Modular Representation Theory of Finite Groups Peter Schneider

Modular Representation Theory of Finite Groups Peter Schneider Department of Mathematics University of Münster Münster Germany

ISBN 978-1-4471-4831-9 ISBN 978-1-4471-4832-6 (eBook) DOI 10.1007/978-1-4471-4832-6 Springer Dordrecht Heidelberg New York London

Library of Congress Control Number: 2012954001

Mathematics Subject Classification: 20C20, 20C05

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Springer is part of Springer Science+Business Media (www.springer.com) Preface

The nature of the representation theory of a finite group G in (finite-dimensional) vector spaces over some field k depends very much on the relation between the order |G| of the group G and the characteristic char(k) of the field k. If char(k) does not divide |G| then all representations are semisimple, i.e. are direct sums of irreducible representations. The reason for this is the semisimplicity of the group al- gebra k[G] in this situation. By the modular representation theory of G one means, on the other hand, the case where char(k) is a divisor of |G| (so that, in particular, char(k) must be a prime number). The group algebra k[G] now may be far from be- ing semisimple. In the extreme case, for example, where |G| is a power of char(k), it is a local ring; there is then a single irreducible representation, which is the trivial one, whereas the structure of a general representation will still be very complicated. As a consequence a whole range of additional tools have to be developed and used in the course of the investigation. To mention some, there is the systematic use of Grothendieck groups (Chap. 2) as well as Green’s direct analysis of indecompos- able representations (Chap. 4). There also is the strategy of writing the category of all k[G]-modules as the direct product of certain subcategories, the so-called blocks of G, by using the action of the primitive idempotents in the center of k[G]. Brauer’s approach then establishes correspondences between the blocks of G and blocks of certain subgroups of G (Chap. 5), the philosophy being that one is thereby reduced to a simpler situation. This allows us, in particular, to measure how nonsemisimple a category a block is by the size and structure of its so-called defect group. Begin- ning in Sect. 4.4 all these concepts are made explicit for the example of the group G = SL2(Fp). The present book is to be thought of as an introduction to the major tools and strategies of modular representation theory. Its content was taught during a course lasting the full academic year 2010/2011 at Münster. Some basic algebra together with the semisimple case were assumed to be known, although all facts to be used are restated (without proofs) in the text. Otherwise the book is entirely self- contained. The references [1Ð10] provide a complete list of the sources I have drawn upon. Of course, there already exist several textbooks on the subject. The older ones like [5] and [6] are written in a mostly group theoretic language. The beautiful

v vi Preface book [1] develops the theory entirely from the module theoretic point of view but leaves out completely the comparison with group theoretic concepts. For example, the concept of defect groups can be introduced either purely group theoretically or purely module theoretically. To my knowledge all existing books essentially restrict themselves to a discussion of one of these approaches only. Although my presenta- tion is strongly biased towards the module theoretic point of view, I make an attempt to strike a certain balance by also showing the reader the other aspect. In particular, in the case of defect groups a detailed proof of the equivalence of the two approaches will be given. This book is not addressed to experts. It does not discuss any very advanced aspects nor any specialized results of the theory. The aim is to familiarize students at the masters level with the basic results, tools, and techniques of a beautiful and important algebraic theory, hopefully enabling them to subsequently pursue their own more specialized problems. I wish to thank T. Schmidt for carefully reading a first draft and I. Reckermann and G. Dierkes for their excellent typesetting of the manuscript. Münster, Germany Peter Schneider Contents

1 Prerequisites in Module Theory ...... 1 1.1ChainConditionsandMore...... 1 1.2Radicals...... 3 1.3 I -Adic Completeness ...... 4 1.4 Unique Decomposition ...... 9 1.5 Idempotents and Blocks ...... 15 1.6 Projective Modules ...... 26 1.7 Grothendieck Groups ...... 34 2 The Cartan–Brauer Triangle ...... 43 2.1 The Setting ...... 43 2.2 The Triangle ...... 46 2.3 The Ring Structure of RF (G), and Induction ...... 54 2.4TheBurnsideRing...... 59 2.5 Clifford Theory ...... 67 2.6 Brauer’s Induction Theorem ...... 71 2.7 Splitting Fields ...... 75 2.8 Properties of the CartanÐBrauer Triangle ...... 78 3 The Brauer Character ...... 87 3.1Definitions...... 87 3.2 Properties ...... 91 4 Green’s Theory of Indecomposable Modules ...... 97 4.1 Relatively Projective Modules ...... 97 4.2 Vertices and Sources ...... 105 4.3 The Green Correspondence ...... 110 4.4 An Example: The Group SL2(Fp) ...... 119 4.5 Green’s Indecomposability Theorem ...... 140 5 Blocks ...... 147 5.1 Blocks and Simple Modules ...... 147

vii viii Contents

5.2CentralCharacters...... 151 5.3 Defect Groups ...... 153 5.4 The Brauer Correspondence ...... 159 5.5 Brauer Homomorphisms ...... 165 References ...... 175 Index ...... 177 Chapter 1 Prerequisites in Module Theory

Let R be an arbitrary (not necessarily commutative) ring (with unit). By an R-module we will always mean a left R-module. All ring homomorphisms respect the unit element, but a subring may have a different unit element.

1.1 Chain Conditions and More

For an R-module M we have the notions of being

finitely generated, artinian, noetherian, simple, and semisimple.

The ring R is called left artinian, resp. left noetherian, resp. semisimple, if it has this property as a left module over itself.

Proposition 1.1.1 i. The R-module M is noetherian if and only if any submodule of M is finitely generated. ii. Let L ⊆ M be a submodule; then M is artinian, resp. noetherian, if and only if L and M/L are artinian, resp. noetherian. iii. If R is left artinian, resp. left noetherian, then every finitely generated R-module M is artinian, resp. noetherian. iv. If R is left noetherian then an R-module M is noetherian if and only if it is finitely generated.

Proposition 1.1.2 (Jordan–Hölder) For any R-module M the following conditions are equivalent: i. M is artinian and noetherian; ii. M has a composition series {0}=M0 ⊆ M1 ⊆ ··· ⊆ Mn = M such that all Mi/Mi−1 are simple R-modules.

P. Schneider, Modular Representation Theory of Finite Groups, 1 DOI 10.1007/978-1-4471-4832-6_1, © Springer-Verlag London 2013 2 1 Prerequisites in Module Theory

In this case two composition series {0}=M0 ⊆ M1 ⊆···⊆Mn = M and {0}= ∼ L0 ⊆ L1 ⊆···⊆Lm = M satisfy n = m and Li/Li−1 = Mσ(i)/Mσ(i)−1, for any 1 ≤ i ≤ m, where σ is an appropriate permutation of {1,...,n}.

An R-module M which satisfies the conditions of Proposition 1.1.2 is called of finite length and the integer l(M) := n is its length.Let

Rˆ := set of all isomorphism classes of simple R-modules.

For τ ∈ Rˆ and an R-module M the τ -isotypic component of M is

M(τ):= sum of all simple submodules of M in τ .

Lemma 1.1.3 For any R-module homomorphism f : L −→ M we have f(L(τ))⊆ M(τ).

Proposition 1.1.4 i. For any R-module M the following conditions are equivalent: a. M is semisimple, i.e. isomorphic to a direct sum of simple R-modules; b. M is the sum of its simple submodules; c. every submodule of M has a complement. ii. Submodules and factor modules of semisimple modules are semisimple. iii. If R is semisimple then any R-module is semisimple. iv. Any τ -isotypic component M(τ) of any R-module M is semisimple. = v. If the R-module M is semisimple then M τ∈Rˆ M(τ). = vi. If R is semisimple then R τ∈Rˆ R(τ) as rings.

Lemma 1.1.5 Any R-module M contains a unique maximal submodule which is semisimple (and which is called the socle soc(M) of M).  := Proof We define soc(M) τ∈Rˆ M(τ). By Proposition 1.1.4.i the submodule soc(M) is semisimple. On the other hand if L ⊆ M is any semisimple submod- = ⊆ ule then L τ L(τ) by Proposition 1.1.4.v. But L(τ) M(τ) by Lemma 1.1.3 and hence L ⊆ soc(M). 

Definition An R-module M is called decomposable if there exist nonzero submod- ules M1,M2 ⊆ M such that M = M1 ⊕ M2; correspondingly, M is called indecom- posable if it is nonzero and not decomposable.

Lemma 1.1.6 If M is artinian or noetherian then M is the direct sum of finitely many indecomposable submodules.

Proof We may assume that M = {0}. Step 1: We claim that M has a nonzero in- decomposable direct summand N.IfM is artinian take a minimal element N of 1.2 Radicals 3 the set of all nonzero direct summands of M.IfM is noetherian let N be a max- imal element of the set of all direct summands = M of M, and take N such that = ⊕ = ⊕ = { } M N N. Step 2: By Step 1 we find M M1 M1 with M1 0 indecom- = { } = ⊕ = { } posable. If M1 0 then similarly M1 M2 M2 with M2 0 indecomposable. Inductively we obtain in this way a strictly increasing sequence

{0} M1 M1 ⊕ M2 ··· as well as a strictly decreasing sequence ··· M1 M2 of submodules of M. Since one of the two (depending on M being artinian or ={ } noetherian) stops after finitely many steps we must have Mn 0 for some n. Then M = M1 ⊕···⊕Mn is the direct sum of the finitely many indecomposable submodules M1,...,Mn. 

Exercise The Z-modules Z and Z/pnZ, for any prime number p and any n ≥ 1, are indecomposable.

1.2 Radicals

The radical of an R-module M is the submodule

rad(M) := intersection of all maximal submodules of M.

The Jacobson radical of R is Jac(R) := rad(R).

Proposition 1.2.1 i. If M = {0} is finitely generated then rad(M) = M. ii. If M is semisimple then rad(M) ={0}. iii. If M is artinian with rad(M) ={0} then M is semisimple. iv. The Jacobson radical

Jac(R) = intersection of all maximal left ideals of R = intersection of all maximal right ideals of R   × = a ∈ R : 1 + Ra ⊆ R

is a two-sided ideal of R. v. Any left ideal which consists of nilpotent elements is contained in Jac(R). vi. If R is left artinian then the ideal Jac(R) is nilpotent and the factor ring R/Jac(R) is semisimple. vii. If R is left artinian then rad(M) = Jac(R)M for any finitely generated R-module M. 4 1 Prerequisites in Module Theory

Corollary 1.2.2 If R is left artinian then any artinian R-module is noetherian and, in particular, R is left noetherian and any finitely generated R-module is of finite length.

Lemma 1.2.3 (Nakayama) If L ⊆ M is a submodule of an R-module M such that M/L is finitely generated, then L + Jac(R)M = M implies that L = M.

Lemma 1.2.4 If R is left noetherian and R/Jac(R) is left artinian then R/Jac(R)m is left noetherian and left artinian for any m ≥ 1.

Proof It suffices to show that R/Jac(R)m is noetherian and artinian as an R- module. By Proposition 1.1.1.ii it further suffices to prove that the R-module Jac(R)m/ Jac(R)m+1, for any m ≥ 0, is artinian. Since R is left noetherian it cer- tainly is a finitely generated R-module and hence a finitely generated R/Jac(R)- module. The claim therefore follows from Proposition 1.1.1.iii. 

Lemma 1.2.5 The Jacobson ideal of the matrix ring Mn×n(R), for any n ∈ N, is the ideal Jac(Mn×n(R)) = Mn×n(Jac(R)) of all matrices with entries in Jac(R).

1.3 I-Adic Completeness

We begin by introducing the following general construction. Let

fn (Mn+1 −−→ Mn)n∈N be a sequence of R-module homomorphisms, usually visualized by the diagram

fn+1 fn fn−1 f1 ···−−−→ Mn+1 −−→ Mn −−−→···−−→ M1.  In the direct product module n∈N Mn we then have the submodule   := ∈ : = ∈ N ←lim− Mn (xn)n Mn fn(xn+1) xn for any n , n which is called the projective limit of the above sequence. Although suppressed in the notation the construction depends crucially, of course, on the homomorphisms fn and not only on the modules Mn. Let us fix now a two-sided ideal I ⊆ R. Its powers form a descending sequence

R ⊇ I ⊇ I 2 ⊇···⊇I n ⊇··· of two-sided ideals. More generally, for any R-module M we have the descending sequence of submodules

M ⊇ IM ⊇ I 2M ⊇···⊇I nM ⊇··· 1.3 I -Adic Completeness 5 and correspondingly the sequence of residue class projections

pr pr ···−→M/In+1M −→ M/InM −→···−→M/I2M −→ M/IM.

We form the projective limit of the latter

n ←lim− M/I M   n n n = + ∈ : + − ∈ ∈ N xn I M n M/I M xn 1 xn I M for any n . n

There is the obvious R-module homomorphism

I : −→ n πM M ←lim− M/I M

−→ + n x x I M n.

Definition The R-module M is called I -adically separated, resp. I -adically com- I n ={ } plete, if πM is injective (i.e. if n I M 0 ), resp. is bijective.

n Exercise If I 0 M ={0} for some n0 ∈ N then M is I -adically complete.

Lemma 1.3.1 Let J ⊆ R be another two-sided ideal such that I n0 ⊆ J ⊆ I for some n0 ∈ N; then M is J -adically separated, resp. complete, if and only if it is I -adically separated, resp. complete.

Proof We have I nn0 M ⊆ J nM ⊆ I nM for any n ∈ N. This makes obvious the separatedness part of the assertion. Furthermore, by the right-hand inclusions we have the well-defined map

: n −→ n α ←lim− M/J M ←lim− M/I M

+ n −→ + n xn J M n xn I M n. ◦ J = I Since α πM πM it suffices for the completeness part of the assertion to show that = + n ∈ n ∈ n α is bijective. That α(ξ) 0for(xn J M)n ←lim− M/J M means that xn I M ∈ N ∈ nn0 ⊆ n − ∈ n for any n . Hence xnn0 I M J M. On the other hand xnn0 xn J M. n It follows that xn ∈ J M for any n, i.e. that ξ = 0. For the surjectivity of α let (yn + n ∈ n := + − = I M)n ←lim− M/I M be any element. We define xn ynn0 . Then xn 1 xn + − ∈ nn0 ⊆ n + n ∈ n ynn0 n0 ynn0 I M J M and hence (xn J M)n ←lim− M/J M. Moreover, + n = + n = + n + n = + xn I M ynn0 I M yn I M which shows that α((xn J M)n) (yn n I M)n. 

Lemma 1.3.2 If R is I -adically complete then I ⊆ Jac(R). 6 1 Prerequisites in Module Theory

Proof Let a ∈ I be any element. Then

+ +···+ n−1 + n ∈ n 1 a a I n ←lim− R/I . Therefore, by assumption, there is an element c ∈ R such that

− c + I n = 1 + a +···+an 1 + I n for any n ∈ N.

It follows that

− (1 − a)c ∈ (1 − a) 1 + a +···+an 1 + I n = 1 − an + I n ={1} n n and similarly that c(1 − a) = 1. This proves that 1 + I ⊆ R×. The assertion now follows from Proposition 1.2.1.iv. 

Lemma 1.3.3 Let f : M −→ N be a surjective R-module homomorphism and sup- pose that M is I -adically complete; if N is I -adically separated then it already is I -adically complete.

Proof The R-module homomorphism : n −→ n φ ←lim− M/I M ←lim− N/I N

+ n −→ + n xn I M n f(xn) I N n fits into the commutative diagram

f M N

I =∼ I αM αN φ n n ←lim− M/I M ←lim− N/I N.

+ n ∈ n Hence it suffices to show that φ is surjective. Let (yn I N)n ←lim− N/I N be an arbitrary element. By the surjectivity of f we find an x1 ∈ M such that f(x1) = y1. We now proceed by induction and assume that elements x1,...,xn ∈ M have been found such that j xj+1 − xj ∈ I M for 1 ≤ j

− ∈ − + n = n f xn+1 xn yn+1 yn I N I N. 1.3 I -Adic Completeness 7

f Since the restricted map I nM −→ I nN still is surjective we find an element z ∈ ker(f ) such that − − ∈ n xn+1 xn z I M. := − − ∈ n = Therefore, if we put xn+1 xn+1 z then xn+1 xn I M and f(xn+1) =  f(xn+1) yn+1.

Proposition 1.3.4 Suppose that R is commutative and noetherian and that I ⊆ Jac(R); then any finitely generated R-module M is I -adically separated.

:= n Proof We have to show that the submodule M0 n I M is zero. Since R is noetherian M0 is finitely generated by Proposition 1.1.1. We therefore may apply the Nakayama lemma 1.2.3 to M0 (and its submodule {0}) and see that it suffices to show that IM0 = M0. Obviously IM0 ⊆ M0. Again since R is noetherian we find a submodule IM0 ⊆ L ⊆ M which is maximal with respect to the property that L ∩ M0 = IM0. In an intermediate step we establish that for any a ∈ I there is an integer n0(a) ≥ 1 such that an0(a)M ⊆ L. j Fixing a we put Mj := {x ∈ M : a x ∈ L} for j ≥ 1. Since R is commutative the Mj ⊆ Mj+1 form an increasing sequence of submodules of M. Since R is noethe- rian this sequence ⊆···⊆ = =··· M1 Mn0(a) Mn0(a)+1 n (a) has to stabilize. We trivially have IM0 ⊆ (a 0 M + L) ∩ M0. Consider any n (a) element x0 = a 0 x + y, with x ∈ M and y ∈ L, in the right-hand side. We n (a)+1 have ax0 ∈ IM0 ⊆ L and ay ∈ L, hence a 0 x = ax0 − ay ∈ L or equiva- ∈ = n0(a) ∈ lently x Mn0(a)+1.ButMn0(a)+1 Mn0(a) so that a x L and consequently n (a) x0 ∈ L ∩ M0 = IM0. This shows that IM0 = (a 0 M + L) ∩ M0 holds true. The maximality of L then implies that an0(a)M ⊆ L. The ideal I in the noetherian ring R can be generated by finitely many elements a1,...,ar . We put n0 := maxi n0(ai) and n1 := rn0. Then

n1 = +···+ rn0 ⊆ n0 +···+ n0 ⊆ n0(a1) +···+ n0(ar ) I (Ra1 Rar ) Ra1 Rar Rar Rar and hence I n1 M ⊆ L which implies

M = I nM ⊆ I n1 M ⊆ L and therefore M = L ∩ M = IM . 0 0 0 0  n

Let R0 be a commutative ring. If α : R0 −→ Z(R) is a ring homomorphism into the center Z(R) of R then we call R an R0-algebra (with respect to α). In particular, R then is an R0-module. More generally, by restriction of scalars, any R-module also is an R0-module. 8 1 Prerequisites in Module Theory

For any R-module M we have the two endomorphism rings EndR(M) ⊆

EndR0 (M).BothareR0-algebras with respect to the homomorphism

R0 −→ EndR(M)

a0 −→ a0 · idM .

Lemma 1.3.5 Suppose that R is an R0-algebra which as an R0-module is finitely generated, and assume R0 to be noetherian; we than have: i. R is left and right noetherian; ii. for any finitely generated R-module M its ring EndR(M) of endomorphisms is left and right noetherian and is finitely generated as an R0-module; iii. Jac(R0)R ⊆ Jac(R).

Proof i. Any left or right ideal of R is a submodule of the noetherian R0-module R and hence is finitely generated (cf. Proposition 1.1.1). ii. Step 1: We claim that, for any finitely generated R0-module M0,theR0- module EndR0 (M0) is finitely generated. Let x1,...,xr be generators of the R0- module M0. Then

EndR (M0) −→ M0 ⊕···⊕M0 0 f −→ f(x1),...,f(xr ) is an injective R0-module homomorphism. The right-hand side is finitely generated by assumption and so is then the left-hand side since R0 is noetherian. Step 2: By assumption M is finitely generated over R, and R is finitely generated over R0.

