aut(Ag). As seen above S can now be chosen so that S = and

Aa n S = Atj , whence Sa 1 satisfies the conclusion of the lemma.

Lemma 2.18. Let z, t, x be as described immediately preceding lemma 1.2. Then N G () < N G () .

Proof. We first claim that is a characteristic sub­ group of Cg (E) where E = . Indeed E is con­ tained in CG(x) by lemma 1.5. We may therefore appeal to the proof of lemma 1.3 to obtain that either is characteristic in c q (e ) or ^q CIS) kas a uni

P/ is the unique Sylow 3-subgroup of CG(E)/, which in turn is isomorphic to C. ((12) (34)(56)(78)). M i i As <(135)(246),(9 10 11)> and <(137)(248),(9 10 11)> are distinct Sylow 3-subgroups of C. ((12)(34)(56)(78)), A i i we conclude that is characteristic in CG(E), so normal in N q (E). The result follows. 110

Proposition 2.19. Let K be a subgroup of LyS iso­ morphic to Ag and suppose K contains a subgroup A

isomorphic to A 5 such that 3 divides |CG(A)|.

Then 3 divides |CG(K)| as well and Nq (K) is not maximal in LyS .

Proof. With K and A as in the proposition statement, we choose S in accordance with lemma 2. 17. Thus S < K,

S = and A n S = A^ . Denote by E the unique nor­ mal fours-group in S. As LyS has a unique class of fours-groups (lemma 1 .1 ) we may apply lemma 2 . 18 to ob­ tain S :£ Nq(E) < N^() for some Si-element y of

LyS. This gives A n S = [S,S] < [NG () ,NQ ()] <

Cq (y ) whence y € Cq (A n S). But as Cq (A n S) < CG (E)

< NG (), this clearly implies -o CG (A n S) .

Now |CG (A nS ) | 3 d= ( Cq(E) | 3 ^ 3 2 from the proof of lemma

1.3. It therefore follows (as no element of order 9 in

LyS centralizes an A^ [13,p.551] ) that a Sylow 3-sub­ group P of CG(An S) is necessarily elementary abelian.

By lemma 2. 6 P must be 3x-pure ; by lemma 1.13 P = E3.

Thus P = and, as is normal in CQ (AnS), is the unique Sylow 3-subgroup of Cg(AnS). Now as 3 divides |CG(A)| by assumption, there exists an element x in Cg(A) order 3 . Therefore = follows since CQ(A) is clearly a subgroup of CG(AnS). As Ill

S < N G () and A < CQ (x) ^ N Q () = NG (), we have

K = < N Q(). Since K is perfect it now follows that K = [K,K] < [Nq (),NQ()] < CG(y), and 3 di­ vides |Cg (K)| as claimed. Moreover is the unique

Sylow 3-subgroup of CG(K) as well, whence is char­ acteristic in Cq (K). We therefore conclude that is normal in NG(K) and the non-maximality of NG(K) is proved. 112

Proposition 2.20. Let L denote an arbitrary subgroup of LyS isomorphic to L 3 (4). Then Nq(L) is not maxi­ mal in L y S .

Proof. It is well known that L = L 3 (4) contains two classes of E 1 6 's, each having normalizer maximal in L and isomorphic to E 1 6 *A5 . Let us denote representatives of each class by and E 2 and their normalizers in L by Nx and N 2 respectively. Now as |EjJ = 16, we see from lemma 1.5 that E^ is contained in CG(xi) for some

3^-element Xi of LyS ( i = 1 , 2 ). Thus lemma 1.3 applies and we conclude that each of Nq(E1) , NG (E2) is contained in a 3-local subgroup of LyS (so certainly in a maximal 3-local subgroup). But then by theorem 1.24, it is immediate that Ng(Ei) < N G(), yi€ 3j (i = l , 2 ), as no other maximal 3-local subgroup of LyS contains an

E16. We now observe from Sylow's theorem (as jNx|2 = |L|2 ) that E 2 may be assumed to be contained in Nj. Thus Ej and E 2 are distinct E 16's in N x ^ N(-,(), and it follows from 1.4 (as NQ() S Aut(Mc) ) that E 2 = E^ for some element n in N<,(). Thus N 2 < N G (E2 ) =

Ng(E?) = NG(E1)n < N Q()n = NQ(). We therefore conclude that L = < NQ() and as L is per­ fect, we have L = [L,L] < [NQ ( ) ,NQ ( ) ] ^ C ^ y j ) .

Now since L contains a subgroup A isomorphic to A5, 113 we see that yx € CQ(L) < CG(A). From the proof of lemma

2.13, cq(a) is therefore isomorphic to either Z 3 or

2 3 whence an identical statement holds for CG(L). Thus

is characteristic in CG (L) and so normal in NQ (L).

Ng(L) ^ NG() now follows, asserting the non-maximality

of Nq(L) as claimed.

Proposition 2.21. Let P be a subgroup of LyS iso­ morphic to PSp(4,3). Then NG(P) is not maximal in LyS.

Proof. It is well known that P contains (as maximal parabolics) the two subgroups H = Cp(z) and L = Np(E), where z is a Sylow central involution of P , E is an

elementary abelian 2-group of order 16 , H is isomorphic

to (SL(2,3)*SL(2,3))•Z 2 , and L is isomorphic to E i6 * ^ 5

[15]. Also from [15] we see that L n H = C^(z) = Njj(E) is

isomorphic to E 1 6 *A4. We denote this latter group by K,

so that K = E*B where B = A^. Now |H:K| = 3, so letting

N denote the core of K in H (i.e. N = n (K*1: h€H} is the

kernel of the standard action 0 :H — *-£3 of H on the cosets

of K in H) we have |K:N| ^ 2. As B has no subgroup

of index 2 and E n N ■« N, we easily see that N = (EnN)*B.

By lemma 1.1, LyS has a unique class of fours-groups. As

E n N is elementary abelian of order at least 8 , we may

assume ^ EnN with t as in lemmas 1.2-1.5. Thus 114

by lemma 1.5 we have E < C G(x), while by lemma 1.2 we get Cg(E) < Cg(E nN) < Ng (). This gives x € CG(E)<

Nq () whence o CG(E). As 32-elements of LyS can­ not centralize an E 16, we see that a Sylow 3-subgroup of

Cg(E) must be 3x-pure. We now conclude from lemma 1.13 that is Sylow in Cq(E) > anc* as we have shown Cg(E), it of course follows that is character­ istic in Cg(E) whence normal in Ng(E). Thus L =

Np(E) ^ N g (E) < NG( ) . As E < CG(x) and L/E = Ag has no subgroup of index 2, we in fact have L ^ CG(x) whence x € Cq (L) < Cg (N) < Cq(E nN) < Ng ().

Now let h be an arbitrary element of H. As N is normal in H, we have x*1 € CG (N^) = Cq (N) ^ Nq () so that x*1 normalizes . Therefore < x , x h > is a 3- group in CG(N). As cq(N) ^ we see from lemma 2 -6 that < x , x h > is 3 j-pure. Thus = E 3 (lemma 1.13) whence = < x > h , i.e. h € NG (). Thus H ^ N G () and we conclude that P = < NG (). As P is per­ fect, P = [P,P] < [Nq (),Nq ()] < C q(x). As P con­ tains a subgroup A isomorphic to A5, we have x € Cq(P)

< Cg(A) and as seen in the proof of lemma 2.13, must be characteristic in CG(P). (Indeed CQ(A) is iso­ morphic to either Z 3 or £3-) Thus NG(P) and the result follows. 115

Lemma 2.22. N^() has a unique class of subgroups isomorphic to Z 4 x .

Proof. We first recall from 1.17 that NG() = RQ 2 where

R= S 51+it and Q x= A{6,..,11}•<(2354)(6798)> under the identification of CG(z) = CG(z)/ with A12.

Thus Q 2 < Cg((25)(34)) where B is the stabilizer in A 21 of the letter 1 .

Now let H = x denote an arbitrary copy of

Z^ x zh in NG(). By Sylow's theorem, we may certainly assume H ^£Q2. Without loss of generality we may assume x is a product of two four-cycles. (Indeed by 1.1, the only other possibility for an element of order 4 in Cq(z) is to map onto a product of two transpositions (modulo ), and hence square to z . As n = 1 , H must contain an element of order 4 of the type asserted above, i.e. map­ ping onto a product of two four-cycles.) Moreover, as x is an element of CB((25)(34)), we may assume x =

(2354)(efgh) for e,f,g,h distinct members of { 6 ,7,...,11}.

We now claim that Cg(x,(25)(34)) = x <(2354)(ij)>

= Z 4 x Z^ where i and j are distinct members of

{2,3,...,11} ^{2,3,4,5,e,f,g,h}. First note by a simple counting argument that B contains precisely 56700 conju­ gates of x whence |Cg(x)| =32. As (2e3f5g4h)(ij) is clearly an element of Cg(x) no'fc contained in Cg((25)(34)), 116 we see that |Cg(x,(25)(34))| ^ 16. Finally as

Cg(x,(25)(34)) certainly contains x <(2354)(ij)>, the result follows. But now |Cn (x-)| ^ 2* |CU(x) I ^ 2*|Cpr(x)| W 1 '961 ^ 2-|Cg(x,(25)(34))| = 32. If |CQ(x)| = 32, then CQ(x)

= x = x Zg where k = (2354)(ij). If Cq (x ) iias order smaller than 32 , then C,~(x) = H easily follows. <9*1 In either case H is the unique x Z^ in Cq (x ) whence

H = x . Finally, as Q is 4-transitive on

{6 ,7,...,1 1 } and fixes {2 ,3,4,5} pointwise, we see that

H is Qj-conjugate to x where u = (2354)(6798).

Thus x represents the unique class of x z^'s in N(,(), and the lemma is proved.

Lemma 2.23. Let ) such that <5X € N^().

Moreover, with notation as described immediately preceding lemma 1.28, A( 5x (u) ,

Proof. The first statement of the lemma is an immediate consequence of Sylow's theorem and lemma 1.31 (ii),(iii).

For the proof of the second statement one first easily verifies that 5^y6 = 3Iy where y represents the bilinear form A with respect to the basis {g2 ,g3 ,1^,g^gs)• Now for any y € X = GSp(4,5) we certainly have y^yy = ily for some integer i , 1 ^ i ^ 4. Let (x- 1 )^yx_1 = jly 117 where x is chosen as in the lemma statement. Then

(6x )ty( 6 x ) = (x“ 1 6x)ty(x“ 1 6x) = x^S^ (x” 1 )’tyx“ 1 6x = x^ 6 ^[(x” 1 )^yx“ 1 ]6 x = x'tfi'^C jly )6 x = jlCx^S^yfix) = jlxt(6ty5)x = jlxt(31y)x = 31x't(jly)x = 31(x^Cx-1)tYx_lx) =

31(x_1x)^y(x” 1x) = 3IY , i.e. (6 x )ty( 6x) = 3Iy. This implies for all u , v € R, A(Sx(u),5x(v)) = (6 x )^y(6x )v = u^SIyv = Su^yv = 3A(u,v), the desired result.

Lemma 2.24. R contains precisely two H-invariant sub­ groups of order 5 3 having center where H is a fixed complement of S 5 in Nq(S5).

Proof. From [13,p.548] we see that N^,(S5) = S 5 *H with

H = Zh x Recall from the previous lemma that <5X is an element of N^() whence 5X normalizes S5. As

N S Qj we may assume

We now claim that P x = and P 2 = are the only H-invariant subgroups of R having the prop­ erties described in the lemma statement. We first show that any such subgroup must in fact equal one of Pj or P2.

Thus let P be an H-invariant subgroup of R with |P| =

5 3 and Z(P) = . In particular P is -invariant, 118

so that P is a 2-dimensional < 6 x>-invariant subspace of

R. Let v = avj + bv 2 + cv 3 + dv^ be an arbitrary vector in P . Then each of av^, bv 2 , cv3 , dv^ is contained in

P as can be verified by the following formulae:

avj = (I - Sx )[£(v + (

b v 2 = (31 - 5x )[£(v - (6x )2v)]

cv 3 = £(v + (6 x )2v)- a?j

dv^ = 4 (v - (6 x )2v) - bv 2

This proves that P is a sum of eigenspaces for 6X.

(Indeed we may apply the above to two independent vectors of P if necessary.) Thus P = x and P =

for some distinct i , j in {1,2,3,4}. Now as [v^Vj] f 1, we have A(Vi,vj) = k f 0. Thus by lemma

2.23, ^i^jk = XiXjA(Vi,vj) = A(XiVi,XjVj) = A(6 xVj_, 6 xV j ) =

3A(vi,Vj) = 3k where Xn is the eigenvalue for the eigen­ vector vn ( n = 1 , 2 ). As k^O, this gives = 3.

Now if i = 1, then Xj^ = 1 whence Xj = 3. This occurs only if j = 4 whence P = P x. Similarly, the assumption that i = 2 (resp. 3 , 4 ) leads to the conclusion P = P 2

(resp. P 2 , Pi). Thus P = Px or P 2 as claimed.

Conversely, it is clear that each of Pi , P 2 is non-abelian whence Z(Pi) = Z(P2) = . (Indeed, given any eigenvector v^ , 1 ^ k ^ 4 , there exists some eigen­ vector vj such that [vk,Vj] f 1. Otherwise we would have 119 vk € Z(R), a contradiction. But now from above, vj is uniquely determined.) Also each of PJf P 2 is H-invari­ ant by the following argument. Each eigenspace is

Sx-invariant. As H is abelian we have for h € H ,

5xh(v]£) = h 6 x (vjt) = hCXjj-Vjj) = ^ M v ^ ) . Thus h(vk) is in the eigenspace for 5X , i.e. is H-invari­ ant whence each of P ^ P 2 is H-invariant as well. The proof of the lemma is now complete.

Lemma 2.25. |NN ():NN () | = 6 where N = Nq(R).

Proof. Clearly NN() permutes the six subgroups of S 2 > 6 3 > which contain . As seen in lemma 1.37 these six subgroups are -conjugate where y maps onto t £t~ 1 modulo R . As < N , Njj() has index 6 = |N | in % ( < f l , S z , S 3 >) as claimed.

Lemma 2.26. With notation as in lemma 2.24 , PiH is LyS- conjugate to P 2H.

Proof. We first observe that without loss of generality, we may assume Vj = g 2 in lemma 2.24. Indeed from lemma

1.29 we see that g 2 is an eigenvector for 5 with eigen­ value 1 . But x is upper triangular (lemmas 1.28, 2.23) whence x preserves the eigenspace f°r 5. It follows that as

follows. Thus P 1 = .

Now as |N^() I2 = 2 5 (e.g. see theorem 1.47), we see that 2 4 = |h| ^ |Nn ( ) |2 < |NN ( ) | 2

^ 2 5 , whence H is Sylow in N^(). (The strict inequality in the above follows from lemma 2.25.) Let K be a Sylow 2-subgroup of N^() containing H.

We claim that Pj and P 2 are K-conjugate. Indeed let k be an element of K H. As k normalizes R , we see that

lr a P1 is a subgroup of R of order 5 3 having center .

Moreover (as |K:H[ = 2) Hk = H so that P^ is H-invari- k ant. By virtue of lemma 2.24 we must therefore have P x =

Px or P2. But Pk = Pj implies k =

( P L n )k = Pk n k = Pl n

~ > a contradiction as k £ NN (). Thus P k

= P 2 as asserted. The lemma now follows as (P 1H)k = P^Hk

= P 2 Hk = P 2H .

Lemma 2.21. Let L denote a subgroup of LyS isomorphic to L 3 (5) and P a Sylow 5-subgroup of L . Then NL(P) is conjugate to a subgroup of NG(S5) .

Proof. It is well known from the structure of L 3 (5) that

NL(P) = P*U with P = 5 1+2 (exponent 5) and U = Z^ x z^.

We first claim that f is of LyS-type 5X where =

Z(P). Indeed as U normalizes Z(P) we see that 16 121 divides the order of N^() . As NG()/CG(f) embeds in Aut() = , we further have that 4 divides the order of CG(f) . But the centralizer of a 52-element has

2-part 2 (see Table 1) whence f must be a 5x-element as asserted. We may therefore assume N^(P) < NG().

Now suppose P £ R ( R as in 1.14 ). By lemma 1.27,

N^(P) normalizes R so that RP is a Sylow 5-subgroup of

NG() with N^(P) < Nq(RP). The lemma easily follows in this case from Sylow1s theorem.

We next assume P ^ R. As = Z(S5), Nq (S5) =

S5*H is a subgroup of NG(). Thus by lemma 2.22,

Ux = H < Ng (S5) for some element x of NG(). But

Px < Rx = R < Ng (S5) as well. This proves NL(P)X = PXUX

< N g (S5), which is the desired result.

Lemma 2.28. With notation as in lemma 2.27, assume NL(P) is contained in NG(S5). Then P :< R.

Proof. As a consequence of [13,p.550] every subgroup of

Sg of exponent 5 is contained in either R or S =

. In particular this applies to the extra­ special group P . We assume by way of contradiction that

P ^ R whence P < S . Now |P:PnR| = |RP:R| = | S 5 : R | = 5 so that |PnR| = 52. We may therefore write P= with x € R and y € S n R. Letting bar denote images 122

modulo we now have [x,yf2] - 1 and, as R is abelian, x and y commute whence [x,?2] ~ ^ - As R is a uniserial -module (1.14), we see that = whence = . Thus P = . But

[fi,f2J = [g2 ,f2] = 1 (from 1.13) while [f^y] = [g2 ,y] =

1 (as y € S and = Z(S) ). Thus P is abelian, a contradiction. We therefore conclude that P < R as claimed.

Lemma 2.29. N^(P) represents the unique class of subgroups of LyS having its isomorphism type.

Proof. As a consequence of Sylow's theorem and lemma 2.27, we may assume NL(P) = P»H < Nq (S5) = Ss'H . It follows from lemma 2.28 that P = 5 1+2 is contained in R . As P is H-invariant with center , the result follows from lemmas 2.24 and 2.26.

Lemma 2.30. Let H denote a subgroup of LyS isomorphic to G 2 (5) and containing S 5 . Then NH(R) = R*K where

K is isomorphic to GL(2,5) and acts irreducibly on the

Frattini quotient R of R .

Proof. It is well known from standard rank-2 Chevalley theory that H contains precisely two maximal parabolics:

= 5 1+ 4 *GL(2,5) and M 2 = 5 2 + 1+ 2 *GL(2,5). As H contains 123

S5, we may assume 0 5 (1^) = R and 0 5 (M2) = S = C ^ f ^ g 2)

(see 1.14, 1.15) whence = NH (R) and M 2 = Ng().

Now suppose R is not K-irreducible , and let V be a proper K-invariant subspace of R . Letting V denote the full inverse image of V in R , we see that

V is clearly Mj-invariant, so in particular -invari- ant. This makes V into an -invariant subspace of E: and, as such, V is uniquely determined by its dimension

(see lemma 1.28). If dim(V) = 1, then V = whence

V = . Thus ^ Njj() = M2, a contradiction.

If dim(V) = 2, then V = whence V = .

As NQ () ^ NQ () was proved in lemma 1.25, we here obtain M 2 = NH () ^ NH (&z> ) whence, by maximality of M2, M 2 = NH () = Nfi( ) .

But then ^ Nfl() = M2 , again a contradiction.

Finally suppose dim(V) = 3 whence V = Sh> and

V = . We claim that NH (V) < NH () .

Indeed for n € NH(V), n is an elementary abel­ ian subgroup of V of order 53. As V < C Q(f1 ,g2 ), we must have < n. (Otherwise V would equal the abelian group n , a contradic­ tion.) But as shown in the proof of lemma 1.26, is the unique 5x-pure E 5 3 in V which contains

Thus n = and the claim NH(V) ^

NH () is proved. As before, this gives rise to 124 the contradiction Mj. :£ ^H (V) ^ M2. Lemma 2.30 is thereby proved.

Proposition 2.31. Let L denote an arbitrary subgroup of

LyS isomorphic to L 3 (5). Then NQ(L) is not maximal in

LyS.

Proof. With H , , and M 2 as in the previous lemma, let X be a subgroup of H isomorphic to L 3 (5). By virtue of lemma 2.29, we may assume that N^(P) is a Sylow normalizer in X where P is Sylow in L . In particular,

P is Sylow in X as well. Now from rank-2 Chevalley theory, it is well known that P is generated by two of its elementary abelian subgroups Ej and E 2 (each of order 25 ) with Ex, E 2 interchanged by an outer automor­ phism of X . In addition N^CEj), NL(E2) are maximal parabolics in L so that L = . Finally, from the structure of H = G 2 (5), we know that NH (X) is isomorphic to Aut(X) [ 8 ].

Now as Nx (E1) ’s E 5 2 *GL(2 ,5 ) is a (maximal) parabol­ ic of X , we can invoke a theorem of Borel and Tits [ 1 ] to embed NX(E1) in a maximal parabolic of H . Without loss of generality, we may therefore assume Nx(Ea) ^ Mx or M 2 . 125

Suppose Nx(Ex) ^ . Then we may assume NX(EX) =

E x-K < M1 = R«K with K SGL(2,5). If Ex < R, then E 1 is a K-invariant suhspace of R, contradicting lemma 2.30.

Thus E 1 ^ R and REX is a Sylow 5-subgroup of LyS normalized by K . As Nq(S 5 )/S5 = x , this is an ob­ vious contradiction, and we conclude that NX(EX) < M2. We show presently that E x = .

First we observe that NX (E1) = E x*K «s M2 = C^(f1,g 2 ) *K.

Thus Ej < CQ

CQ), we see that K normal­ izes E1. Clearly |E1| ^ 54. If the order of E 1 is 5 4 , then E 1 is normal in C ^ f ^ g ^ whence normal in M2. By [13,p.552] this is a contradic­ tion, as the maximal such subgroup of CG(f1 ,g2) has order

53. Thus E 1 has order at most 53. If this order is exactly 5 3 then n E x is a K-invariant subgroup of of order 5. But n E! is central, so certainly normal, in CG(fi,g2). Thus n Ex is M2- invariant, a contradiction as is minimal normal in

M 2 by [13,p.552]. We conclude that |E1 | = 5 2 whence = Ex as desired.

Now from [13,p.552] we see that N(,() is contained in H . As discussed earlier, there exists an element x of NH(X) = Aut(X) such that x = E 2 . 126

Thus Nq(E2 ) = NG ()x ^ H. We may now conclude that

L - ),NL(E2)> < ),NG (E2 )> < H .

Finally, let n € NG(L). Then L = Ln ^ Hn whence

L 5 H n H n , By [13, p. 565] , the only double-point stabil­

izer (in the action of LyS on the cosets of H in LyS) which contains L is H itself. Thus H = Hn and n is

an element of NG(H) = H, i.e. NG(L) ^ H . The proof of the proposition is now complete.

Theorem 2.32. Let M be a maximal non-local subgroup of

LyS. Then M = NG(K) where K is isomorphic to one of the following groups: L 2 (7), L 2 (8 ), L 2 (9), L 2 (ll), U 3 (3),

G 2 (5), Mu. (Of these groups, G 2 (5) is the only one pres­ ently known to occur as a maximal subgroup of LyS. )

Proof. The theorem is an immediate consequence of theorem

2.12 and propositions 2.16, 2.20, 2.21, and 2.31. 127

2. Conclusions

For convenience we begin with a restatement of the following theorem from Chapter I, Section 5.