Hence M is finitely generated over R0, Step 1 applies, and EndR0 (M) is a finitely generated R0-module. Since R0 is noetherian the submodule EndR(M) is finitely generated as well. Furthermore, since any left or right ideal of EndR(M) is an R0- submodule the ring EndR(M) is left and right noetherian. iii. Set L ⊆ R be a maximal left ideal. Then M := R/L is a simple R-module. By ii. the R0-module EndR(M) is finitely generated. It is nonzero since M is nonzero. The Nakayama lemma 1.2.3 therefore implies that Jac(R0) EndR(M) = EndR(M).ButEndR(M), by Schur’s lemma, is a skew field. It follows that Jac(R0) EndR(M) ={0}. Hence

Jac(R0)M = idM Jac(R0)M = Jac(R0) idM (M) ={0} or equivalently Jac(R0)R ⊆ L. Since L was arbitrary we obtain the assertion. 

For simplicity we call an R-module complete if it is Jac(R)-adically complete.

Proposition 1.3.6 Suppose that R is an R0-algebra which as an R0-module is finitely generated, and assume that R0 is noetherian and complete and that R0/ Jac(R0) is artinian; for any of the rings S = R or S = EndR(M), where M is a finitely generated R-module, we then have: 1.4 Unique Decomposition 9

i. S is left and right noetherian; ii. S/Jac(S) is left and right artinian; iii. any finitely generated S-module is complete.

Proof By Lemma 1.3.5.ii the case S = R contains the case S = EndR(M). Lemma 1.3.5.i says that R is left and right noetherian. Since R0 maps to the cen- ter of R the right ideal Jac(R0)R in fact is two-sided. Being finitely generated as an R0/ Jac(R0)-module the ring R/Jac(R0)R is left and right artinian by Proposi- tion 1.1.1.iii. According to Lemma 1.3.5.iii the ring R/Jac(R) is a factor ring of R/Jac(R0)R and therefore, by Proposition 1.1.1.ii, is left and right artinian as well. Using Proposition 1.2.1.vi it also follows that

n0 Jac(R) ⊆ Jac(R0)R ⊆ Jac(R) for some n0 ∈ N. Because of Lemma 1.3.1 it therefore remains to show that any finitely generated R-module N is Jac(R0)R-adically complete. Since

n n Jac(R0)R N = Jac(R0) N for any n ≥ 1 this further reduces to the statement that any finitely generated R0-module N is com- plete. We know from Proposition 1.3.4 that the R0-module N is Jac(R0)-adically separated. On the other hand, by finite generation we find, for some m ∈ N,asur- m  m jective R0-module homomorphism R0 N. With R0 also R0 is complete by as- sumption. We therefore may apply Lemma 1.3.3 and obtain that N is complete. 

1.4 Unique Decomposition

We first introduce the following concept.

Definition 1 AringA is called local if A \ A× is a two-sided ideal of A.

We note that a local ring A is nonzero since 1 ∈ A× whereas 0 must lie in the ideal A \ A×.

Proposition 1.4.1 For any nonzero ring A the following conditions are equivalent: i. A is local; ii. A \ A× is additively closed; iii. A \ Jac(A) ⊆ A×; iv. A/Jac(A) is a skew field; v. A contains a unique maximal left ideal; vi. A contains a unique maximal right ideal.

Proof We note that A = {0} implies that Jac(A) = A is a proper ideal. 10 1 Prerequisites in Module Theory

i. =⇒ ii. This is obvious. ii. =⇒ iii. Let b ∈ A \ A× be any element. Then also −b/∈ A×. Suppose that 1 + ab∈ / A× for some a ∈ A. Using that A \ A× is additively closed we first obtain that −ab ∈ A× and in particular that −a,a∈ / A×. It then follows that a + b/∈ A× and that 1 + a,1+ b ∈ A×. But in the identity

(1 + ab) + (a + b) = (1 + a)(1 + b) now the right-hand side is contained in A× whereas the left-hand side is not. This contradiction proves that 1 + Ab ⊆ A× and therefore, by Proposition 1.2.1.iv, that b ∈ Jac(A). We conclude that A ⊆ A× ∪ Jac(A). iii. =⇒ iv. It immediately follows from iii. that any nonzero element in A/Jac(A) is a unit. iv. =⇒ v., vi. Let L ⊆ A be any maximal left, resp. right, ideal. Then Jac(A) ⊆ L A by Proposition 1.2.1.iv. Since the only proper left, resp. right, ideal of a skew field is the zero ideal we obtain L = Jac(A). v., vi. =⇒ i. The unique maximal left (right) ideal, by Proposition 1.2.1.iv, is necessarily equal to Jac(A).Letb ∈ A \ Jac(A) be any element. Then Ab (bA)is not contained in any maximal left (right) ideal and hence Ab = A (bA = A). Let a ∈ A such that ab = 1(ba = 1). Since a/∈ Jac(A) we may repeat this reasoning and find an element c ∈ A such that ca = 1(ac = 1). But c = c(ab) = (ca)b = b (c = (ba)c = b(ac) = b). It follows that a ∈ A× and then also that b ∈ A×.This shows that A = Jac(A) ∪ A×. But the union is disjoint. We finally obtain that A \ A× = Jac(A) is a two-sided ideal. 

We see that in a local ring A the Jacobson radical Jac(A) is the unique maximal left (right, two-sided) ideal and that A \ Jac(A) = A×.

Lemma 1.4.2 If A is nonzero and any element in A \ A× is nilpotent then the ring A is local.

Proof Let b ∈ A \ A× and let n ≥ 1 be minimal such that bn = 0. For any a ∈ A we then have (ab)bn−1 = abn = 0. Since bn−1 = 0 the element ab cannot be a unit in A. It follows that Ab ⊆ A \ A× and hence, by Proposition 1.2.1.v, that Ab ⊆ Jac(A). We thus have shown that A \ A× ⊆ Jac(A) which, by Proposition 1.4.1.iii, implies that A is local. 

Lemma 1.4.3 (Fitting) For any R-module M and any f ∈ EndR(M) we have: i. If M is noetherian and f is injective then f is bijective; ii. if M is noetherian and f is surjective then f is bijective; iii. if M is of finite length then there exists an integer n ≥ 1 such that a. ker(f n) = ker(f n+j ) for any j ≥ 0, b. im(f n) = im(f n+j ) for any j ≥ 0, c. M = ker(f n) ⊕ im(f n), and ∼ ∼ n = n n = d. the induced maps idM −f : ker(f ) −→ ker(f ) and f : im(f ) −→ im(f n) are bijective. 1.4 Unique Decomposition 11

Proof We have the increasing sequence of submodules

ker(f ) ⊆ ker f 2 ⊆···⊆ker f n ⊆··· as well as the decreasing sequence of submodules

im(f ) ⊇ im f 2 ⊇···⊇im f n ⊇··· .

If M is artinian there must exist an n ≥ 1 such that

+ + im f n = im f n 1 =···=im f n j =··· .

For any x ∈ M we then find a y ∈ M such that f n(x) = f n+1(y). Hence f n(x − f(y))= 0. If f is injective it follows that x = f(y). This proves M = f(M)under the assumptions in i. If M is noetherian there exists an n ≥ 1 such that

+ + ker f n = ker f n 1 =···=ker f n j =··· .

Let x ∈ ker(f ).Iff , and hence f n, is surjective we find a y ∈ M such that x = f n(y). Then f n+1(y) = f(x)= 0, i.e. y ∈ ker(f n+1) = ker(f n). It follows that x = f n(y) = 0. This proves ker(f ) ={0} under the assumptions in ii. Assuming that M is of finite length, i.e. artinian and noetherian, we at least know the existence of an n ≥ 1 satisfying a. and b. To establish c. (for any such n)we first consider any x ∈ ker(f n) ∩ im(f n). Then f n(x) = 0 and x = f n(y) for some y ∈ M. Hence y ∈ ker(f 2n) = ker(f n) which implies x = 0. This shows that

ker f n ∩ im f n ={0}.

Secondly let x ∈ M be arbitrary. Then f n(x) ∈ im(f n) = im(f 2n),i.e.f n(x) = f 2n(y) for some y ∈ M. We obtain f n(x − f n(y)) = f n(x) − f 2n(y) = 0. Hence

x = x − f n(y) + f n(y) ∈ ker f n + im f n .

n n For d. we note that idM −f : ker(f ) −→ ker(f ) has the inverse idm +f +···+ f n−1. Since ker(f ) ∩ im(f n) ={0} by c., the restriction f | im(f n) is injective. Let x ∈ im(f n). Because of b we find a y ∈ M such that x = f n+1(y) = f(fn(y)) ∈ f(im(f n)). 

Proposition 1.4.4 For any indecomposable R-module M of finite length the ring EndR(M) is local.

Proof Since M = {0} we have idM = 0 and hence EndR(M) = {0}.Letf ∈ EndR(M) be any element which is not a unit, i.e. is not bijective. According to Lemma 1.4.3.iii we have

M = ker f n ⊕ im f n for some n ≥ 1. 12 1 Prerequisites in Module Theory

But M is indecomposable. Hence ker(f n) = M or im(f n) = M. In the latter case f would be bijective by Lemma 1.4.3.iii.d which leads to a contradiction. It follows that f n = 0. This shows that the assumption in Lemma 1.4.2 is satisfied so that EndR(M) is local. 

Proposition 1.4.5 Suppose that R is left noetherian, R/Jac(R) is left artinian, and any finitely generated R-module is complete; then EndR(M), for any finitely gener- ated indecomposable R-module M, is a local ring.

Proof We abbreviate J := Jac(R). By assumption the map

∼ J : −→= m πM M ←lim− M/J M is an isomorphism. Each M/JmM is a finitely generated module over the factor ring R/J m which, by Lemma 1.2.4, is left noetherian and left artinian. Hence M/JmM is a module of finite length by Proposition 1.1.1.iii. On the other hand for any f ∈ EndR(M) the R-module homomorphisms

m m fm: M/J M −→ M/J M x + J mM −→ f(x)+ J mM are well defined. The diagrams

fm+1 M/Jm+1M M/Jm+1M

pr pr (1.4.1)

fm M/JmM M/JmM obviously are commutative so that in the projective limit we obtain the R-module homomorphism

: m −→ m f∞ ←lim− M/J M ←lim− M/J M

+ m −→ + m xm J M m f(xm) J M m. Clearly the diagram

f M M

J =∼ =∼ J πM πM f∞ m m ←lim− M/J M ←lim− M/J M 1.4 Unique Decomposition 13 is commutative. If follows, for example, that f is bijective if and only if f∞ is bijective. We now apply Fitting’s lemma 1.4.3.iii to each module M/JmM and obtain an increasing sequence of integers 1 ≤ n(1) ≤ ··· ≤ n(m) ≤ ··· such that the triple m (M/J M,fm, n(m)) satisfies the conditions a.Ðd. in that lemma. In particular, we have

m = n(m) ⊕ n(m) ≥ M/J M ker fm im fm for any m 1. The commutativity of (1.4.1) easily implies that we have the sequences of surjective (!) R-module homomorphisms

···−pr→ n(m+1) −pr→ n(m) −pr→···−pr→ n(1) ker fm+1 ker fm ker f1 and

···−pr→ n(m+1) −pr→ n(m) −pr→···−pr→ n(1) im fm+1 im fm im f1 . Be defining

:= n(m) := n(m) X ←lim− ker fm and Y ←lim− im fm we obtain the decomposition into R-submodules =∼ m = ⊕ M ←lim− M/J M X Y. But M is indecomposable. Hence X ={0} or Y ={0}. Suppose first that X = {0}. By the surjectivity of the maps in the corresponding sequence this implies n(m) ker(fm ) ={0} for any m ≥ 1. The condition d then says that all the fm, hence f∞, and therefore f are bijective. If, on the other hand, Y ={0} then analogously n(m) im(fm ) ={0} for all m ≥ 1, and conditions d. implies that idM −f is bijec- tive. So far we have shown that for any f ∈ EndR(M) either f or idM −f is a unit. To prove our assertion it suffices, by Proposition 1.4.1.ii, to verify that the nonunits in EndR(M) are additively closed. Suppose therefore that f,g ∈ × × EndR(M) \ EndR(M) are such that h := f + g ∈ EndR(M) . By multiplying −1 by h we reduce to the case that h = idM . Then the left-hand side in the iden- tity

g = idM −f is not a unit, but the right-hand side is by what we have shown above. This is a × contradiction, and hence EndR(M) \ EndR(M) is additively closed. 

Proposition 1.4.6 Let

M = M1 ⊕···⊕Mr = N1 ⊕···⊕Ns be two decompositions of the R-module M into indecomposable R-modules Mi and Nj ; if EndR(Mi), for any 1 ≤ i ≤ r, is a local ring we have: 14 1 Prerequisites in Module Theory i. r = s; ∼ ii. there is a permutation σ of {1,...,r} such that Nj = Mσ(j) for any 1 ≤ j ≤ r.

Proof The proof is by induction with respect to r. The case r = 1 is trivial since M then is indecomposable. In general we have the R-module homomorphisms

pr pr Mi ⊆ Nj ⊆ fi: M −−−→ Mi −→ M and gj : M −−−→ Nj −→ M in EndR(M) satisfying the equation

idM = f1 +···+fr = g1 +···+gs.

In particular, f1 = f1g1 +···+f1gs , which restricts to the equation = ( g )|M +···+( g )|M idM1 prM1 1 1 prM1 s 1 in EndR(M1).ButEndR(M1) is local. Hence at least one of the summands must be a unit. By renumbering we may assume that the composed map

pr pr ⊆ N1 ⊆ M1 M1 −→ M −−−→ N1 −→ M −−−→ M1 is an automorphism of M1. This implies that N =∼ M ⊕ ( |N ). 1 1 ker prM1 1 =∼ Since N1 is indecomposable we obtain that g1 : M1 −→ N1 is an isomorphism. In particular

M1 ∩ ker(g1) = M1 ∩ (N2 ⊕···⊕Ns) ={0}.

On the other hand let x ∈ N1 and write x = g1(y) with y ∈ M1. Then

g1(x − y) = x − g1(y) = 0 and hence

x = y + (x − y) ∈ M1 + ker(g1) = M1 + (N2 ⊕···⊕Ns).

This shows that N1 ⊆ M1 + (N2 ⊕···⊕Ns), and hence

M = N1 +···+Ns = M1 + (N2 ⊕···⊕Ns). Together we obtain

M = M1 ⊕ N2 ⊕···⊕Ns and therefore ∼ ∼ M/M1 = M2 ⊕···⊕Mr = N2 ⊕···⊕Ns. We now apply the induction hypothesis to these two decompositions of the R- module M/M1.  1.5 Idempotents and Blocks 15

Theorem 1.4.7 (KrullÐRemakÐSchmidt) The assumptions of Proposition 1.4.6 are satisfied in any of the following cases:

i. M is of finite length; ii. R is left artinian and M is finitely generated; iii. R is left noetherian, R/Jac(R) is left artinian, any finitely generated R-module is complete, and M is finitely generated; iv. R is an R0-algebra, which is finitely generated as an R0-module, over a noethe- rian complete commutative ring R0 such that R0/ Jac(R0) is artinian, and M is finitely generated.

Proof i. Use Proposition 1.4.4. ii. This reduces to i. by Proposition 1.1.1.iii. iii. Use Proposition 1.4.5. iv. This reduces to iii. by Proposition 1.3.6. 

Example Let K be a field and G be a finite group. The group ring K[G] is left artinian. Hence Proposition 1.4.6 applies: Any finitely generated K[G]-module has a “unique” decomposition into indecomposable modules.

1.5 Idempotents and Blocks

Of primary interest to us is the decomposition of the ring R itself into indecompos- able submodules. This is closely connected to the existence of idempotents in R.

Definition

i. An element e ∈ R is called an idempotent if e2 = e = 0. ii. Two idempotents e1,e2 ∈ R are called orthogonal if e1e2 = 0 = e2e1. ii. An idempotent e ∈ R is called primitive if e is not equal to the sum of two orthogonal idempotents. iv. The idempotents in the center Z(R) of R are called central idempotents in R.

We note that eRe, for any idempotent e ∈ R, is a subring of R with unit ele- ment e. We also note that for any idempotent 1 = e ∈ R the element 1 − e is another idempotent, and e,1 − e are orthogonal.

Exercise If e1,...,er ∈ R are idempotents which are pairwise orthogonal then e1 + ···+er is an idempotent as well. 16 1 Prerequisites in Module Theory

Proposition 1.5.1 Let L = Re ⊆ R be a left ideal generated by an idempotent e; the map

setofallsets{e1,...,er } of pair- set of all decompositions ∼ wise orthogonal idempotents ei ∈ R −→ L = L1 ⊕···⊕Lr of L

such that e1 +···+er = e into nonzero left ideals Li

{e1,...,er } −→L = Re1 ⊕···⊕Rer is bijective.

Proof Suppose given a set {e1,...,er } in the left-hand side. We have L = Re = R(e1 +···+er ) ⊆ Re1 +···+Rer . On the other hand L ⊇ Reie = R(eie1 +···+ eier ) = Rei . Hence L = Re1 +···+Rer . To see that the sum is direct let    aej = aiei ∈ Rej ∩ Rei with a,ai ∈ R i=j i=j be an arbitrary element. Then    aej = (aej )ej = aiei ej = aieiej = 0. i=j i=j

It follows that the asserted map is well defined. { } To establish its injectivity let e1,...,er be another set in the left-hand side such that the two decompositions ⊕···⊕ = = ⊕···⊕ Re1 Rer L Re1 Rer { } = coincide. This means that there is a permutation σ of 1,...,r such that Rei Reσ(i) for any i. The identity +···+ = = +···+ eσ(1) eσ(r) e e1 er = { }={ } then implies that ei eσ(i) for any i or, equivalently, that e1,...,er e1,...,er . We prove the surjectivity of the asserted map in two steps. Step 1: We assume that e = 1 and hence L = R. Suppose that R = L1 ⊕···⊕Lr is a decomposition as a direct sum of nonzero left ideals Li . We then have 1 = e1 +···+er for appropriate elements ei ∈ Li and, in particular, Rei ⊆ Li and R = R · 1 = R(e1 +···+er ) ⊆ Re1 +···+Rer . It follows that Rei = Li . Since Li = {0} we must have ei = 0. Furthermore,

ei = ei · 1 = ei(e1 +···+er ) = eie1 +···+eier for any 1 ≤ i ≤ r.

Since eiej ∈ Lj we obtain 2 = = = ei ei and eiej 0forj i. 1.5 Idempotents and Blocks 17

We conclude that the set {e1,...,er } is a preimage of the given decomposition un- der the map in the assertion. Step 2: For a general e = 1 we first observe that the elements 1 − e and e are orthogonal idempotents in R. Hence, by what we have shown already, we have the decomposition R = R(1 − e) ⊕ Re = R(1 − e) ⊕ L. Suppose now that L = L1 ⊕···⊕Lr is a direct sum decomposition into nonzero left ideals Li . Then

R = R(1 − e) ⊕ L1 ⊕···⊕Lr is a decomposition into nonzero left ideals as well. By Step 1 we find pairwise orthogonal idempotents e0,e1,...,er in R such that

1 = e0 + e1 +···+er ,Re0 = R(1 − e), and Rei = Li for 1 ≤ i ≤ r.

Comparing this with the identity 1 = (1 − e) + e where 1 − e ∈ Re0 and e ∈ L = L1 ⊕ ··· ⊕ Lr we see that 1 − e = e0 and e = e1 + ··· + er . Therefore the set {e1,...,er } is a preimage of the decomposition L = L1 ⊕···⊕Lr under the asserted map. 