Theorem.1.47. With ,M2 ,...,MX Q as described below,

{Mx,M2 ,...,MX0> constitutes a complete set of represen­ tatives of the maximal local subgroups of LyS . This is to say every local subgroup of LyS is conjugate to a subgroup of some Mj_ and {Mj,M2 ,...,M10} with respect to this property.

local subgroup P isomorphism type order /s

M x=Ng () 2 A n 28-31*-52-7-ll

M 2=Ng ( ) 2 E 2 3\GL( 3,2) 26•3 • 7

3 Z 3\ Aut ( Me ) 28 -37 -53-7-11 M 3=NG (:)

M 4 =Ng (V) 3 E 3 5‘(Mj! X Z2 ) 25 • 37 • 5 • 11

3 2 +^.(SL(2,5)*Zq)• , • M 5=NG (

;)

M 7 ~Nq( S 3 >) 5 E 5 3\SL(3,5) 25• 3 • 58•31

m 8 =ng (s3 i) 31 Frob(31,6) 2 - 3*31

M 9 =Ng (S37) 37 Frob(37,18) 2 - 32-37

Frob(67,22) 2 - 11-67 M 10_NG^®67^ 67 128

Propos it ion 2.33. M2 and M0 are not maximal in LyS.

Proof. By Sylow's theorem and the fact that G 2 (5) pos­ sesses 5 classes of elements of order 31 , it is easy to see that M 0 is contained in a copy of G 2 (5) in LyS.

As for M2, we first observe [ 7 ] that G 2 (5 ) contains a group E = E 8 whose normalizer N(E) in G 2 (5) is iso­ morphic to Eg\GL(3,2). Identifying G 2 (5) with a sub­ group H of LyS, we claim that N(E) = N^(E) is conju­ gate to M 2 . Suppose not. Then E is not conjugate to

, so by lemma 1.6 E is centralized by some 3X- element. By lemma 1.3 we now have that Ng(E) is contained in some 3-local subgroup of LyS, and it therefore follows that Nh (E) < Nq () where x has LyS-type 3j. (Indeed, no other class of maximal 3-locals in LyS has represen­ tatives whose order is divisible by 7 .) But now as NH(E) is perfect we have Nh (E) = [NH (E),NH (E)] < [Nq ( ),N^( )]

= CG (x).

We next show that NG(Ng(E)) :£ H. Let n be an arbi­ trary element of NG(NH(E)). Then NH(E) = NH (E)n < Hn so that Njj(E) < H n H n . By [13,p.565] the double-point stabilizers (in the action of LyS on the cosets of H in

LyS) are isomorphic to G 2 (5), (SL(2,3)*SL(2,5))•Z2,

5 1+lf *SL(2,3)•Z ^ , U 3 (3), and Z 3 »PGL(2,7). As |HnHn |2^

|NH (E)|2 = 2 6 , the only possibilities for the isomorphism 129

type of H n H n are G 2 (5) and (SL(2,5)*SL(2,5))*Z2 . As the latter group has order not divisible by 7 , we see that

H nH n & G2 (5) whence H = Hn . Thus n € NQ(H) = H and the inclusion Nq (Nh (E)) ^ H is proved. In particular x€H whence x € N H(E) < CG (x). Thus is central in N^(E) .

Modulo E , this implies that Z 3 = is central in

Njj(E) = GL(3,2), a contradiction. We therefore conclude that Njj(E) is indeed conjugate to M2, which proves the non-maximality of M 2 in LyS.

Lemma 2.34. Let M be a maximal non-local subgroup of

LyS. Then M embeds in Aut(K) where K is isomorphic to one of L 2 (7), L 2 (8 ), L 2 (9), L 2 (ll), U 3 (3), G 2 (5), M u .

Furthermore, if p divides |M| for p f 2 , 3 , then p divides |K|.

Proof. Let F*(M) denote the generalized Fitting subgroup of M . Then F*(M) = F(M)E(M) where F(M) is the Fitting subgroup of M and E(M) has the property:

E(M)/Z(E(M)) s Ki x K 2 x • " x Kr with each simple, non- solvable. Now if p is a prime divisor of the order of

F(M), then a Sylow p-subgroup of F(M) is necessarily nor­ mal in M whence M is p-local. As M is non-local by assumption, we conclude that F(M) = 1 whence it follows that F*(M) = E(M) and (as Z(E(M)) < F(M) ) Z(E(M)) = 1. 130

Therefore F*(M)=E(M) is itself a direct product of sim­ ple non-solvable subgroups of M . We conclude from proposi­ tion 2.1 that F*(M) is in fact simple whence by theorem

2.32, M = Nq(F*(M)) with F*(M) isomorphic to one of

L 2 (7), L 2 (8 ), L 2 (9), L 2 (ll), U 3 (3), G 2 (5), Mxx. As

Cq(F*(M)) = Cm(F*(M)> = Z(F*(M))‘ = Z(E (M )) = 1 , we may now conclude that M embeds in Aut(F*(M)) as desired.

Finally suppose p divides |M| where p is a prime distinct from 2 or 3. Then by the previous para­ graph, p divides |Aut(K)| = |K||0ut(K)| where K is as described in the lemma statement. As it is easily checked

[9] that Out(K) is a y-group for y a subset of

{2,3}, the result follows at once. 131

Proposition 2.35. Mi is maximal in LyS for i f 2,8.

Thus {Mj,M3 ,M^,M5 ,M6 ,M7 ,M9 ,Mio} is a complete set of representatives for the maximal subgroups of LyS which are local.

Proof. As {Mj,M2 ,...,Mjo} is already known to be a com­ plete set of representatives for the maximal local sub­ groups of LyS (theorem 1.47), it suffices to show that Mi is not contained in any maximal non-local subgroup of LyS for i f 2 , 8 . We accomplish this presently.

Suppose then, by way of contradiction, that Mi is contained in some maximal non-local subgroup of LyS for i f 2 , 8 . By lemma 2.34 Mi embeds in Aut(K) where

K is isomorphic to one of L 2 (q) ( q = 7,8,9,11), U 3 <3 ),

G 2 (5), MX1. As K has order not divisible by 37, 67, and

77, we see that i cannot be 1, 3, 9, or 10 (lemma 2.34).

Moreover as |Aut(K) | 3 — 33, i ^ 4 , 5 . Finally suppose i is equal to either 6 or 7 . Then 5 s divides |Aut(K)| whence it is easily verified that K = G 2 (5) [ 9]. But as each of Me and M 7 is 5-local, each must surely be con­ tained in a maximal parabolic of K for the prime 5. As the maximal parabolics for G 2 (5) both have order 2S*3*56, we obtain the final contradiction. This proves that each of

Mi, M 3, Mt+, Ms, M g , M 7 , Mg, Mio is indeed maximal in LyS as claimed. The proof of the proposition is now complete. 132

Theorem 2.36. Let M i , M 2 ,•••, M 10 be as in theorem 1.47 and let H be a subgroup of LyS isomorphic to G 2 (5). If

M is conjugate to a member of {Mx ,M3 ,Mi* ,M5 ,M6 ,M7 ,M9 ,M10 ,H} then M is maximal in LyS. Conversely, if M is an arbi­ trary maximal subgroup of LyS, then either M is conjugate to a member of the above set or M = N^(K) where K is isomorphic to one of the following groups: L 2 (7), L 2 (8 ),

L 2 (9), L 2 (11), U 3 (3), MX1.

Proof. From theorems 1.47 and 2.32 it is immediate that H is maximal in LyS; by proposition 2.35, each of M x, M3,

Mi*, M 5 , Me, M 7 , Mg, M x 0 is maximal in LyS as well. This proves the first statement of the theorem.

For the converse statement, let M denote an arbitrary maximal subgroup of LyS. If M is local, then by proposi­ tion 2.33 M is conjugate to one of the M.^ where i f 2 ,8 .

If M is non-local, we conclude from theorem 2.32 that M =

Nq(K) for K isomorphic to one of L 2 (7), L 2 (8 ), L 2 (9),

L 2 (ll), U 3 (3), G 2 (5), Mu. Tlle theorem now follows from the fact that LyS contains a unique class of subgroups isomor­ phic to G 2 (5) [13,p.565,14-29]. 133

Proposition 2.37. Let H be any subgroup of LyS such that H n Hx contains a full Sylow p-normalizer for some prime p . Then H = Hx.

Proof. Clearly Sp < H n Hx whence Sp and SpX 1 are

Sylow p-subgroups of H . Thus there exists an element h in H such that SpX = Sp . But then x_1h € NG (Sp), and as Ng(Sp) < H by assumption, we obtain x € H. The result clearly follows.

Lemma 2.38. Nq(S2) = S 2 .

Proof. It naturally suffices to show that Nq (S2 ) is a

2-group. Suppose, by way of contradiction, that x € N G(S2) has order p , p an odd prime. Clearly x centralizes

= Z(S2) so that x is an element of odd prime order in CQ(z) = CQ(z)/. Moreover x normalizes S”2 which is a Sylow 2-subgroup of CG(z). Identifying CQ(z) with

A,i we may therefore assume x is an element of N.(P) Au. where P is the Sylow 2-subgroup of Ax x given as follows:

P = ( D x D' )*E with D = (<(12)(34)> x <(13)( 24 ) > ) • < (12 )( 9 10) >,

D'= (<(56)(78)> x < (57) (68)> ) • < ( 56) (9 10)> , and E =

<(15)(26)(37)(48)>. (Thus D = D' = D 8 , and P = Da \Z 2 =

( D a x d 8 )*Z2.) We now show that Z(P) = where t =

(12)(34)(56)(78). First we claim that Z(P) < DxD'. If not, let y € Z(P) with y = gf , g€D xD' and f = (15) (26) (37) (48) . 134

Then for u = (12)(34) we have 1 = [u,y] = [u,gf] = [u,f] , a

contradiction. (Note that we have used the fact that u and g commute as u rs Z(DxD'). ) Thus Z(P) < Dx D 1 as claimed, whence Z(P) < Z(DxD') = Z(D) x Z(D') =

<(12)(34)>x <(56)(78)>. But f interchanges (12)(34) and (56)(78) whence Z(P) = as originally asserted.

As Z(P) is characteristic in P, x normalizes so centralizes t. As C.(t) is a (2,3)-group (e.g. see H xi proof of lemma 1.3), x (of odd prime order) must be of order 3 . Clearly x fixes {9,10,11) setwise; as x has order 3 we see that x either fixes {9,10,11} point- wise or permutes 9, 10, and 11 cyclicly. But the latter cannot occur as then we would have (12)(9 10)x = (12) (30 31), a contradiction as (12)(1011) is not in P . We conclude that x € C_(t) where B = A{1,2,...,8), and we may now 15 assume that x = (135)(246). But then (13)(24)x = (35)(46) must be an element of P . As DxD' fixes each of {1,2,3,4) and {5,6 ,7,8) setwise, we easily see that (35)(46) is not contained in D X D', Thus (35) (46) = d( 15) (26) (37) (48) for some d in D*D'. But then d = (1573) (2684), giving the same contradiction as above. We conclude that Nq(S2 )

is a 2-group and the proof of the lemma is now complete. 135

Theorem 2.39. The lattice of maximal subgroups of LyS which contain NG(S2 ) is given by

LyS

Ng (S2 )

where A = A 1]L and B = Z3\Aut(Mc).

Proof. From theorem 1.47 and lemma 2.38, we see that Ng (S2) is indeed contained in maximal subgroups A and B as de­ scribed in the theorem statement. Theorem 1.47 also shows that N q (S2) is not contained in any maximal local which is not conjugate to A or B. It now follows from proposition

2.37 that A and B are the only maximal local subgroups of LyS which contain NG(S2).

Suppose now that Nq(S2) ^ M where M is a maximal non-local of LyS. By lemma 2.34, M embeds into Aut(K) where K is isomorphic to one of L 2(q) (q=7,8 ,9,11), U 3(3), G 2(5), Mu. As |Aut(K)|2 ^ 26 , we obtain a con­ tradiction. We therefore conclude that NG(S2) is not contained in any maximal non-local subgroup of LyS and the theorem is proved. 137

Theorem 2.40. The lattice of maximal subgroups of LyS which contain NG(S3) is given, by

LyS

Ng (S3 )

where A = 32+l* • (SL(2 ,5)*Z8 ) *Z2 and B = E 35•(Mii * Z2 ).

Proof. As each of Z(S3) and V = is characteristic in S3, we certainly have that NG(S3) is contained in both NG (Z(S3 )) = 32+lf • (SL(2 , 5)*Z8 ) • Z2 and

N g (V) = E 35*(Mh x Z2). Theorem 1.47 and proposition 2.37 together imply that no other maximal local contains NG(S3).

As it was earlier shown (in the proof of proposition 2.35) that no maximal non-local subgroup of LyS contains a Sylow

3-subgroup, the lattice is now complete. 138

Theorem 2.41. The lattice of maximal subgroups of LyS which contain N G (Ss) is given by

LyS

where A S 5 1 + lt-SL(2,9) , B = E 53\SL(3,5) , and

C = G 2(5).

Proof. Using the presentation for S 5 (1.13) it is easy to verify that E = [S,S] where E = anc*

S = • By [13, p. 550,11.7-11] S is a characteristic subgroup of S5, so by the above it follows that E is characteristic in S 5 as well. This proves

Ng(S5) < Nq(E) s E 53\SL(3 ,5). As Z(S5) = , we also see that NG (S5 ) < NQ () = 5l + lt . S L ( 2 , 9 ) . Finally from theorem 1.47 and proposition 2.37, we conclude that

NG() and NG(E) are the only maximal locals of LyS which contain NG(S5).

Turning now to the non-locals, we recall from the proof of proposition 2.35 that K = G 2(5) is the only 139 maximal non-local subgroup of LyS which contains a Sylow

5-subgroup. But NK(S5)/S5 = Z 4. * Zi* = NG (Ss)/S5 where

K is a copy of G 2(5) in LyS containing S 5. Thus

Nq(Ss) is contained in K, and by proposition 2.37, K is the unique maximal non-local in LyS with this property.

The theorem is thereby proved. 140

Theorem 2.42. The lattice of maximal subgroups of LyS

which contain N q (S7) is given by

LyS

A

n g (s 7>

where A = Aj j.

Proof. By lemma 1.44, NG(S7) is indeed contained in a

copy A of Ajj in LyS. By Lagrange's theorem, theorem

1.47, and propositions 2.33 and 2.37, the only other maxi­ mal local subgroup of LyS that could possibly contain

Nq(S7) is a copy of Z 3\Aut(Mc). Suppose NG(S7) < B where B is such a copy. Then Nq(S7) = Ng(S7) and by

Sylow’s theorem 2i*»3s*56*ll = |B:Ng(S7)| is congruent to

1 mod 7 . As this is an obvious contradiction, we conclude

that A is the unique maximal local subgroup of LyS which

contains NG(S7).

We show presently that no maximal non-local of LyS

contains NG (S7 ). Indeed, let N q (K) be a maximal non­

local containing NG(S7). By lemma 2.34, 7 divides the 141

order of K whence K is isomorphic to one of L 2(7),

L 2(8 ), U 3(3), G 2(5). But Ng(S7) contains elements of

order 14 (e.g. see Table 1), while Aut(K) does not.

This is a contradiction as NG(Sy) embeds in Aut(K) by

lemma 2.34. The theorem is thereby proved. 142

Theorem 2.43. The lattice of maximal subgroups of LyS which contain Nq CSjj) is given by

LyS

A

n g (s h )

where A = Z 3\Aut(Mc).

Proof. That such a group A exists as in the theorem statement follows immediately from lemma 1.45. From theorem 1.47 and proposition 2.37 , we are able to elimi­ nate all remaining maximal locals except those isomorphic

A to A n or E 35 - (Mi i xZ2) as possible overgroups of

NqCSh). But neither of these contains elements of order

33, while NG ( S n ) in fact does. (An element of order 33 in E 35 *(M1 1 xZ2) must lie in E 35 *MX1. By Table 6 , no such element exists.) We therefore conclude that A is the unique maximal local subgroup of LyS which contains

Ng(Sh), the desired result. 143

Suppose next that Nq (Six) contained in N^(K),

a maximal non-local subgroup of LyS . By lemma 2.34, K must be isomorphic to either L 2(ll) or M ^ . But NqCSjx)

contains elements of order 33 , while each of Aut(L2(ll))

and A u t ( M u ) do not. As Ng(Sxi) embeds in Aut(K) by

lemma 2.34, we obtain a contradiction. The proof of the

theorem is now complete. 144

Theorem 2.44. The lattice of maximal subgroups of LyS which contain NG(S31) is given by

LyS

A

Ng (S31)

where A = G 2(5).

Proof. As evidenced in proposition 2.33, Ng(S3i) is contained in a subgroup A of LyS isomorphic to G 2(5).

By theorem 1.47 and Lagrange's theorem, no maximal local subgroup of LyS contains Ng(S3i); by theorem 2.32, lemma

2.34, and proposition 2.37,. A is the unique maximal non­ local containing it. The proof of the theorem is now com­ plete. 145

Theorem 2.45. The lattice of maximal subgroups of LyS which contain Nq(S3 7) is given by

LyS

Ng (S37) .

Proof. The theorem is an immediate consequence of theorem

2. 36.

Theorem 2.46. The lattice of maximal subgroups of LyS which contain Nq(S67) is given by

LyS

N g (S67)

Proof. The theorem is an immediate consequence of theorem

2.36. CHAPTER III

EVIDENCE FOR A 111-DIMENSIONAL IRREDUCIBLE

LyS-MODULE OVER A FIELD OF CHARACTERISTIC 5

Our goal in this chapter will be to supply evidence for the existence of a 111-dimensional irreducible

F[LyS]-module, where F is a field of characteristic 5.

The inspiration for this work stems from research recently conducted at Cambridge University and communicated to me by David Benson. Although the existence of such a module remains an open question at the time of this writing,

Conway et. al. have developed a plausible probablistic argument that such a module exists by way of constructing and analyzing representing matrices for what appear to be generators of Lyons' group inside GL(111,F), F an appropriate finite field of characteristic 5.

Returning then to the task in question, we first perform a rather indepth study of the 5-modular structure of various distinguished subgroups of LyS. We then show that a hypothetical irreducible 5-Brauer character of

LyS of degree 111 withstands many compatability checks

146 147

involving restriction to subgroups and values on 5-

regular LyS-classes.

Faithful Brauer irreducibles of smaller degree simply do not exist. By appropriate ordinary and modular charac­ ter theoretic arguments, we subsequently prove that 111

is indeed a lower bound for the degree of any faithful F- character of LyS, F any field, and we show that this

lower bound is in fact much greater if we further impose that the characteristic of F be different from 5 .

We list below, for easy reference, various facts from elementary modular character theory which will be used freely throughout the chapter. G denotes a finite group, p a prime.

3.1. Let ip € Irr(G). ip has p-defect zero if and

only if | Gj p divides ijKl) •

3.2. Let x>V € Irr(G). x » ^ belong to the same

p-block of G if and only if

|K|X (x) | K | t|j(x ) ------= (mod p) X d ) « K D

holds for every p-regular class K of G .

( x here is a representative of K .) 148

3.3. Let IBrp(G) denote the set of all irreducible

p-Brauer characters of G. Then |lBrp(G)|

equals the number of p-regular classes of G.

3.4. If x is a p-projective Brauer character

of G, then |G|p divides x(l)- Moreover,

X vanishes on all p-singular classes of G.

3.5. A projective character can be uniquely expressed

as a non-negative integral sum of projective

indecomposable characters.

3.6. Let H and K denote subgroups of G with

H < K. Let i|; be a projective character of

K. Then i|j+G and iJ>+H are each projective.

3.7. Let be a p-projective character of G

and 0 any p-Brauer character of G. Then

ip ® 0 is p-projective.

3.8. The dual ^ of a projective (resp. indecompo­

sable, irreducible) character ip is projective

(resp. indecomposable, irreducible). Let be a Brauer character of G, a an automorphism of G. Then

<{>a(x) = (xa) is a Brauer character of G , and is.

Let x = 2ae0 be a p-projective character of

G ( 0 € Irr(G) ), and B a p-block of G.

Define b 0 as follows:

a 0 , 0 € B

0 , 0 i B

Then Eb 06 is p-projective. (We call Eb00 the block projective of x with respect to

B and denote it by {x^b > or simPly by {X}, when the block B is understood.)

Let € IBrp(G), ip an arbitrary p-Brauer character of G . Then the multiplicity of

^ in ip equals the multiplicity of in ip.

Let B be a p-block of G having cyclic defect group. Then a tree can be drawn with vertices the ordinary irreducibles of G oc­ curring in B and edges the modular irreduc- 150

ibles which correspond to B such that two

vertices Xi and Xj ar© joined by the edge if and only if tf> is a common constituent

of Xi anc* Xj • This tree (called the Brauer

tree of B ) uniquely determines the decompo­

sition of the block. Moreover, if all the

characters in the block are real-valued on p-

regular elements, then the tree is an open

polygon.

3.13. Let B be a p-block of G having cyclic

defect group. Then all entries in the decom­

position matrix for the block B are from

{0 ,1}.

3.14. Let $ = Zaxx be a projective character of

G ( X € Irr(G) ) with d a divisor of each

of the (integral) coefficients ax. Then

(l/d)$ is projective.

As is customary in the literature, we shall gen­

erally denote an irreducible Brauer character \p by its

degree n = ip(l), using primes to distinguish amomg char­

acters of the same degree (e.g. n , n' , n" , ...). 151

n = n x +n2 + ••• + nr shall represent a decomposi­ tion of n into constituents at the character theoretic level and is in no way to be interpreted as a statement about the splitting of modules.

Our use of the term ’’block" will be somewhat non­ standard. To us, a block shall consist only of ordinary irreducible characters and we shall speak of irreducible

Brauer characters corresponding to a given block. (Thus in the text we shall denote by B what is commonly denoted by Bnlrr(G) in the literature, where B is a block for a fixed prime and the group G .)

Finally we remark that blank entries occurring in all induction-restriction tables and decomposition matrices which appear in this chapter are to be interpreted as zeros. 152

1. 5-Modular Analysis of Me

In this section we attempt to derive the 5-decompo­ sition matrix for Me, McLaughlin’s sporadic simple group.

An interesting problem in its own right, its relevance here stems from the fact that Me occurs as a section in LyS.

Only partial success is achieved.