There is a completely analogous right ideal version, sending {e1,...,er } to e1R ⊕···⊕er R, of the above proposition.

Corollary 1.5.2 For any idempotent e ∈ R the following conditions are equivalent:

i. The R-module Re is indecomposable; ii. e is primitive; iii. the right R-module eR is indecomposable; iv. the ring eRe contains no idempotent other than e.

Proof The equivalence of i., ii., and iii. follows immediately from Proposition 1.5.1 and its right ideal version. ii. =⇒ iv. Suppose that e = f ∈ eRe is an idempotent. Then ef = fe= f , and e = (e − f)+ f is the sum of the orthogonal idempotents e − f and f .Thisisa contradiction to the primitivity of e. iv. =⇒ ii. Suppose that e = e1 + e2 is the sum of the orthogonal idempotents ∈ = 2 + = = 2 + = = e1,e2 R. Then ee1 e1 e2e1 e1 and e1e e1 e1e2 e1 and hence e1 ee1e ∈ eRe. Since e1 = e this again is a contradiction. 

Proposition 1.5.3 Let I = Re = eR be a two-sided ideal generated by a central idempotent e; we then have: i. I is a subring of R with unit element e; 18 1 Prerequisites in Module Theory ii. the map

set of all sets {e1,...,er } set of all decompositions of pairwise orthogonal I = I ⊕···⊕I −→∼ 1 r idempotents ei ∈ Z(R) such that of I into nonzero two-sided

e1 +···+er = e ideals Ii ⊆ R

{e1,...,er } −→I = Re1 ⊕···⊕Rer

is bijective; moreover, the multiplication in I = Re1 ⊕···⊕Rer can be carried out componentwise.

Proof i. We have I = eRe. ii. Obviously, since ei is central the ideal Rei = eiR is two-sided. Taking Proposition 1.5.1 into account it therefore remains to show that, if in the decomposition I = Re1 ⊕···⊕Rer with e = e1 +···+er the Rei are two-sided ideals, then the ei necessarily are central. But for any a ∈ R we have

ae1 +···+aer = ae = ea = a(e1 +···+er )

= (e1 +···+er )a

= e1a +···+er a where aei and eia both lie in Rei . Hence aei = eia since the summands are uniquely determined in Rei . For the second part of the assertion let a = a1 +···+ar and b = b1 +···+br with ai,bi ∈ Ii be arbitrary elements. Then ai = aiei and bi = eibi and hence

ab = (a1 +···+ar )(b1 +···+br )

= (a1e1 +···+ar er )(e1b1 +···+er br )   = aieiej bj = aieibi i,j i  = a b . i i  i

Corollary 1.5.4 A central idempotent e ∈ R is primitive in Z(R) if and only if Re is not the direct sum of two nonzero two-sided ideals of R.

Proposition 1.5.5 If R is left noetherian then we have: i. 1 ∈ R can be written as a sum of pairwise orthogonal primitive idempotents; ii. R contains only finitely many central idempotents; iii. any two different central idempotents which are primitive in Z(R) are orthogo- nal; 1.5 Idempotents and Blocks 19 iv. if {e1,...,en} is the set of all central idempotents which are primitive in Z(R) then e1 +···+en = 1.

Proof i. By Lemma 1.1.6 we have a direct sum decomposition R = L1 ⊕ ··· ⊕ Lr of R into indecomposable left ideals Li . According to Proposition 1.5.1 this corresponds to a decomposition of the unit element

1 = f1 +···+fr (1.5.1) as a sum of pairwise orthogonal idempotents fi such that Li = Rfi . Moreover, Corollary 1.5.2 says that each fi is primitive. ii. We keep the decomposition (1.5.1). Let e ∈ Z(R) be any idempotent. Then e = ef 1 +···+ef r and fi = ef i + (1 − e)fi .Wehave

2 = 2 2 = − 2 = − 2 2 = − (efi) e fi ef i, (1 e)fi (1 e) fi (1 e)fi, − = − 2 = − = − 2 = ef i(1 e)fi e(1 e)fi 0, and (1 e)fief i (1 e)efi 0.

But fi is primitive. Hence either ef i = 0or(1 − e)fi = 0, i.e. ef i = fi . This shows that there is a subset S ⊆{1,...,r} such that  e = fi, i∈S and we see that there are only finitely many possibilities for e. iii. Let e1 = e2 be two primitive idempotents in the ring Z(R). We then have

e1 = e1e2 + e1(1 − e2)

2 2 where the summands satisfy (e1e2) = e1e2,(e1(1 − e2)) = e1(1 − e2), and e1e2e1(1 − e2) = 0. Hence e1e2 = 0ore1 = e1e2. By symmetry we also obtain e2e1 = 0ore2 = e2e1. It follows that e1e2 = 0ore1 = e1e2 = e2. The latter case being excluded by assumption we conclude that e1e2 = 0. iv. First we consider any idempotent f ∈ Z(R).Iff is not primitive in Z(R) then we can write it as the sum of two orthogonal idempotents in Z(R).Anyofthe two summands either is primitive or again can be written as the sum of two new (exercise!) orthogonal idempotents in Z(R). Proceeding in this way we must arrive, because of ii., after finitely many steps at an expression of f as a sum of pairwise orthogonal idempotents which are primitive in Z(R). This, first of all, shows that the set {e1,...,en} is nonempty. By iii. the sum e := e1 + ··· + en is a central idempotent. Suppose that e = 1. Then 1 − e is a central idempotent. By the initial observation we find a subset S ⊆{1,...,n} such that  1 − e = ei. i∈S 20 1 Prerequisites in Module Theory

For i ∈ S we then compute   ei = ei ej = ei(1 − e) = ei − ei(e1 +···+en) = ei − ei = 0 j∈S which is a contradiction. 

Let e ∈ R be a central idempotent which is primitive in Z(R).AnR-module M is said to belong to the e-block of R if eM = M holds true.

Exercise If M belongs to the e-block then we have: a. ex = x for any x ∈ M; b. every submodule and every factor module of M also belongs to the e-block.

We now suppose that R is left noetherian. Let {e1,...,en} be the set of all cen- tral idempotents which are primitive in Z(R). By Proposition 1.5.5.iii/iv the ei are pairwise orthogonal and satisfy e1 +···+en = 1. Let M be any R-module. Since ei is central eiM ⊆ M is a submodule which obviously belongs to the ei -block. We also have M = 1 · M = (e1 + ··· + en)M ⊆ e1M + ··· + enM and hence M = e1M +···+enM.For    eix = ej xj ∈ eiM ∩ ej M with x,xj ∈ M j=i j=i we compute   eix = eieix = ei ej xj = eiej xj = 0. j=i j=i Hence the decomposition

M = e1M ⊕···⊕enM is direct. It is called the block decomposition of M.

Remark 1.5.6

i. For any R-module homomorphism f : M −→ N we have f(eiM)⊆ eiN. ii. If the submodule N ⊆ M lies in the ei -block then N ⊆ eiM. iii. If M is indecomposable then there is a unique 1 ≤ i ≤ n such that M lies in the ei -block.

Proof i. Since f is R-linear we have f(eiM) = eif(M)⊆ eiN. ii. Apply i. to the inclusion eiN = N ⊆ M. iii. In this case the block decomposition can have only a single nonzero summand.  1.5 Idempotents and Blocks 21

As a consequence of Proposition 1.5.3 the map

n =∼ R −→ Rei i=1

a −→ (ae1,...,aen) is an isomorphism of rings. If M lies in the ei -block then

ax = a(eix) = (aei)x for any a ∈ R and x ∈ M.

This means that M comes, by restriction of scalars along the projection map R −→ Rei , from an Rei -module (with the same underlying additive group). In this sense the ei -block coincides with the class of all Rei -modules. We next discuss, for arbitrary R, the relationship between idempotents in the ring R and in a factor ring R/I.

Proposition 1.5.7 Let I ⊆ R be a two-sided ideal and suppose that either every element in I is nilpotent or R is I -adically complete; then for any idempotent ε ∈ R/I there is an idempotent e ∈ R such that e + I = ε.

Proof Case 1: We assume that every element in I is nilpotent. Let ε = a + I and put b := 1 − a. Then ab = ba = a − a2 ∈ I , and hence (ab)m = 0forsomem ≥ 1. Since a and b commute the binomial theorem gives   2m 2m − 1 = (a + b)2m = a2m ibi = e + f i i=0 with     m 2m 2m − 2m − e := a2m ibi ∈ aR and f := a2m j bj . i j i=0 j=m+1

For any 0 ≤ i ≤ m and m

− − − − + − − − + − a2m ibia2m j bj = ambma3m i j bi j m = (ab)ma3m i j bi j m = 0 and hence ef = 0. It follows that e = e(e + f)= e2. Moreover, ab ∈ I implies   m    2m e + I = a2m + a2m−i−1bi−1 ab + I = a2m + I = ε2m = ε. i i=1 ∼ −=→ n Case 2: We assume that R is I -adically complete. Since R ←lim− R/I it suffices n to construct a sequence of idempotents εn ∈ R/I ,forn ≥ 2, such that pr(εn+1) = 2n n+1 n n+1 n+1 εn and ε1 = ε. Because of I ⊆ I the ideal I /I in the ring R/I is 22 1 Prerequisites in Module Theory nilpotent. Hence we may, inductively, apply the first case to the idempotent εn in n n+1 the factor ring R/I of the ring R/I in order to obtain εn+1. 

We point out that, by Proposition 1.2.1.v and Lemma 1.3.2, the assumptions in Proposition 1.5.7 imply that I ⊆ Jac(R).

Remark 1.5.8 i. Jac(R) does not contain any idempotent. ii. Let I ⊆ Jac(R) be a two-sided ideal and e ∈ R be an idempotent; we then have: a. e + I is an idempotent in R/I; b. if e + I ∈ R/I is primitive then e is primitive.

Proof i. Suppose that e ∈ Jac(R) is an idempotent. Then 1 − e is an idempotent as well as a unit (cf. Proposition 1.2.1.iv). Multiplying the equation (1 − e)2 = 1 − e by the inverse of 1 − e shows that 1 − e = 1. This leads to the contradiction that e = 0. ii. The assertion a. is immediate from i. For b. let e = e1 + e2 with orthogonal idempotents e1,e2 ∈ R. Then e + I = (e1 + I)+ (e2 + I) with (e1 + I)(e2 + I)= e1e2 + I = I . Therefore, by a., e1 + I and e2 + I are orthogonal idempotents in R/I. This, again, is a contradiction. 

Lemma 1.5.9 i. Let e,f ∈ R be two idempotents such that e + Jac(R) = f + Jac(R); then Re =∼ Rf as R-modules. ii. Let I ⊆ Jac(R) be a two-sided ideal and e,f ∈ R be idempotents such that the idempotents e + I,f + I ∈ R/I are orthogonal; then there exists an idempotent f ∈ R such that f + I = f + I and e,f are orthogonal.

Proof i. We consider the pair of R-modules L := Rf e ⊆ M := Re. Since f − e ∈ Jac(R) we have

L + Jac(R)M = Rf e + Jac(R)e = R e + (f − e) e + Jac(R)e = Re + Jac(R)e = Re = M.

The factor module M/L being generated by a single element e+L we may apply the Nakayama lemma 1.2.3 and obtain Rf e = Re. On the other hand the decomposition R = Rf ⊕ R(1 − f)leads to Re = Rf e ⊕ R(1 − f)e= Re ⊕ R(1 − f)e. It follows that R(1 − f)e={0} and, in particular, (1 − f)e= 0. We obtain that

e = fe and, by symmetry, also f = ef. 1.5 Idempotents and Blocks 23

This easily implies that the R-module homomorphisms of right multiplication by f and e, respectively,

Re  Rf re → ref r fe←[ r f are inverse to each other. ii. We have fe∈ I ⊆ Jac(R), hence 1 − fe∈ R×, and we may introduce the idempotent −1 f0 := (1 − fe) f(1 − fe). −1 −1 Obviously f0 + I = f + I and f0e = (1 − fe) f(e− fe)= (1 − fe) (f e − fe)= 0. We put

f := (1 − e)f0. Then

f + I = f0 − ef 0 + I = (f + I)− (e + I)(f + I)= f + I. Moreover,

f e = (1 − e)f0e = 0 and ef = e(1 − e)f0 = 0. Finally

2 = − − = − 2 − = − = f (1 e)f0(1 e)f0 (1 e) f0 f0ef 0 (1 e)f0 f . 

Proposition 1.5.10 Under the assumptions of Proposition 1.5.7 we have: i. If e ∈ R is a primitive idempotent then the idempotent e + I ∈ R/I is primitive as well; ii. if ε1,...,εr ∈ R/I are pairwise orthogonal idempotents then there are pairwise orthogonal idempotents e1,...,er ∈ R such that εi = ei + I for any 1 ≤ i ≤ r.

Proof i. Let e + I = ε1 + ε2 with orthogonal idempotents ε1,ε2 ∈ R/I. By Propo- sition 1.5.7 we find idempotents ei ∈ R such that εi = ei + I ,fori = 1, 2. By

Lemma 1.5.9.ii there is an idempotent e ∈ R such that e + I = e2 + I = ε2 and 2 2 e1,e are orthogonal. The latter implies that f := e1 + e is an idempotent as well. 2 ∼ 2 It satisfies f + I = ε1 + ε2 = e + I . Hence Rf = Re by Lemma 1.5.9.i. Applying ⊕ =∼ Proposition 1.5.1 we obtain that Re1 Re2 Re and we see that e is not primitive. This is a contradiction. ii. The proof is by induction with respect to r. We assume that the idempo- tents e1,...,er−1 have been constructed already. On the one hand we then have the idempotent e := e1 +···+er−1. On the other hand we find, by Proposition 1.5.7, 24 1 Prerequisites in Module Theory an idempotent f ∈ R such that f + I = εr . Since e + I = ε1 +···+εr−1 and εr are orthogonal there exists, by Lemma 1.5.9.ii, an idempotent er ∈ R such that er + I = f + I = εr and e,er are orthogonal. It remains to observe that

er ei = er (eei) = (er e)ei = 0 and eier = eieer = 0 for any 1 ≤ i

Proposition 1.5.11 Suppose that R is complete and that R/Jac(R) is left artinian; then R is local if and only if 1 is the only idempotent in R.

Proof We first assume that R is local. Let e ∈ R be any idempotent. Since e/∈ Jac(R) by Remark 1.5.8.i we must have e ∈ R×. Multiplying the identity e2 = e by e−1 gives e = 1. Now let us assume, vice versa, that 1 is the only idempotent in R. As a consequence of Proposition 1.5.7, the factor ring R := R/Jac(R) also has no other idempotent than 1. Therefore, by Proposition 1.5.1,theR-module L := R is indecomposable. On the other hand, the ring R being left artinian the module L,by Corollary 1.2.2, is of finite length. Hence Proposition 1.4.4 implies that EndR(L) is a local ring. But the map ∼ op −→= R EndR(R) c −→[a → ac]

op is an isomorphism of rings (exercise!). We obtain that R and R are local rings. Since Jac(R) = Jac(R/ Jac(R)) ={0} the ring R in fact is a skew field. Now Propo- sition 1.4.1 implies that R is local. 

Proposition 1.5.12 Suppose that R is an R0-algebra, which is finitely generated as an R0-module, over a noetherian complete commutative ring R0 such that R0/ Jac(R0) is artinian; then the map set of all central set of all central idem- −→ idempotents in R potents in R/Jac(R0)R

e −→ e := e + Jac(R0)R is bijective; moreover, this bijection satisfies: Ð e,f are orthogonal if and only if e,f are orthogonal; Ð e is primitive in Z(R) if and only if e is primitive in Z(R/Jac(R0)R).

Proof In Lemma 1.3.5.iii we have seen that Jac(R0)R ⊆ Jac(R). Hence Jac(R0)R, by Remark 1.5.8.i, does not contain any idempotent. Therefore e = 0 which says that the map in the assertion is well defined. To establish its injectivity let us assume that e1 = e2. Then

ei − e1 e2 = 0 and hence ei − e1e2 ∈ Jac(R0)R. 1.5 Idempotents and Blocks 25

But since e1 and e2 commute we have − 2 = 2 − + 2 2 = − + = − (ei e1e2) ei 2eie1e2 e1e2 ei 2e1e2 e1e2 ei e1e2.

It follows that ei − e1e2 = 0 which implies e1 = e1e2 = e2. For the surjectivity let ε ∈ R := R/Jac(R0)R be any central idempotent. In the proof of Proposition 1.3.6 we have seen that R is Jac(R0)R-adically complete. Hence we may apply Proposition 1.5.7 and obtain an idempotent e ∈ R such that e = ε. We, in fact, claim the stronger statement that any such e necessarily lies in the center Z(R) of R.Wehave

R = Re + R(1 − e) = eRe + (1 − e)Re + eR(1 − e) + (1 − e)R(1 − e) and correspondingly

R = εRε + (1 − ε)Rε + εR(1 − ε) + (1 − ε)R(1 − ε).

But

(1 − ε)Rε = (1 − ε)εR ={0} and εR(1 − ε) = Rε(1 − ε) ={0} since ε is central in R. It follows that (1 − e)Re and eR(1 − e) both are contained in Jac(R0)R. We see that 2 2 (1 − e)Re = (1 − e) Re ⊆ (1 − e)Jac(R0)Re = Jac(R0)(1 − e)Re and similarly eR(1 − e) ⊆ Jac(R0)eR(1 − e). This means that for the two finitely generated (as submodules of R) R0-modules M := (1 − e)Re and M := eR(1 − e) we have Jac(R0)M = M. The Nakayama lemma 1.2.3 therefore implies (1−e)Re = eR(1 − e) ={0} and consequently that

R = eRe + (1 − e)R(1 − e).

If we write an arbitrary element a ∈ R as a = ebe + (1 − e)c(1 − e) with b,c ∈ R then we obtain ea = e(ebe) = ebe = (ebe)e = ae. This shows that e ∈ Z(R). For the second half of the assertion we first note that with e,f also e,f are orthogonal for trivial reasons. Let us suppose, vice versa, that e,f are orthogonal. By Lemma 1.5.9.ii we find an idempotent f ∈ R such that f = f and e,f are orthogonal. According to what we have established above f necessarily is central. Hence the injectivity of our map forces f = f .Soe,f are orthogonal. Finally, if e = e1 + e2 with orthogonal central idempotents e1,e2 then obviously e = e1 + e2 with orthogonal idempotents e1, e2 ∈ Z(R). Vice versa, let e = ε1 + ε2 with orthogonal idempotents ε1,ε2 ∈ Z(R). By the surjectivity of our map we find idempotents ei ∈ Z(R) such that ei = εi . We have shown already that e1,e2 necessarily are orthogonal. Since e1 +e2 then is a central idempotent with e1 + e2 = e1 + e2 = ε1 + ε2 = e the injectivity of our map forces e1 + e2 = e.  26 1 Prerequisites in Module Theory

Remark 1.5.13 Let e ∈ R be any idempotent, and let L ⊆ M be R-modules; then eM/eL =∼ e(M/L) as Z(R)-modules.

Proof We have the obviously well-defined and surjective Z(R)-module homomor- phism

eM/eL −→ e(M/L) ex + eL −→ e(x + L) = ex + L.

If ex + L = L then ex ∈ L and hence ex = e(ex) ∈ eL. This shows that the map is injective as well. 