As Ui*(3) and M 22 are prominent subgroups of Me, it is natural to appeal to their respective 5-modular structures for information. With this purpose in mind, we begin by reproducing the 5-decomposition matrices for

Ui*C3) and M 22> as well as the induction-restriction tables for the pairs (Me,11^(3)) and (Mc,M22) (Table

2 and Table 3 respectively). These were determined by J. Thackray in his Ph.D. thesis, in which (among other things) he derived the decomposition matrices for Me for all relevant primes p , except for p = 3 and p = 5

[18] • 153

(i) The 5-decomposition matrix for 1^(3) is given by

1 5 D = D,

where 1 188 708 21

1 1

189 1

D, 896 1 1

729 1 1

21 1

(ii) The 5-decomposition matrix for M 22 is given by

Di

where 1 98 133 21

1 1

99 1

D, 231 1 1

154 1 1

21 1 TABLE 2. The Character Induction-Restriction Table for (110,11^(3)) U„(3) - 1 21 35 35/ 90 140 189 210 280 280 280/ 280/315 315/420 560 640 640 729 891 UMe1U + 1 1 22 1 1 231 1 1 252 1 1 1 1 770 1 1 770 1 1 896 1 896 1 1750 1 1 2 1 1 1 3520 1 1 1 1 1 1 1 1 1 1 3520 1 1 1 1 2 1 1 4500 1 1 1 1 1 1 2 4752 1 1 1 1 1 1 1 2 5103 1 1 1 1 1 1 1 1 2 1 5544 1 1 1 1 1 1 2 1 1 1 1 8019 1 1 1 1 1 1 2 1 2 2 2 8019 1 1 1 1 1 1 2 2 1 2 2 8250 1 1 1 1 1 1 1 1 1 1 1 2 2 2 8250 1 1 1 1 1 1 1 1 1 1 2 1 2 2 9625 1 1 1 1 1 2 2 1 2 2 3 3 9856 1 1 1 1 2 2 2 2 2 3 9856 1 1 1 1 2 2 2 2 2 3 10395 1 1 1 1 1 1 1 1 1 2 2 2 2 3 10395 1 1 1 1 1 1 1 1 1 2 2 2 2 3 154 TABLE 3

The Character Induction-Restriction Table for (Mc,M22

1 21 45 45 55 99 154 210 231 280 280 385 c 1 1

22 1 1

231 1 1

252 1 2 1 1

770 1 1 1

770 1 1 1

896 1 1 1

896 1 1 1

1750 1 2 2 1 3 2 1 1

3520 2 2 1 3 3 2 1 1 3

352o' 1 1 2 4 4 2

4500 1 1 1 1 2 1 3 2 2 5

4752 1 1 2 2 4 4 4

5103 1 2 2 4 2 3 2 2 5

5544 1 1 1 1 3 1 5 5 4

8019 1 2 2 2 3 3 6 6 7

8019 2 1 2 2 3 3 6 6 7

8250 1 1 1 2 3 5 4 5 5 7

8250 1 1 1 2 3 5 4 5 5 7

9625 1 3 3 6 5 8 4 4 8

9856 1 1 1 2 3 4 6 6 6 9

9856 1 1 1 2 3 4 6 6 6 9

10395 1 1 1 2 3 5 5 7 7 9

10395 1 1 1 2 3 5 5 7 7 9 156

Theorem 3.1. The 5-decomposition matrix for Me is

I.

Di where Dx is given below.

1 21 210 230 56( 896 896 8 1200 1200 4>ii W i 1 1

22 1 1

231 1 1

252 1 1 1

770 1 1

770 1 1 896 1 896 1

3520 a 1 1 1

3520/ 2 1 1

4752 1 11 11

5103 1 a 1 1 1 1

5544 1 1 1 2 11 11

8019 1 3 e 1 1 5 1 1

8019 1 3 e II 15 1 9856 Y e 1115 ? 1-S 9856 Y e I I I 5 1-C ?

10395 1 a 1 2 2 2 111 10395 1 a 1 2 2 2 111

In the above, a , 8 > Y > 5 , e , ? are non-negative integers subject to the following conditions: c t ^ 2 , 8 ^ 2 , y ^ 2 , 157

6^1, e ^ 2, 5 ^ 1 , y = a + $ - 2. Moreover, if C = 0 then

6 = 1. Finally <(>8 and ii have degrees 3080 - 21a

( ^ 3038) and 5026 - 12006 - 560e - 21p ( ^ 2664) re­ spectively .

Proof. It is immediate from 3.1 that {1750,4500,8250,

8250,9625} is precisely the set of defect zero characters of Me. (We suppress the prime p from our notation as p = 5 is understood throughout.) It is also immediate

(from 3.2 ) that all characters of non-zero defect lie in a common block (the principal block), whence the matrix

Dj consists of 19 rows. Finally by 3.3, Dx consists of 12 columns as Me possesses 17 5-regular classes, and 5 irreducible Brauer characters (those of defect zero) have already been accounted for. Thus the decomposition matrix for Me is indeed of the form

D = D i where D: is a 19 x 12 matrix which shall be described shortly. Following convention, we index the columns of D1 by the irreducible Brauer characters of Me occurring in the principal block: ! , ej>2 , ... , 12 . we denote by the projective cover of (j>^, 1 — i — 12, so that $i,$2 > • •>^12 158

are precisely the projective indecomposable characters of

Me which are in natural correspondence with the columns of Dj . We also index the rows of D1 by the ordinary irreducible characters of Me of non-zero defect: 1, 22,

231, 252, 770, 770, 896, 896, 3520, 3520', 4752, 5103, 5544,

8019, 8019, 9856, 9856, 10395, 10395. We shall refer to rows and columns of D x by the appropriate index (e.g. row 770, column <{j3, etc.)

We now proceed with the proof:

(1) Without loss of generality, set x =1. (Our double

use of the symbol 1 should cause no confusion.

Here 1 denotes the principal character (also 1)

restricted to the 5-regular classes of Me.

Clearly 1 6 IBr(Mc).) Completing row 1 is a

triviality.

(2) From the decomposition matrix for 1^(3), it is im­

mediate that 1 + 189 is a projective character of

U 4(3). Thus {(1+189)+Mc} = 1 + 22 + 252 + 5103 +

5544 + 8019 + 8019 + 10395 + 10395 is projective

as well (see 3.6, 3.10). As is self-dual,

8019 < if and only if 8019 < . A similar

statement holds for 10395 and 10395. Using this

fact along with 3.4, one readily concludes that

{(1+189)+Mc} = $!• (Although 5 3 = |Mc|s divides 159

the degree of 22 + 5103, this character fails to

vanish on 5-singular classes.) Thus we obtain

column 1.

(3) From column 1, 1 < 22 (as Brauer characters) so 22

is not Brauer irreducible. As 22+M22 = 1 + 21

where 1,21 € IBr(Mc), we see that 21 = 22 - 1 is

an irreducible Brauer character of Me. Set

21. Row 22 can now be completed.

(4) {210+Mc} = 231 + 770 + 770 + 3520 + 5544 + 10395 +

10395 where 210 € IBr(U4(3)). As 210 has defect

zero (so is projective), {210+Mc} is projective

(3.6). By 3.4, {210+Mc} is indecomposable.

Column (p3 is thereby obtained.

(5) 770+Uu(3) = 210 + 560, so either 770 € IBr(Mc) or

770 = 210 + 560 with 210, 560€IBr(Mc). As

3 < 770 as Brauer characters, we have 3 = 210,

560, or 770. But 3 < 231 (column 3), so we

conclude that <)>3 = 210 and 560 € IBr(Mc). Set

<(>5 = 560. As 770 = 770 as Brauer characters,

rows 770, 770 may now be completed. 160

(6) Computation of 22® 22 gives the following:

1 + 252 + 231 = 22 ® 22 = (21+1) ® (21+1) =

2 1 0 21 + 2-21 + 1. Thus 21 ® 21 = 252 + 231 - 2-21.

As 210 < 231, we cannot have 2*21 < 231. Thus

21 < 252. From column 1, 1 < 252 as well. We

claim 230 = 252 - 1 - 21 is Brauer irreducible.

Indeed no other possibility is compatible with the

two restrictions of 252 into irreducible Brauer

constituents:

252+U4(3) = 1 + 21 + 90 + 140

252+M22 = 1 + 3*21 + 55 + 133

We set 4^ = 230. As we now have 2*21 ^ 252, it

follows that 21 < 231, and rows 231, 252 are

each easily completed.

(7) Observe that 896+11^(3) = 108 + 788 and 896+M22 =

98 + 133 + 280 + 385, where all constituents appear­

ing are Brauer irreducible. As no proper decompo­

sition of 896 in Me is compatable with both re­

strictions, we conclude that 896 € IBr(Mc). By

3.8, 896 € IBr(Mc) as well. Set 6 = 896,

<|>7 = 896. Rows 896, 896 are trivially completed. 161

(8) By 3.4, 3.6, and 3.10, {90+Mc} = 252 + 3520 + 5103

is trivially projective indecomposable. (Here 90

represents the unique irreducible of U4(3) of that

degree.)

(9) {55+Mc} = 252 + 2*3520 + 2*5103 + 9856 + 9856 +

10395 + 10395 is projective (3.6,3.10) where 55

is the unique irreducible of M22 that degree.

As 252 < {55+Mc} but 22 ^ {55+Mc}, we see that

must occur precisely once as a constituent of

{55+Mc}. But then by 3.4, {55+Mc} - ^ =

3520 + 5103 + 9856 + 9856 + 10395 + 10395 must be

indecomposable. Without loss of generality we

assume $8 = {55+Mc} - and we complete

column cj> 8.

(10) Consider {280+Mc} = 352o'+ 4752 + 5544 + 8019 +

8019 + 9856 + 10395 + 10395 where 280 € IBr(Utf(3)).

Let be the unique projective indecomposable

constituent of {280+Mc} satisfying 3520/< <5-^.

By inspection, i ^ 9; so without loss of generality

i = 9. Now applying 3.4, we see that either

$9 = {280+Mc} or $9 = {280+Mc} - (8019 + 9856).

(Indeed, if $9 = {280+Mc} - (8019 + 9856), the 162

latter obtains as a consequence of relabeling.)

This determines column 9 as indicated in the

theorem statement, with 6 = 0 or 1. In any

case $ 9 is not self-dual, so we set ¥ 7 o = $ 9 >

and column 10 is determined as well.

(11) We now show that cf>12 = TTi • Suppose not. Then

we must have , c|>12 = * From the

character table for Aut(Mc) it is readily seen

that there exists an outer automorphism a of Me

such that 896° = 896, 896a = 896, 8019° = 8019,

and 9856a = 9856. Thus, as Brauer characters,

the multiplicity of 896 in 8019 is equal to

its multiplicity in 8019. A similar statement

holds if we replace 8019 with 9856 in the above.

But then 6 = 0 implies 9856 = 9856 as Brauer

characters, a contradiction as 9856, 9856 disagree

on elements of order 9. Thus 6=1. But this

implies 8019 = 8019 as Brauer characters. As

8019, 8019 disagree on elements of order 7, we

obtain our final contradiction and 4>12 = as

claimed. 163

(12) As 4500 6 IBr(Mc) is of defect zero (and so is

projective) we see from 3.7 (and 3.10) that

{218 4500} = 3520 + 5103 + 8019 + 8019 + 2*9856 +

2*9856 + 10395 + 10395 is projective.

If $ 8 £ {21®4500}, then by necessity

3520 < $u < {218 4500}. As each of {21® 4500}

and 3520 is self-dual, we obtain

3520 < $ 12 < {21.® 4500} as well. But then

2*3520 < {21«4500}, a contradiction. Thus

$ 8 < {21®4500} and {218.4500} - $ 8 = 8019 +

8019 + 9856 + 9856 is projective. As = $ 12

we conclude without loss of generality that

$1X = 8019 + 9856 or $n = 8019 + 9856. As we

cannot eliminate this ambiguity, column <|}x x is

completed as indicated in the theorem statement.

If $u = 8019 + 9856 then 5 = 1; if * lx =

8019 + 9856 then C = 0. We finally observe

that the latter possibility ( ? = 0 ) implies that

the character {280+Mc} of (1 0 ) must be indecom­

posable, whence 6 = 1. As = $n, column 12

is also obtained.

(13) By the usual arguments, {560+Mc} = 770 + 770 +

2*3520' + 4752 + 2*5544 + 2*8019 + 2*8019 + 2*9856 +

2•9856 + 2*10395 + 2*10395 is projective, as 164

560 € IBrCU^CS)) has defect zero.

As 231 4 {560+Mc} we must have $s < {560fMc},

whence {560+Mc} - $ 5 is projective. It is now

easily argued that 4 {560+Mc} - $ 5 for all

i, 1 ^ i ^ 10 . As nothing conclusive can be

said about the cases i = 1 1 , 1 2 (except that the

multiplicities of $1X and $ 12 in {560+Mc} - $ 5

are equal), we can only determine column 560 to

within three possibilities. These are reflected

in the three choices for e appearing in column 560

in the theorem statement: e = 0 , 1 , or 2 .

(14) From the decomposition matrix for 1^(3), it is

apparent that 189 + 896 is projective, where

189, 896 6 IBr(Ult(3)). Therefore, so is

{(189+896)+Mc} = 896 + 896 + 2*4752 + 2*5103 +

2*5544 + 3*8019 + 3*8019 + 3*9856 + 3*9856 +

4*10395 + 4*10395. An easy calculation yields

21 ® 896 * 9856 + 9856 - 896, whence 896 < 9856.

Recalling our work in (11) , this implies

896 < 9856 as well. Furthermore 2*896 4 9856.

Indeed 2*896 < 9856 implies 2*9856 < $6.

As 2*896 < 9856 also follows (from (11)), we

see that 2•9856 < $ 6 as well, whence 2*9856 < $ 7 . 165

But this gives 4*9856 < $ 6 + $ 7 < {(189+896)tMe},

a contradiction. Thus 896 occurs in 9856 (and

so also in 9856 ) with multiplicity one. We are

now able to conclude that {(189+896)+Mc} = $ 6 + $ 7 +

$ ii + $ 1 2 - Columns 896, 896 are thereby determined.

(15) From the decomposition matrix for M22, we readily

see that 21 + 154 is projective, where

21, 154 € IBr(M22). Our usual arguments show that

{(21+154)+Mc>=22 + 231 + 3*252 + 5*3520 + 5*5103 +

5544 + 2*8019 + 2*8019 + 3*9856 + 3*9856 + 3*10395 +

3• 10395 is projective. As 1 j- {(21+154)+Mc}, we

immediately observe that $ 2 < {(21+154)+Mc}. This

proves at once that neither of 3520/, 4752 can be

a constituent of $ 2 . As i- {(21+154)+Mc} - 2

for all values of i for which 5544 < we Set

that the multiplicity of 5544 in $ 2 is precisely

one. As a consequence, we obtain rows 3520/, 4752,

and 5544, and 9 = 1200,

follows.

(16) The only possibility for the decomposition of

{(21+154)+Me} (as in (15) ) into a sum of inde-

composables is the following: 166

{(21+154)tMc} = $ 2 + 2®* + a® 8 + b(#ir + ® 12)

where a , b are non-negative integers. But

then, regardless of the values of a and b,

the multiplicities of 3520, 5103, 10395, and

10395 in $ 2 are seen to be equal. We denote

this common value by a. We also denote by 3

and y the respective multiplicities of 8019

and 9856 in $2. Column 21 is now completed

as indicated in the theorem statement.

(17) A trivial check (Table 2 ) of the restrictions

3520+1^(3) , 8019+1^(3) , 9856+1^(3) now yields

the respective inequalities:

* a ^ 2 , $ ^ 2 , and y ^ 2 .

By virtue of the fact that $ 2 must vanish on

5-singular classes (3.4), we obtain a further

relation:

y = a + 8 - 2 .

This follows from calculating $ 2 on the class

of 5-elements of Me which is not Sylow-central.

Finally we routinely compute 8 = 3080 - 21a and

12 = n where n = 5026-12006-560e-213• 167

Thus lower bounds for the degrees of lx

are respectively 3038 and 2664. The proof of

the theorem is now complete.

Remark. In an attempt to secure more precise results in the preceding theorem, I wrote several computer programs.

The following three are particularly noteworthy:

Program A decomposes into irreducible constituents

the restrictions to Me of each ordinary irreducible

character of Conway1s sporadic group .3 .

Program B computes and then decomposes into irredu­

cible constituents all second order tensors of

ordinary irreducibles of Me.

Program C finds dependence relations among the

natural Brauer characters of Me (i.e. the ordinary

irreducibles of Me restricted to 5-regular classes).

Although no further information could be gathered from these programs, they did nonetheless provide many valuable checks on information deduced from other sources. 168

We close this section with the following two propo­ sitions which, although not germane to the central theme of the text, still provide facts of general interest concern­ ing Conway's group .3 .

Proposition 3.2. The character induction-restriction table for the pair (.3,Me) is as appears in Table 4 below.

Proof. Program A (Fortran; Amdahl 740). 169

TABLE 4

The Character Induction-Restriction Table for (.3,Me)

o O o o 0 3 CO C l 0 1 o o m CO CO Me H N O o C O CO i n C 3 0 3 o m o r H H m i n 0 3 i n in 00 0 0 r H C O CO in m 00 00 00 0 0 0 3 0 1 ci 10395

1 1 + 23 1 1 253 1 1 253' 1 1 275 1 1 1 896 896 1771 1 1 1 2024 1 1 3520 1 3520 1 4025 1 1 1 1 . 5544 1 1 1 1 7084 1 1 8855 1 9625 1 9625 1 20608 1 1 20608 1 1 23000 1 1 1 1 1 26082 1 1 1 1 31625 1 1 1 1 1 1 1 31625' 1 1 1 1 1 1 1 31625" 1 2 1 1 2 31878 1 1 1 1 1 1 1 40250 1 1 1 1 1 57960 1 1 1 1 1 1 1 1 1 63250 1 1 1 1 1 1 1 1 73600 1 1 1 1 1 1 1 1 1 80960 2 2 2 1 1 1 1 1 1 91125 1 1 1 1 1 1 2 1 1 1 1 93312 1 1 H 1 1 1 1 1 1 1 129536 1 1 2 2 2 2 1 1 1 1 2 2 129536' 2 2 2 1 1 2 2 2 1 1 1 1 177100 1 2 2 1 2 2 1 1 2 2 2 2 2 184437 1 1 1 2 1 1 1. 2 2 3 2 2 2 2 221375 1 2 1 2 1 1 2 2 5 3 3 2 2 226688 1 1 2 2 1 1 2 2 2 2 3 2 2 3 3 246400 1 1 2 2 1 1 2 2 2 2 3 3 3 3 3 249480 2 1 2 1 2 3 3 2 2 1 3 3 3 3 253000 1 2 1 2 1 2 3 3 2 2 1 3 3 3 3 255024 1 1 1 2 2 2 2 2 3 3 2 3 3 3 3 170

Proposition 3.3. Conway's sporadic group .3 has two

5-blocks of non-zero defect: the principal block and the unique block of defect one given by

B = {275,4025,73600,177100,246400}.

The Brauer tree for the block B is given by

o------x------o------x------o 275 73600 246400 177100 4025

Proof. The first statement of the theorem is immediate from 3.2 . As each ordinary irreducible of B is real on 5-regular classes of .3, we may apply 3.12 to conclude that the Brauer tree for B is an open polygon.

By 3.4, the only possibilities for this tree are

o------x------o------x------o 275 73600 246400 177100 4025

and

o------x------o------x------o 275 177100 246400 73600 4025 .

We eliminate the second possibility presently.

Suppose then, by way of contradiction, that 4025 and 73600 are connected in the tree for B . This implies

69575 = 73600 - 4025 is Brauer irreducible, and thus

{69575+Mc} is a non-negative integral combination of ele­ ments of IBr(Mc). But {69575+Mc} = 5103 + 5544 + 8019 + 171

+ 10395 + 10395 - (22 + 231 + 252 + 3520) = 4*1 +

(2a + 20 - 1 ) • 21 + 210 - 230 + (6 - 2e)*560 + 8*896 +

+ 3*

As 230 has a negative coefficient in this equation, we have the desired contradiction. 172

2. The 5-Decomposition Matrices for (5a£n^Tl) and An 1—

Our main objective in this section is to derive the

5-decomposition matrix for A11, which occurs as a section of LyS. This is most easily accomplished by first deriv­

ing the decomposition matrices for £n , 5 ^ n ^ 11; the matrix for A 1]L is then easily obtained from that of I x x by restricting projectives.

Before proceeding with our program, we discuss first some of the prominent features of the ordinary and modular character theory of the symmetric group which will be used

in this section. A general reference for this material is the book by James and Kerber [12] . We adopt the notation found there.

* (character-partition correspondence) To each

partition a = (nj,n2,...,nr) of n there cor­

responds a unique irreducible character of En

denoted by [a] or [nj,n2 ,...,np ] ..

* (character induction and restriction) Let [a] be

be the irreducible character of 2 n which corre­

sponds to the partition a of n. Then

(i) [a]+En + 1 = I [8 ] , where the sum ranges

over all partitions B of n+ 1 whose

diagrams are obtained from that of a by 173

adding a single node.

(ii) = £ [3 ] , where the sum ranges

over all partitions 8 of n - 1 whose

diagrams are obtained from that of a

by removing a single node.

* (p-cores) Fix a partition a and a prime p .

From the diagram of a successively remove (for

as long as possible) rim hooks of length p. The

partition corresponding to the resulting diagram

is called the p-core of a. If p is under­

stood, we denote the p-core of a by a.

* (Nakayama’s Conjecture) [a] , [8 ] e Irr(En) lie

in the same p-block if and only if a = B .

(We note that despite its designation, Nakayama's

Conjecture is a proved result.)

Table 5, which follows, gives the 5-core a of each partition a of n (5 ^ n ^ 11) and the degree of the corresponding irreducible character [a]. The construction of the table is time-consuming but certainly not conceptual­ ly difficult. 174

TABLE 5

5-Cores of Partitions of Size n (5^n^ll)

n_ a a deg[a] n a a deg

5 (5) 0 1 7 (3,2 2 ) a 2) 2 1

5 (4,1) 0 4 7 (3,2,12 )(3,2,12 ) 35 5 (3,2) (3,2) 5 7 (3,1*) (3,1*) 15

5 (3,l2 ) 0 6 7 (23,1) a 2) 14 5 (2 2 ,1 ) (2 2 ,1 ) 5 7 (22 ,13) (2 ) 14

5 (2,13) 0 4 7 (2,15) (2 ) 6

5 (l5) 0 1 7 (l7) a 2) 1

6 (6 ) (1 ) 1 8 (8) (3) 1

6 (5,1) (5,1) 5 8 (7,1) (2 ,1 ) 7

6 (4,2) (1 ) 9 8 (6 ,2 )(6 ,2 ) 2 0

6 (4,l2) (4,l2) 1 0 8 (6 ,l2) (I3) 2 1

6 ( 3 2 ) ( 3 2 ) 5 8 (5,3) (2 ,1 ) 28

6 (3,2,1) (1 ) 16 8 (5,2,1) a 3) 64 6 (3,13) (3,13) 1 0 8 (5,13) (5,13) 35

6 (2 3) (23) 5 8 (42 ) (3) 14

6 (2 2 ,l2 ) (1 ) 9 8 (4,3,1) (4,3,1) 70 6 (2 ,1 **) (2 ,1 *) 5 8 (4,2 2 ) d 3) 56

6 a 6) (1 ) 1 8 (4,2,12 )(4,2,12 ) 90 7 (7) (2 ) 1 8 (4,1*) (4,1*) 35

7 (6 ,1 ) (l2 ) 6 8 (3 2 ,2 ) (2 ,1 ) 42 7 (5,2) a 2) 14 8 (3 2 ,l2 ) (3) 56 7 (5,l2 ) (5,l 2 ) 15 8 (3,22 ,1)(3,22 ,1) 70

7 (4,3) (2 ) 14 8 (3,2,13) (3) 64 7 (4,2,1) (4,2,1) 35 . ' 8 (3,15 ) (3) 2 1 7 (4,13) (4,13) 2 0 8 (2 *) a 3) 14 7 (3 2 ,1 )(2 ) 2 1 8 (23,12) (2 ,1 ) 28 175

TABLE 5 (cont inued)

n a a deg[a] n a a deg[a] 8 (22, l1*) (22 ,14) 20 9 (24 ,1) (2,I2) 42 8 (2,l6) (2,1) 7 9 (23,l3) ( 22 ) 48 8 (I8) (l3) 1 9 (22 ,13) ( 22 ) 27 9 (9) (4) 1 9 (2,l7) (2,l2) 8 9 (8,1) (3,1) 8 9 (l9) (l4) 1 9 (7,2) ( 22 ) 27 10 (10) 0 1 9 (7,l2) (2,l2) 28 10 (9,1) 0 9 9 (6,3) ( 22 ) 48 10 (8,2) (3,2) 35 9 (6,2,1) (6,2,1) 105 10 (8,l2) 0 36 9 (6,l 3) (1*) 56 10 (7,3) (7,3) 75 9 (5,4) (3,1) 42 10 (7,2,1) (22 ,1) 160 9 (5,3,1) (2,l2) 162 10 (7,l3) 0 84 9 (5,22 ) (5,22) 120 10 (6,4) (3,2) 90 9 (5,2,l2) (I4) 189 10 (6,3,1) (22 ,1) 315 9 (5,1**) (5,l4) 70 10 (6,22)(6,22) 225 9 (42 ,1) (4) 84 10 (6,2,l2) (6,2,l2) 350 9 (4,3,2) (2,l2) 168 10 (6,l4) 0 126 9 (4,3,l2) (4) 216 10 ( 52 ) 0 42 9 (4,22 ,1) (I1*) 216 10 (5,4,1) 0 288 9 (4,2,l3) (4) 189 10 (5,3,2) (5,3,2) 450 9 (4,1s) (4) 56 10 (5,3,l2) 0 567 9 ( 3 3 ) ( 22 ) 42 10 (5,22 ,1) (5,22 ,1) 525

9 (32 ,2,1) (3,1) 168 10 (5,2,l3) 0 448

9 (32 ,l3) (32,l3) 120 10 (5,l5) 0 126

9 (3,23 ) (l4) 84 10 (42 ,2 ) 0 252

9 (3,22 ,l2) (3,1) 162 10 (42 ,l2) (42 ,l2 ) 300

9 (3,2 ,l4 ) (3,2,l4 ) 105 10 (4,32)(2 2 ,1) 210

9 (3,l6) (3,1) 28 10 (4,3,2,1) 0 768 176

TABLE 5 (continued)

n a a deg[a] n a a deg[a

10 (4,3,l 3) (4,3,l3) 525 11 (7,22) (23) 385 10 (4,23) (4,2 3) 300 11 (7,2,l2) (1) 594 10 (4,22,l2) P 567 11 (7,1**) (2,1**) 210 10 (4,2,l4 ) (4,2,1**) 350 11 (6,5) (1) 132 10 (4,l6) P 84 11 (6,4,1) (1) 693 10 (33,1) (3,2) 210 11 (6,3,2) (23 ) 990 10 (32,22) P 252 11 (6,3,l2) (1) 1232 10 (32,2,l2) (32,2,l2) 450 11 (6,22 ,1) (6,22 ,1) 1100 10 (32,1**) (32,1**) 225 11 (6,2,l3) (1) 924 10 ( 3,2 3 ,1) P 288 11 (6,1s) (1) 252 10 (3,22,13) (3,2) 315 11 (52 ,1) (4,l2) 330 10 (3,2,l 5) (3,2) 160 11 (5,4,2) (5,1) 990 10 (3,l7) P 36 11 (5,4,l2) (3,l3) 1155 10 (25) P 42 11 ' (5,32) (23 ) 660 10 ( 2 M 2 ) (22 ,1) 90 11 (5,3,2,1) (5,1) 2310 10 (23,1**) (23,1**) 75 11 (5,3,l3) (2,1**) 1540 10 (22 ,16) (22 ,1) 35 11 (5,23) (5,2 3) 825 10 (2,I8) P 9 11 (5,22,l2) (5,1) 1540 10 (l10) P 1 11 (5,2,1**) (1) 924 11 (11) (1) 1 11 (5,l6) (5,1) 210 11 (10,1) (5,1) 10 11 (42 ,3) (1) 462 11 (9,2) (1) 44 11 (42,2,1) (3,l 3) 1320 11 (9,I2) (4,l2) 45 11 (42 ,l3) (42 ,l3 ) 825 11 (8,3) (32) 110 11 (4,32,1) (1) 1188 11 (8,2,1) (1) 231 11 (4,3,22) (4,12) 1320 11 (8,l 3) (3,l3) 120 11 (4,3,2,12)(2,1**) 2310 11 (7,4) (32 ) 165 11 (4 , 3 , l1* ) (4 , 3 ,14 ) 1100 11 (7,3,1) (7,3,1) 550 11 (4,2 3,1) (4,l2) 1155 TABLE 5 (continued)

n a______a deg[a] n______a______a deg[a]

11 (4,22 ,l 3) (1 ) 1232 1 1 (3,22 ,14 )(3,22 ,14 ) 550

1 1 (4,2,15) (1 ) 594 1 1 (3,2,l5 ) (1 ) 231 1 1 (4,l 7) (4,l 7) 1 2 0 1 1 (3,l 8) (3,l 3) 45

1 1 (33 ,2) (1 ) 462 1 1 (2 5 ,1 ) (1 ) 132

1 1 (33 ,12 ) ( 32 ) 660 1 1 (2 4 ,l 3) (2 3 ) 165

1 1 (3 2 ,2 2 ,1 ) (2 ,l4) 990 1 1 (2 3 ,l5) (2 3 ) 1 1 0

1 1 (3 2 ,2 ,1 3 ) ( 3 2 ) 990 1 1 (2 2 ,1 7) (1 ) 44

1 1 (32 ,l 5 ) (32 ) 385 1 1 (2 ,l9) (2 ,l4 ) 1 0

1 1 (3,24 ) (3,l 3) 330 1 1 d 11) (1 ) 1

1 1 (3,2 3 ,l2 ) (1 ) 693 178

Proposition 3.4. (Decomposition matrix for Eg)

(i) Z 5 has two characters of defect zero: [3,2 ],[2Z,1 ].

(ii) All remaining irreducibles of Eg lie in a single

block B.

(iii) The decomposition matrix for Eg is given by

D = D,

where

3' l7 [5] 1

[4,1] 1 1

D, = [3.12] 1 1

[2 .13] 1 1

[l5] 1

Proof. (i) and (ii) follow immediately from Table 5 and Nakayama's Conjecture. As all irreducible characters of Eg are rational integer valued (so certainly real on

5-regular elements), we conclude from 3.12 that the

Brauer tree for B is an open polygon. By 3.4 (and

Table 5), we easily see that [5] is connected to either

[4,1] or [2,l3]. But [4]+E5 = [5] + [4,1] is pro­ jective. ( [4] € Irr (2^) is projective as E1| is a

5 7 -group.) Thus [5] is connected to [4,1] and a 179

complete determination of the Brauer tree follows from this

(and 3.4):

-x------o------x- [5] [4,1] [3,12] [2 j1 3] [l5]

(iii) now follows.

Proposition 3.5. (Decomposition matrix for Zg)

(i) Zg has six characters of defect zero: [5,1 ],[4,l 2] ,

[32] , [3,l 3] , [23] , [2,ll+] .

(ii) All remaining irreducibles of Zg lie in a single

block B.

(iii) The decomposition matrix for Z g is given by

D = D.

where 1 8 8 ' l'

[6] 1

[4,2] 1 1

[3,2,1] 1 1

[22,l2] 1 1 [I6] 1 180

Proof. (1) and (ii) follow from Table 5 and Nakayama1s

Conjecture. By the previous proposition, [5] + [4,1] is a projective character of E5. Thus {([5]+[4,1]) + £6) =

[6 ] + [4,2] is also projective (3.6, 3.10), and it is trivially indecomposable. As a consequence, [6 ] is connected to [4,2] in the Brauer tree for B which, as

in proposition 3.4, is an open polygon. The entire tree

is now uniquely determined (from 3.4, Table 5) as follows:

o------x------o------x------o [6] [4,2] [3,2,1] [22,12] [16]

(iii) follows at once. 181

Proposition 3.6. (Decomposition matrix for Ey)

(i) Ey has five characters of defect zero: [5,l2],

[4,2,1], [4,13], [3,2,12], [3,1**].

(ii) The remaining irreducibles of E7 fall into two

blocks B x , B 2 as follows:

B x = {[7],[4,3],[32 ,1], [22 ,l 3 ],[2,l 5] >

B 2 = {[6,1],[5,2],[3,22 ] ,[23,1],[l7] }

(iii) The decomposition matrix for Ey is given by

where 1 13 8 6

[7] 1

[4,3] 1 1

E 3 2,1 ] 1 1

[2 2,13] 1 1 [2,15] 1

1 ' 13' 8 ' 6 '

[1 ?] 1

[ 2 3,1 ] 1 1

[3,2 2] 1 1

[5,2] 1 1

[6 ,1 ] 1 182

Proof. (i), (ii) follow from Table 5 and Nakayama's Con­ jecture. As we are still in the cyclic defect group situ­ ation, we get open polygons for the Brauer trees corre­ sponding to Bj , B 2 (3.12). As the trees are dual, it suffices to determine the tree for Bj. Clearly by 3.4,

[7] (of degree 1) must connect to either of [4,3] or

[22 ,13] (each of degree 14), and the complete tree is uniquely determined from this. From proposition 3.5,

[6 ] + [4,2] is projective in S6. Thus we have that

{ ( [6]+[4,2])+Z?}Bi = [7] + [4,3] is projective (3.6, 3.10) and [7] is connected to [4,3] . We now conclude that the tree for Bx is given by

o------x - - — o------x------o [7] [4,3] [32,1] [22,13] [2,15]

Taking the dual now determines the tree for B 2 :

o------x------o------x------o [17] [23,1] [3,22] [5,2] [6,1] and (iii) follows. 183

Proposition 3.7. (Decomposition matrix for 2 )

(i) has seven characters of defect zero: [6,2],[5,l3],

[4,3,1] , [4,2, l2] , [4,14] , [3,22 ,1] , [22 ,1"1 .

(ii) The remaining irreducibles fall into three blocks

as follows:

- {[8],[42 ],[32 ,12 ],[3,2,13 1,[3,15]}

B 2 = {[7,1],[5,3],[32 ,2 ] , [ 2 3 , l 3 ] ,[2 ,l6]}

B 3 = ([6 ,12],[5,2,1],[4,22],[2-],[Is]}

(iii) The decomposition matrix for 1Q is given by

I, D, D =

D.

where D x, D 2 , and D 3 are given sequentially

as follows:

1 13 43 21 7 21 21' 7* [8] 1 [7,1] 1

[42] 1 1 [5,3] 1 1

[32 ,I2] 1 1 [32 ,2] 1 1

[3,2,13] 1 1 [23,l2] 1 1

[3,l5] 1 [2 ,1s] 1 184

1' 13' 43' 2l' [l8] 1

[2 “] 1 1

[4,22] 1 1

[5,2,1] 1 1

[6 ,1 2 ] 1

Proof. Once again Table 5 and Nakayama's Conjecture give the complete block structure of Eg , proving (i) and (ii). As in each of the preceding propositions, the blocks Bj, B2, B 3 are associated with Brauer trees. As

Bj and B 3 are dual blocks, it suffices to determine the tree for each of Bj and B2. To accomplish this, we first derive some projectives:

{([7]+[4,3])+E8}Bi = [8 ] + [42]

{([4,3]+[32 ,1])+E8}Bi = [42] + [32,12]

{([7]+[4,3])+E8 >B 2 = [7,1] + [5,3]

{([4,3] + [32 ,l]) + S8>B;i = [5,3] + [32 ,2]

(Note [7] + [4,3] and [4,3] + [32 ,1] are each pro­ jective by proposition 3.6.) The first two projectives show that [4 2] is connected to each of [8 ] and [3 2,1 2] in the tree for Bj. This determines the complete tree as follows: 185

[8 ] [42] [32?12] [3,2?aJ] ^ 1 5 ]

The dual tree (that for B3) is now determined:

[18] [2*T [4?22] [5,2,1] • [6?12]

We finally see from the last two projectives com­ puted above that [5,3] is connected to each of [7,1] and [32 ,2]. This uniquely determines the Brauer tree for the block B 2 :

[7?1] [5*3] [32,2] [2**1*] [2?l6] #

(iii) now follows. 186

Proposition 3.8. (Decomposition matrix for Z 9)

(i) Eg has five characters of defect zero: [6 ,2 ,1 ],

[5,22],[5,I*],[32,13],[3,2,1*].

(ii) The remaining irreducibles fall into five blocks

as follows:

= {[9],[ 4 M ] , [4,3,12],[4,2,13],[4,15]}

B 2 = {[8 ,1 ],[5,4],[32,2,1],[3,22,12],[3,1 6 ]}

B 3 = {[7,2],[6,3],[33 ] ,[23 ,13 ] ,[22,1s]}

B^ = {[7 ,l2],[5,3,1],[4,3,2],[2^,1],[2,l7]}

B s = {[6 ,l 3] ,[5,2,l2] ,[4,22 ,1],[3,2 3 ],[l9]}

(iii) The decomposition matrix for Zg is given by

D =

where Dj, D2, D3, D [|, and D 5 are given

sequentially as follows:

1 83 133 56 8 34 134 28 [9] 1 [8,1] 1

[42 ,1] 1 1 [5,4] 1 1

[4.3.12] 1 1 [32,2,1] 1 1

[4.2.13] 1 1 [3,22,12] 1 1

[4,15] 1 [3,16] 1 187

27 21 21' 27' 8' 34' 134' 28' [7,2] 1 [2,l7] 1

[6,3] 1 1 [2\1] 1 1

[3 3 ] 1 1 [4,3,2] 1 1

[ 2 3 ,13 ] 1 1 [5,3,1] 1 1

[S^,!3] 1 [7,l2] 1

83 133 56 [I9] 1

[3,23] 1 1

[4,22,1] 1 1

[5,2,12] 1 1

[6 ,1 3] 1

Proof. As usual, (i) and (ii) are consequences of

Nakayama’s Conjecture and Table 5. We next derive the

Brauer trees for the blocks Bj , B2, and B3. (Those for Blf and B 5 are dual respectively to the trees for

B 2 and Bj.) These trees are open polygons by 3.12.

As {( [8 ] + [42 ])tZ9}Bi = [9] + [42 ,1] is projective (by

3.6, 3.10, and proposition 3.7), we see that [9] is connected to [42 ,1] . This fact enables us to complete the tree for Bx :

o- — o [9] [42 *L] [4^3,I2] [4,2,13] [4,15] 188

Similarly, {([8 ] + [42 ])+E9 >g2 = [8,1] + [5,4] is projec­ tive (and trivially indecomposable) so [8 ,1 ] is connected to [5,4] in the Brauer tree for B2. 3.4 now allows us to complete the tree:

[8?1] [5*4] [S^B.l] [ 3 , 2 ^ i [3 ? 1 6]

Finally, from proposition 3.7, each of [5,3]+[32 ,2] and [32 ,2]+[23 ,l2] is projective in 2 0 . Thus (by 3.6,

3.10, and Table 5) we obtain the following projectives of

2 g :

{([5,3]+[32 ,2])+E9}B = [6 ,3]+[33] 3

{( [32 ,2] + [23 ,12] )+S9 )b 3 - [33] + [ 2 M 3 ]

This shows [33] is connected to each of [6,3] and

[2 3 ,1 3] in the tree for B3, which can now be uniquely determined from 3.4:

[7?2] [6?3] [33] [2**1*] [??1S] .

Taking duals of the appropriate trees now gives the remain­ ing trees (those for B^ and B5):

[2?l7] [2\T] [4^3,2] [5,3~] [7?l2 ]

(B4) 189

[19] [3?23] [4^2,1] [ 5 , 2 ~ L * ] [6?13]

(Bs) and (iii) follows.

Proposition 3.9. (Decomposition matrix for 210)

(i) E 10 has twelve characters of defect zero:

[7,3],[6,22],[6 ,2,l2] ,[5,3,2],[5,22 ,1],[42 ,12] ,

[4,3,l 3] , [4,23 ] ,[4,2,1^], [32 ,2,12] , [32 ,1^],

[2 3 , l1* ] .

(ii) The remaining irreducihles of S 10 fall into three

blocks, Bx, B 2 , and B3. Bx and B 2 are given

below. B 3 contains all non-zero defect characters

not in Bj or B 2 .

Bj - t [8,2],[6,4],[33 ,1],[3,22 ,13 ],[3,2,l5]}

B2 = {[7.2,1],[0,3,1],[4,32],[2",1»],[22,16]}

(iii) The decomposition matrix for Z 1Q is given by

1 1 2

D 1

where Dx, D2, and D 3 are given sequentially as : 190

160 155 55 35 160' 155' 55' 35' [3,2,15] ]_ [7,2,1] 1

[3,22,13] 1 1 [6,3,1] 1 1

[33 ,1] 1 1 [4,3 2] 1 1

[6,4] 1 1 [ 2 M 2] 1 1

[8 ,2] 1 [22 ,l6] 1

28 56 70 34 217 266 56' 34' 217' 28' l' 8' [10

[9,1 1

[8,l2 1

[7,13 1

[ 6 , 1 ** 1

[52 [5,4,1

[5,3,l 2 [5,2,13 [5,15

[42 ,2 [4,3,2,1

[4,2 2 ,l 2

[4,l 6

[32 ,2 2 1 [3,23,1 I’

[3,l 7 [25 [2 ,1® [110 191

Proof. Nakayama's Conjecture and Table 5 imply both (i) and (ii). As Bj and B 2 are blocks of defect one (so have cyclic defect groups) we have by 3.12 (and the fact that all irreducibles in these blocks are real on 5-regu­ lar elements) that their respective Brauer trees are open polygons. As these blocks are dual to one another, we have only to determine the tree for one of these blocks, say B x.

Now from the previous proposition, [8,1] + [5,4] and [5,4] + [32 ,2,1] are both projectives of S9. Thus

{([8,1] + [5,4])+Z10}b = [8,2] + [6,4] and

{ ( [5,4] + [32 ,2,1] )+S10}g2 = [6,4] + [33 ,1] are projectives, and we conclude from this that [6,4] is connected to each of [8,2] and [33 ,1]. This information (along with 3.4) provides a complete determination of the tree for B x :

[8?2] [6*4] [3*,1] [3,22 ,l3] [3,2,l5]

The tree for B 2 is now obtained by dualizing that for

B i *•

o ------X ------o------X ------o

[22 ,Is] [ 2 \ 1 2] [4,3 2] [6,3,1] [7,2,1] 192

We now proceed to derive D3. Twenty-two of the forty-two irreducibles of E 10 have already been accounted for (twelve of defect zero; ten of defect one). Thus D 3 consists of 20 rows. As £ 10 contains precisely eight

5-regular classes, we have |lBr(S10)| = 3 4 by 3.3 .

Since twenty of these Brauer irreducibles have already been accounted for, D 3 consists of 14 columns, and

D 3 is a 2 0 x 1 4 matrix as claimed in (iii).

From proposition 3.8, the characters [9] + [42 ,1],

[8,1] + [5,4], [7,l2] + [5,3,1], [6 ,l3] + [5,2,l2], [5,1*],

[5,4] + [32 ,2,1], [5,3,1] + [4,3,2], [5,2,l2] + [4,22 ,1] are all projectives in Eg. The following computations supply us with related projectives (for E10) and we indi­ cate, after each computation, the column of D 3 which is determined as a consequence. The braces ({ }) indicate block reduction with respect to B3.

{([9]+[42 ,l])+Z10> = [10] + [9,1] + [5,4,1] + [42 ,l2]

gives the first column (by 3.4) and the 13th column

(by 3.8).

{([8,l]+[5,4])+Sl0> = [9,1] + [8 ,l2]+ [52] + [5,4,1]

gives the second column (by 3.4) and the 14th (by

3.8) .

{([7,12 ]+[5,3,1])+Sio} = [8 ,l2] + [7,13] + [5,4,1] +

[5,3,l2] gives the 3rd column (by 3.4) and the 12th 193

(by 3.8).

{([6,13] + [5,2,12]) + Z10} = [7,l 3] + [6 ,l1*] + [5,3,l2] +

[5,2,I3] gives the 4th column (by 3.4) and the 9th

(by 3.8).

[5,lIf]tS 10 = [6,1^] + [5,2,l3] + [5,1s] gives the

5th column (3.4).

{([5,4]+[32 ,2,l])+E10} = [52] + [5,4,1] + [4,3,2,1] +

[32 ,22] gives the 6 th column (3.4) and the 10th

column (3.8).

{([5,3,l]+[4,3,2])+EX0} = [5,4,1] + [5,3,l2] + [42 ,2]

+ [4,3,2,1] gives the 7th column (3.4) and the 11th

column (3.8).

{([5,2,12] + [4,22 ,1] ) + E10} = [5,3,l2] :+ [5 , 2 , l3] +

[4,3,2,1] + [4,22 ,12] gives the 8 th column (3.4).

With little effort (although more so than necessary in previous propositions) the degrees of the irreducible

Brauer characters of B 3 can be computed from the matrix entries in D3. The proof is now complete. 194

Proposition 3.10. (Decomposition matrix for '2^)

(i) S X1 has six characters of defect zero: [7,3,1],

[6,22,1], [5 ,2 3 ], [4 2 ,1 3 ], [4,3,1*], [3,22,1*].

(ii) The remaining irreducibles for Z X1 fall into seven

blocks, Bj, B2,..., By. We list below the charac­

ters which occur in Bx through Bg. The charac­

ters of the block By are precisely those which

have non-zero defect and do not belong to any of the

blocks listed below.

B x = {[1 0 ], [5,4,2 ],[5,3,2 ,1 ], [5,22,12],[5 ,1 6 ]}

B 2 = {[9,l2],[52 ,1],[4,3,2 2 ],[4,23,1],[4,17]>

B 3 = {[8 ,3],[7,4],[33,1 2 ],[3 2 ,2 ,1 3 ],[3 2 ,i5]}

Bk = {[8 ,13],[5,4,12],[42,2,1],[3,2*],[3,18]}

B 5 = {[7,22], [6,3,2], [5,32] ,[2*,13] , [23,15]}

B 6 = {[7,1*],[5,3,13],[4,3,2,12],[32,22,1],[2,19]}

(iii) The decomposition matrix for Eu is given by

D,

D. D =

D c

D. 195

where Dx , D2 ,..., D? are given sequentially as follows

10 980 1330 210 45 285 1035 120 [10,1] 1 [9,l2] 1

[5.4.2] 1 1 [52 ,1] 1 1

[5,3,2,1] 1 1 [4,3,22] 1 1 .

[5,2 2 ,l2] 1 1 [4,23,1] 1 1

[5,l6] 1 [4,l7] 1

(Dj) (D2 >

385 605 55 110 45' 285'1035'120' [32 ,Is] 1 [3,l 8] 1

[32 ,2, l3] 1 1 [3,2**-] 1 1

[33 ,l2] 1 1 [42 ,2,1] 1 1

[7,4] 1 1 [5,4,l2] 1 1

[8 ,3 ] 1 [8 ,l3] 1

(D3) (D^)

385' 605' 55' 110' 10' 980'1330'210' [7,22] 1 [2,l9] 1

[6.3.2] 1 1 [32 ,2 2 ,1] 1 1

[5,32] 1 1 [4,3,2,l2] 1 1

[2\13] 1 1 [5,3,l3] 1 1

[23 ,l5] 1 [7,1^] 1

(Dr ) (D e) 196

1 43 188 406 89 372 266 252 406' 89' 372' 188' 1' 43' [11 1 [9,2 1 1

[8 ,2,1 1 1

[7,2,l2 1 1

[6,5 1 1

[6,4,1 1 1 1 1 1

[6,3,l 2 1 1 1 1

[6,2 ,l 3 1 1 1

[6,15 1

[5,2,1*+ 1 1 1

[42 ,3 1

[4,32 ,1 1 1 1 1 1

[4,2 2 ,l 3 1 1 1

[4,2,l 5 1

[33 ,2 1 1

[3,2 3 ,l2 11111

[3,2,l 6 1 1

[2 s, 1 1 1

[22,l 7 1 1

[I11

(Dy) 197

Proof. By Table 5, Nakayama's Conjecture, and 3.12, the block structure for E X1 is determined, and the blocks of defect one, namely Bj, B2, B3,' B^, Bg, Bg, have Brauer trees which are open polygons.