Keeping the assumptions of Proposition 1.5.12 we consider the block decompo- sition

M = e1M ⊕···⊕enM of any R-module M. It follows that

M/Jac(R0)M = e1M/Jac(R0)e1M ⊕···⊕enM/Jac(R0)enM

= e1M/e1 Jac(R0)M ⊕···⊕enM/en Jac(R0)M

= e1 M/Jac(R0)M ⊕···⊕en M/Jac(R0)M is the block decomposition of the R/Jac(R0)R-module M/Jac(R0)M.IfeiM = {0} then obviously ei(M/ Jac(R0)M) ={0}. Vice versa, let us suppose that ei(M/ Jac(R0)M) ={0}. Then eiM ⊆ Jac(R0)M and hence = 2 ⊆ eiM ei M Jac(R0)eiM.

If M is finitely generated as an R-module then eiM is finitely generated as an R0- module and the Nakayama lemma 1.2.3 implies that eiM ={0}. We, in particular, obtain that a finitely generated R-module M belongs to the ei -block if and only if M/Jac(R0)M belongs to the ei -block.

1.6 Projective Modules

We fix an R-module X. For any R-module M, resp. for any R-module homomor- phism g : L −→ M,wehavetheZ(R)-module

HomR(X, M), resp. the Z(R)-module homomorphism

HomR(X, g): HomR(X, L) −→ HomR(X, M) f −→ g ◦ f. 1.6 Projective Modules 27

h g Lemma 1.6.1 For any exact sequence 0 −→ L −→ M −→ N of R-modules the se- quence

HomR(X,h) HomR(X,g) 0 −→ HomR(X, L) −−−−−−−→ HomR(X, M) −−−−−−−→ HomR(X, N) is exact as well.

Proof Whenever a composite

f h X −→ L −→ M is the zero map we must have f = 0 since h is injective. This shows the injectivity of HomR(X, h).Wehave

HomR(X, g) ◦ HomR(X, h) = HomR(X, g ◦ h) = HomR(X, 0) = 0.

Hence the image of HomR(X, h) is contained in the kernel of HomR(X, g). Let now f : X −→ M be an R-module homomorphism such that g ◦ f = 0. Then im(f ) ⊆ ker(g). Hence for any x ∈ X we find, by the exactness of the original sequence, a unique fL(x) ∈ L such that

f(x)= h fL(x) .

This fL : X −→ L is an R-module homomorphism such that h ◦ fL = f . Hence

image of HomR(X, h) = kernel of HomR(X, g). 

Example Let R = Z,X = Z/nZ for some n ≥ 2, and g : Z −→ Z/nZ be the sur- jective projection map. Then HomZ(X, Z) ={0} but HomZ(X, Z/nZ)  idX = 0. Hence the map HomZ(X, g) : HomZ(X, Z) −→ HomZ(X, Z/nZ) cannot be surjec- tive.

Definition An R-module P is called projective if, for any surjective R-module ho- momorphism g : M −→ N,themap

HomR(P, g): HomR(P, M) −→ HomR(P, N) is surjective.

In slightly more explicit terms, an R-module P is projective if and only if any exact diagram of the form

P f f g M N 0 can be completed, by an oblique arrow f , to a commutative diagram. 28 1 Prerequisites in Module Theory

Lemma 1.6.2 For an R-module P the following conditions are equivalent: i. P is projective; ii. for any surjective R-module homomorphism h : M −→ P there exists an R-module homomorphism s : P −→ M such that h ◦ s = idP .

Proof i. =⇒ ii. We obtain s by contemplating the diagram

P s idP h M P 0.

ii. =⇒ i. Let P

f g M N 0 be any exact “test diagram”. In the direct sum M ⊕ P we have the submodule   M := (x, y) ∈ M ⊕ P : g(x) = f (y) .

The diagram

h((x,y)):=y M P

f ((x,y)):=x f g M N is commutative. We claim that the map h is surjective. Let y ∈ P be an arbitrary element. Since g is surjective we find an x ∈ M such that g(x) = f (y). Then (x, y) ∈ M with h((x, y)) = y. Hence, by assumption, there is an s : P −→ M such that h ◦ s = idP . We define f := f ◦ s and have

g ◦ f = g ◦ f ◦ s = f ◦ h ◦ s = f ◦ idP = f . 

In the situation of Lemma 1.6.2.ii we see that P is isomorphic to a direct sum- mand of M via the R-module isomorphism

=∼ ker(h) ⊕ P −→ M (x, y) −→ x + s(y)

(exercise!). 1.6 Projective Modules 29

Definition An R-module F is called free if there exists an R-module isomorphism ∼ F = ⊕i∈I R for some index set I .

Example If R = K is a field then, by the existence of bases for vector spaces, any K-module is free. On the other hand for R = Z the modules Z/nZ, for any n ≥ 2, are neither free nor projective.

Lemma 1.6.3 Any free R-module F is projective. ∼ Proof If F = ⊕i∈I R then let ei ∈ F be the element which corresponds to the tuple (...,0, 1, 0,...) with 1 in the ith place and zeros elsewhere. The set {ei}i∈I is an “R-basis” of the module F . In particular, for any R-module M,themap  =∼ HomR(F, M) −→ M i∈I

−→ f f(ei) i

g is bijective. For any surjective R-module homomorphism M −→ N the lower hori- zontal map in the commutative diagram

HomR(F,g) HomR(F, M) HomR(F, N)

=∼ =∼

 (xi )i −→(g(xi ))i  i∈I M i∈I N obviously is surjective. Hence the upper one is surjective, too. 

Proposition 1.6.4 An R-module P is projective if and only if it isomorphic to a direct summand of a free module.

Proof First we suppose that P is projective. For a sufficiently large index set I we find a surjective R-module homomorphism  h: F := R −→ P. i∈I

By Lemma 1.6.2 there exists an R-module homomorphism s : P −→ F such that h ◦ s = idP , and ker(h) ⊕ P =∼ F. 30 1 Prerequisites in Module Theory

Vice versa, let P be isomorphic to a direct summand of a free R-module F .Any module isomorphic to a projective module itself is projective (exercise!). Hence we may assume that F = P ⊕ P ⊆ for some submodule P ⊆ F . We then have the inclusion map i : P −→ F as well as the projection map pr : F −→ P . We consider any exact “test diagram”

P

f g M N 0.

By Lemma 1.6.3 the extended diagram

F

pr ˜ f P

f g M N 0 can be completed to a commutative diagram by an oblique arrow f˜. Then

g ◦ (f˜ ◦ i) = (g ◦ f)˜ ◦ i = f ◦ pr ◦i = f which means that the diagram

P f :=f˜◦i f g M N is commutative. Hence P is projective. 

Example Let e ∈ R be an idempotent. Then R = Re ⊕ R(1 − e). Hence Re is a projective R-module.

Corollary 1.6.5 If P1,P2 are two R-modules then the direct sum P1 ⊕ P2 is pro- jective if and only if P1 and P2 are projective. 1.6 Projective Modules 31

Proof If ∼ (P1 ⊕ P2) ⊕ P = F for some free R-module F then visibly P1 and P2 both are isomorphic to direct summands of F as well and hence are projective. If on the other hand   ⊕ =∼ ⊕ =∼ P1 P1 R and P2 P2 R i∈I1 i∈I2 then  ⊕ ⊕ ⊕ =∼ (P1 P2) P1 P2 R.  i∈I1∪I2 Lemma 1.6.6 (Schanuel) Let

h1 g1 h2 g2 0 −→ L1 −−→ P1 −→ N −→ 0 and 0 −→ L2 −−→ P2 −→ N −→ 0 be two short exact sequences of R-modules with the same right-hand term N; if P1 ∼ and P2 are projective then L2 ⊕ P1 = L1 ⊕ P2.

Proof The R-module   M := (x1,x2) ∈ P1 ⊕ P2 : g1(x1) = g2(x2) sets in the two short exact sequences

y −→(0,h2(y)) (x1,x2) −→x1 0 −→ L2 −−−−−−−−−→ M −−−−−−−−→ P1 −→ 0 and

y −→(h1(y),0) (x1,x2) −→x2 0 −→ L1 −−−−−−−−−→ M −−−−−−−−→ P2 −→ 0. By applying Lemma 1.6.2 we obtain ∼ ∼ L2 ⊕ P1 = M = L1 ⊕ P2. 

Lemma 1.6.7 Let R −→ R be any ring homomorphism; if P is a projective

R-module then R ⊗R P is a projective R -module.  ⊕ =∼ Proof Using Proposition 1.6.4 we write P Q i∈I R and obtain    ∼ R ⊗R P ⊕ R ⊗R Q = R ⊗R (P ⊕ Q) = R ⊗R R i∈I  

= R ⊗R R = R .  i∈I i∈I 32 1 Prerequisites in Module Theory

Definition i. An R-module homomorphism f : M −→ N is called essential if it is surjective but f(L)= N for any proper submodule L M. ii. A projective cover of an R-module M is an essential R-module homomorphism f : P −→ M where P is projective.

Lemma 1.6.8 Let f : P → M and f : P → M be two projective covers; then =∼ there exists an R-module isomorphism g : P −→ P such that f = f ◦ g.

Proof The “test diagram”

P g f

P M 0 f shows the existence of a homomorphism g such that f = f ◦ g. Since f is sur- jective we have f(g(P )) = M, and since f is essential we deduce that g(P ) = P . This shows that g is surjective. Then, by Lemma 1.6.2, there exists an R-module homomorphism s : P −→ P such that g ◦ s = idP .Wehave

f s(P) = f g s(P) = f(P)= M.

Since f is essential this implies s(P) = P . Hence s and g are isomorphisms. 

Remark 1.6.9 Let f : M −→ N be a surjective R-module homomorphism between finitely generated R-modules; if

ker(f ) ⊆ Jac(R)M then f is essential.

Proof Let L ⊆ M be a submodule such that f(L)= N. Then

M = L + ker(f ) = L + Jac(R)M.

Hence L = M by the Nakayama lemma 1.2.3. 

Proposition 1.6.10 Suppose that R is complete and that R/Jac(R) is left artinian; then any finitely generated R-module M has a projective cover; more precisely, there =∼ exists a projective cover f : P −→ M such that the induced map P/Jac(R)P −→ M/Jac(R)M is an isomorphism. 1.6 Projective Modules 33

Proof By Corollary 1.2.2 and Proposition 1.5.5.i we may write 1 + Jac(R) = ε1 + ···+εr as a sum of pairwise orthogonal primitive idempotents εi ∈ R := R/Jac(R). According to Proposition 1.5.1 and Corollary 1.5.2 we then have

R = Rε1 ⊕···⊕Rεr where the R-modules Rεi are indecomposable. But R is semisimple by Proposi- tion 1.2.1.vi. Hence the indecomposable R-modules Rεi in fact are simple, and all simple R-modules occur, up to isomorphism, among the Rεi . On the other hand, M/Jac(R)M is a semisimple R-module by Proposition 1.1.4.iii and as such is a direct sum

M/Jac(R)M = L1 ⊕···⊕Ls of simple submodules Lj . For any 1 ≤ j ≤ s we find an 1 ≤ i(j) ≤ r such that ∼ Lj = Rεi(j).

By Proposition 1.5.7 there exist idempotents e1,...,er ∈ R such that ei + Jac(R) = εi for any 1 ≤ i ≤ r.UsingRemark1.5.13 we now consider the composed R-module isomorphism     s s s ∼ f : Rei(j) Jac(R) Rei(j) = Rεi(j) j=1 j=1 j=1 s ∼ ∼ = Lj = M/Jac(R)M. j=1 It sits in the “test diagram”  := s P j=1 Rei(j)

pr f   s s ( j=1 Rei(j))/ Jac(R)( j=1 Rei(j))

=∼ f

M M/Jac(R)M 0. pr

Each Rei is a projective R-module. Hence P is a projective R-module by Corol- lary 1.6.5. We therefore find an R-module homomorphism f : P −→ M which makes the above diagram commutative. By construction it induces the isomorphism f modulo Jac(R). It remains to show that f is essential. Since f ◦ pr is surjective we have f(P)+ Jac(R)M = M, and the Nakayama lemma 1.2.3 implies that f is surjective. Moreover, ker(f ) ⊆ Jac(R)P by construction. Hence f is essential by Remark 1.6.9.  34 1 Prerequisites in Module Theory

1.7 Grothendieck Groups

We first recall that over any set S one has the free abelian group Z[S] with basis S given by  Z[S]= mss : ms ∈ Z, all but finitely many ms are equal to zero s∈S and      mss + nss = (ms + ns)s. s∈S s∈S s∈S Its universal property is the following: For any map of sets α : S −→ B from S into some abelian group B there is a unique homomorphism of abelian groups α˜ : Z[S]−→B such that α˜ |S = α. Let now A be any ring and let M be some class of A-modules. It is partitioned into isomorphism classes where the isomorphism class {M} of a module M in M consists of all modules in M which are isomorphic to M. We assume that the iso- morphism classes of modules in M form a set M/ =∼, and we introduce the free abelian group Z[M]:=Z[M/ =∼].InZ[M] we consider the subgroup Rel gener- ated by all elements of the form

{M}−{L}−{N} whenever there is a short exact sequence of A-module homomorphisms

0 −→ L −→ M −→ N −→ 0 with L,M,N in M.

The corresponding factor group

G0(M) := Z[M]/ Rel is called the Grothendieck group of M. We define [M]∈G0(M) to be the image of {M}. The elements [M] are generators of the abelian group G0(M), and for any short exact sequence as above one has the identity

[M]=[L]+[N] in G0(M).

Remark For any A-modules L,N we have the short exact sequence 0 → L → L ⊕ N → N → 0 and therefore the identity [L ⊕ N]=[L]+[N] in G0(M) provided L,N, and L ⊕ N lie in the class M.

For us two particular cases of this construction will be most important. In the first case we take MA to be the class of all A-modules of finite length, and we define

R(A) := G0(MA). 1.7 Grothendieck Groups 35

The set Aˆ of isomorphism classes of simple A-modules is obviously a subset of ∼ ˆ MA/ =. Hence Z[A]⊆Z[MA] is a subgroup, and we have the composed map

ˆ ⊆ pr Z[A] −→ Z[MA] −→ R(A).

=∼ Proposition 1.7.1 The above map Z[Aˆ] −→ R(A) is an isomorphism.

Proof We define an endomorphism π of Z[MA] as follows. By the universal prop- erty of Z[MA] we only need to define π({M}) ∈ Z[MA] for any A-module M of finite length. Let {0}=M0 M1 ··· Mn = M be a composition series of M. According to the Jordan–Hölder Proposition 1.1.2 the isomorphism classes {M1}, {M2/M1},...,{M/Mn−1} do not depend on the choice of the series. We put

π {M} := {M1}+{M2/M1}+···+{M/Mn−1} , and we observe:

Ð The modules M1,M2/M1,...,M/Mn−1 are simple. Hence we have im(π) ⊆ Z[Aˆ]. ÐIfM is simple then obviously π({M}) ={M}. It follows that the endomorphism ˆ π of Z[MA] is an idempotent with image equal to Z[A], and therefore ˆ Z[MA]=im(π) ⊕ ker(π) = Z[A]⊕ker(π). (1.7.1)

Ð The exact sequences

0 −→ M1 −→ M −→ M/M1 −→ 0

0 −→ M2/M1 −→ M/M1 −→ M/M2 −→ 0 . .

0 −→ Mn−2/Mn−1 −→ M/Mn−2 −→ M/Mn−1 −→ 0

imply the identities

[M]=[M1]+[M/M1], [M/M1]=[M2/M1]+[M/M2],...,

[M/Mn−2]=[Mn−1/Mn−2]+[M/Mn−1]

in R(A). It follows that

[M]=[M1]+[M2/M1]+···+[M/Mn−1]

and therefore {M}−π({M}) ∈ Rel. We obtain ˆ Z[MA]=im(π) + Rel = Z[A]+Rel . (1.7.2) 36 1 Prerequisites in Module Theory

f g Ð Finally, let 0 −→ L −→ M −→ N −→ 0 be a short exact sequence with L,M,N in MA.Let{0}=L0 ... Lr = L and {0}=N0 ... Ns = N be composi- tion series. Then

{0}=M0 M1 := f(L1) ··· Mr := f(L) −1 −1 Mr+1 := g (N1) ··· Mr+s−1 := g (Ns−1) Mr+s := M

is a composition series of M with  ∼ Li/Li−1 for 1 ≤ i ≤ r, Mi/Mi−1 = Ni−r /Ni−r−1 for r

Hence r+s r s π {M} = {Mi/Mi−1}= {Li/Li−1}+ {Nj /Nj−1} i=1 i=1 j=1

= π {L} + π {N}

which shows that π({M}−{L}−{N}) = 0. It follows that

Rel ⊆ ker(π). (1.7.3)

The formulae (1.7.1)Ð(1.7.3) together imply ˆ Z[MA]=Z[A]⊕Rel which is our assertion. 

In the second case we take MA to be the class of all finitely generated projective A-modules, and we define

K0(A) := G0(MA).

Remark As a consequence of Lemma 1.6.2.ii the subgroup Rel ⊆ Z[MA] in this case is generated by all elements of the form

{P ⊕ Q}−{P }−{Q} where P and Q are arbitrary finitely generated projective A-modules.

Let A˜ denote the set of isomorphism classes of finitely generated indecomposable ˜ projective A-modules. Then Z[A] is a subgroup of Z[MA].

Lemma 1.7.2 If A is left noetherian then the classes [P ] for {P }∈A˜ are generators of the abelian group K0(A). 1.7 Grothendieck Groups 37

Proof Let P be any finitely generated projective A-module. By Lemma 1.1.6 we have

P = P1 ⊕···⊕Ps with finitely generated indecomposable A-modules Pi . The Proposition 1.6.4 im- ˜ plies that the Pi are projective. Hence {Pi}∈A. As remarked earlier we have

[P ]=[P1]+···+[Ps] in K0(A). 

Remark Under the assumptions of the KrullÐRemakÐSchmidt Theorem 1.4.7 (e.g., if A is left artinian) an argument completely analogous to the proof of Proposi- tion 1.7.1 shows that the map

˜ =∼ Z[A] −→ K0(A) {P } −→[P ] is an isomorphism.

But we will see later that, since the unique decomposition into indecomposable modules is needed only for projective modules, weaker assumptions suffice for this isomorphism.

Remark 1.7.3 If A is semisimple we have: i. A˜ = Aˆ; ii. any A-module is projective; iii. K0(A) = R(A).

Proof By Proposition 1.1.4.iii any A-module is semisimple. In particular, any indecomposable A-module is simple, which proves i. Furthermore, any simple A-module is isomorphic to a module Ae for some idempotent e ∈ A and hence is projective (compare the proof of Proposition 1.6.10). It follows that any A-module is a direct sum of projective A-modules and therefore is projective (extend the proof of Corollary 1.6.5 to arbitrarily many summands!). This establishes ii. For iii. it re- mains to note that by Corollary 1.2.2 the class of all finitely generated (projective) A-modules coincides with the class of all A-modules of finite length. 

Suppose that A is left artinian. Then, by Corollary 1.2.2, any finitely generated A-module is of finite length. Hence the homomorphism

K0(A) −→ R(A) [P ] −→[P ] 38 1 Prerequisites in Module Theory is well defined. Using the above remark as well as Proposition 1.7.1 we may intro- duce the composed homomorphism

˜ =∼ =∼ ˆ cA: Z[A] −→ K0(A) −→ R(A) −→ Z[A].

It is called the Cartan homomorphism of the left artinian ring A.If  cA {P } = n{M}{M} {M}∈Aˆ then the integer n{M} is the multiplicity with which the simple A-module M oc- curs, up to isomorphism, as a subquotient in any composition series of the finitely generated indecomposable projective module P . Let A −→ B be a ring homomorphism between arbitrary rings A and B.By Lemma 1.6.7 the map ∼ ∼ MA/ = −→ MB / =

{P } −→{B ⊗A P } and hence the homomorphism

Z[MA]−→Z[MB ]

{P } −→{B ⊗A P } ∼ are well defined. Because of B ⊗A (P ⊕ Q) = (B ⊗A P)⊕ (B ⊗A Q) the latter map respects the subgroups Rel in both sides. We therefore obtain a well-defined homomorphism

K0(A) −→ K0(B)

[P ] −→[B ⊗A P ].

Exercise Let I ⊆ A be a two-sided ideal. For the projection homomorphism A −→ A/I and any A-module M we have

A/I ⊗A M = M/IM.

In particular, the homomorphism

K0(A) −→ K0(A/I) [P ] −→[P/IP] is well defined.

Proposition 1.7.4 Suppose that A is complete and that A := A/Jac(A) is left ar- tinian; we then have: 1.7 Grothendieck Groups 39

i. The maps ˜ ˆ =∼ A −→ A and K0(A) −→ K0(A) = R(A)     {P } −→ P/Jac(A)P [P ] −→ P/Jac(A)P

are bijective; ii. the inverses of the maps in i. are given by sending the isomorphism class {M} of an A-module M of finite length to the isomorphism class of a projective cover of M as an A-module; ˜ =∼ iii. Z[A] −→ K0(A).

Proof First of all we note that A is semisimple by Proposition 1.2.1.iii. We already have seen that the map

α: K0(A) −→ K0(A) = R(A)   [P ] −→ P/Jac(A)P is well defined. If M is an arbitrary A-module of finite length then by Proposi- −→f tion 1.6.10 we find a projective cover P M M of M as an A-module such that ∼ −→= P M / Jac(A)P M M. (1.7.4)

The proof of Proposition 1.6.10 shows that P M in fact is a finitely generated { } A-module. According to Lemma 1.6.8 the isomorphism class P M only depends on the isomorphism class {M}. We conclude that Z[M ]−→Z[M ] A A { } −→{ } M P M is a well-defined homomorphism. If N is a second A-module of finite length with −→g projective cover P N N as above then ⊕ ⊕ −f−−→g ⊕ P M P N M N is surjective with ⊕ ⊕ = ⊕ (P M P N )/ Jac(A)(P M P N ) P M / Jac(A)P M P N / Jac(A)P N =∼ M ⊕ N. ⊕ = ⊕ It follows that ker(f g) Jac(A)(P M P N ). This, by Remark 1.6.9, implies that f ⊕ g is essential. Using Corollary 1.6.5 we see that f ⊕ g is a projective cover of M ⊕ N as an A-module. Hence we have { ⊕ }={ } P M P N PM⊕N . 40 1 Prerequisites in Module Theory

This means that the above map respects the subgroups Rel in both sides and conse- quently induces a homomorphism

β: K0(A) −→ K0(A) [ ] −→[ ] M P M .

But it also shows that M is a simple A-module if P M is an indecomposable A-module. The isomorphism (1.7.4) says that

α ◦ β = id . K0(A) On the other hand, let P be any finitely generated projective A-module. As a con- sequence of Remark 1.6.9 the projection map P −→ M := P/Jac(A)P is essential { }={ } and hence a projective cover. We then deduce from Lemma 1.6.8 that P P M which means that ◦ = β α idK0(A) .

It follows that α is an isomorphism with inverse β. We also see that if P = P1 ⊕ P2 is decomposable then M = P1/ Jac(A)P1 ⊕ P2/ Jac(A)P2 is decomposable as well. This establishes the assertions i. and ii. For iii. we consider the commutative diagram

=∼ K0(A) R(A)

=∼ ∼ = ˆ Z[A˜] Z[A] where the horizontal isomorphisms come from i. and the right vertical isomorphism was shown in Proposition 1.7.1. Hence the left vertical arrow is an isomorphism as well. 

Corollary 1.7.5 Suppose that A is complete and that A/Jac(A) is left artinian; then

A = P1 ⊕···⊕Pr decomposes into a direct sum of finitely many finitely generated indecomposable projective A-modules Pj , and any finitely generated indecomposable projective A-module is isomorphic to one of the Pj .

Proof The projection map A −→ A/Jac(A) is a projective cover. We now decom- pose the semisimple ring

A/Jac(A) = M1 ⊕···⊕Mr 1.7 Grothendieck Groups 41 as a direct sum of finitely many simple modules Mj , and we choose projective covers Pj −→ Mj . Then P1 ⊕···⊕Pr is a projective cover of A/Jac(A) and con- sequently is isomorphic to A.IfP is an arbitrary finitely generated indecomposable projective A-module then P is a projective cover of the simple module P/Jac(A)P . ∼ The latter has to be isomorphic to some Mj . Hence P = Pj . 

Assuming that A is left artinian let us go back to the Cartan homomorphism ˆ ∼ ∼ ˆ cA: Z[A] = K0(A) −→ R(A) = Z[A]. ˜ By Corollary 1.7.5 the set A ={{P1},...,{Pt }} is finite. We put

Mj := Pj / Jac(A)Pj .

Then, by Proposition 1.7.4.i, {M1},...,{Mt } are exactly the isomorphism classes of simple A/Jac(A)-modules. But due to the definition of the Jacobson radical the simple A/Jac(A)-modules coincide with the simple A-modules, i.e.   ˆ A = {M1},...,{Mt } .

The Cartan homomorphism therefore is given by an integral matrix

CA = (cij )1≤i,j≤t ∈ Mt×t (Z) defined by the equations

cA {Pj } = c1j {M1}+···+ctj {Mt }.

The matrix CA is called the Cartan matrix of A. Chapter 2 The Cartan–Brauer Triangle

Let G be a finite group. Over any commutative ring R we have the group ring   R[G]= agg : ag ∈ R g∈G with addition      agg + bgg = (ag + bg)g g∈G g∈G g∈G and multiplication      = agg bgg ahbh−1g g. g∈G g∈G g∈G h∈G

We fix an algebraically closed field k of characteristic p>0. The representation theory of G over k is the module theory of the group ring k[G]. This is our primary object of study in the following.

2.1 The Setting

The main technical tool of our investigation will be a (0,p)-ring R for k which is a complete local commutative integral domain R such that

Ð the maximal ideal mR ⊆ R is principal, Ð R/mR = k, and Ð the field of fractions of R has characteristic zero.

j ≥ { } Exercise The only ideals of R are mR for j 0 and 0 .

P. Schneider, Modular Representation Theory of Finite Groups, 43 DOI 10.1007/978-1-4471-4832-6_2, © Springer-Verlag London 2013 44 2 The CartanÐBrauer Triangle

We note that there must exist an integer e ≥ 1—the ramification index of R— = e such that Rp mR. There is, in fact, a canonical (0,p)-ring W(k)for k—its ring of Witt vectors—with the additional property that mW(k) = W(k)p.LetK/K0 be any finite extension of the field of fractions K0 of W(k). Then   := ∈ : ∈ R a K NormK/K0 (a) W(k) is a (0,p)-ring for k with ramification index equal to [K : K0]. Proofs for all of this can be found in ¤3Ð6 of [9]. In the following we fix a (0,p)-ring R for k. We denote by K the field of fractions of R and by πR a choice of generator of mR,i.e.mR = RπR. The following three group rings are now at our disposal:

K[G]

⊆ pr R[G] k[G] such that

K[G]=K ⊗R R[G] and k[G]=R[G]/πRR[G]. As explained before Proposition 1.7.4 there are the corresponding homomorphisms between Grothendieck groups

K0(K[G])

[P ]−→[K⊗RP ] κ ρ K0(R[G]) K0(k[G]). [P ]−→[P/πRP ]

For the vertical arrow observe that, quite generally for any R[G]-module M,we have

K[G]⊗R[G] M = K ⊗R R[G]⊗R[G] M = K ⊗R M. We put

RK (G) := R K[G] and Rk(G) := R k[G] . Since K has characteristic zero the group ring K[G] is semisimple, and we have

RK (G) = K0 K[G] by Remark 1.7.3.iii. On the other hand, as a finite-dimensional k-vector space the group ring k[G] of course is left and right artinian. In particular we have the Cartan 2.1 The Setting 45 homomorphism

cG: K0 k[G] −→ Rk(G) [P ]−→[P ].

Hence, so far, there is the diagram of homomorphisms

RK (G) Rk(G)

κ cG ρ K0(R[G]) K0(k[G]).

Clearly, R[G] is an R-algebra which is finitely generated as an R-module. Let us collect some of what we know in this situation. Ð (Proposition 1.3.6) R[G] is left and right noetherian, and any finitely generated R[G]-module is complete as well as R[G]πR-adically complete. Ð (Theorem 1.4.7) The KrullÐRemakÐSchmidt theorem holds for any finitely gen- erated R[G]-module. Ð (Proposition 1.5.5)1∈ R[G] can be written as a sum of pairwise orthogonal primitive idempotents; the set of all central idempotents in R[G] is finite; 1 is equal to the sum of all primitive idempotents in Z(R[G]);anyR[G]-module has a block decomposition. Ð (Proposition 1.5.7) For any idempotent ε ∈ k[G] there is an idempotent e ∈ R[G] such that ε = e + R[G]πR. Ð (Proposition 1.5.12) The projection map R[G]−→k[G] restricts to a bijection between the set of all central idempotents in R[G] and the set of all central idem- potents in k[G]; in particular, the block decomposition of an R[G]-module M re- duces modulo R[G]πR to the block decomposition of the k[G]-module M/πRM. Ð (Proposition 1.6.10) Any finitely generated R[G]-module M has a projective =∼ cover P −→ M such that P/Jac(R[G])P −→ M/Jac(R[G])M is an isomor- phism.

We emphasize that R[G]πR ⊆ Jac(R[G]) by Lemma 1.3.5.iii. Moreover, the ideal Jac(k[G]) = Jac(R[G])/R[G]πR in the left artinian ring k[G] is nilpotent by Proposition 1.2.1.vi. We apply Proposition 1.7.4 to the rings R[G] and k[G] and we see that the maps {P }−→{P/πRP }−→{P/Jac(R[G])P } induce the commuta- tive diagram of bijections between finite sets

R[G] k [G]

R[G]=k [G] 46 2 The CartanÐBrauer Triangle as well as the commutative diagram of isomorphisms

=∼ Z[R[G]] Z[k [G]]

=∼ =∼ =∼ K0(R[G]) K0(k[G]). ρ

For purposes of reference we state the last fact as a proposition.

=∼ Proposition 2.1.1 The map ρ : K0(R[G]) −→ K0(k[G]) is an isomorphism; its in- verse is given by sending [M] to the class of a projective cover of M as an R[G]- module.

We define the composed map

− ρ 1 κ eG: K0 k[G] −−→ K0 R[G] −→ K0 K[G] = RK (G).

Remark 2.1.2 Any finitely generated projective R-module is free.

Proof Since R is an integral domain 1 is the only idempotent in R. Hence the free R-module R is indecomposable by Corollary 1.5.2. On the other hand, according to Proposition 1.7.4.i the map

R˜ −→ kˆ

{P }−→{P/πRP } is bijective. Obviously, k is up to isomorphism the only simple k-module. Hence R is up to isomorphism the only finitely generated indecomposable projective R-module. An arbitrary finitely generated projective R-module P , by Lemma 1.1.6,isafi- nite direct sum of indecomposable ones. It follows that P must be isomorphic to some Rn. 

2.2 The Triangle We already have the two sides

RK (G) Rk(G)

eG cG

K0(k[G]) of the triangle. To construct the third side we first introduce the following notion. 2.2 The Triangle 47

Definition Let V be a finite-dimensional K-vector space; a lattice L in V is an R-submodule L ⊆ V for which there exists a K-basis e1,...,ed of V such that

L = Re1 +···+Red .

Obviously, any lattice is free as an R-module. Furthermore, with L also aL,for any a ∈ K×, is a lattice in V .

Lemma 2.2.1 i. Let L be an R-submodule of a K-vector space V ; if L is finitely generated then L is free. ii. Let L ⊆ V be an R-submodule of a finite-dimensional K-vector space V ; if L is finitely generated as an R-module and L generates V as a K-vector space then L is a lattice in V . iii. For any two lattices L and L in V there is an integer m ≥ 0 such that m ⊆ πR L L .

Proof i. and ii. Let d ≥ 0 be the smallest integer such that the R-module L has d generators e1,...,ed .TheR-module homomorphism

Rd −→ L

(a1,...,ad ) −→ a1e1 +···+ad ed is surjective. Suppose that (a1,...,ad ) = 0 is an element in its kernel. Since at least one ai is nonzero the integer   := ≥ : ∈ j  max j 0 a1,...,ad mR =  ∈ ∈ × is defined. Then ai πRbi with bi R, and bi0 R for at least one index 1 ≤ i0 ≤ d. Computing in the vector space V we have = +···+ =  +···+ 0 a1e1 ad ed πR(b1e1 bd ed ) and hence

b1e1 +···+bd ed = 0. − But the latter equation implies e =− b 1b e ∈ Re , which is a con- i0 i =i0 i0 i i i =i0 i tradiction to the minimality of d. It follows that the above map is an isomorphism. This proves i. and, in particular, that

L = Re1 +···+Red .

For ii. it therefore suffices to show that e1,...,ed , under the additional assumption that L generates V ,isaK-basis of V . This assumption immediately guarantees 48 2 The CartanÐBrauer Triangle that the e1,...,ed generate the K-vector space V . To show that they are K-linearly independent let

c1e1 +···+cd ed = 0 with c1,...,cd ∈ K. ≥ := j ∈ ≤ ≤ We find a sufficiently large j 0 such that ai πRci R for any 1 i d. Then =  · =  +···+ = +···+ 0 πR 0 πR(c1e1 cd ed ) a1e1 ad ed .

By what we have shown above we must have ai = 0 and hence ci = 0 for any 1 ≤ i ≤ d. iii. Let e1,...,ed and f1,...,fd be K-bases of V such that

L = Re1 +···+Red and L = Rf1 +···+Rfd .

We write

ej = c1j f1 +···+cdj fd with cij ∈ K, and we choose an integer m ≥ 0 such that

m ∈ ≤ ≤ πR cij R for any 1 i, j d. m ∈ +···+ = ≤ ≤ It follows that πR ej Rf1 Rfd L for any 1 j d and hence that m ⊆  πR L L .

Suppose that V is a finitely generated K[G]-module. Then V is finite-dimension- al as a K-vector space. A lattice L in V is called G-invariant if we have gL ⊆ L for any g ∈ G. In particular, L is a finitely generated R[G]-submodule of V , and L/πRL is a k[G]-module of finite length.

Lemma 2.2.2 Any finitely generated K[G]-module V contains a lattice which is G-invariant.

Proof We choose a basis e1,...,ed of the K-vector space V . Then L := Re1 + ···+Red is a lattice in V . We define the R[G]-submodule  L := gL g∈G of V . With L also L generates V as a K-vector space. Moreover, L is finitely generated by the set {gei : 1 ≤ i ≤ d,g ∈ G} as an R-module. Therefore, by Lemma 2.2.1.ii, L is a G-invariant lattice in V . 

Whereas the K[G]-module V always is projective by Remark 1.7.3 a G-invariant lattice L in V need not to be projective as an R[G]-module. We will encounter an example of this later on. 2.2 The Triangle 49

Theorem 2.2.3 Let L and L be two G-invariant lattices in the finitely generated K[G]-module V ; we then have  

[L/πRL]= L /πRL in Rk(G).

Proof We begin by observing that, for any a ∈ K×,themap

=∼ L/πRL −→ (aL)/πR(aL)

x + πRL −→ ax + πR(aL) is an isomorphism of k[G]-modules, and hence   [L/πRL]= (aL)/πR(aL) in Rk(G).

m By applying Lemma 2.2.1.iii (and replacing L by πR L for some sufficiently large m ≥ 0) we therefore may assume that L ⊆ L . By applying Lemma 2.2.1.iii again to L and L (while interchanging their roles) we find an integer n ≥ 0 such that

n ⊆ ⊆ πRL L L .

We now proceed by induction with respect to n.Ifn = 1 we have the two exact sequences of k[G]-modules

0 −→ L/πRL −→ L /πRL −→ L /L −→ 0 and

0 −→ πRL /πRL −→ L/πRL −→ L/πRL −→ 0. It follows that          

L /πRL = L/πRL + L /L = L/πRL + πRL /πRL    

= L/πRL +[L/πRL]− L/πRL

=[L/πRL] in Rk(G).Forn ≥ 2 we consider the R[G]-submodule

:= n−1 + M πR L L.

It is a G-invariant lattice in V by Lemma 2.2.1.ii and satisfies

n−1 ⊆ ⊆ ⊆ ⊆ πR L M L and πRM L M.

Applying the case n = 1toL and M we obtain [M/πRM]=[L/πRL]. The induc- tion hypothesis for M and L gives [L /πRL ]=[M/πRM].  50 2 The CartanÐBrauer Triangle

The above lemma and theorem imply that

Z[MK[G]]−→Rk(G)

{V }−→[L/πRL], where L is any G-invariant lattice in V , is a well-defined homomorphism. If V1 and V2 are two finitely generated K[G]-modules and L1 ⊆ V1 and L2 ⊆ V2 are G-invariant lattices then L1 ⊕ L2 is a G-invariant lattice in V1 ⊕ V2 and   (L1 ⊕ L2)/πR(L1 ⊕ L2) =[L1/πRL1 ⊕ L2/πRL2]

=[L1/πRL1]+[L2/πRL2].

It follows that the subgroup Rel ⊆ Z[MK[G]] lies in the kernel of the above map so that we obtain the homomorphism

dG: RK (G) −→ Rk(G)

[V ]−→[L/πRL]. It is called the decomposition homomorphism of G.TheCartanÐBrauer triangle is the diagram

dG RK (G) Rk(G)

eG cG

K0(k[G]).

Lemma 2.2.4 The CartanÐBrauer triangle is commutative.

Proof Let P be a finitely generated projective R[G]-module. We have to show that

dG κ [P ] = cG ρ [P ] holds true. By definition the right-hand side is equal to [P/πRP ]∈Rk(G). More- over, κ([P ]) =[K ⊗R P ]∈RK (G). According to Proposition 1.6.4 the R(G)- module P is a direct summand of a free R[G]-module. But R[G] and hence any free R[G]-module also is free as an R-module. We see that P as an R-module is finitely generated projective and hence free by Remark 2.1.2. We conclude that P =∼ Rd is a ∼ d d d G-invariant lattice in the K[G]-module K ⊗R P = K ⊗R R = (K ⊗R R) = K , and we obtain dG(κ([P ])) = dG([K ⊗R P ]) =[P/πRP ]. 

Let us consider two “extreme” situations where the maps in the CartanÐBrauer triangle can be determined completely. First we look at the case where p does not divide the order |G| of the group G. Then k[G] is semisimple. Hence we have k[G]=k[G] and K0(k[G]) = Rk(G) by Remark 1.7.3.ThemapcG, in particular, is the identity. 2.2 The Triangle 51

Proposition 2.2.5 If p |G| then any R[G]-module M whichisprojectiveasan R-module also is projective as an R[G]-module.

Proof We consider any “test diagram” of R[G]-modules

M

α β L N 0.

Viewing this as a “test diagram” of R-modules our second assumption ensures the existence of an R-module homomorphism α0 : M −→ L such that β ◦α0 = α. Since |G| is a unit in R by our first assumption, we may define a new R-module homo- morphism α˜ : M −→ L by  −1 −1 α(x)˜ := |G| gα0 g x for any x ∈ M. g∈G

One easily checks that α˜ satisfies

α(hx)˜ = hα(x)˜ for any h ∈ G and any x ∈ M.