By the previous proposition, the following characters of E 10 are projective: [10] + [9,1] + [5,4,1] .+ [42 ,2] ,

[6 ,4] + [33 ,1] , [7,3] . We therefore obtain that

{([10]+[9,l]+[5,4,l]+[42 ,2])+E11}Bi = [10,1] + [10,1] +

[5,4,2] + [5,4,2] = 2*[10,1] + 2*[5,4] is projective.

By either 3.13 or 3.14 we are led to conclude that

[10.1] + [5,4,2] is projective indecomposable, and [10,1] is connected to [5,4,2] in the tree for B x. By 3.4, it is immediate that precisely two of the characters [1 0 ,1 ],

[5.3.2.1], [5,l6] must have valency one in this tree.

As deg([5,3,2,1]) > deg(0) for all ordinary irreducibles 0 in Bj, we see that [5,3,2,1] must have valency two.

Thus [10,1] has valency one and the tree is uniquely determined:

[10?1] [5,4~2] [5,3?2,1] [5,2 2 ,l2] [5?16]

Its dual tree (that for Bg) is also determined:

[2?l9] [32 , 2 2 ,1] [4,3?2,12] [5 , 3 ~ ] [7?14] 198

We next observe that

{([10]+[9,1]+[5,4,1]+[42,2])+S11}b = [9,12] + [5 2 ,1] 2 is projective indecomposable, whence [9,l2] is connected

to [52 ,1] in the tree for B2. By a similar argument

as that used above for Bj, [9,l2] has valency one in

the Brauer tree for B 2 and the tree follows:

[9?l2] [52*1] [4^3,22] [4,2371] [ I V ]

Dualizing this tree now gives the tree for B, :

[3?18] [3?2»] [4272,1] [5,4*12] [ T V ]

Finally, {([6,4]+[33 ,1])+z,,}_ = [7,4] + [3 3 ,1 2 ] 3 and {[7,3]+Z1;l}B = [8,3] + [7,4] are trivially projec- 3 tive indecomposable, whence [7,4] is connected to each of [8,3] and [33 ,12] in the Brauer tree for B 3 .

From this information we see that [33,!2] cannot have valency one and the tree for B 3 follows:

[8?3] [7?4] [33?I2] [32 ,2?!^] [ ^ l 3]

Dualizing this tree gives that for B 5 :

— o------x--- [23?15] [ 2 \ l 3] [5,32] [6,3,2] [7?22] 199

We now attack the problem of determining Dy. First we observe that |lrr(Exl)| = 56 and E X1 contains twelve

5-regular classes. As a consequence |lBr(Exx)| =56-12

=44. As 36 irreducibles and 30 Brauer irreducibles have already been accounted for, we conclude that the block

B? contains 20 ordinary irreducibles and 14 Brauer ir­ reducibles, whence D 7 is a 20 x 14 matrix as indicated in the proposition statement.

The projectivity of each of the characters we induce below can be easily checked from the decomposition matrix for £ 10 (proposition 3.9). We indicate, following each character induction, the column of D 7 obtained as a con­ sequence. Braces ({ }) denote block reduction with respect to B 7 .

{([10]+[9,l]+[5,4,l]+[42 ,2])+Exx} = [11] + [9,2] +

[6,4,1] + [42 ,3] gives the first column (by 3.4)

and the 13th column (by 3.8).

{([9,l]+[8,l2 ]+[52 ]+[5,4,l])+Exx} = [9,2] + [8,2,1]

+ [6,5] + [6,4,1] gives the second column (by 3.4)

and the 14th column (by 3.8).

{([8,12 ]+[7,13]+[5,4,1]+[5,3,12 ])+E11} = [8,2,1] +

[7,2,l2] + [6,4,1] + [6 ,3,l2] gives the 3rd column

(3.4) and the 12th column (3.8). 200

{ [6,2,l2 ']t21 = [7 , 2 , l2] + [6 ,3,l2] + [6 ,2,l 3]

gives the 4th column (3.4) and the 9th column

(3.8).

{([6,4]+[33 ,l])+Z:x} = [6,5] + [6,4,1] + [4,32 ,1]

+ [33 ,2] gives the 5th column (3.4) and the 10th

column (3.8).

{([6,3,l]+[4,32] ) + £ x = [6,4,1] + [6 ,3,l2] +

[42 ,3] + [4., 3 2 ,1] gives the 6 th column (3.4) and

the 11th (3.8).

{([5,3,l2 ]+[5,2,l 3 ]+[4,3,2,1]+[4,2 2 ,l2 ])+Exl} =

[6 ,3,l2] + [6,2,l3] + [5,2,l4] + [4,32 ,1] +

[4,22 ,13] gives the 7th column (by 3.4).

{([e.l^ + ts^^sj + ^. l 5] ) ! ^ ^ = 2.[6,2,l3] +

2* [6 ,l5] + 2*[5,2,l4] gives the 8 th column (by

3.4 and 3.14).

As in the previous proposition, the irreducible

Brauer degrees are now easily obtained. This completes the proof of the proposition.

Before proceeding to a derivation of the decomposi­ tion matrix for A11? we state without proof the follow­ ing useful lemma. (Its proof can be found in [12]). 201

Lemma 3.11. Let n be a fixed positive integer, p a fixed prime, a a partition of n, and a/ its dual partition. Then if a = a, every irreducible constituent of [a]+An forms its own p-block and each modular representation associated with such a constituent is irreducible. If a f a, then to each p-block of an irreducible constituent of [a]+An belong exactly the irreducible constituents of restrictions [0]+An of such irreducible characters [3] of En for which 3 = a or

3 = a'.

Proposition 3.12. (Decomposition matrix for A11)

(i) A 1X has three characters of defect zero: [7,3,1]*,

[6 ,2 2 ,1 ]*, [5,23]*. (ii) The remaining irreducibles of An fall into four

blocks as follows:

B1 = {[10,1]*, [5,4,2]*,[5,3,2 ,1 ]*, [5,2 2 ,l2]*, [5, l6]*}

B 2 = {[9,l 2 ]*,[52 ,1]*,[4,3,22 ]*,[4,23 ,1]*,[4,l 7]*}

B 3 = {[8,3]*,[7,4]*,[32 ,l2 ]*,[32 ,2,l 3 ]*,[3,l5]*}

B 4 = {[1 1 ]*,[9 ,2]*,[8,2,1]*,[7,2,l2]*, [6 ,5]*,[6 ,4,1]*,.

[6 ,3,l 2 ]*,[6 ,2,l 3]*,[6,15 ]+ ,[6 ,I 5 ] ' , [ 4 2 ,3]*,

[4,32 ,1]+ ,[4,32 ,1]-}

In the above, if [3]+Ai;l € Irr(Alx) we define

[8 ]* = [31+Aj!. If [8 ]+A1]L $ Irr(Alx) we denote 202

by [0 ]+ , [3 l~ the irreducible constituents of

[3]+Aj1.

(iii) The decomposition matrix for All is given by

where D1, D2, Dg , and are given sequentially

as follows:

10 980 1330 210 45 285 1035 120

[1 0 ,1 ]* 1 [9,I2]* 1

[5,4,2]* 1 1 [52 ,1]* 1 1

[5,3,2,1]* 1 1 [4,3,22]* 1 1

[5,2 2 ,l2]* 1 ‘ 1 [4,23,1]* 1 1

[5,l 6]* 1 [4,l 7]* 1

(Dx) (D2)

385 605 55 110 [3,l 5]* 1

[32 ,2,13]* 1 1

[32 ,l2]* 1 1 [7,4]* 1 1 [8,3]* 1

0>3> 203

1 43 188 406 89 372 133 133 126 126 [11]* 1 [9,2]* 1 1 [8,2,1]* 1 1

[7.2.12]* 1 1

[6,5]* 1 1

[6,4,1]* 1 1 1 1 1

[6.3.12]* 11 1 1 1 [6.2.13]* 1 11.11

[42 ,3]* 1 1 1 [6 , l5 ] + 1

[6,I5]" 1

[4.32 .1]+ 1 1 1

[4.32.1]~ 1 1 1

Proof. (i) and (ii) follow immediately from lemma 3.11.

The determinations of D 1, D2, and D 3 follow from the block decompositions matrices, D 1? D2, and D 3 for £ X1 by mere character restriction. As llrrCA^)! = 31 and

A 1X contains 25 5-regular classes, we readily conclude from 3.3 that is a 13 * 10 matrix as indicated in

(iii) of the proposition statement. 204

From proposition 3.10, each of the characters which have been restricted below is indeed projective. We indi­ cate after each restriction the corresponding column of which is determined as a result.

([ll]+[9,2]+[6,4,l]+[42 ,3])+A11 = [11]* + [9,2]* +

[6,4,1]* + [42 ,3]* gives the first column (by 3.4).

([9,2]+[8,2,l]+[0,5]+[6,4,l])+An = [9,2]* + [8,2,1]*

+ [6,5]* + [6,4,1]* gives the second column (by 3.4).

([8,2,1]+[7,2,12 ]+[6,4,1]+[6,3,12 ])+A11 = [8,2,1]* +

[7.2.12]*+ [6,4,1]* + [6 ,3,l2]* gives the 3rd column

(by 3.4).

([7,2,l 2 ]+[6 ,3,I2] + [6 ,2,l 3])+Ax! = [7,2,l2]* +

[6 .3.12]* + [6 ,2,l 3]* gives the 4th column (3.4).

([6,5]+[6,4,l]+[4,32 ,l]+[33 ,2])+A11 = [6,5]* + [6,4,1]*

+ [4,32 ,1]+ + [4,32 ,1]“ + [33 ,2]* gives the 5th

column (from 3.4 and the fact that [33 ,2]* = [42 ,3]*).

([6,4,l] + [6,3,l2] + [42 ,3] + [4,32 ,l] )4-Axl = [6,4,1]* +

[6.3.12]* + [42 ,3]* + [4,32 ,1]+ + [4,3 2 ,1]" gives the

sixth column (3.4).

It remains only to determine the final four columns of . From proposition 3.10 we easily see that [6 ,3,l2]

4- [6,2,l 3] + [5,2,1**] + [4, 3 2 ,1] + [4 , 2 2 , l 3 ] and [6 ,2,l 3]

+ [6 , l5] + [5,2,1**] are projective characters of Z l x . 205

Denoting these projectives by x and respectively, we compute their restrictions to AX1:

X+Axl = 2 » [6 ,3,l2]* + 2* [ 6 ,2,l3]* + [4 ,32 ,1]+ + [4, 32 ,1] "

l|i+Axl = 2» [6 ,2, l3]* + [6 , l5 ] + + [6,15]"

We assume X+Ax x a n d ij;+A1 x are both indecomposable and derive a contradiction. Without loss of generality, we allow X+Aj 1 and i^+Ax x to correspond to the 7th and

8 th columns of , and we reproduce below the final four rows of •

*2 ♦3 ♦3 *5 *6 * 7 4>r 8 T 9 [6,l 5]+ 0 0 0 0 0 0 0 1 a b

[6,15]“ 0 0 0 0 0 0 0 1 b a

[4,32 ,1]+ 0 0 0 0 1 1 1 0 c d

[4,32 ,1]“ 0 0 0 0 1 1 1 0 d c

Here a, b, c, d denote respectively the multiplicities of [6,15]+ in * 9 , [6,15]+ in ® 10, [4,32 ,1]+ in

$9, [4,32,l]+ in $10, where $9 and $10 are the respective indecomposables of A1X corresponding to the

9th and 10th columns of . Now ¥g = $10 (indeed

? 9 f $1(j implies ¥ g = 3>g , ¥70 = $10 whence [6,15]+ =

[6,l5]- as Brauer characters, a contradiction), so the

multiplicity of [6,l 5]" = [6,1^]+ in $ 9 equals that

of [6,15]+ in $10 (and that of [6,15]- in $10 is 206

equal to the multiplicity of [6,15]+ in $g). A similar statement obtains if we replace [6 ,l5] with

[4,32 ,1] in the above. This justifies the repetition of the symbols a, b, c, and d which occur in the final two columns of our partial decomposition matrix above.

We next make the following important observations:

(1) 16 + 4 4 is a projective indecomposable of M :1

(16, 44 € Irr(Mn )>.

(2) [4,3 2 ,1]++M1x = [4,32 ,1]-+M11

(3) [6,15 ]++M 11 = 11 + 44 +55+16 (where

11, 44 , 55 , 16 € Irr(Mi;i)

( (1 ) is immediate from the decomposition matrix for M X1

(see lemma 3.24) while (2) and (3) can be verified by computing with the character tables for M1X and AX1.)

By (2), (3), and Frobenius reciprocity, we obtain the following.

(a) ([6,15]+,(16+44)+A11)Aii = ( [ 6 ,l 5 ]++M1:,16+44)^

= 2

(b) ([6,15]-,(16+44)+A11)A ii = ([6,l5]+ ,(16+44)+A11)Aii

* ([6,15]+ ,(16+44)+A11)A ii =

( [6,15 ]+ +M,.,16+44)m = 1 , as 16 f 16 but 11 m i i 44 = 44. 207

(c) ([4,32,1]+ ,(16+44)+A11)A ii =

([4,32 ,1]++M1x ,16+44)M ii = ([4,32 ,1]-+M11,16+44)M ii

= ([4,32 ,1]“ ,(16+44)+A11)a . "11

Now (16+44)+Axx is projective by (1). Let $ be the projective constituent of (16+44)+AJ1 corresponding to the block of A X1, i.e. $ = { (16+44 ) + Ax x . We may therefore write $ = where the a^’s are non­ negative integers, and denotes the projective indecom­ posable corresponding to the i-th column of , 1 ^ i ^ 1 0 .

But then from (a) above, 2 = ([6,15]+ ,(16+44)+Axx)Aj =

([6 ,1 5]+ ,$)Aii = a 8 + a ga + a 10b, while from (b) we ob­ tain 1 = ([6,15]-,(16+44)+A1x)Aii = ([6,15]-,$)Aii = a 8 + a 9b + a 10a. Thus clearly a9a + axob f a 9b + a 10 a, from which we conclude a 9 f a x 0.

We next observe from (c) that a 5 + a 6 + a 7 + a 9c

+ a xod = ([4,32 ,1]+ ,0)Aii = ([4,32 ,1]+ ,(16+44)+Axx)Aii =

([4,32,l]-,(16+44)tAlx)Aii = ([4,32 ,1]",$)Aii = a 5 + a 6 + a 7 + a 9d + a 10c, whence a gc + a 1(Jd = agd + a 1 (Jc.

Therefore (ag-axo)c = (a9-aX0 )d, and as we have already shown a g f ax , we conclude that c = d. But this implies

[4,32 ,1]+ = [4,32 ,1]" as Brauer characters, an obvious con­ tradiction as these characters differ on elements of order

21. Thus we have successfully shown that at least one of

X+Ax x , t(>+Axx is decomposable. We show presently that in 208

fact both are.

Assume first that X+A 1^ alone splits. Then

X+Axl = ([6,3,l2 ]*+[6,2,l3]*+[4,32 ,1]+ ) +

([0 ,3 ,12]*+[6 ,2 ,13]*+[4 ,32,1]-), the characters in parentheses being projective indecom- posables. Without loss of generality, let these indecom- posables correspond to the 7th and 8 th columns of D 4 respectively, and let 1 (indecomposable by assumption) correspond to the 9th column of . We therefore have the following portion of the block decomposition matrix

*1 $2 * 3 * 5 *6 *7 + 8 v 9 4»

[ 6 ,Is] + 0 0 0 0 0 0 0 0 1 a

[6 ,1 5]- 0 0 0 0 0 0 0 0 1 b

where a , b denote respectively the multiplicities of

[6 ,15]+ and [6,1s]" in $ 10. But $ 10 must be self­ dual (indeed we have run out of columns!). We must there­ fore have a = b whence [6,15]+ = [6,15]- as Brauer characters, a contradiction. The assumption that i^+A11 alone splits leads to the contradiction [4,32 ,1]+ =

[4,32 ,1]~ by an almost identical argument. We therefore conclude that each of x+Ajx, iHAx! splits, and by 3.4 the decomposition of each is uniquely determined. As a 209

consequence is determined and the proposition is proved. 210

3. The Character Tables and 5-Decomposition Matrices for

N 3 E 35 »M11 and S a 3z + l * • (SL(2,5)*2g)

Each of the groups N = EgS'M^ and

S = 32+l* • (SL(2, 5)*Zq ) occurs in LyS with index two in a maximal local: |Ng(E):N| = 2 with E = E 3 5 , and

|N g (< u 1 , u2 > ) :S| = 2 with = E 32 . I have chosen to work with the groups N and S rather than NG(E) and Nq() because the construction of the character tables for N and S is far easier than that for the latter groups, and there is no great loss of information involved. After the far more difficult problem of con­ structing the character tables comes the simple task of determining the respective 5-decomposition matrices for

N and S, which is in truth the real goal of this section.

We turn now to the construction of the character tables. No attempt is made to supply rigorous proofs; the details are far too numerous (and the computations too laborious) to mention. We are content to merely provide a brief account of the more salient aspects of their construction. First we state without proof the following result from elementary character theory which is attributed to

Clifford. 211

Theorem 3.13. Let G be a finite group, N

X€Irr(G), and 6 an irreducible constituent of x+N *

Then there exists an irreducible character ip of

Iq (6 ) = {g€G:0g=0} such that = e 0 for some

positive integer e, and i|;tG = x*' ^q^0) is called t*10 inertia group of 0 in G.)

This theorem suggests a general strategy for acquir­

ing irreducible characters of a group from those of cer­

tain subgroups. In fact, an actual algorithm is obtained

if we are willing to endure the following sacrifice of

generality:

Wigner's Method. Let G = A*H be a split extension of

A by H, with A abelian and G finite. Then the ordinary irreducible characters of G are obtained as

follows.

Step 1. Determine the orbits 0X, 02,..., of G on

Irr(A). (The action of G on Irr(A) is contragredient

to that of G on A . )

Step 2. For each i, 1 ^ i ^ t, determine Ijj(^ ± ) where ^ is a fixed representative of 0-^. 212

Step 3. Extend to € Irr(lQ(^-j_)) by

$jL(ah) = ipi(a), a € A, h € H.

Step 4. Form the distinct irreducible characters $ 1 ? ^ of Iq(^i) where runs over Irr(IjjCi^) ).

Step 5. Induce each i to G.

Remarks. Clearly, by theorem 3.13, $i?ijtG e Irr(G).

In actuality, every irreducible of G is obtained in this manner for some i and some € Irr( Ijj(if'i)) •

We also remark that only step 3 uses the full strength of our hypotheses ( A abelian, G splits over A ). 213

The Character Table for N

Clearly Wigner’s Method applies as N is a split extension of the abelian group E = xxxx

S E 0 5 by M 11 . There are three orbits of N on Irr(E), having respective representatives 3-E » ip , and x • Here lj, denotes the principal character of E, and ip and x are defined as follows:

ip(uf lu|2uf 3x?ixf*) = r)*3

x ( u ^ i u2u|3xJix^2 f ) = nfil (n = eZVi/3)

The inertia groups (in H = Mn ) of these characters are given by Ih (1e) = H > ZjjC'/') = E32*D8’ and IH^X^ s A 5 ’ The complete character table for N appears in Table 6 .

Irrational entries are given as follows:

1 + i/ll _ 1 + i/27 -1 + i/15 2

-1 + i/3 e = -3 + i/27 , X = 1 + i/27 , o = i/3 , n 2 TABLE 6

The Character Table for N = EjS-Mjj

|Cent|: |.N| 1296 87480 8748 162 81 24 15 324 324 216 108 18 8 8 54 54 11 11 12 15 15 18 18

Class: 1A 2A 3A 3B 3C 3D 4A 5A 6A 6B 6C 6D 6E 8A 8B 9A 9B 11A11B12A15A15B18A18B

1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1

x 2 10 2 10 10 1 1 2 0 2 2 2 2 -1 0 0 1 1 -1 -1 2 0 0 -1 -1

x 3 11 3 11 11 2 2 -1 1 3 3 3 3 0 -1 -1 2 2 0 0 -1 1 1 0 0

Xu 10 -2 10 10 1 1 0 0 -2 -2 -2 -2 1 a a 1 1 -1 -1 0 0 0 1 1

X, 10 -2 10 10 1 1 0 0 -2 -2 -2 -2 1 a a 1 1 -1 -1 0 0 0 1 1

X 5 16 0 16 16 -2 -2 0 1 0 0 0 0 0 0 0 -2 -2 3 3 0 1 1 0 0

x 5 16 0 16 16 -2 -2 0 1 0 0 0 0 0 0 0 -2 -2 3 3 0 1 1 0 0

0 X 6 44 4 44 44 -1-10-1 4 4 4 4 1 0 -1 -1 0 0 0 -1 -1 1 1

x7 45 -3 45 45 0 0 1 0 -3 -3 -3 -3 0 -1 -1 0 0 1 l 1 0 0 0 0

-1 -1 1 1 1 0 0 -1 0 0 -1 -1 X 8 55 -1 55 55 1 1 -1 0 -1 -1 -1 1

2 2 0 -1 -1 0 0 -1 0 0 -1 -1 = (fe)+N 110 14 -25 2 2 2 2 0 -4 -4 -1 0 214

=($c 2)+n 110 -2 -25 2 2 2 2 0 -2 -2 7 -2 -2 0 0 -1 -1 0 0 -1 0 0 1 1 TABLE 6 (continued)

|C e n t|: |H| 1296 87480 8748 162 81 24 15 324 324 216 108 18 8 8 54 54 11 11 12 15 15 18 18

C lass: 1A 2A 3A 3B 3C 3D 4A 5A 6A 6B 6C 6D 6E 8A 8B 9A 9B 11A11B12A15A15B18A18B

$ 3“ ( f e 3)+N 110 6 -25 2 2 2 -2 0 "e e 3 0 0 0 0 -1 -1 0 0 1 0 0 a 0

110 6 -25 2 2 2 -2 0 e e- 3 0 000-1-100 100 a o

(M C ksM N 220 -12 -50 4 4 4 0 0 6 6 -6 0 0 0 0-2-20 0 0 0 0 0 0

4>5“ ($C6)+N 440 8 -100 8 -1 -1 0 0 -X -X -4 2 -1 0 0 Y Y 0 0 0 0 0 -o -n2

6= (fe 7)+N 440 -8 -100 8 -1 -1 0 0 XX 4 -2 1 0 0 YY 0 0 0 0 0 n n2

440 8 -100 8 -1 -1 0 0 -X -X -4 2 -1 0 0 YY 0 0 0 0 0 -o2-n

*6 440 -8 -100 8 -1 -1 0 0 I X 4 -2 1 0 0 YY 0 0 0 0 0 o2 T1

Gj-CXB^+N 132 12 24 -3 6 -3 0 2 3 3 0 -3 0 0 0 0 0 0 0 0 -1 -1 0 0

62=(x B2)+N 528 0 96 -12 6 -3 0 -2 0 0 0 0 0 0 0 0 0 0 0 0 1 1 0 0 e 3=(xB3)+N 660 12 120 -15 -6 3 0 0 3 3 0 -3 0 0 0 0 0 0 0 0 0 0 0 0

64=(X3,t)+N 396 -12 72 -9 0 0 0 1 -3 -3 0 3 0 0 0 0 0 0 0 0 6 6 0 0

396 -12 72 -9 0 0 0 1 -3 -3 0 3 0 0 0 0 0 0 0 0 6 6 0 0

o. 215 216

The Character Table for S

Here Wigner's Method has limited application. Still,

39 of the 57 irreducibles of S can be obtained from it. We describe presently how this is accomplished.