This means that α˜ is,infact,anR[G]-module homomorphism. Moreover, we com- pute   −1 −1 −1 −1 β α(x)˜ =|G| gβ α0 g x =|G| gα g x g∈G g∈G  =|G|−1 gg−1α(x) = α(x). g∈G 

Corollary 2.2.6 If p |G| then all three maps in the CartanÐBrauer triangle are isomorphisms; more precisely, we have the triangle of bijections

K[G] k [G]

{P }−→{K⊗RP } {P }−→{P/πRP } R[G].

Proof We already have remarked that cG is the identity. Hence it suffices to show that the map κ : K0(R[G]) −→ RK (G) is surjective. Let V be any finitely generated K[G]-module. By Lemma 2.2.2 we find a G-invariant lattice L in V . It satisfies V = K ⊗R L by definition. Proposition 2.2.5 implies that L is a finitely generated 52 2 The CartanÐBrauer Triangle projective R[G]-module. We conclude that [L]∈K0(R[G]) with κ([L]) =[V ]. This argument in fact shows that the map ∼ ∼ MR[G]/ = −→ MK[G]/ =

{P }−→{K ⊗R P } is surjective. Let P and Q be two finitely generated projective R[G]-modules such ∼ that K ⊗R P = K ⊗R Q as K[G]-modules. The commutativity of the CartanÐBrauer triangle then implies that

[P/πRP ]=dG [K ⊗R P ] = dG [K ⊗R Q] =[Q/πRQ].

Let P = P1 ⊕···⊕Ps and Q = Q1 ⊕···⊕Qt be decompositions into indecom- posable submodules. Then

s t P/πRP = Pi/πRPi and Q/πRQ = Qj /πRQj i=1 j=1 are decompositions into simple submodules. Using Proposition 1.7.1 the identity

s t [Pi/πRPi]=[P/πRP ]=[Q/πRQ]= [Qj /πRQj ] i=1 j=1 implies that s = t and that there is a permutation σ of {1,...,s} such that ∼ Qj /πRQj = Pσ(j)/πRPσ(j) for any 1 ≤ j ≤ s as k[G]-modules. Applying Proposition 1.7.4.i we obtain ∼ Qj = Pσ(j) for any 1 ≤ j ≤ s as R[G]-modules. It follows that P =∼ Q as R[G]-modules. Hence the above map between sets of isomorphism classes is bijective. Obviously, if K ⊗R P is in- decomposable (i.e. simple) then P was indecomposable. Vice versa, if P is in- decomposable then the above reasoning says that P/πRP is simple. Because of [P/πRP ]=dG([K ⊗R P ]) it follows that K ⊗R P must be indecomposable. 

The second case is where G is a p-group. For a general group G we have the ring homomorphism

k[G]−→k   agg −→ ag g∈G g∈G 2.2 The Triangle 53 which is called the augmentation of k[G]. It makes k into a simple k[G]-module which is called the trivial k[G]-module. Its kernel is the augmentation ideal    Ik[G]:= agg ∈ k[G]: ag = 0 . g∈G g∈G

Proposition 2.2.7 If G is a p-group then we have Jac(k[G]) = Ik[G]; in particular, k[G] is a local ring and the trivial k[G]-module is, up to isomorphism, the only simple k[G]-module.

Proof We will prove by induction with respect to the order |G|=pn of G that the trivial module, up to isomorphism, is the only simple k[G]-module. There is nothing to prove if n = 0. We therefore suppose that n ≥ 1. Note that we have

n n n gp = 1 and hence (g − 1)p = gp − 1 = 1 − 1 = 0 for any g ∈ G. The center of a nontrivial p-group is nontrivial. Let g0 = 1bea central element in G. We now consider any simple k[G]-module M and we denote by π : k[G]−→Endk(M) the corresponding ring homomorphism. Then

pn pn pn π(g0) − idM = π(g0 − 1) = π (g0 − 1) = π(0) = 0.

Since g0 is central (π(g0) − idM )(M) is a k[G]-submodule of M.ButM is simple. Hence (π(g0) − idM )(M) ={0} or = M. The latter would inductively imply that pn (π(g0) − idM ) (M) = M = {0} which is contradiction. We obtain π(g0) = idM which means that the cyclic subgroup g0 is contained in the kernel of the group homomorphism π : G −→ Autk(M). Hence we have a commutative diagram of group homomorphisms

π G Autk(M)

pr π

G/g0.

We conclude that M already is a simple k[G/g0]-module and therefore is the trivial module by the induction hypothesis. The identity Jac(k[G]) = Ik[G] now follows from the definition of the Jacobson radical, and Proposition 1.4.1 implies that k[G] is a local ring. 

Suppose that G is a p-group. Then Propositions 2.2.7 and 1.7.1 together imply that the map =∼ Z −→ Rk(G) m −→ m[k] 54 2 The CartanÐBrauer Triangle is an isomorphism. To compute the inverse map let M be a k[G]-module of finite length, and let {0}=M M ··· M = M be a composition series. We must ∼ 0 1 t have Mi/Mi−1 = k for any 1 ≤ i ≤ t. It follows that t t [M]= [Mi/Mi−1]=t ·[k] and dimk M = dimk Mi/Mi−1 = t. i=1 i=1 Hence the inverse map is given by

[M]−→dimk M. Furthermore, from Proposition 1.7.4 we obtain that

=∼ Z −→ K0 R[G]   m −→ m R[G] is an isomorphism. Because of dimk k[G]=|G| the Cartan homomorphism, under these identifications, becomes the map

cG: Z −→ Z m −→ m ·|G|. For any finitely generated K[G]-module V and any G-invariant lattice L in V we have dimK V = dimk L/πRL. Hence the decomposition homomorphism becomes

dG: RK (G) −→ Z

[V ]−→dimK V. Altogether the CartanÐBrauer triangle of a p-group G is of the form

[V ]−→dimK V RK (G) Z

1−→[K[G]] ·|G| Z.

We also see that the trivial K[G]-module K has the G-invariant lattice R which cannot be projective as an R[G]-module if G = {1}. Before we can establish the finer properties of the CartanÐBrauer triangle we need to develop the theory of induction.

2.3 The Ring Structure of RF (G), and Induction In this section we let F be an arbitrary field, and we consider the group ring F [G] and its Grothendieck group RF (G) := R(F[G]). 2.3 The Ring Structure of RF (G), and Induction 55

Let V and W be two (finitely generated) F [G]-modules. The group G acts on the tensor product V ⊗F W by

g(v ⊗ w) := gv ⊗ gw for v ∈ V and w ∈ W.

In this way V ⊗F W becomes a (finitely generated) F [G]-module, and we obtain the multiplication map

Z[MF [G]]×Z[MF [G]]−→Z[MF [G]]

{V }, {W} −→{V ⊗F W}.

Since the tensor product, up to isomorphism, is associative and commutative this multiplication makes Z[MF [G]] into a commutative ring. Its unit element is the isomorphism class {F } of the trivial F [G]-module.

Remark 2.3.1 The subgroup Rel is an ideal in the ring Z[MF [G]].

Proof Let V be a (finitely generated) F [G]-module and let

α β 0 −→ L −→ M −→ N −→ 0 be a short exact sequence of F [G]-modules. We claim that the sequence

idV ⊗α idV ⊗β 0 −→ V ⊗F L −−−−→ V ⊗F M −−−−→ V ⊗F N −→ 0 is exact as well. This shows that the subgroup Rel is preserved under multiplication by {V }. The exactness in question is purely a problem about F -vector spaces. But as ∼ ∼ vector spaces we have M = L ⊕ N and hence V ⊗F M = (V ⊗F L) ⊕ (V ⊗F N). 

It follows that RF (G) naturally is a commutative ring with unit element [F ] such that

[V ]·[W]=[V ⊗F W]. Let H ⊆ G be a subgroup. Then F [H]⊆F [G] is a subring (with the same unit element). Any F [G]-module V , by restriction of scalars, can be viewed as an F [H ]- module. If V is finitely generated as an F [G]-module then V is a finite-dimensional F -vector space and, in particular, is finitely generated as an F [H ]-module. Hence we have the ring homomorphism

G : −→ resH RF (G) RF (H ) [V ]−→[V ].

On the other hand, for any F [H]-module W we have, by base extension, the F [G]- module F [G]⊗F [H ] W . Obviously, the latter is finitely generated over F [G] if the 56 2 The CartanÐBrauer Triangle former was finitely generated over F [H].Thefirst Frobenius reciprocity says that

=∼ HomF [G] F [G]⊗F [H ] W,V −→ HomF [H ](W, V )   α −→ w −→ α(1 ⊗ w) is an F -linear isomorphism for any F [H]-module W and any F [G]-module V .

α β Remark 2.3.2 For any short exact sequence 0 −→ L −→ M −→ N −→ 0ofF [H ]- modules the sequence of F [G]-modules

idF [G] ⊗α idF [G] ⊗β 0 → F [G]⊗F [H ] L −−−−−−→ F [G]⊗F [H ] M −−−−−−→ F [G]⊗F [H ] N → 0 is exact as well.

Proof Let g1,...,gr ∈ G be a set of representatives for the left cosets of H in G. Then g1,...,gr also is a basis of the free right F [H ]-module F [G]. It follows that, for any F [H ]-module W ,themap

∼ r = W −→ F [G]⊗F [H ] W

(w1,...,wr ) −→ g1 ⊗ w1 +···+gr ⊗ wr is an F -linear isomorphism. We see that, as a sequence of F -vector spaces, the sequence in question is just the r-fold direct sum of the original exact sequence with itself. 

Remark 2.3.3 For any F [H]-module W ,iftheF [G]-module F [G]⊗F [H ] W is simple then W is a simple F [H]-module.

Proof Let W ⊆ W be any F [H]-submodule. Then F [G]⊗F [H ] W is an F [G]- submodule of F [G]⊗F [H ] W by Remark 2.3.2. Since the latter is simple we must have F [G]⊗F [H ] W ={0} or = F [G]⊗F [H ] W . Comparing dimensions using the argument in the proof of Remark 2.3.2 we obtain

[G : H ]·dimF W = dimF F [G]⊗F [H ] W = 0or

= dimF F [G]⊗F [H ] W =[G : H ]·dimF W.

We see that dimF W = 0or= dimF W and therefore that W ={0} or = W .This proves that W is a simple F [H]-module. 

It follows that the map

Z[MF [H ]]−→Z[MF [G]]   {W}−→ F [G]⊗F [H ] W 2.3 The Ring Structure of RF (G), and Induction 57 preserves the subgroups Rel in both sides and therefore induces an additive homo- morphism indG : R (H ) −→ R (G) H F  F  [W]−→ F [G]⊗F [H ] W

(which is not multiplicative!).

Proposition 2.3.4 We have

G · = G · G ∈ ∈ indH (y) x indH y resH (x) for any x RF (G) and y RF (H ).

Proof It suffices to show that, for any F [G]-module V and any F [H ]-module W , we have an isomorphism of F [G]-modules

∼ F [G]⊗F [H ] W ⊗F V = F [G]⊗F [H ] (W ⊗F V).

One checks (exercise!) that such an isomorphism is given by

(g ⊗ w) ⊗ v −→ g ⊗ w ⊗ g−1v . 

G : −→ Corollary 2.3.5 The image of indH RF (H ) RF (G) is an ideal in RF (G).

We also mention the obvious transitivity relations H ◦ G = G G ◦ H = G resH resH resH and indH indH indH for any chain of subgroups H ⊆ H ⊆ G. An alternative way to look at induction is the following. Let W be any F [H ]- module. Then   G := : −→ : = −1 ∈ ∈ IndH (W) φ G W φ(gh) h φ(g) for any g G, h H equipped with the left translation action of G given by

gφ g := φ g−1g is an F [G]-module called the module induced from W . But, in fact, the map ∼ = G F [G]⊗ [ ] W −→ Ind (W)  FH H  

agg ⊗ w −→ φ g := ag hhw g∈G h∈H is an isomorphism of F [G]-modules. This leads to the second Frobenius reciprocity isomorphism ∼ G −→= HomF [G] V,IndH (W) HomF [H ](V, W)   α −→ v −→ α(v)(1) . 58 2 The CartanÐBrauer Triangle

We also need to recall the character theory of G in the semisimple case. For this we assume for the rest of this section that the order of G is prime to the characteristic of the field F . Any finitely generated F [G]-module V is a finite-dimensional F - vector space. Hence we may introduce the function

χV : G −→ F g g −→ tr(g; V)= trace of V −→ V which is called the character of V . It depends only on the isomorphism class of V . Characters are class functions on G, i.e. they are constant on each conjugacy class of G. For any two finitely generated F [G]-modules V1 and V2 we have

= + = · χV1⊕V2 χV1 χV2 and χV1⊗F V2 χV1 χV2 .

Let Cl(G, F ) denote the F -vector space of all class functions G −→ F . By point- wise multiplication of functions it is a commutative F -algebra. The above identities imply that the map

Tr : RF (G) −→ Cl(G, F )

[V ]−→χV is a ring homomorphism.

Definition The field F is called a splitting field for G if, for any simple F [G]- module V ,wehaveEndF [G](V ) = F .

If F is algebraically closed then it is a splitting field for G.

Theorem 2.3.6 i. If the field F has characteristic zero then the characters {χV :{V }∈F [G]} are F -linearly independent. ii. If F is a splitting field for G then the characters {χV :{V }∈F [G]} form a basis of the F -vector space Cl(G, F ). iii. If F has characteristic zero then two finitely generated F [G]-modules V1 and = V2 are isomorphic if and only if χV1 χV2 holds true.

Corollary 2.3.7 i. If F has characteristic zero then the map Tr is injective. ii. If F is a splitting field for G then the map Tr induces an isomorphism of F -algebras =∼ F ⊗Z RF (G) −→ Cl(G, F ). 2.4 The Burnside Ring 59 iii. If F has characteristic zero then the map ∼ MF [G]/ = −→ Cl(G, F )

{V }−→χV

is injective.

Proof For i. and ii. use Proposition 1.7.1. 

2.4 The Burnside Ring

A G-set X is a set equipped with a G-action

G × X −→ X (g, x) −→ gx such that

1x = x and g(hx) = (gh)x for any g,h ∈ G and any x ∈ X.

Let X and Y be two G-sets. Their disjoint union X ∪ Y is a G-set in an obvious way. But also their cartesian product X × Y is a G-set with respect to

g(x,y) := (gx, gy) for (x, y) ∈ X × Y.

We will call X and Y isomorphic if there is a bijective map α : X −→ Y such that α(gx) = gα(x) for any g ∈ G and x ∈ X. Let SG denote the set of all isomorphism classes {X} of finite G-sets X.Inthe free abelian group Z[SG] we consider the subgroup Rel generated by all elements of the form

{X ∪ Y }−{X}−{Y } for any two finite G-sets X and Y.

We define the factor group

B(G) := Z[SG]/ Rel, and we let [X]∈B(G) denote the image of the isomorphism class {X}. In fact, the map

Z[SG]×Z[SG]−→Z[SG]

{X}, {Y } −→{X × Y } makes Z[SG] into a commutative ring in which the unit element is the isomorphism class of the G-set with one point. Because of (X1 ∪ X2) × Y = (X1 × Y)∪ (X2 × Y) 60 2 The CartanÐBrauer Triangle the subgroup Rel is an ideal in Z[SG]. We see that B(G) is a commutative ring. It is called the Burnside ring of G. Two elements x,y in a G-set X are called equivalent if there is a g ∈ G such that y = gx. This defines an equivalence relation on X. The equivalence classes are called G-orbits. They are of the form Gx ={gx : g ∈ G} for some x ∈ X. A nonempty G-set which consists of a single G-orbit is called simple (or transi- tive or a principal homogeneous space). The decomposition of an arbitrary G-set X into its G-orbits is the unique decomposition of X into simple G-sets. In particular, the only G-subsets of a simple G-set Y are Y and ∅.WeletSG denote the set of isomorphism classes of simple G-sets.

=∼ Lemma 2.4.1 Z[SG] −→ B(G).

Proof If X = Y1 ∪···∪Yn is the decomposition of X into its G-orbits then we put

π {X} := {Y1}+···+{Yn}.

This defines an endomorphism π of Z[SG] which is idempotent with im(π) = Z[SG]. It is rather clear that Rel ⊆ ker(π). Moreover, using the identities

[Y1]+[Y2]=[Y1 ∪ Y2], [Y1 ∪ Y2]+[Y3]=[Y1 ∪ Y2 ∪ Y3],...,

[Y1 ∪···∪Yn−1]+[Yn]=[X] we see that

[X]=[Y1]+···+[Yn] and hence that {X}−π({X}) ∈ Rel. As in the proof of Proposition 1.7.1 these three facts together imply  Z[SG]=Z[SG]⊕Rel . For any subgroup H ⊆ G the coset space G/H is a simple G-set with respect to G × G/H −→ G/H

g,g H −→ gg H.

Simple G-sets of this form are called standard G-sets.

Remark 2.4.2 Each simple G-set X is isomorphic to some standard G-set G/H .

Proof We fix a point x ∈ X.LetGx := {g ∈ G : gx = x} be the stabilizer of x in G. Then

G/Gx −→ X

gGx −→ gx is an isomorphism.  2.4 The Burnside Ring 61

It follows that the set SG is finite.

Lemma 2.4.3 Two standard G-sets G/H1 and G/H2 are isomorphic if and only if ∈ −1 = there is a g0 G such that g0 H1g0 H2. −1 = Proof If g0 H1g0 H2 then

G/H1 −→ G/H2

gH1 −→ gg0H2 is an isomorphism of G-sets. Vice versa, let

α: G/H1 −→ G/H2 be an isomorphism of G-sets. We have α(1H1) = g0H2 for some g0 ∈ G and then

g0H2 = α(1H1) = α(h1H1) = h1α(1H1) = h1g0H2 ∈ −1 ⊆ for any h1 H1. This implies g0 H1g0 H2. On the other hand −1 = −1 = −1 = −1 = −1 g0 H1 α (1H2) α (h2H2) h2α (1H2) h2g0 H1 ∈ −1 ⊆  for any h2 H2 which implies g0H2g0 H1.

Exercise 2.4.4 Let G/H1 and G/H2 be two standard G-sets; we then have    −1 [G/H1]·[G/H2]= G/H1 ∩ gH2g in B(G) g∈H1\G/H2 where H1\G/H2 denotes the space of double cosets H1gH2 in G.

Let F again be an arbitrary field. For any finite set X we have the finite- dimensional F -vector space   F [X]:= axx : ax ∈ F x∈X “with basis X”. Suppose that X is a finite G-set. Then the group G acts on F [X] by     := = g axx axgx ag−1xx. x∈X x∈X x∈X In this way F [X] becomes a finitely generated F [G]-module (called a permutation module). If α : X −→ Y is an isomorphism of finite G-sets then =∼ α˜ : F [X] −→ F [Y ]    −→ = axx axα(x) aα−1(y)y x∈X x∈X y∈Y 62 2 The CartanÐBrauer Triangle is an isomorphism of F [G]-modules. It follows that the map ∼ SG −→ MF [G]/ =   {X}−→ F [X] is well defined. We obviously have

F [X1 ∪ X2]=F [X1]⊕F [X2] for any two finite G-sets X1 and X2. Hence the above map respects the subgroups Rel in both sides and induces a group homomorphism

b: B(G) −→ RF (G)   [X]−→ F [X] .

Remark There is a third interesting Grothendieck group for the ring F [G] which is the factor group

AF (G) := Z[MF [G]]/ Rel⊕ with respect to the subgroup Rel⊕ generated by all elements of the form

{M ⊕ N}−{M}−{N} where M and N are arbitrary finitely generated F [G]-modules. We note that Rel⊕ ⊆ Rel. The above map b is the composite of the maps

pr B(G) −→ AF (G) −→ RF (G) [X]−→F [X] −→ F [X] .