Let P = 0 3(S). Thus P = =

32+lt. Also let Z = Z(P) = and = , i = 1,2. Now S/Z s Egif-(SL(2,5)*Z8) , so that Wigner’s

Method applies directly to the group S/Z. There are two orbits of S/Z on Irr(P/Z) with respective representa­ tives lpyz , y and corresponding inertia groups ( in

H = S/P ) given by ^(Ip/z^ = H > IrCY) = Eg. The sub­ sequent inductions of products of characters yield 39 irreducibles for S, each of which has Z in its kernel.

(These are the first 39 characters occurring in the finished table.)

We next observe that S/Qj = S/Q2 S 31+I+• (SL(2 , 5)*ZQ ) with P/Q1 = P/Q2 = 31+It (extraspecial with exponent 3).

Now 31+4 possesses 81 linear characters and 2 faith­ ful irreducibles of degree 9. Thus for i = 1,2 , we can choose € Irr(P/Q^) with ip^(l) = 9. Regarding

as an irreducible character of P having in its kernel, we find that Ig(^i) = 32+1* *SL(2,5) (i = 1,2).

In general, there is no guarantee that 0 extends to

Ig(0); but as SL(2,5) has trivial multiplier, it follows 217

from [11,p.178,£&.30-36] that ^ indeed extends to

IgC^) for i =1,2. Each therefore gives rise to

9 distinct irreducibles of S, yielding 18 in all.

The character table for S can now be completed. (It appears in Table 7. ) Irrational entries are given as

follows:

1 + /5 1-/5 e e l 2 2 2 218

TABLE 7

The Character Table fo r S = 32+t* •(SL(2,5)*Z 8^

|C e n t |: |s| 4320 144 87480 87480 1944 486 486 486 480 480 C la s s : 1A 2A 2B 3A 3B 3C 3D 3E 3F 4A 4B

Rep: 1 z t 2a U1 u 2 b2 Ujb2 u 2b 2 X1 t 2 t 2z 1 1 1 1 XS 1 1 1 1 1 1 1 5 5 1 5 5 -1 -1 -1 5 5 5

X 6 -6 0 6 6 0 0 0 6 . 61 -61

0x 4 -4 0 4 4 1 1 1 4 41 -41 4 0 02 4 4 4 1 1 1 4 4 4

5x 3 3 -1 3 3 0 0 0 3 3 3 3 3 -1 3 3 0 S2 0 0 3 3 3 2 -2 0 2 2 -1 -1 -1 2 21 - 2 i

n2 2 -2 0 2 2 -1 -1 -1 2 2 i -21 i7 1 1 -1 1 1 1 1 1 1 -1 -1

< / 5 5 -1 5 5 -1 -1 -1 5 -5 -5

x / 6 -6 0 6 6 0 0 0 6 -6 1 61 4 -4 0 4 4 1 1 1 4 -41 41 ei 4 4 0 4 4 1 1 1 4 -4 -4 < 3 3 1 3 3 0 0 0 3 -3 It -3 3 3 1 3 3 0 0 0 5 / 3 -3 -3

n / 2 -2 0 2 2 -1 -1 -1 2 -21 21

n / 2 -2 0 2 2 -1 -1 -1 2 -21 21 i" 1 1 1 1 1 1 1 1 1 1 1 219

TABLE 7 • (Continued)

| Cent | : |s | 4320 144 87480 87480 1944 486 486 486 480 480

C la s s : 1A 2A 2B 3A 3B 3C 3D 3E 3F 4A 4B

R ep: 1 z t 2a u l U2 b 2 Ujb2 u 2b 2 X1 t 2 t 2z

♦ " 5 5 1 5 5 -1 -1 -1 5 5 5

X* 6 -6 0 6 6 0 0 0 6 6 i - 6 i 1 4 41 - 4 i < 4 -4 0 4 4 1 i 1 4 4 4 < 4 4 0 4 4 1 l 3 3 -1 3 3 0 0' 0 3 3 3 ef 3 -1 3 3 0 0 0 3 3 3 < 3 2 -1 - l -1 2 21 -21 i f 2 -2 0 2 2 21 -21 i 2" 2 -2 0 2 2 -1 - l -1 i"' 1 1 -1 1 1 1 i 1 1 -1 -1

♦ * 5 5 -1 5 5 -1 - l -1 5 -5 -5

X* 6 -6 0 6 6 0 0 0 6 -61 61 4 -4 0 4 4 1 1 1 4 - 4 i 41 • r CD 4 4 0 4 4 1 1 1 4 -4 -4

3 3 1 3 3 0 0 0 3 -3 -3 e f c f 3 3 1 3 3 0 0 0 3 -3 -3 /// 2 -1 -1 -1 2 -21 21 n i 2 -2 0 2 /// 2 -2 0 2 2 -1 -1 -1 2 -21 21

Y+S 80 0 8 80 80 8 8 8 -1 0 0

Y3+S 80 0 -8 80 80 8 8 8 -1 0 0 TABLE 7

(Continued)

|S | 4320 144 87480 87480 1944 486 486 486 480

1A 2A 2B 3A 3B 3C 3D 3E 3F 4A

1 z t 2a U1 u 2 b 2 Ujb2u 2b 2 Xj t 2 t* 160 0 0 160 160 -8 -8 -8 -2 0 0 I 36 4 0 9 00 12 3 -6 0 0 0

180 20 0 45 -90 -12 -3 6 0 0 0

216 -24 0 54 -108 0 0 0 0 0 0

144 -16 0 36 -72 12 3 -6 0 0 0

144 16 0 36 -72 12 3 -6 0 0 0

108 12 0 27 -54 0 0 0 0 0 0

108 12 0 27 -54 0 0 0 0 0 0

72 -8 0 18 -36 -12 -3 6 0 0 0

72 -8 0 18 -36 -12 -3 6 0 0 0

36 4 0 -18 9 12 -6 3 0 0 0

180 20 0 -90 45 -12 6 -3 0 0 0

216 -24 0 -108 54 0 0 0 0 0 0

144 -16 0 -72 36 12 -6 3 0 0 0

144 16 0 -72 36 12 -6 3 0 0 0

108 12 0 -54 27 0 0 0 0 0 0

108 12 0 -54 27 0 0 0 0 0 0

72 -8 0 -36 18 -12 6 -3 0 0 0

72 -8 0 -36 18 -12 6 -3 0 0 0 221

TABLE 7

(Continued)

Sent |: 144 360 360 1080 1080 216 54 54 18 480 480 480

Lass: 4C 5A 5B 6A 6B 6C 6D 6E 6F 8A 8B 8C sp: a c d zu^ zu 2 b u^b u2b t2ax t tz t?

XS 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 5 5 -1 -1 -1 1 5 5 5

X 0 1 1 -6 -6 0 0 0 0 6(0 -6co 6(03

0 1 0 -1 -1 -4 -4 -1 -1 -1 0 4(0 -4(0 4(0 3

02 0 -1 -1 4 4 1 1 1 0 4 4 4

-1 El e 2 3 3 0 0 0 -1 3 3 3

-1 e2 El 3 3 0 0 0 -1 3 3 3 -2 1 "l 0 -e2 “6i -2 1 1 0 2(0 -2(0 2(0 3

n2 0 -£i -e2 -2 -2 1 1 1 0 2(0 -2(0 2(0 3

i' 1 1 1 1 1 1 1 1 -1 i 1 -1

< / 1 0 0 5 5 -1 -1 -1 -1 5 i 51 -51

x' • 0 1 1 -6 -6 0 0 0 0 6(03 -6(o 3 6(0 CO 3 0 -1 -4 -1 -1 4(0 3 1 < -1 -4 -1 0 4(0

CD 0 -1 -1 4 4 1 1 1 0 41 41 -4 1

-1 3 3 0 0 0 1 3i 31 -3 1 *1 e i

- 1 0 0 0 1 e 2 e i 3 3 31 31 -3 1

0 - 2 - 2 1 1 1 0 2(0 -e2 “ e i 2(0 3 - 2 ( 0 3

0 -e2 - 2 - 2 1 1 1 0 2(0 3 -2(0 3 2(0 n 2 " e i

I* 1 1 1 1 1 1 1 1 -1 1 -1 222

TABLE 7

(Continued)

| Cent] : 144 360 360 1080 1080 216 54 54 18 480 480 480

C la ss: 4C 5A 5B 6A 6B 6C 6D 6E 6F' 8A 8B 8C

Rep: a c d zu^ zu 2 bUjb u 2b t 2ax t tz t 3 -e- 1 0 0 5 5 -1 -1. -1 1 -5 -5 -5

x " 0 1 1 -6 -6 0 0 0 0 - 6 oj 6(i) -6(0 3 ^ * CD 0 -1 -1 -4 -4 -1 -1 -1 0 -4w 4(i) -4(0 3 CM ^ CD 0 -1 -1 4 4 1 1 1 0 -4 -4 -4

5 f -1 E2 3 3 0 0 0 -1 -3 -3 -3

-1 e 2 E i 3 3 0 0 0 -1 -3 -3 -3

< 0 ~e z -E l -2 -2 1 1 1 0 —2w 2d) -2(0 3

_ // n 2 0 -E2 -2 -2 1 1 1 0 —2(i) 2(i) ‘ -2(0 3

i ' " 1 1 1 1 1 1 1 1 -1 - i - 1 i

1 0 0 5 5 -1 - 1 -1 -1 - 5 i - 5 i 51

x ' " 0 1 1 -6 -6 0 0 0 0 -6o)3 6(i)3 -6(0

CD 0 -1 -1 -4 -4 -1 -1 -1 0 —4(i)3 4(i)3 -4(0 CD 0 -1 -1 4 4 1 1 1 0 -41 - 4 i 41

e f -1 El £2 3 3 0 0 0 1 -3 1 -31 31

-1 £ 2 El 3 3 0 0 0 1 -31 -31 31 /// Hi 0 - £ 2 —Ej. -2 -2 1 1 1 0 -2(i)3 2w 3 -2(0

n f 0 "E l - e 2 -2 -2 1 1 1 0 -2d)3 2(0 3 -2co Y+S 0 0 0 0 0 0 0 0 -1 0 0 0 y 3+S 0 0 0 0 0 0 0 0 1 0 0 0 223

TABLE 7

(Continued)

|C e n t |: 144 360 360 1080 1080 216 54 54 18 480 480 480

Class: 4C 5A 5B 6A 6B 6C 6D 6E 6F 8A 8B 8C

Rep: a •c d z u x zu2 b Ujb U2b t2ax t tz t 3

y6+S 0 0 0 0 0 0 0 0 0 0 0 0

1|>!+S 4 -4 -4 1 -2 4 1 -2 0 0 0 0

4 0 0 5 -10 -4 -1 2 0 0 0 0

4»lX+s 0 -4 -4 -6 12 0 0 0 0 0 0 0

^10!+S 0 4 4 -4 8 -4 -1 2 0 0 0 0

^102+s 0 4 4 4 -8 4 1 -2 0 0 0 0

'f'lSx+s -4 -4ex -4e2 3 -6 0 0 0 0 0 0 0

< M 2+S -4 -4 e 2 -4 e ! 3 -6 0 0 0 0 0 0 0

0 4 e 2 4 ei -2 4 4 1 -2 0 0 0 0 ip1n 2i‘S 0 4 ei 4e2 -2 4 4 1 -2 0 0 0 0

1^2+S 4 -4 -4 -2 1 4 -2 1 0 0 0 0

^2<|)+S 4 0 0 -1 0 5 -4 2 -1 0 0 0 0

ip2X+S 0 -4 -4 12 -6 0 0 0 0 0 0 0

^ 29x+S 0 4 4 8 -4 -4 2 -1 0 0 0 0

^ 202+S 0 4 4 -8 4 4 -2 1 0 0 0 0

^2Cx+S -4 -4ex - 4 e 2 -6 3 0 0 0 0 0 0 0

11^2+S -4 -4 e 2 -4 e ! -6 3 0 0 0 0 0 0 0

^ rij+ S 0 4 e2 4ei 4 -2 4 -2 1 0 0 0 0

0 4ei 4e2 4 -2 4 -2 1 0 0 0 0 224

TABLE 7

(Continued)

|C e n t |: 480 16 16 27 360 360 36 36 24 24 90 90 90 90

C la ss: 8D 8E 8F 9A 10A 10B 12A 12B 12C 12D 15A 15B 15C 15D

Rep: t 3z ta t3a u3Wi cz dz auj au2 t 2b t 2b2 cu^ CU2 d u j du2

1S 1 1 1 1 1 1 1 1 1 1 1 1 1 1 4> 5 1 1 -1 0 0 1 1 -1 -1 0 0 0 0

X -6u)3 0 0 0 -1 -1 0 0 0 0 1 1 1 1

0 i -4(0 3 0 0 1 1 1 0 0 - 1 1 - 1 -1 -1 -1

02 4 0 0 1 -1 -1 0 0 1 1 -1 -1 -1 -1

Ci 3 -1 -1 0 El £ 2 -1 -1 0 0 £i £1 £2 £ 2

C2 3 -1 -1 0 £2 El -1 - 1 0 0 £2 £ 2 £1 El

Hi -2(0 3 0 0 -1 E2 El 0 0 1 - 1 - £ 2 - £ 2 -£i -El

2 3 Tl2 - co 0 0 -1 El £ 2 0 0 1 -i - £ 1 - £ 1 - £ 2 - £ 2

1/ -i i -i 1 1 1 1 1 -1 -1 1 1 1 1

- 5 i 1 -1 -1 0 0 1 1 1 1 0 0 0 0

x/ -6(0 0 0 0 -1 -1 0 0 0 0 1 1 1 1

*( -4(0 0 0 1 1 1 0 0 i - 1 - 1 - 1 - 1 - 1

Qz - 4 i 0 0 1 - 1 - 1 0 0 - 1 - 1 - 1 - 1 - 1 - 1

S( -3 1 -i i 0 El £2 - 1 - 1 0 0 El El £2 £ 2

- 1 1 0 d -3 1 £2 £1 -1 -1 0 0 £2 £2 £1 El

nl -2(0 0 0 -1 £2 £1 0 0 -i 1 - £ 2 - £ 2 -El -El

ri2 -2(0 0 0 - 1 El £2 0 0 - 1 1 “El -El -E2 - £ 2

1 M -1 -1 -1 1 1 1 1 1 1 1 1 1 1 1 225

TABLE 7

(Continued)

|C e n t]: 480 16 16 27 360 360 36 36 24 24 90 90 90 90

C la s s : 8D 8E 8E. 9A 10A 10B 12A 12B 12C 12D 15A 15B 15C 15D

Rep: t 3z ta t 3a UgWi cz dz a u i au£t 2b t 2b 2 cucu j 2duj du2

*" -5 -1 -1 -1 0 0 1 1 -1 -1 0 0 0 0

X" 6co3 0 0 0 -1 -1 0 0 0 0 1 1 1 1

e i' 4(d 3 0 0 1 1 1 0 0 - i 1 -1 -1 -1 -1 ^CM CD -4 0 0 1 -1 -1 0 0 1 1 -1 -1 -1 -1

-3 1 1 0 Ei £2 -1 -1 0 0 El ei £2 £2

-3 1 1 0 £2 £i -1 -1 0 0 £2 £2 El El

n r 2to3 0 0 -1 £2 El 0 0 1 -1 -£2 -£2 — El -E l // ri2 2 w 3 0 0 -1 £i £2 0 0 1 -1 -E l -E l — £2 -£2

l " ' i -1 1 1 1 1 1 1 -1 -1 1 1 1 1

/" 5 i -1 1 -1 0 0 1 1 1 1 0 0 0 0

x"' 6co 0 0 0 -1 -1 0 0 0 0 1 1 1 1 CD 4 a) 0 0 1 1 1 0 0 i -1 -1 -1 -1 -1

CD 41 0 0 1 -1 -1 0 0 -1 -1 -1 -1 -1 -1

31 i -i 0 £l £2 -1 -1 0 0 El £1 £2 £2

d " 31 i -i 0 £2 El -1 -1 0 0 £2 £2 £1 £1

n'" 2d) 0 0 -1 £2 El 0 0 -1 i -£2 -E2 -El -El ^/// ri2 2d) 0 0 -1 El £2 0 0 -1 i -E l -El —£2 -£2

Y+S 0 0 0 -1 0 0 0 0 0 0 0 0 0 0

Y3+S 0 0 0 -1 0 0 0 0 0 0 0 0 0 0 226

TABLE 7

(Continued)

| Cent |: 480 16 16 27 360 360 36 36 24 24 90 90 90 90

C la ss : 8D 8E 8F 9A 10A 10B 12A 12B 12C 12D 15A 15B 15C 15D

t2b t2b 2 cuj CU2 duj du2 Rep: t3Z ta t3a U 3W1 cz dz a m au2

y6+S 0 0 0 1 0 0 0 0 0 0 0 0 0 0

i^l+S 0 0 0 0 4 4 1 -2 0 0 -1 2 -1 2

lpi<|>+S 0 0 0 0 0 0 1 -2 0 0 0 0 0 0

•Ihx+S 0 0 0 0 -4 -4 0 0 0 0 -1 2 -1 2 i M i + S 0 0 0 0 4 4, 0 0 0 0 1 -2 1 -2

^102't'S 0 0 0 0 -4 -4 0 0 0 0 1 -2 1 -2 ihSi+S 0 0 0 0 4£ i 4£2 -1 2 0 0 -El 2ei -£2 2e2 0 0 0 0 4e 2 4£x -1 2 0 0 - e 2 2e2 -El 2Ei 0 0 if'xni+s 0 0 4e 2 46 ! 0 0 0 0 £2 “2s2 £l ■-2£i

0 0 0 0 4e x 4£2 0 0 0 0 ei -2 S i £2 1-2e2

0 0 0 0 4 4 -2 1 0 0 2 -1 2 -1

^2+S 0 0 0 0 0 0 -2 1 0 0 0 0 0 0

^2X+S 0 0 0 0 -4 -4 0 0 0 0 2 -1 2 -1

I(i201+S 0 0 0 0 4 4 0 0 0 0 -2 1 -2 1

4,202't>S 0 0 0 0 -4 -4 0 0 0 0 -2 1 -2 1

^2?1+S 0 0 0 0 4£i 4e 2 2 -1 0 0 2ei -Ei 2e 2 -£2

4ei ^2?2+S 0 0 0 0 4e2 2 -1 0 0 2e2 -£2 2Ei “El 4£i ihni+S 0 0 0 0 4e 2 0 0 0 0 -2e2 £2 -2ei £l

^2H2+S 0 0 0 0 4s i 4e 2 0 0 0 0 -2s i £i -2e 2 £2 227

TABLE 7

(Continued)

jCent |: 40 40 40 40 24 24 24 24 90 90 90 90

C la ss : 20A 20B 20C 20D 24A 24B 24C 24D 30A 30B 30C 30D N>

ip: rt O t2d t2CZ t2dz tb tb2 t3b t3b2 CZUj CZU2 dzuj dzu2

1S 1 1 1 1 1 1 1 1 1 1 1 1

4> 0 0 0 0 -1 -1 -1 -1 0 0 0 0

X i i -i -i 0 0 0 0 -1 -1 -1 -1

9i -i -i i i -00 0) -033 O)3 1 1 1 1

02 -1 -1 -1 -1 1 1 1 1 -1 -1 -1 -1

Cl El £2 El £2 0 0 0 0 e i El e 2 £2

C 2 £2 El E2 El 0 0 0 0 £2 e2 El El

Til -i£2 -i£i ±£2 i£i 00 -oo O)3 -O)3 £2 e 2 El El •rl (J 1 ri2 -i£i i£i i£2 0) -00 O)3 -oo 3 El El e2 £2

1' -1 -1 -1 -1 i i -i -i 1 1 1 1

4>/ 0 0 0 0 -i -i i i 0 0 0 0

X/ -i -i i i 0 0 0 0 -1 -1 -1 -1

CD i 1 -i -i -003 OO3 -0) 0) 1 1 1 1 CD roN 1 1 1 1 i i -i -i -1 -1 -1 -1

cr -El -e2 “El -E2 0 0 0 0 El El £2 £2

Ki -e2 -El —£2 "El 0 0 0 0 e2 e2 El El

ni i£2 i£i -±£2 -iEi U)3 -00 3 0) -00 £2 e2 El El

ri2 i£i i£2 -i£i -ie2 003 —OJ3 OJ -00 El El £2 £2 l/y 1 1 1 1 -1 -1 -1 -1 1 1 1 1 228

TABLE 7

(Continued)

|Cent|: 40 40 40 40 24 24 24 24 90 90 90 90

Class: 20A 20B 20C 20D 24A 24B 24C 24D • 30A 30B 30C 30D

Rep: t2c t2d t2CZ t2dz tb tb2 t3b t3b2 CZUj czu2 dzuj dzu2

" 0 0 0 0 1 1 1 1 0 0 0 0

X " i i -i -i 0 0 0 0 -1 -1 -1 -1 CD -i -i i i to -to to3 -to3 1 1 1 1

e f -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1

K f El £2 El £2 0 0 0 0 El El £2 e 0 0 £ £2 El £2 El 0 0 £2 £2 £1 e

nf “iE2 -iEl ±E 2 iEl -to to -to3 to3 £2 £2 El e

nf -i£i -ie2 iEl i£ 2 -to to -to3 to3 el El £2 e

i * -1 -1 -1 -1 -i -i i i 1 1 1 1

0 0 0 0 i i -i -i 0 0 0 0

xf" -i -i i i 0 0 0 0 -1 -1 -1 -1

e f i i -i -i to3 -to3 to -to 1 1 1 1

e f 1 1 1 1 - i -i i i -1 -1 -1 -1

5 f “El “£2 “El -£2 0 0 0 0 El £1 £2 £2

s f -£2 -El “£2 “El 0 0 0 0 £2 £2 El £1

n f i£2 iEl -ie2 -iEl -to3 to3 -to to £2 £2 El £1

n2 iEl i£2 “i£ i -is 2 -to3 to3 -to to El El £2 £2

yts 0 0 0 0 0 0 0 0 0 0 0 0

y3+s 0 0 0 0 0 0 0 0 0 0 0 0 229

TABLE 7

(Continued)

| Cent | : 40 40 40 40 24 24 24 24 90 90 90 90

C la s s : 20A 20B 20C 20D 24A 24B 24C 24D 30A 30B 30C 30D ro

Rep: ft o t2d t2CZ t2dz tb tb2 t3b t3b2 CZUj CZU2 dzui dzu£

y6+S 0 0 0 0 0 0 0 0 0 0 0 0

ipi+S 0 0 0 0 0 0 0 0 1 -2 1 -2

ipi+S 0 0 0 0 0 0 0 0 0 0 0 0

'hx+s 0 0 0 0 0 0 0 0 -1 2 -1 2 i M i + S 0 0 0 0 0 0 0 0 1 -2 1 -2

Ipl02+S 0 0 0 0 0 0 0 0 -1 2 -1 2

0 0 0 0 0 0 0 0 El -2ei £2 -2e2

^l?2+s 0 0 0 0 0 0 0 0 £2 -2e2 El -2ei thni+s 0 0 0 0 0 0 0 0 e2 -2e2 El -2ei

^ iH2+S 0 0 0 0 0 0 0 0 El - 2ei £2 -2e2

0 0 0 0 0 0 0 0 -2 1 -2 1

Ip2+S 0 0 0 0 0 0 0 0 0 0 0 0

^2X^S 0 0 0 0 0 0 0 0 2 -1 2 -1

^201+S 0 0 0 0 0 0 0 0 -2 1 -2 1

^202+S 0 0 0 0 0 0 0 0 2 -1 2 7-1

^2?1+S 0 0 0 0 0 0 0 0 -2ei El -2e2 £2

^2?2+S 0 0 0 0 0 0 0 0 -2e2 e 2 -2ei El

^201+S 0 0 0 0 0 0 0 0 -2e2 e 2 -2ei El

^202+S 0 0 0 0 0 0 0 0 -2ei El -2e2 £2 230

TABLE 7 (Continued)

|C e n t |: 40 40 40 40 40 40 40 40

C la s s : 40A 40B 40C 40D 40E 40F 40G 40H

Rep: tc td tcz td z t 3e t 3d t 3cz t 3dz

XS 1 1 1 1 1 1 1 1

0 0 0 0 0 0 0 0

X W W -w -w w 3 w 3 -w3 -w3

01 -W —to w w -w3 -w3 w 3 w 3

02 -1 -1 -1 -1 -1 -1 -1 -1

Ci El £2 £x £2 El £2 £1 £2

Cz £2 El £2 £1 £2 £1 £2 El

Oi -we 2 -wei we 2 wei -w3e2 -w3ei w 3e2 w 3ei

02 —we i -we 2 wei we 2 -w3ei —w 3e2 w 3ei w 3e2

l' i i i i -i -i -i -i

4>' 0 0 0 0 0 0 0 0

X' w3 w3 -w3 -w3 w w -w -w N> r*4 CD -u>3 - w 3 w 3 w 3 -w -w w w

02 -i -1 -i -i i i i i

ci lei ie2 lei ±£2 -lei -ie2 -i£i -ie2

Cz ie2 i£i i£2 i£i -ie2 -iei -ie2 “i£i

-w3e2 -w3ei w 3e2 w 3ei —we 2 -wei we 2 wei

ri2 —w 3 e i —w 3e2 w 3ei w 3e2 -wei -we2 wei we 2

l' -1 -1 -1 -1 -1 -1 -1 -1 231

TABLE 7

(Continued)

| Cent | : 40 40 40 40 40 40 40 40

C la s s : 40A 40B 40C 40D 40E 40F 40G 40H

Rep: tc td tc z td z t 3c t 3d t 3cz t 3dz * -e- 0 0 0 0 0 0 0 0

X" -01 -01 0) 0) -O)3 -O)3 O)3 O)3 CD 0) 0) -0) -0) O)3 O)3 -O)3 -O)3

CD 1 1 1 1 1 1 1 1

ef “ El “ 6 2 -E l — £ 2 -E l - £ 2 "El - £ 2

e ? - 6 2 -E l -e2 -E l —e2 -E l - £ 2 -E l

n f 0)82 0)81 - 0)8 2 - 0)8 1 0)3 8 2 0)38 i —0)38 2 -0)38 i n f 018l 0)82 - 0)81 - 0)82 0)38 i 0)38 2 —O)3 81 —O)3 8 2 i'" - i - 1 - 1 - 1 1 i i i

0 0 0 0 0 0 0 0

x ' " -o>3 - 0)3 O)3 O)3 - 0) - 0) 0) 0) CD 0)3 0)3 -O)3 -O)3 0) 0) - 0) - 0) ^ CD CD N N 1 1 1 1 - 1 - i - i - 1

€ f - 1 8 1 - 1 8 2 —iE i - 1 6 2 i£ i 1E2 iE l i £ 2

e f - 1 8 2 - iE l - 1 8 2 - iE l i£2 i£ l i £ 2 i£ i

Hi 0)3e2 0)38 i O)3 82 —0)38 i 0)82 0)81 —0)82 -0)8 1

n 2 0)38 i o)3 e 2 O)3 81 - 0)3 8 2 0)81 0)82 -0)81 - 0)82

Y+S 0 0 0 0 0 0 0 0 y3+s 0 0 0 0 0 0 0 0 TABLE 7

(Continued)

|C e n t |: 40 40 40 40 40 40 40 40

C la ss: 40A 40B 40C 40D 40E 40F 40G 40H

Rep: t c td tc z td z t 3c t 3d t 3cz t 3dz

0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0

'J'lX+S 0 0 0 0 0 0 0 0 lM i+ S 0 0 0 0 0 0 0 0

1M 2 +S 0 0 0 0 0 0 0 0

^iS i+ S 0 0 0 0 0 0 0 0 i p i ^ S 0 0 0 0 0 0 0 0 th o i+ s 0 0 0 0 0 0 0 0

^iil2+S 0 0 0 0 0 0 0 0

ipz+S 0 0 0 0 0 0 0 0

ip2+S 0 0 0 0 0 0 0 0

^2X+S 0 0 0 0 0 0 0 0

^201+S 0 0 0 0 0 0 0 0

1^202+S 0 0 0 0 0 0 0 0

^2?1+S 0 0 0 0 0 0 0 0

^2?2+S 0 0 0 0 0 0 0 0

^2 n i+ s 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 233

Propos it ion 3.14. (Decomposition matrix for N)

(i) N has fifteen characters of defect zero: X k > X 2

x7 > x8» 3 ’ $3’ ^1* ^2’ ^5* ^5’ ^6’ ^6’ 63* (ii) The remaining irreducibles of N fall into two

blocks:

— •X 3 > X 5 > X 5 > X 6 3"

(iii) The decomposition matrix for N is given by:

1 5 D, D =

where and D 2 are given as follows:

1 1 1 16 16

lN 1 132 396

1 X 3 9 i 1

X s 1 9, !

1 X 5 9. 1

1 1 1 0 2 1 1 x 6 1 (Dx) (d 2 )

(i) and (ii) follow from 3.1, 3.2. As N = 5, CO follows from 3 CO and 3.13 234

Proposition 3.15. (Decomposition matrix for S)

(i) S has nine characters of defect zero: , ', ",

Y+S, Yf5+S, y 6 +S, ^ 2 +s*

(ii) The remaining irreducibles fall into twelve blocks

as follows:

B 5 = (b x ,n 2 » »x >

2 B = {1' .C'a.e'a} B6 = tn'i» ^ 2 > 9i »x' >

b 3 = b 7 = {r\l ,n^,e",X"}

b^ = b 8 = {nr.n^e'i'.x'"}

b 9 = {i^+s^^+s^^+s.^e^s}

B 10 = 2 ^ 2 ? X ^ 2 ^2 ^ 2 ® 1 ^ Bn = ^iB1+s,ijJ1n2+s,ipie2+s,^1x+s> Bi2 = {^2n1+s,ip2B2ts,ijj2e2+s,^2x/t-s}

(iii) The decomposition matrix for S is given by

D, D =

12

where D j , D 2 ,..., D 1 2 are given sequentially

as follows: 235

l 3 i' 3' 1 " 3' ^■s l 1 ' 1 1 " 1 5i 1 51 1 51 1

1 1 1 ^ 2 5i 5;

l 1 1 1 1 1 0 2 «; 0 2

(Dx ) (d 2 )

1 "' 3'" 2 4 2 ' 4'

1 1 "' 1 Hi 1 ^'1

1 1 1 57 ^ 2 n'2

1 1 1 57 e l 0 'i e'" 1 1 X 1 1 X' 1 1

(D„) ) < V < D 6

2 " 4" 2 '" 4 '" 36 lOi

1 i+s 1 n i 1 Tl7 n" 1 n"' 1 1

1 e"' 1 ^ S 2+s 1 0 "

X" 1 1 X'" 1 1 1 1

(D? ) (Dq ) (Dg) 236

36' 108' 72 144 72' 144'

ip2+s 1 ipxDx+S 1 1 ^ 2 711 'hS

1 il^nz+s 1 i|>2n 2+s 1

* 2 52+s 1 ^ e 4+s 1 ip2e2+s 1

i M i + s 1 1 ^iX + S 1 1 ^ 2 x+s 1 1

(Dio)

Proof. (i) and (ii) follow at once from 3.1 and 3.2.

As |S| 5 = 5, (iii) follows from 3.3,3.4, and 3.13. 237

4. Evidence

As proposed in the introductory statements of the

chapter, we let x denote a hypothetical irreducible

5-Brauer character of LyS of degree 111. We then

systematically restrict x to certain distinguished sub­

groups of LyS in an effort to obtain values for x on

5-regular LyS-classes. For some of these subgroups the

5-modular structure is well understood, and a compatabil-

ity check is immediately provided via the corresponding

restrictions. Even when this is not the case, it will

frequently be true that a pair of restrictions still

yields a compatability check as follows. Let A and

B be subgroups of LyS, and let , Kg be conjugacy

classes of A , B respectively such that K^ C K and

Kg C K where K is a LyS-class. Then the restriction

X+A conveys information regarding the value of x on

K , which must be consistent with that obtained via the

restriction x+B « To prevent the ensuing proofs from

becoming too cumbersome, we remark here that all compata­

bility checks of the latter type have been carried out whenever possible, even when not explicitly stated.

We begin by focussing attention on the groups N

and S of the previous section. 238

Proposition 3.16. Let N = E 3 5 *MX 1 be the subgroup of

LyS whose 5-structure was determined in proposition 3.14.

Then x+N = 1N + $2.

Proof. As all Brauer characters of degree less than 110 have E = E 3 5 in their kernels, and as must be faith­ ful, we see that x+N = 1 + 0 where 0 € { x, 2 , cf> 3 , ^ 3 } .

Furthermore, as LyS has a unique class of elements of order 18, we certainly must have x+N(a) = x+N(a-1) where a is an element of order 18 in N. By inspec­ tion of the character table for N, this yields

0€{1 ,x presently.

Suppose then, by way of contradiction, that x^N =

1N + cf> x . Let U be a subgroup of LyS isomorphic to

U 3 (3). Our first task is to determine the LyS-classes into which certain of the U-classes fall. The classes of U with which we are concerned are given as follows, along with the corresponding centralizer orders.

U-class of x 1A 2A 3A 3B 4A 4C 6 A

I^Cx)! 6048 96 108 9 96 16 12

As LyS has a unique class of elements of each of the orders 1, 2, 4, the only difficulty is to determine the corresponding LyS-classes for the 3-elements listed above. Now if 3A U 3B C 3 X, then all elements of order 239

3 in U would be of LyS-type 3j, and LyS would have a 3 -pure Eg, contradicting lemma 1.13. Thus at least one of the classes 3A , 3B must be of LyS-type 3 2 .

But under the assumption 3A C 32, it is readily seen that

(i|>+U,lu) is not a non-negative integer, where ipelrr(LyS) with iKl) = 45694. This contradiction leads us to con­ clude that 3A C 3 1 and 3B C 32. As an element of 6 A squares to an element of 3A (so to an element of the LyS- class 3.!), we further conclude from the structure of LyS that 6 A C 6 1. The LyS-classes corresponding to 1A, 2A,

3A, 3B, 4A, 4C, and 6 A have now been determined.

Under our assumption that x+N = ^-n + we easily obtain the following values for x on t*1® LyS-classes as indicated:

LyS-class ofx 1 2 3X 3 2 4 6 X

X(x) 111 15 -24 3 3 0

In terms of U-classes this becomes:

U-class of x 1A 2A 3A 3B 4A 4C 6 A

X ( x ) 111 15 -24 3 3 3 0

Now consider x^U. As U is a 5'-group, IBr(U) =

Irr(U) = {1,6,7,7',7',14,21,21',21',27,28,28,32,32} , and as

X(a) = x(a_1) = x(a ) f°r a € 4A , we see that 9 and 9 have the same multiplicity in x^U for every 9 eiBr(U).

Thus all possibilities for the (unique) decomposition of 240

X+U as a non-negative integral sum of irreducible Brauer characters of U can be determined from the matrix equa­ tion :

1 6 7 14 14 21 42 27 56 64 111 x.

1 -2 -1 -2 5 2 3 -8 0 15 X, 1 -3 -2 -4 5 3 6 0 2 -8 -24 X, 1 0 1 2 - 1 0 0 0 2 -2 3

1 -2 3 -2 2 1 -6 3 0 0 3 X, 1 2 -1 2 2 1 -2 - 1 0 0 3

1 1 2 0 1 - 1 2 0 -2 0 0 X 10

Indeed this equation corresponds to a system of linear equations that must be satisfied by the character x+u

(and is obtained from the character table for U). As the general solution to this equation is found to be

(-9+a+2b+6c,-1+a+b,4-b-2c,x— - a-b-2c,3-a-b-2c,a,-£,b,0,c)'fc, we conclude that -£ must equal the multiplicity of 2 1 in x+u * This gives the desired contradiction and proves

X+N = 1N + (j)2 . 241

Proposition 3.17. With notation as in the previous proposition, x+U =2*6 +3*7 +7' + 1' + 32 + 32 .

Proof. As x+N = lpj + 2 , we obtain the following values for x on the LyS-classes as indicated:

LyS-class of x 1 2 3 1 3 2 4

X(x) 111 -1 -24 8

In terms of U-classes this becomes:

U-class of x 1A 2A 3A 3B 4A 4C 6 A

X(x) 111 -1 -24 3 3 3 8

The relevant matrix equation for computing x+U as a non-negative integral sum of irreducible Brauer charac­ ters of U is given as follows:

f X 1 > 1 6 7 14 14 21 42 27 56 64 1 1 1 X 2

1 -2 -1 -2 5 2 3 - 8 0 X 3 - 1

1 -3 -2 -4 5 3 6 0 2 - 8 X* -24 X 5 1 0 1 2 -1 0 0 0 2 -2 = 3 X 6

1 -2 3 -2 2 1 - 6 3 0 0 X 7 3

1 2 -1 2 2 1 - 2-1 0 0 X 8 3 X 9 1 1 2 0 1-1 2 0-2 0 8 X 10 » .

The general solution to this equation is given by

(-6 +a+2 b+6 c ,2 +a+b,5-b-2c,3-a-b-2c,2 -a-b-2 c,a,0 ,b,0 ,c)t 242

As all entries (in particular 2-a-b-2c, a, b, and c) must be non-negative integers, we easily see that c = 0 or 1. But c = 0 implies a + b ^ 2, which contradicts

-6+a+2b+6c ^ 0. Thus c = 1, whence -a-b = 2-a-b-2c ^ 0 and a = b = 0 follows. Thus the above matrix equation has a unique solution of (0 ,2 ,3,1 ,0 ,0 ,0 ,0 ,0 ,I)1- and the proposition is proved.

Corollary 3.18. X assumes the values as indicated on the

LyS-classes below:

L y S - c l a s s o f x 1 2 4 31 32 61 9 18 7 11 2

X(x) 111 -1 3 -24 3 8 0 2 -1 1

LyS-class ofx ll2 12 1

X(x) 1 0

Proof. All values are obvious from proposition 3.16 and the character table for N, except for those on the class­ es 7 and 12x. But the former follows readily from proposition 3.17, and the latter is obtained from the cal­ culation #(3^4,12) f 0 in N which shows that the IT- class 12 is contained in the LyS-class 121. 243

Proposition 3.19. Let S = 32+If • (SL(2,5.) *Zg ) be the sub­ group of LyS whose 5-raodular characters were determined in proposition 3.15. Then x +s = 3" + 36 + 72' .

Proof. We first observe from the character table for S that the element u x is in the kernel of all irreducible

Brauer characters of S except for 36, 36', 108, 108',

72, 72', 144, 144', ipjCfr+S, ^ 2

LyS-type 3a, we have -24 = x(ua) = X+S(ui) = ©(u^+^Uj) = 0(1) + s(u:) = 111 - c(l) + £(Uj) whence r,(1)-?(Uj )=135.

From this it is not hard to show (again from the character table for S) that there are only two possibilities for

? as a non-negative integral sum of irreducible Brauer characters, namely ? = 36 + 72' and z, = 36 + 2*36'. Now

(36+72')(z) = -4 and (36+2»36')(z) = 12, z an involu­ tion in S. .As |0(z)|=0(l)=3 (from the character table for S) and as x(z) = -1 (from corollary 3.18), we see that the only possibility is S(z) = -4 and 0(z) = 3.

Thus x+S =0+36+72' where 0 is equal to one of

3*1, 3*1" , 3, or 3", as 0 must be real on elements of order 8 by the structure of LyS. We next consider

X(a), x("t2) where a, t 2 are elements, of order 4 as 244

indicated in the character table for S. We easily

compute:

3 = x(a) = x+S(a) = 9(a) + £(a) = 0(a) + 4

3 = x(t2) = X+S(t2) = 0(t2) + ?(t2) = 0(t2)

Thus 0 (a) = -1 and 0 (t2) = 3 whence 0 = 3 or

3". Finally, we consider X(t) = X+S(t) = 0(t)+?(t) = 0(t) where t is of order 8 as indicated in the character table for S. Thus x(t)=3 or -3 corresponding to the cases 0 = 3

or 3", respectively. By propositions 3.16 and 3.17, x(e ) = 1

and xCe')3' -3 for elements e and e' of order 8 in N and U

respectively. We therefore conclude that e and e' repre­

sent the distinct LyS-classes of elements of order 8 , whence

X(t)=3 is impossible. Thus x('t) = ~3 and 0 = 3” as claimed.

Corollary 3.20. x assumes the following values on LyS- classes as indicated:

LyS-class of x 6 2 6 3 8 X 8 2 12 2 24x

X(x) -1 -1 1-3-3 0

Proof. As < S, S contains representatives from every LyS-class of elements of order 6 . We list below in tabular form representatives for all S-classes of elements of order 6 , their centralizer orders, and their values under x : 245

X b z u r ZU 2 t 2ax \.\b T^b

Cs (x)| 216 1080 1080 18 54 54

X(x ) 8 8 - 1 - 1 - 1 - 1

As zuj has LyS-type 6 ,,, xCzi^) = 8 agrees with our former calculations. Now zu 2 has LyS-type 6 2 [13], whence x assumes the value -1 on this class. Clearly for x of LyS-type 6 3, either x(x ) = 8 or -1- But

|Cq(x) | 2 = 4 while |Cg(b) | 2 = 8 . Thus x(x ) = as claimed.

From proposition 3.19, x(w) = ® where w € S is an element of order 24. We show w has LyS-type 24

Let w = ue with [u,e] = 1, u of order 3 and e of order 8 . As u € Cg(e) we must have |Cg(e)| = 480.

(Indeed for x of order 8 in S, the only possibilities for |Cg(x)| are 480 and 16.) But then e must be an

8 2-element of LyS as j CQ(e x)| = 96 for ex of LyS- type 8 a. This proves w is in the LyS-class 241 [13], and x assumes the value 0 on this class as claimed.

This argument also shows t must be of LyS-type 8 2

(as |Cg(t)| = 480); therefore x takes on the value -3 on the class 8 2. As discussed in the previous proposi­ tion, the elements of order 8 in U are of different

LyS-type from that of t, hence of type 8 X . Thus x assumes the value 1 on the LyS-class 8 j . 246

It remains only to determine x on 122-elements of LyS. Consider the element au 2 of order 12 (as indicated in the character table for S). As a is an element of Cg(u2), it is immediate from [13] that au 2 is of LyS-type 122. We easily calculate x(au 2 ) = from proposition 3.19 and the proof is complete.

Lemma 3.21. (5-decomposition matrix for L 3 (4) )

(i) L 3 (4) has six characters of defect zero: 20, 35, 35',

35", 45, 45.

(ii) All remaining irreducibles of L 3 (4) lie in a unique

block.

(iii) The decomposition matrix for L 3 (4) is given by

1 63 1 1 D = where D, = 6 3 D, 1 63 1 64 1 1

Proof. (i) and (ii) follow from 3.1 and 3.2. (iii) is immediate from 3.13.

Proposition 3.22. Let L S L 3 (4) be a subgroup of LyS.

Then x+L = 1 + 20 + 45 + 45 with constituents as in lemma 3.21. 247

Proof. Let 3A denote the unique L-class of elements of or­

der 3. As a Sylow 3-subgroup of L is elementary abelian of

order 9 and no such subgroup of LyS is 32-pure (lemma

1.13), we conclude that 3A C 32 . The remaining 5-regular

classes of L are easily identified with their respective

LyS-classes, so that (as in proposition 3.17) we can de­

termine the decomposition of x+L from matrix computation:

* > • 1 1 1 1 2 0 35 35 35 90 63 X 1

1 4 3 3 3 6 1 x 2 - 1

1 - 1 - 1 - 1 2 0 0 X 3 3

1 0 3 - 1 - 1 2 - 1 x * — 3

1 - 1 - 1 2 - 1 0 3 X 5 3

1 0 - 1 - 1 3 2 - 1 X 6 3

1 - 1 0 0 0 - 1 0 - 1 X 7 » * * 4

The unique solution to this equation is given by

(l,l,0,0,0,l,0)t whence x+L = 1 + 20 + 45 + 45 as

claimed. (Note that the 6 th column of our coefficient matrix is obtained by virtue of the fact that 45 and

45 occur with the same multiplicity in x+L since LyS

has a unique class of elements of order 7.) The proposi­

tion is thereby proved. 248

Proposition 3.23. Let Cu = > 't^ie centralizer in

LyS of an element u of Sj-type. (Thus Cu = Me.)

Then X ^ ^ = 21 + 45 + 45 where 21, 45, 45 € IBr(Cu ) .

Proof. Let ip denote any faithful irreducible constitu­ ent of X+Cu, and let n be an element of Nq() which inverts u. Then ipn (defined by ipn(g) = iKs11)) is a faithful irreducible constituent of X+Cu as well.

Indeed < (x*Cu)n = xn+Cu = X+Cu* the last equality following from xn= X as n is an element of LyS.

Moreover, ip and ipn are necessarily distinct as can be seen by evaluation at u. This proves that faithful ir­ reducible constituents of X+Cu occur in pairs. As X't'Cu is faithful, it must contain at least one such pair of constituents.

Now Cu contains a subgroup L isomorphic to

L 3 (4). Thus by the previous proposition, (x+Cu)+L =

X+L = 1 + 20 + 45 + 45 with all constituents being irreducible Brauer characters of L. It is now im­ mediate from our earlier discussion that there exist pre­ cisely two faithful irreducible constituents of x+Cu which restrict irreducibly to 45 , 45 € IBr(L) respective­ ly. We denote these by 45 , 45 as well. As all remain­

ing constituents of X+Cu must be faithless (and as such, can be regarded as Brauer characters of Cu/ ~ Me ) we 249

see that theorem 3.1 applies. Since Me has no Brauer irreducible of degree 20, the proposition follows.

Lemma 3.24. (5-decomposition matrix for

(i) 1 has five characters of defect zero: 10, 10' , 10

45, 55.

(ii) All remaining irreducibles of Mjj lie in a unique

block.

(iii) The decomposition matrix for Mj j is given by

1 11 16 16 1 1 D = where D = 11 1 16 1 16 1 44 1 1 1 1

Proof. The proof is an easy consequence of 3.1, 3.2,

3.3, and 3.13.

Proposition 3.25. Let M < Cu , M = M ^ . Then =

10 + 11 + 2*45 with 10 , 11 , 45 € IBr(M). In particular,

45 6 IBr(Cu) restricts irreducibly to M.

Proof. If j € LyS inverts u, then CG(u,j) = MX1

[13,p.544,AS. .12-15] , so such an M as indicated in the proposition statement exists. As LyS contains no 3j- pure E 9's (lemma 1.13), the elements of order 3 in M 250

must be of type 32. Now by [l3,p.551] Sj-elements and

8 l-elements of LyS fail to commute. Thus (as u is of type 3X) all elements of order 8 in Cu must be of type 82. As the latter statement must certainly hold for M, we are now able to formulate the usual matrix equation which will disclose all possibilities for x+M:

1 • * 10 11 20 32 45 55 X 1 111

2 3 -4 0 -3 - 1 x 2 -1

0 1 - 1 2 -1 0 X 3 3 0 -1 0 0 -1 1 x 4 = -3

0 1 2 2 -4 1 X 5 3

0 0 - 1 -1 -1 0 2 X 6

-1 - 2 -1 1 0 1 0 X 7 > > 4 « 4

(The 4th and 5th columns of the coefficient matrix follow from the fact that each of 10 , 16 € Irr(M) occurs with the same multiplicity in x^M as does TU , respec­ tively.) The unique solution to this equation is given by (0,l,l,0,0,2,0)'fc whence the proposition follows. 251

Corollary 3.26. x assumes the values as indicated on the

LyS-classes below.