Remark 2.4.5 For any two finite G-sets X1 and X2 we have ∼ F [X1 × X2] = F [X1]⊗F F [X2] as F [G]-modules.

Proof The vectors (x1,x2), resp. x1 ⊗ x2,forx1 ∈ X1 and x2 ∈ X2,formanF -basis of the left-, resp. right-, hand side. Hence there is a unique F -linear isomorphism mapping (x1,x2) to x1 ⊗ x2. Because of

g(x1,x2) = (gx1,gx2) −→ gx1 ⊗ gx2 = g(x1 ⊗ x2), for any g ∈ G, this map is an F [G]-module isomorphism. 

It follows that the map

b: B(G) −→ RF (G) 2.4 The Burnside Ring 63 is a ring homomorphism. Note that the unit element [G/G] in B(G) is mapped to the class [F ] of the trivial module F which is the unit element in RF (G).

Lemma 2.4.6 i. For any standard G-set G/H we have

[ ] = G b G/H indH (1)

where 1 on the right-hand side denotes the unit element of RF (H ). ii. For any finite G-set X we have   tr g; F [X] = {x ∈ X : gx = x} ∈ F for any g ∈ G.

Proof i. Let F be the trivial F [H]-module. It suffices to establish an isomorphism ∼ F [G/H ] = F [G]⊗F [H ] F.

For this purpose we consider the F -bilinear map

β: F [G]×F −→ F [G/H ]    agg,a −→ aaggH. g∈G g∈G

Because of β(gh,a) = ghH = gH = β(g,a) = β(g,ha) for any h ∈ H the map β is F [H]-balanced and therefore induces a well-defined F -linear map ˜ β: F [G]⊗F [H ] F −→ F [G/H ]    agg ⊗ a −→ aaggH. g∈G g∈G

∼ [G:H ] As discussed in the proof of Remark 2.3.2 we have F [G]⊗F [H ] F = F as F - vector spaces. Hence β˜ is a map between F -vector spaces of the same dimension. It obviously is surjective and therefore bijective. Finally the identity         ˜ ˜ β g agg ⊗ a = β agg g ⊗ a = aagg gH g∈G g∈G g∈G      ˜ = g aaggH = g β agg ⊗ a g∈G g∈G for any g ∈ G shows that β˜ is an isomorphism of F [G]-modules. 64 2 The CartanÐBrauer Triangle

g· ii. The matrix (ax,y)x,y of the F -linear map F [X] −→ F [X] with respect to the basis X is given by the equations  gy = ax,yx. x∈X But gy ∈ X and hence  1ifx = gy ax,y = 0 otherwise. It follows that       tr g; F [X] = ax,x = 1 = {x ∈ X : gx = x} ∈ F.  x∈X gx=x

Remark

1. The map b rarely is injective. Let G = S3 be the symmetric group on three letters. It has four conjugacy classes of subgroups. Using Lemma 2.4.1, Remark 2.4.2, ∼ 4 and Lemma 2.4.3 it therefore follows that B(S3) = Z . On the other hand, S3 has only three conjugacy classes of elements. Hence Proposition 1.7.1 and The- ∼ 3 orem 2.3.6.ii imply that RC(S3) = Z . 2. In general the map b is not surjective either. But there are many structural results about its cokernel. For example, the Artin induction theorem implies that

|G|·RQ(G) ⊆ im(b).

We therefore introduce the subring

PF (G) := im(b) ⊆ RF (G).

Let H be a family of subgroups of G with the property that if H ⊆ H is a subgroup of some H ∈ H then also H ∈ H. We introduce the subgroup B(G,H) ⊆ B(G) generated by all [G/H ] for H ∈ H as well as its image PF (G, H) ⊆ PF (G) under the map b.

Lemma 2.4.7 B(G,H) is an ideal in B(G), and hence PF (G, H) is an ideal in PF (G).

Proof We have to show that, for any H1 ∈ H and any subgroup H2 ⊆ G, the element [G/H1]·[G/H2] lies in B(G,H). This is immediately clear from Exercise 2.4.4. But a less detailed argument suffices. Obviously [G/H1]·[G/H2] is the sum of the × = −1 classes of the G-orbits in G/H1 G/H2.LetG(g1H1,g2H2) G(H1,g1 g2H2) ⊆ −1 be such a G-orbit. The stabilizer H G of the element (H1,g1 g2H2) is contained in H1 and therefore belongs to H. It follows that    

G(g1H1,g2H2) = G/H ∈ B(G,H).  2.4 The Burnside Ring 65

Definition i. Let  be a prime number. A finite group H is called -hyper-elementary if it contains a cyclic normal subgroup C such that  |C| and H/C is an -group. ii. A finite group is called hyper-elementary if it is -hyper-elementary for some prime number .

Exercise Let H be an -hyper-elementary group. Then: i. Any subgroup of H is -hyper-elementary; ii. let C ⊆ H be a cyclic normal subgroup as in the definition, and let L ⊆ H be any

-Sylow subgroup; then the map C × L −→ H sending (c, g) to cg is a bijection of sets.

Let Hhe denote the family of hyper-elementary subgroups of G. By the exercise Lemma 2.4.7 is applicable to Hhe.

Theorem 2.4.8 (Solomon) Suppose that F has characteristic zero; then

PF (G, Hhe) = PF (G).

Proof Because of Lemma 2.4.7 it suffices to show that the unit element 1 ∈ PF (G) already lies in PF (G, Hhe). According to Lemma 2.4.6.ii the characters     χF [X](g) = {x ∈ X : gx = x} ∈ Z, for any g ∈ G and any finite G-set X, have integral values. Hence we have the well-defined ring homomorphisms

tg: PF (G) −→ Z z −→ Tr(z)(g) for g ∈ G. On the one hand, by Corollary 2.3.7.i, they satisfy  ker(tg) = ker Tr |PF (G) ={0}. (2.4.1) g∈G

On the other hand, we claim that

tg PF (G, Hhe) = Z (2.4.2) holds true for any g ∈ G.Wefixag0 ∈ G in the following. Since the image H Z Z tg0 (PF (G, he)) is an additive subgroup of and hence is of the form n for some n ≥ 0 it suffices to find, for any prime number ,an-hyper-elementary subgroup H ⊆ G such that the integer 66 2 The CartanÐBrauer Triangle     [ ] = = { ∈ : = } tg0 F G/H χF [G/H ](g0) x G/H g0x x    −1  = gH ∈ G/H : g g0g ∈ H is not contained in Z.Wealsofix. The wanted -hyper-elementary subgroup H will be found in a chain of sub- groups

C ⊆g0⊆H ⊆ N with C being normal in N which is constructed as follows. Let n ≥ 1 be the order s of g0, and write n =  m with l m. The cyclic subgroup g0⊆G generated by g0 then is the direct product      = s × m g0 g0 g0  m :=  s  where g0 is an -group and C g0 is a cyclic group of order prime to . We define N := {g ∈ G : gCg−1 = C} to be the normalizer of C in G. It contains g0, of course. Finally, we choose H ⊆ N in such a way that H/C is an -Sylow subgroup of N/C which contains the -subgroup g0/C. By construction H is -hyper-elementary. In the next step we study the cardinality of the set   −1 {gH ∈ G/H : g0gH = gH}= gH ∈ G/H : g g0g ∈ H .

−1 −1 −1 −1 Suppose that g g0g ∈ H . Then g Cg ⊆ g g0g =g g0g⊆H . But, the two sides having coprime orders, the projection map g−1Cg −→ H/C has to be the trivial map. It follows that g−1Cg = C which means that g ∈ N. This shows that

{gH ∈ G/H : g0gH = gH}={gH ∈ N/H : g0gH = gH}.

The cardinality of the right-hand side is the number of g0-orbits in N/H which consist of one point only. We note that the subgroup C, being normal in N and contained in H , acts trivially on N/H. Hence the g0-orbits coincide with the orbits of the -group g0/C. But, quite generally, the cardinality of an orbit, being the index of the stabilizer of any point in the orbit, divides the order of the acting group. It follows that the cardinality of any g0-orbit in N/H is a power of .We conclude that     {gH ∈ N/H : g0gH = gH} ≡|N/H|=[N : H ] mod .

But by the choice of H we have  [N : H]. This establishes our claim. By (2.4.2) we now may choose an element zg ∈ PF (G, Hhe), for any g ∈ G, such that tg(zg) = 1. We then have    tg (zg − 1) = 0 for any g ∈ G, g ∈G 2.5 Clifford Theory 67 and (2.4.1)impliesthat  (zg − 1) = 0. g ∈G

Multiplying out the left-hand side and using that PF (G, Hhe) is additively and mul- tiplicatively closed easily shows that 1 ∈ PF (G, Hhe). 

2.5 Clifford Theory

As before F is an arbitrary field. We fix a normal subgroup N in our finite group G. Let W be an F [N]-module. It is given by a homomorphism of F -algebras

π: F [N]−→EndF (W).

For any g ∈ G we now define a new F [N]-module g∗(W) by the composite homo- morphism π F [N]−→ F [N] −→ EndF (W) h −→ ghg−1 or equivalently by

F [N]×g∗(W) −→ g∗(W) (h, w) −→ ghg−1w.

Remark 2.5.1 ∗ i. dimF g (W) = dimF W . ii. The map U → g∗(U) is a bijection between the set of F [N]-submodules of W and the set of F [N]-submodules of g∗(W). iii. W is simple if and only if g∗(W) is simple. ∗ ∗ = ∗ ∈ iv. g1 (g2 (W)) (g2g1) (W) for any g1,g2 G. v. Any F [N]-module homomorphism α : W1 −→ W2 also is a homomorphism of ∗ ∗ F [N]-modules α : g (W1) −→ g (W2).

Proof Trivial or straightforward. 

Suppose that g ∈ N. One checks that then

=∼ W −→ g∗(W) w −→ gw is an isomorphism of F [N]-modules. Together with Remark 2.5.1.iv/v this implies that 68 2 The CartanÐBrauer Triangle ∼ ∼ G/N × (MF [N]/ =) −→ MF [N]/ =   gN,{W} −→ g∗(W) ∼ is a well-defined action of the group G/N on the set MF [N]/ =. By Remark 2.5.1.iii this action respects the subset F[N]. For any {W} in F[N] we put     ∗ IG(W) := g ∈ G : g (W) ={W} .

As a consequence of Remark 2.5.1.iv this is a subgroup of G, which contains N of course.

Remark 2.5.2 Let V be an F [G]-module, and let g ∈ G be any element; then the map

set of all F [N]- set of all F [N]- − → submodules of V submodules of V W −→ gW is an inclusion preserving bijection; moreover, for any F [N]-submodule W ⊆ V we have: i. The map

=∼ g∗(W) −→ g−1W w −→ g−1w

is an isomorphism of F [N]-modules; ii. gW is a simple F [N]-module if and only if W is a simple F [N]-module; ∼ ∼ iii. if W1 = W2 are isomorphic F [N]-submodules of V then also gW1 = gW2 are isomorphic as F [N]-modules.

Proof For h ∈ N we have

h(gW) = g g−1hg W = gW since g−1hg ∈ N. Hence gW indeed is an F [N]-submodule, and the map in the assertion is well defined. It obviously is inclusion preserving. Its bijectivity is im- mediate from g−1(gW) = W = g(g−1W). The assertion ii. is a direct consequence. The map in i. clearly is an F -linear isomorphism. Because of

− − − g 1 ghg 1w = h g 1w for any h ∈ N and w ∈ W it is an F [N]-module isomorphism. The last assertion iii. follows from i. and Re- mark 2.5.1.v.  2.5 Clifford Theory 69

Theorem 2.5.3 (Clifford) Let V be a simple F [G]-module; we then have: i. V is semisimple as an F [N]-module; ii. let W ⊆ V be a simple F [N]-submodule, and let W˜ ⊆ V be the {W}-isotypic component; then ˜ a. W is a simple F [IG(W)]-module, and b. V =∼ IndG (W)˜ as F [G]-modules. IG(W) Proof Since V is of finite length as an F [N]-module we find a simple F [N]- submodule W ⊆ V . Then gW, for any g ∈ G, is another simple F [N]-submodule := by Remark 2.5.2.ii. Therefore V0 g∈G gW is, on the one hand, a semisimple F [N]-module by Proposition 1.1.4. On the other hand it is, by definition, a nonzero F [G]-submodule of V . Since V is simple we must have V0 = V , which proves the assertion i. As an F [N]-submodule the {W}-isotypic component W˜ of V is of the form W˜ = W ⊕···⊕W with simple F [N]-submodules W =∼ W . Let first g be 1 m i ∼ ∼ an element in IG(W).UsingRemark2.5.2.i/iii we obtain gWi = gW = W for any ˜ ˜ ˜ 1 ≤ i ≤ m. It follows that gWi ⊆ W for any 1 ≤ i ≤ m and hence gW ⊆ W .Wesee ˜ that W is an F [IG(W)]-submodule of V . For a general g ∈ G we conclude from Remark 2.5.2 that gW˜ is the {gW}-isotypic component of V . We certainly have  V = gW.˜

g∈G/IG(W) ˜ ˜ Two such submodules g1W and g2W , being isotypic components, either are equal ˜ = ˜ −1 ˜ = ˜ −1 =∼ or have zero intersection. If g1W g2W then g2 g1W W , hence g2 g1W W , −1 ∈ and therefore g2 g1 IG(W). We see that in fact  V = gW.˜ (2.5.1)

g∈G/IG(W)

The inclusion W˜ ⊆ V induces, by the first Frobenius reciprocity, the F [G]-module homomorphism G ˜ ∼ ˜ Ind (W)= F [G]⊗ [ ] W −→ V IG(W) F IG(W)    agg ⊗˜w −→ aggw.˜ g∈G g∈G Since V is simple it must be surjective. But both sides have the same dimension ˜ as F -vector spaces [G : IG(W)]·dimF W , the left-hand side by the argument in the proof of Remark 2.3.2 and the right-hand side by (2.5.1). Hence this map is an [ ]⊗ ˜ =∼ isomorphism which proves ii.b. Finally, since F G F [IG(W)] W V is a sim- ˜ ple F [G]-module it follows from Remark 2.3.3 that W is a simple F [IG(W)]- module. 

In the next section we will need the following particular consequence of this result. But first we point out that for an F [N]-module W of dimension dimF W = 1 70 2 The CartanÐBrauer Triangle the describing algebra homomorphism π is of the form

π: F [N]−→F.

The corresponding homomorphism for g∗(W) then is x → π(gxg−1). Since an endomorphism of a one-dimensional F -vector space is given by multiplication by a × scalar we have g ∈ IG(W) if and only if there is a scalar a ∈ F such that

− aπ(h)w = π ghg 1 aw for any h ∈ N and w ∈ W.

It follows that   −1 IG(W) = g ∈ G : π ghg = π(h) for any h ∈ N (2.5.2) if dimF W = 1.

Remark 2.5.4 Suppose that N is abelian and that F is a splitting field for N; then any simple F [N]-module W has dimension dimF W = 1.

Proof Let π : F [N]−→EndF (W) be the algebra homomorphism describing W . Since N is abelian we have im(π) ⊆ EndF [N](W). By our assumption on the field F the latter is equal to F . This means that any element in F [N] acts on W by multiplication by a scalar in F . Since W is simple this forces it to be one- dimensional. 

Proposition 2.5.5 Let H be an -hyper-elementary group with cyclic normal sub- group C such that  |C| and H/C is an -group, and let V be a simple F [H ]- module; we suppose that a. F is a splitting field for C, b. V does not contain the trivial F [C]-module, and c. the subgroup C0 := {c ∈ C : cg = gc for any g ∈ H } acts trivially on V ; then there exists a proper subgroup H H and an F [H ]-module V such that =∼ H [ ] V IndH (V ) as F H -modules.

Proof We pick any simple F [C]-submodule W ⊆ V . By applying Clifford’s Theo- rem 2.5.3 to the normal subgroup C and the module W it suffices to show that

IH (W) = H.

According to Remark 2.5.4 the module W is one-dimensional and given by an alge- bra homomorphism π : F [C]−→F , and by (2.5.2)wehave   −1 IH (W) = g ∈ H : π gcg = π(c) for any c ∈ C .

The assumption c. means that

C0 ⊆ C1 := ker(π|C). 2.6 Brauer’s Induction Theorem 71

We immediately note that any subgroup of the cyclic normal subgroup C also is normal in H . By assumption b. we find an element c2 ∈ C \ C1 so that π(c2) = 1. Let L ⊆ H be an -Sylow subgroup. We claim that we find an element g0 ∈ L such −1 = ∈ that π(g0c2g0 ) π(c2). Then g0 IH (W) which establishes what we wanted. We point out that, since H = C × L as sets, we have

C0 ={c ∈ C : cg = gc for any g ∈ L}.

−1 Arguing by contradiction we assume that π(gc2g ) = π(c2) for any g ∈ L. Then

−1 −1 gc2C1g = gc2g C1 = c2C1 for any g ∈ L.

This means that we may consider c2C1 as an L-set with respect to the conjugation action. Since L is an -group the cardinality of any L-orbit in c2C1 is a power of . On the other hand we have  |C1|=|c2C1|. There must therefore exist an element c0 ∈ c2C1 such that −1 gc0g = c0 for any g ∈ L

(i.e. an L-orbit of cardinality one). We conclude that c0 ∈ C0 ⊆ C1 and hence c2C1 = c0C1 = C1. This is in contradiction to c2 ∈ C1. 

2.6 Brauer’s Induction Theorem

In this section F is a field of characteristic zero, and G continues to be any finite group.

Definition i. Let  be a prime number. A finite group H is called -elementary if it is a direct product H = C × L of a cyclic group C and an -group L. ii. A finite group is called elementary if it is -elementary for some prime number .

Exercise i. If H is -elementary then H = C × L is the direct product of a cyclic group C of order prime to  and an -group L. ii. Any -elementary group is -hyper-elementary. iii. Any subgroup of an -elementary group is -elementary.

Let He denote the family of elementary subgroups of G.

Theorem 2.6.1 (Brauer) Suppose that F is a splitting field for every subgroup of G; then  G = indH RF (H ) RF (G). H ∈He 72 2 The CartanÐBrauer Triangle

G Proof By Corollary 2.3.5 each indH (RF (H )) is an ideal in the ring RF (G). Hence the left-hand side of the asserted identity is an ideal in the right-hand side. To obtain equality we therefore need only to show that the unit element 1G ∈ RF (G) lies in the left-hand side. According to Solomon’s Theorem 2.4.8 together with Lemma 2.4.6.i we have      ∈ Z [ ] = Z [ ] = Z G 1G F G/H b G/H indH (1H ) H ∈Hhe H ∈Hhe H ∈Hhe  ⊆ G indH RF (H ) . H ∈Hhe By the transitivity of induction this reduces us to the case that G is -hyper- elementary for some prime number . We now proceed by induction with respect to the order of G and assume that our assertion holds for all proper subgroups H G. We also may assume, of course, that G is not elementary. Using the transitivity of induction again it then suffices to show that  ∈ G 1G indH RF H . H G

Let C ⊆ G be the cyclic normal subgroup of order prime to  such that G/C is an -group. We fix an -Sylow subgroup L ⊆ G. Then G = C × L as sets. In C we have the (cyclic) subgroup

C0 := {c ∈ C : cg = gc for any g ∈ G}. Then

H0 := C0 × L is an -elementary subgroup of G. Since G is not elementary we must have H0 G. We consider the induction IndG (F ) of the trivial F [H ]-module F . By semisim- H0 0 plicity it decomposes into a direct sum

IndG (F ) = V ⊕ V ⊕···⊕V H0 0 1 r of simple F [G]-modules V . We recall that IndG (F ) is the F -vector space of all i H0 functions φ : G/H0 −→ F with the G-action given by

g −1 φ g H0 = φ g g H0 .