LyS-class of x 21 1 212 3 3 1 3 3 2

X(x) -2 -2

Proof. We first observe from [13,p.552] that 211 = 3 X*7,

212 = 3^7, 33x = Sj'llj, 332 = 3 l ’llz . Thus all of the above classes of LyS in fact meet Cu . Now as =Z(Cu) acts as scalar matrices via any irreducible representation of Cu , we obtain the following (where 21 , 45 € IBr(Cu ) ):

21(ug) = 21(g), 45(ug) = o)45(g), 45(ug)=w245(g) for all g € Cu . Here to denotes e2iri^3 = (_i + i/3)/2.

Now let y = us with s 6 Cu of order 7. Then

X (y) = 21(y) + 45(y) + 45(y) = 21(s) + to45(s) + oj24 5 ( s ) .

We obtain 21(s) = 0 directly from the character table for Cu S Me (as 21 = 22 - 1 where 22 € Irr(Cu ) ).

The value 45(s), however, depends on the Cu-class to which s belongs. Thus there are two possibilities for

45(s), each of which is obtained via restriction to L:

45+L(s) = (-l±i/7)/2 ; and correspondingly two possibili­ ties for x(y): X(y) = (l±/21)/2 (easily computed from above). Without loss of generality we may assume 252

1 + /21 y e 21x 2

1 - V21 y € 212 2

Now let H € M < Cu , M = M]_ 1 , Z of order 11 .

Then x(u£) = 21(u£) + 45(u£) + 45(u£) = 21(£) + w45(£)

+ w245(&). From the character table for Me we have

21(&) = -1, while restriction to M (proposition 3.25) gives

45(Jt) = 1. Thus we get x(u^) = -1 + w + = “2, irre­ spective of the LyS-class to which Z belongs. The proof is now complete.

Proposition 3.27. Let K < C Z, K = SL(2,8). Then

X+K = 7 + 2*(7'+ 7"+ 7'") + 8 + 2* (9 + 9'+ 9") where

7, 7' , 7 " , 7"', 8, 9, 9' , 9* are irreducible (Brauer) characters of K and {7' ,7", 7"'}, {9,9',9"} are each rational classes of characters.

Proof. SL(2,8) embeds into Ag via its action on the projective line of order 9. As SL(2,8) has trivial multiplier, such a K as appears in the statement of the proposition exists. The identification of K-classes with those of LyS, and the solving of the resulting matrix equation below, are both routine. (Column 4 of the coef­ ficient matrix corresponds to the character 7' + 7"+ 7'" ; 253

column 5 corresponds to 9 + 9'+ 9". We may group them because - as is revealed by the character table for SL(2,8)

7',7",7"' all occur with the same multiplicity in x+K > as do 9, 9' , 9 ". )

1 8 7 21 27 X 1 111 1 0-1-3 3 x 2 -1 = 1-12 3 0 X 3 3

1-1 10 0 X 4 0

110 0 - 1 X 5 -1

This matrix equation has unique solution (0,1,1,2,2)"^ whence x+K decomposes as claimed. Since K is a 5 -group, all its irreducible characters are Brauer irreducible as well.

Proposition 3.28. Let Cz = CG (z), the centralizer in

LyS of an involution (so Cz = A 1X). Then X + Cz = 55 + 56 where 55 , 56 € IBr(Cz ).

Proof. Consider CG(z,v) where v is a 3x-element in­ verted by z. Then CG (z,v) = M 1]L (as noted in the proof of proposition 3.25) and x+CG (z,v) = 10 + 11 + 2*45

(again proposition 3.25). As CG(z,v) < Cz , the restric­ tion to CG (z,v) gives much information about the charac­ ter X't'Cz- 254

Let X+Cjg = 9 + ^ > where 8 denotes the constituent of x+Cz ojf maximal degree such that ker8 >. (Thus

if s < tp is irreducible, then it must be faithful.)

Evaluation at 1 and z gives:

111 = x(l) = X+Cz (l) = 6(1) + i|>( 1)

-1 = x(z) = X+Cz(z) = 0(z) + ij;(z) = 0(1) - i|;(l)-

The last equality follows as z is represented by -I under the representation of Cz affording 1p. We thereby obtain 0(1) = 55 , ^ (1) = 56.

But from above we must have 0+CG (z,v) =10+45 and

\jj+CG(z,v) = 11 + 45. Comparing this with proposition 3.27, it is easy to see that each of 0 , ip must be Brauer irredu­ cible. Indeed no proper decomposition of 0 (respectively ip ) admits both the restrictions 0+CG(z,v) , 0+K (respec­ tively ^4-Cg (z ,v ) , iHK )• The proposition follows with

55 denoting 0 and 56 denoting

Proposition 3.29. Let 55 € IBr(Cz ) be as in proposition

3.28. Then regarding 55 as a Brauer character of

Cz/ s A xj, we have 55 = [7,4]* - [8,3]*.

Proof. By proposition 3.28, ker(55) = . The proof now follows directly from proposition 3.12. 255

Corollary 3.30. x assumes the values as indicated on the

LyS-classes below:

LyS-class of x 14 4 2 1 422 22 1 222 28 , „ , -5-/21 -5+/21 H X(x) -1 — ------1 -1 3

Proof. Let g . € Cz. As z € ker(55) (55 € IBr(Cz) ), we have 55(zg) = 55(g), and as z acts as -I under the representation of Cz affording 56 € IBr(Cz) we have 56(zg) = -56(g). Thus we are able to compute:

X(zg) * X+Cz(zg) = 55(zg) + 56(zg) = 55(g) - 56(g) =

55(g) - (X (g) - 55(g)) = 2•55(g) - X(g). But the value 55(g) can be obtained using proposi­ tion 3.29 : 55(g) = [7,4]*(g)•- [8,3]*(g), where [7,4]*,

[8,3]* are here regarded as faithless characters of

A Cz = An . We therefore have a useful formula for estab­ lishing the value x(zg) when x(g) is given.

Let s € 7. We have already seen that x(s) = -1

(corollary 3.18). We readily compute:

55(s) = [7,4]*(s) - [8,3]*(s) = -3 - (-2) = -1. Thus

X(zs) = 2* 55(s) - x(s) = 2(-1) - (-1) = -1. As zg € 14,

X assumes -1 on the LyS-class 14.

We summarize the calculations for the LyS-classes

4 2 1 , 422 , 22},222 below. 256

LyS-class of g 2 1 i 212 1 1 1 11 1 + /21 1 - V21 x(g) 1 1 2 2 [7,4]*(g) 0 0 0 0

[8,3]*(g> 1 1 0 0 -5 - /21 -5 + V21 x(zs) -1 -1 2 2 LyS-class of zg 42 x 422 22l 22

Finally, we observe that x(x) ^or x 6 28 cannot be obtained in this exact manner. Indeed for g € 14 we also have zg € 14. But zx € 28 for x € 28, so we have

X(zx) = x(x )- Therefore x(zx) = 2»55(x) - x(x) =

2»55(x) - x(z x ) whence x(zx) = 55(x) = [7 ,4] *(x)-[8,3] *(x)

= l-(-2) = 3. The corollary is proved.

Proposition 3.31. The degrees of the irreducible 5-Brauer characters of G 2 (5) are given as follows: 1 , 7 , 14 , 27 ,

64 , 77 , 97 , 182 , 189 , 196 , 371 , 469 , 483 , 721 , 792 , 1344 ,

1715 , 2008 , 2247 , 2380 , 2667 , 4830 , 8456 , 15625. For n f

77, there is a unique irreducible of degree n which we shall denote also by n. There are two irreducibles of degree 77 , and we shall denote these by 77 and 77' , respectively.

Proof. [10]. I am indebted to Peter Landrock for bring­ ing this result to my attention. 257

Lemma 3.32. Let U =U.(3) be a subgroup of G„(5).

Then 14 € IBr(G2(5>) restricts irreducibly to U .

Proof. The proof is divided into two cases which reflect the two possibilities for 7+U , 7 € IBr(G2(5)>. To bet­ ter emphasize the distinction between Brauer irreducibles of U and those of G2(5), we adopt the following con­ vention: n will continue to denote a Brauer irreducible of G2(5) of degree n, while such a character of U will henceforth be denoted by n .

Case one: 7+11 is irreducible. As 7 € IBr(G2(5)) is self-dual there is a unique possibility for 7+U under our present assumption of irreducibility, namely 7+U = 7

(with notation as in the proof of proposition 3.16). We calculate in U: 7®7 = l + 7 + 14 + 27. Thus

1 + 7 + 14 + 27 = 7+U®7+U = (7®7)+U. The lemma follows in this case from proposition 3.31.

Case two: 7+U splits. As 7 € IBr(G2(5)) is faithful, the only possibility is clearly 7+U =1+6.

Here we have 7 « 7+U = 74-U ® 7+U = (1+6) ® (1+6) =

2*1 + 2*6 + 14 + 21^. Now from proposition 3.31 it is easily seen that 27 is the only possibility for an ir­ reducible constituent of 7® 7 having the property

21 < 27+U. Thus ( 7 ® 7 - 27)+U =2*1+6+14. But now 258 it follows (again from proposition 3.31) that 14 < 7a7 - 27 with 14+U = 14. (Indeed 14 can he obtained in no other way.) The proof is now complete.

Proposition 3.33. Let H = G2(5) be a subgroup of LyS.

Then x+H = 2»7 + 97.

Proof. By proposition 3.17, x+u = 2*6 + 3-7 + 7/ + 7' +

32 + 32. (We continue our practice of underlining the

Brauer irreducibles of u to contrast them from those of

H.) As we may assume U ^ H, the above gives information regarding X+H.

Now 32 < x+U and x(l) = H I , so we may conclude that (counting multiplicities) precisely one of 64 , 77 ,

77', 97 is an irreducible constituent of x+H. As there do not exist non-negative integers a , b , c such that

7a + 14b + 27c = 47 = 111 - 64 , it is immediate that 64 cannot be a constituent of x+H. It is not hard to show at this point that the only possibilities for x+H are given as follows:

(a) x+H = 7 + 2 7 + 7 7

(b) x+H =7+27+77'

(c) x+H =14+97

(d) x+H = 2*7 + 97 259

Suppose (a) or (b) holds. As 1 <(. x+U we conclude that 7+U = 1_. It is now easily seen from lemma 3.32

(case one) that 27+U = 27. But as 27 < x+H we must have 27 < x+U, a contradiction. This disposes of (a) and (b). We next assume (c) holds. Thus 14 < x+H whence 14 < x+U (by lemma 3.32). This is again contra­ diction and x+H =2*7+97 is proved.

Corollary 3.34. X(x) = 1 f°r x € 2 4 2 U 243 .

A Proof. From the character table for Aj2 we see that gz € 243 for g 6 Cz of type 242. The character table for LyS reveals that g-1 € 243 for g € 242. Thus for g € 242, x(zg) = X(g“l) = X(g)• Combining this with the formula x(zg) = 2*55(g) - x(g) of corollary

3.30 we obtain: x(g) + X(g) = 2*55(g).

Now x+H is self-dual by proposition 3.33. As 242 , 243 each meet H (see for example the character table for G2(5)) we have x(g) = X+H(g) = X+H(g) = x(g) for g € 242. We conclude from above x(g) = 55(g) = [7 »4]*(g) - [8,3]*(g)

= 1 for g € 242 and hence also for g € 243. 260

The following result from modular character theory

is well-known and appears without proof [11].

Proposition 3.35. Let G be a finite group and let p

be a prime. If 0, € IBr(G) lie in different p-blocks,

then Z 0(x)

Corollary 3.36.

(1) X ( W x +w 2 +10 3 + 0) 4 +U) 5 ) = -3 (2) x(Xi+x2) = 0

(3) xCyx+Yz+Ya) = -2 where to^ € 31i (1 ^ i ^ 5), x^€37jL(l^i— 2), and

y i € ®7i — 1 — 3). Moreover x(xi) = X(x 2 ^ ~

Proof. By previous corollaries we have determined the

value x(x) l°r x any 5-regular element of LyS of

order n f 31 , 37 , 67. Applying proposition 3.35 suc­

cessively to x an

X9 of LyS [13,p.557] (of respective degrees 38734375,

64906250, 53765625), we get the values of x 011 I*1® rep­

resentative sums as indicated. From the character table

for LyS it is immediate that ^(x^ = i|j(x2) whenever

i|Kl) t 21312500. As 56 divides 21312500, all ordinary

irreducibles of this degree have 5-defect zero. Thus iKXj)^ ^(x2) for every ordinary irreducible ip in the

5-block containing x* As x is necessarily an integral

linear combination of such irreducibles, x(x i) = X(x2)

follows whence x(x i) = x(x 2 ^ =

In the following theorem we summarize what is now known regarding the values of x on 5-regular elements.

Theorem 3.37. Let x denote a hypothetical irreducible

5-Brauer character of LyS of degree 111. Then X assumes the values on LyS-classes as indicated below.

LyS-class of x 1 2 4 8X 82 3X 32

x (x) 111 -1 3 1 - 3 -24 3

LyS-class of x 9 7 llx 112 62 63

X ( x ) 0 - 1 1 1 8-1-1

LyS-class of x 18 14 42 x 42 2 „ -5-/21 -5 + /21 X(x) 2 -1 ----- j----

LyS-class of x 22x 222 12x 122 28 24x 242

X(x) -1 -1 0 -3 3 0 1

LyS-class of x 243 21x 212 33a 332 1 + /21 1 - /21 262

LyS-class of x 37j 372

X(x) 0 0

Moreover x(w i+aJ2 +t03 +a)4 +a)5 ) = where € 31^, and X( y1+yz+y 3) = ”2 where yj € 67j. Modulo our lack of

success in determining the precise values for x(wi) and

X(yj) (1 ^ i ^ 5 , 1 i j £ 3), our list is exhaustive

(i.e. every 5-regular class of LyS appears). In particu­

lar x is self-dual.

Proof. This is just a restatement of corollaries 3.18,

3.20, 3.26, 3.30, 3.34, and 3.36. As g ~ g -1 for any g

in LyS of order 31 or 67 , we clearly have x(S) = xTiT

for all g € LyS, and x is self-dual.

Before proceeding, we remark that proposition 3.35

supplies us with additional compatability checks which

further support the existence of x> For example, we can

check that 2 x(x )X2i(x) = 0 where Xji is either of the x€S two defect zero characters of LyS of degree 21312500

[13]. In this manner we can apply proposition 3.35 to various restrictions X-*"^- even when the 5-modular

structure of A is not known. 263

5. Minimal Degree for a FaitTiiul Character of LyS

Maintaining earlier established notation, let S5 denote a Sylow 5-subgroup of LyS generated by elements

, g 2 » S 3 , gk , S 5 , f2 of order 5 subject to defining relations [g2,g5] = [g3,S5] = = f!« [£i*f2J = Si-i for i = 3,4,5, and other commutators of generators trivial. Let E denote the unique (up to conjugacy) 5 X- pure E53 of LyS, i.e. E = (lemma 1.37). NQ shall denote the normalizer Nq (E) of E in LyS .

Finally, S shall denote the 5-group and Z its subgroup . We are now ready to pro­ ceed .

Lemma 3.38. S/Z is extraspecial of order 5 3 .

Proof. From the defining relations for S5 it is immedi­ ate that Z is normal (indeed central) in S. Letting bar denote images modulo Z, we get the defining relations for 13: [gi*,?^] = ^ 3 with all other commutators of gen­ erators ( g3 , g4 , f2 each of order 5 ) being trivial.

Thus we clearly have Z(S) = [S,S] = - As S/Z(S) is isomorphic to Z5 *Zg, it easily follows that S is extra­ special, and its order is given by |s| = |S/Z(S)||Z(S)| =

5 3 . 264

Lemma 3.39. N q acts transitively on Irr(E)^.

Proof. By [13,p.552,2.£.24-27], NQ (Z) is a split faith­ ful extension of S by GL(2,5). Now N0 permutes the

31 subgroups of E isomorphic to E52 . As |z^o| —

lNo 1/ |Nq (Z)| = 2 5*3*56*31/25*3*56 = 31, the action is transitive. We next show that Nj^Z) acts transitively on E •— Z. Let 0 denote the action of N0 on E via conjugation: 0:NO ► Aut(E). By lemma 1.38, No is a faithful (non-split) extension of E by SL(3,5); thus identifying Aut(E) with GL(3,5) (relative to the basis

{f1,g2,g3}), we see that 0 maps onto SL(3,5). Now let x = ffg^gf be an arbitrary element of E ^ Z. Letting n be a preimage under 0 of the matrix

• \ 1 a

1/c b

c > we see immediately that n € Njj£Z) and g]1 = x as de­ sired. (c f 0 because x g Z.)

Now let X 6 Irr(E) be defined by X(f 2) = X(g2) = 1,

X(g3) = ri (where n is a primitive 5th root of unity), and let be an arbitrary element of Irr(E)^. We have already established that N acts transitively on the

E ^-subgroups of E ; thus there exists n € N Q such that 265

(kerip)n = kerX = Z. Ghoose y € E such that

Clearly yn i Z; indeed yn € Z implies y 6 Zn_1 = kerij;, a contradiction. As Nj^CZ) is transitive on E ^ Z, there exists w € NN q (Z) such that ynw = g3. We now show ipnw = X, proving the lemma. First ker(ipnw) =

{x € E :ipnw (x)=l} = {x € E :^(x(nw )_1 )=1} = {x 6 E :xCnw)_1 g kerij;}

= {x € E:x € (keri|i)nw} = (kerijj)nw = (kerX)w = kerX. Thus it suffices to show ipnw and- X agree on g 3 . Indeed

^nw(g3) = ^(S 3nw^— 1) = ^(y) = n = X(g3). The proof is com­ plete .

Proposition 3.40. Let ? be a faithful irreducible F- character of N0 where F is a field of characteristic different from 5. Then deg(?) ^ 620.

Proof. As ? is faithful, there exists X € Irr(E)^ such that X < s+E. (Otherwise ?4-E = deg(?)*lg whence E is contained in ker£, a contradiction.) As a consequence of lemma 3.39, we have the following two facts:

(i) We may assume X(f x) = X(g2) = 1, X(g3)=n,

i.e. X(f^g^g^) = nc .

(ii) |N0 :IN o (X)| = |XN o| = |lrr(E)*| = 124.

By Clifford theory, there exists i|> € IrrF( (X)) such that iJj+E = eX ( e positive integral) and ^+N0 = S

[2]. Thus by (ii) above we have deg(C) = 124*deg(^). We 266

show presently that deg(iJO Ss= 5 which gives the desired

result.

First we observe from the defining relations for S 5 that g|‘t=g3(mod Z) and g^2=g3(mod Z). Thus Xe^1(g3) =

*(g3)> Xf^1(g3) =. X(g3) and it follows that S ^ I N (X).

We may therefore consider the restriction !|>+S. As ip + E = eX we see that Z = kerX ^ kerip, whence ^4-S can be con­ strued as a character of S = S/Z having Z in its ker­ nel. But S is extraspecial of order 5 3 (lemma 3.38).

The ordinary character theory of extraspecial groups is well understood (see for example [2 ]) and as char(F) f 5, the F-characters are precisely the ordinary characters.

Thus S has 25 linear characters, each having Z(S) in

its kernel, and 4 faithful irreducible characters, each of degree 5. Therefore if deg(ijj) — 4, we would have ip+S = Oj + a2 + a3 + where each is a linear char­ acter of S . But then = Z(S) ^ ker(if)+S) whence

E < ker (i^+S). This is an obvious contradiction as it was established earlier that = eX. Thus deg(^) ^5, and deg(?) ^ 620 as desired. 267

Theorem 3.41. Let C be a faithful F-character of LyS

where F is any field. Then deg(£) ^ 111. Moreover,

this lower bound is achieved if and only if (of the pre­

vious section) exists.

Proof. As ? is faithful, C+Nq possesses a faithful ir­

reducible constituent. Thus, if char(F) f 5 , we automat­

ically have deg(s) ^ 620 (corollary 3.40). Thus we may

assume char(F) = 5 in what follows. From the decomposi­

tion matrix for N = E 35 *MX1 (proposition 3.14) the mini­ mal degree of a faithful 5-Brauer irreducible of N is

110. Arguing as above, we therefore have deg(£) ^110.

Thus to prove the theorem we need only show deg(c) f 110,

so we assume by way of contradiction that deg(?) = 110.

The only possibilities for ?+N are <{>x ,3 ,^3 , all other 5-Brauer characters of N being faithless. As

LyS has a unique class of elements of order 18, £+N must agree on the two N-classes of elements of that or­ der; thus ?4-N cannot be either of

leaves s+N = cj>! or <|)2. We dispose of each of these

cases presently.

Suppose C+N = 4» j_ . Then exactly as in the proof of proposition 3.16, we obtain a matrix equation which deter­ mines all possibilities for the (unique) decomposition of

S+U as a non-negative integral sum of 5-Brauer irreduc- 268

ibles of U = U 3(3). This matrix equation is given by

6 7 14 14 21 42 27 56 64 x 1 110 x 2 -2 -1 6 -2 5 2 3 -8 0 14 x 3 -3 -2 -4 5 3 6 0 2 -8 x * -25 0 1 2 -1 0 0 0 2 -2 *5 = 2 X 6 -2 3 -2 2 1 -6 3 0 0 2 X 7 2 -1 2 2 1 -2 -1 0 0 X 8 2 X 9 1 2 0 1 -1 2 0 -2 0 / -1 . xio .

and has general solution:

( -8+a+2b+6c, -1+a+b, 4-b-2c, JgL - a-b-2c ,a. ,-i ,b ,0,0 )*.

We conclude from this that the multiplicity of 21' € IBr(U)

in £+U is -£ , a contradiction.

Finally suppose £+N = 2 . This time we obtain a

matrix equation which reflects the decomposition of S+L,

L s L 3(4):

« f \ 1 20 35 35 35 90 63 X 1 110

1 4 3 3 3 -6 -1 X 2 -2

1 2 -1 -1 -1 0 0 X 3 2

1 0 3 -1 -1 2 -1 X* = 2 1 0 -1 3 -1 2 -1 2 X 5 1 0 -1 -1 3 2 -1 2 X 6

1 -1 0 0 0 -1 0 j> . X 7 .-2 j 269

As the unique solution to this equation is given by

(0,1,0,0,0,1,0)^ we have as a result ?+L = 20 + 45 + 45, where 20 , 45 , 45 € IBr(L). We now observe that L can be chosen to be a subgroup of Cu (the centralizer in LyS of a Sj-element u ). Thus £+L = (?'t-Cu)+L. Now express

as 0 + £ where 0 (respectively £ ) denotes the sum of all faithful (respectively faithless) irreducible constituents of S+Cu . Thus we have 0+L + £+L =

(0+£)4-L = 20 + 45 + 45. Now precisely as in the proof of proposition 3.23, the faithful irreducible constituents of

S+L (i.e. the irreducible constituents of 0+L ) occur in pairs. Thus we have 0+L = 45 + 45 and £4-L = 20, and we conclude that 5 is a Brauer irreducible of Cu / = Me of degree 20. But by theorem 3.1 no such character exists.

This being our final contradiction, the theorem is proved. LIST OF REFERENCES

1 . A. Borel and J. Tits, "Elements unipotents et sous- groupes paraboliques de groupes reductifs":I , Invent. Math., 12 (1971), 97-104.

2 . L. Dornhoff, Group Representation Theory, Marcel Dekker, Inc., New York, 1972.

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