This G-action fixes a function φ if and only if φ(gg H0) = φ(g H0) for any g,g ∈ G, i.e. if and only if φ is constant. It follows that the one-dimensional subspace of constant functions is the only simple F [G]-submodule in IndG (F ) which is H0 isomorphic to the trivial module. We may assume that V0 is this trivial submodule. In RF (G) we then have the equation 1 = indG (1 ) −[V ]−···−[V ]. G H0 H0 1 r 2.6 Brauer’s Induction Theorem 73

This reduces us further to showing that, for any 1 ≤ i ≤ r,wehave

[V ]∈indG R (H ) i Hi F i for some proper subgroup Hi G. This will be achieved by applying the criterion in Proposition 2.5.5. By our assumption on F it remains to verify the conditions b. and c. in that proposition for each V1,...,Vr . Since C0 is central in G and is contained in H it acts trivially on IndG (F ) and a fortiori on any V . This is condition c. For 0 H0 i b. we note that CH0 = G and C ∩ H0 = C0. Hence the inclusion C ⊆ G induces a bijection C/C0 −→ G/H0. It follows that the map

=∼ IndG (F ) −→ IndC (F ) H0 C0

φ −→ φ|(C/C0) is an isomorphism of F [C]-modules. It maps constant functions to constant functions and hence the unique trivial F [G]-submodule V0 to the unique trivial F [C]-submodule. Therefore Vi ,for1≤ i ≤ r, cannot contain any trivial F [C]- submodule. 

Lemma 2.6.2 Let H be an elementary group, and let N0 ⊆ H be a normal sub- group such that H/N0 is not abelian; then there exists a normal subgroup N0 ⊆ N ⊆ H such that N/N0 is abelian but is not contained in the center Z(H/N0) of H/N0.

Proof With H also H/N0 is elementary (if H = C × L with C and L having co- ∼ prime orders then H/N0 = C/C ∩ N0 × L/L ∩ N0). We therefore may assume without loss of generality that N0 ={1}. Step 1: We assume that H is an -group for some prime number . By assumption we have Z(H) = H so that H/Z(H) is an -group ={1}. We pick a cyclic normal subgroup {1} =N =g⊆H/Z(H). Let Z(H) ⊆ N ⊆ H be the normal subgroup such that N/Z(H)= N and let g ∈ N be a preimage of g. Clearly N =Z(H),g is abelian. But Z(H) N since g = 1. Step 2: In general let H = C ×L where C is cyclic and L is an -group. With H also L is not abelian. Applying Step 1 to L we find a normal abelian subgroup NL ⊆ L such that NL Z(L). Then N := C × NL is a normal abelian subgroup of H such that N Z(H) = C × Z(L). 

Lemma 2.6.3 Let H be an elementary group, and let W be a simple F [H ]-module; we suppose that F is a splitting field for all subgroups of H ; then there exists a subgroup H ⊆ H and a one-dimensional F [H ]-module W such that

=∼ H W IndH W as F [H ]-modules. 74 2 The CartanÐBrauer Triangle

Proof We choose H ⊆ H to be a minimal subgroup (possibly equal to H ) such [ ] =∼ H that there exists an F H -module W with W IndH (W ), and we observe that W necessarily is a simple F [H ]-module by Remark 2.3.3.Let

π : H −→ EndF W

be the corresponding algebra homomorphism, and put N0 := ker(π ). We claim that H /N0 is abelian. Suppose otherwise. Then, by Lemma 2.6.2, there exists a normal subgroup N0 ⊆ N ⊆ H such that N/N0 is abelian but is not contained in ¯ Z(H /N0).LetW ⊆ W be a simple F [N]-submodule. By Clifford’s Theorem 2.5.3 we have ∼ H ¯˜ W = Ind ¯ (W) IH (W) ˜ where W¯ denotes the {W¯ }-isotypic component of W . Transitivity of induction im- plies

=∼ H H ¯˜ =∼ H ¯˜ W IndH Ind ¯ (W) Ind ¯ (W). IH (W) IH (W) ¯ By the minimality of H we therefore must have IH (W)= H which means that ˜ W = W¯ is {W¯ }-isotypic. ¯ On the other hand, W is an F [H /N0]-module. Hence W is a simple F [N/N0]- module for the abelian group N/N0. Remark 2.5.4 then implies (note that ¯ = ¯ = ¯ EndF [N/N0](W) EndF [N](W) F ) that W is one-dimensional given by an al- gebra homomorphism

χ: F [N/N0]−→F. It follows that any h ∈ N acts on the {W¯ }-isotypic module W by multiplication by the scalar χ(hN0). In other words the injective homomorphism

H /N0 −→ EndF W

hN0 −→ π (h) satisfies

π (h) = χ(hN0) · idW for any h ∈ N.

But χ(hN0) · idW lies in the center of EndF (W ). The injectivity of the homomor- phism therefore implies that N/N0 lies in the center of H /N0. This is a contradic- tion.

We thus have established that H /N0 is abelian. Applying Remark 2.5.4 to W viewed as a simple F [H /N0]-module we conclude that W is one-dimensional. 

Theorem 2.6.4 (Brauer) Suppose that F is a splitting field for any subgroup of G, and let x ∈ RF (G) be any element; then there exist integers m1,...,mr , elementary 2.7 Splitting Fields 75 subgroups H1,...,Hr , and one-dimensional F [Hi]-modules Wi such that r x = m indG [W ] . i Hi i i=1

Proof Combine Theorem 2.6.1, Proposition 1.7.1, Lemma 2.6.3, and the transitivity of induction. 

2.7 Splitting Fields

Again F is a field of characteristic zero.

Lemma 2.7.1 Let E/F be any extension field, and let V and W be two finitely generated F [G]-modules; we then have

HomE[G](E ⊗F V,E⊗F W)= E ⊗F HomF [G](V, W).

Proof First of all we observe, by comparing dimensions, that

HomE(E ⊗F V,E⊗F W)= E ⊗F HomF (V, W) holds true. We now consider U := HomF (V, W) as an F [G]-module via

G × U −→ U

(g, f ) −→ gf := gf g−1 .

G g Then HomF [G](V, W) = U := {f ∈ U : f = f for any g ∈ G} is the {F }- isotypic component of U for the trivial F [G]-module F . Correspondingly we obtain

G G HomE[G](E ⊗F V,E⊗F W)= HomE(E ⊗F V,E⊗F W) = (E ⊗F U) .

This reduces us to proving that

G G (E ⊗F U) = E ⊗F U for any F [G]-module U. The element

1  ε := g ∈ F [G]⊆E[G] G |G| g∈G is an idempotent with the property that

G U = εG · U, 76 2 The CartanÐBrauer Triangle and hence

G G (E ⊗F U) = εG · (E ⊗F U)= E ⊗F εG · U = E ⊗F U .



Theorem 2.7.2 (Brauer) Let e be the exponent of G, and suppose that F contains a primitive eth root of unity; then F is a splitting field for any subgroup of G.

Proof We fix an algebraic closure F¯ of F . Step 1: We show that, for any finitely generated F¯[G]-module V¯ , there is an F [G]-module V such that ¯ ∼ ¯ ¯ V = F ⊗F V as F [G]-modules.

According to Brauer’s Theorem 2.6.4 we find integers m1,...,mr , subgroups ¯ ¯ H1,...,Hr of G, and one-dimensional F [Hi]-modules Wi such that

r   ¯ ¯ ¯ [V ]= m F [G]⊗¯ W i F [Hi ] i i=1      −  ¯ ¯ mi ¯ ¯ mi = F [G]⊗¯ W − F [G]⊗¯ W F [Hi ] i F [Hi ] i mi >0 mi <0       − ¯ ¯ mi ¯ ¯ mi = F [G]⊗¯ W − F [G]⊗¯ W . F [Hi ] i F [Hi ] i mi >0 mi <0 ¯ ¯ ¯ ¯ Let πi : F [Hi]−→F denote the F -algebra homomorphism describing Wi .Wehave e e πi(h) = πi(h ) = πi(1) = 1 for any h ∈ Hi . Our assumption on F therefore implies that πi(F [Hi ]) ⊆ F . Hence the restriction πi|F [Hi] describes a one-dimensional F [Hi]-module Wi such that ¯ ∼ ¯ ¯ Wi = F ⊗F Wi as F [Hi]-modules.

We define the F [G]-modules   − := [ ]⊗ mi := [ ]⊗ mi V+ F G F [Hi ] Wi and V− F G F [Hi ] Wi . mi >0 mi <0

Then   ¯ ⊗ = ¯ ⊗ [ ]⊗ mi = ¯[ ]⊗ mi F F V+ F F F G F [Hi ] Wi F G F [Hi ] Wi mi >0 mi >0  ¯ ¯ mi = F [G]⊗¯ F [H ]⊗ [ ] W F [Hi ] i F Hi i mi >0 2.7 Splitting Fields 77  ¯ ¯ mi = F [G]⊗¯ F ⊗ W F [Hi ] F i mi >0  ∼ ¯ ¯ mi = F [G]⊗¯ W F [Hi ] i mi >0 and similarly  − ¯ ∼ ¯ ¯ mi F ⊗ V− = F [G]⊗¯ W . F F [Hi ] i mi <0 It follows that ¯ ¯ ¯ [V ]=[F ⊗F V+]−[F ⊗F V−] or, equivalently, that   ¯ ¯ ¯ V ⊕ (F ⊗F V−) =[F ⊗F V+].

Using Corollary 2.3.7.i/iii we deduce that ¯ ¯ ∼ ¯ ¯ V ⊕ (F ⊗F V−) = F ⊗F V+ as F [G]-modules.

If V− is nonzero then V− = U ⊕ V− is a direct sum of F [G]-modules where U is simple. On the other hand let V+ = U1 ⊕ ··· ⊕ Um be a decomposition ¯ ¯ into simple F [G]-modules. Then F ⊗F U is a direct summand of F ⊗F V− and ¯ hence is isomorphic to a direct summand of F ⊗F V+. We therefore must have ¯ ¯ Hom ¯[ ](F ⊗F U,F ⊗F Uj ) = {0} for some 1 ≤ j ≤ m. Lemma 2.7.1 implies F G ∼ that HomF [G](U, Uj ) = {0}. Hence U = Uj as F [G]-modules. We conclude that =∼ ⊕ := V+ U V+ with V+ i =j Ui , and we obtain

¯ ¯ ¯ ∼ ¯ ¯ V ⊕ F ⊗F V− ⊕ (F ⊗F U)= F ⊗F V+ ⊕ (F ⊗F U).

The Jordan–Hölder Proposition 1.1.2 then implies that

¯ ¯ ∼ ¯ V ⊕ F ⊗F V− = F ⊗F V+.

By repeating this argument we arrive after finitely many steps at an F [G]-module V such that ¯ ∼ ¯ V = F ⊗F V. ¯ ¯ ¯ Step 2: Let now V beasimpleF [G]-module, and let F ⊗F V = V1 ⊕···⊕Vm be ¯ a decomposition into simple F [G]-modules. By Step 1 we find F [G]-modules Vi such that ¯ ∼ ¯ Vi = F ⊗F Vi.

For any 1 ≤ i ≤ m,theF [G]-module Vi necessarily is simple, and we have ¯ ¯ Hom ¯[ ](F ⊗F Vi, F ⊗F V) = {0}. Hence HomF [G](Vi,V) = {0} by Lemma 2.7.1. F G ∼ It follows that Vi = V for any 1 ≤ i ≤ m. By comparing dimensions we conclude 78 2 The CartanÐBrauer Triangle

¯ ¯ that m = 1. This means that F ⊗F V is a simple F [G]-module. Using Lemma 2.7.1 again we obtain ¯ ⊗ = ¯ ⊗ = ¯ F F EndF [G](V ) EndF¯ [G](F F V) F.

Hence EndF [G](V ) = F must be one-dimensional. This shows that F is a splitting field for G. Step 3: Let H ⊆ G be any subgroup with exponent eH . Since eH divides e the field F a fortiori contains a primitive eH th root of unity. Hence F also is a splitting field for H . 

2.8 Properties of the Cartan–Brauer Triangle

We go back to the setting from the beginning of this chapter: k is an algebraically closed field of characteristic p>0, R is a (0,p)-ring for k with maximal ideal mR = RπR, and K denotes the field of fractions of R. The ring R will be called splitting for our finite group G if K contains a primitive eth root of unity ζ where e is the exponent of G. By Theorem 2.7.2 the field K,in this case, is a splitting field for any subgroup of G. This additional condition can easily be achieved by defining K := K(ζ) and R := {a ∈ K : NormK /K (a) ∈ R}; then R is a (0,p)-ring for k which is splitting for G. It is our goal in this section to establish the deeper properties of the CartanÐ Brauer triangle

dG RK (G) Rk(G)

eG cG

K0(k[G]).

Lemma 2.8.1 For any subgroup H ⊆ G the diagram

dH RK (H ) Rk(H )

G G indH indH dG RK (G) Rk(G) is commutative.

Proof Let W be a finitely generated K[H]-module. We choose a lattice L ⊆ W which is H -invariant. Then   G [ ] = G [ ] = [ ]⊗ indH dH W indH L/πRL k G k[H ] (L/πRL) . 2.8 Properties of the CartanÐBrauer Triangle 79

∼ [G:H ] ∼ Moreover, R[G]⊗R[H ] L = L is a G-invariant lattice in K[G]⊗K[H ] W = W [G:H ] (compare the proof of Remark 2.3.2). Hence   G [ ] = [ ]⊗ dG indH W dG K G K[H ] W   = R[G]⊗R[H ] L /πR R[G]⊗R[H ] L   = k[G]⊗k[H ] (L/πRL) . 

Lemma 2.8.2 We have  = G Rk(G) indH Rk(H ) . H ∈He

G Proof Since the indH (Rk(H )) are ideals in Rk(G) it suffices to show that the unit element 1k[G] ∈ Rk(G) lies in the right-hand side. We choose R to be splitting for G. By Brauer’s induction Theorem 2.6.1 we have  ∈ G 1K[G] indH RK (H ) H ∈He where 1K[G] is the unit element in RK (G). Using Lemma 2.8.1 we obtain   ∈ G = G dG(1K[G]) dG indH RK (H ) indH dH RK (H ) H ∈He H ∈He  ⊆ G indH Rk(H ) . H ∈He

It is trivial to see that dG(1K[G]) = 1k[G]. 

Theorem 2.8.3 The decomposition homomorphism

dG: RK (G) −→ Rk(G) is surjective.

Proof By Lemma 2.8.1 we have   ⊇ G = G dG RK (G) dG indH RK (H ) indH dH RK (H ) . H ∈He H ∈He

Because of Lemma 2.8.2 it therefore suffices to show that dH (RK (H )) = Rk(H ) for any H ∈ He. This means we are reduced to proving our assertion in the case where the group G is elementary. Then G = H × P is the direct product of a group H of order prime to p and a p-group P . By Proposition 1.7.1 it suffices to show that the class [W]∈Rk(G), for any simple k[G]-module W , lies in the image of dG.Viewed 80 2 The CartanÐBrauer Triangle as a k[P ]-module W must contain the trivial k[P ]-module by Proposition 2.2.7.We deduce that W P := {w ∈ W : gw = w for any g ∈ P } ={0}. Since P is a normal subgroup of G the k[P ]-submodule W P in fact is a k[G]- submodule of W .ButW is simple. Hence W P = W which means that k[G] acts on W through the projection map k[G]−→k[H]. According to Corollary 2.2.6 we find a simple K[H]-module V together with a G-invariant lattice L ⊆ V such ∼ that L/πRL = W as k[H]-modules. Viewing V as a K[G]-module through the projection map K[G]−→K[H] we obtain [V ]∈RK (G) and dG([V ]) =[W]. 

Theorem 2.8.4 Let pm be the largest power of p which divides the order of G; the Cartan homomorphism cG : K0(k[G]) −→ Rk(G) is injective, its cokernel is finite, m and p Rk(G) ⊆ im(cG).

m Proof Step 1: We show that p Rk(G) ⊆ im(cG) holds true. It is trivial from the definition of the Cartan homomorphism that, for any subgroup H ⊆ G, the diagram

cH K0(k[H]) Rk(H )

[ ]−→[ [ ]⊗ [ ] ] G P k G k H P indH cG K0(k[G]) Rk(G) is commutative. It follows that

G ⊆ indH im(cH ) im(cG). Lemma 2.8.2 therefore reduces us to the case that G is an elementary group. Let W be any simple k[G]-module. With the notations of the proof of Theorem 2.8.3 we have seen there that k[G] acts on W through the projection map k[G]−→k[H ]. Viewed as a k[H ]-module W is projective by Remark 1.7.3. Hence k[G]⊗k[H ] W is a finitely generated projective k[G]-module. We claim that   cG k[G]⊗k[H ] W =|G/H |·[W] holds true. Using the above commutative diagram as well as Proposition 2.3.4 we obtain     [ ]⊗ = G [ ] = [ ]⊗ ·[ ] cG k G k[H ] W indH W k G k[H ] k W .

In order to analyze the k[G]-module k[G]⊗k[H ] k let h, h ∈ H , g ∈ P , and a ∈ k. Then

h gh ⊗ a = ghh ⊗ a = g ⊗ a = gh ⊗ a.

This shows that H acts trivially on k[G]⊗k[H ] k. In other words, k[G] acts on k[G]⊗k[H ] k through the projection map k[G]−→k[P ]. It then follows from Proposition 2.2.7 that all simple subquotients in a composition series of the k[G]- 2.8 Properties of the CartanÐBrauer Triangle 81 module k[G]⊗k[H ] k are trivial k[G]-modules. We conclude that   k[G]⊗k[H ] k = dimk k[G]⊗k[H ] k · 1 =|G/H |·1

(where 1 ∈ Rk(G) is the unit element). ∼ r Step 2: We know from Proposition 1.7.1 that, as an abelian group, Rk(G) = Z for some r ≥ 1. It therefore follows from Step 1 that Rk(G)/ im(cG) is isomorphic to a factor group of the finite group Zr /pmZr . Step 3: It is a consequence of Proposition 1.7.4 that K0(k[G]) and Rk(G) are isomorphic to Zr for the same integer r ≥ 1. Hence

id ⊗cG: Q ⊗Z K0 k[G] −→ Q ⊗Z Rk(G) is a linear map between two Q-vector spaces of the same finite dimension r.Its injectivity is equivalent to its surjectivity. Let a ∈ Q and x ∈ Rk(G).ByStep1we m find an element y ∈ K0(k[G]) such that cG(y) = p x. Then   a a (id ⊗c ) ⊗ y = ⊗ pmx = a ⊗ x. G pm pm

This shows that id ⊗cG and consequently cG are injective. 

In order to discuss the third homomorphism eG we first introduce two bilinear forms. We start from the maps ∼ ∼ (MK[G]/ =) × (MK[G]/ =) −→ Z

{V }, {W} −→ dimK HomK[G](V, W) and ∼ ∼ (Mk[G]/ =) × (Mk[G]/ =) −→ Z

{P }, {V } −→ dimk Homk[G](P, V ).

They extend to Z-bilinear maps

Z[MK[G]]×Z[MK[G]]−→Z and

Z[Mk[G]]×Z[Mk[G]]−→Z.

Since K[G] is semisimple we have, for any exact sequence 0 → V1 → V → V2 → 0 ∼ in MK[G], that V = V1 ⊕ V2 and hence that

dimK HomK[G](V, W) − dimK HomK[G](V1,W)− dimK HomK[G](V2,W)= 0. 82 2 The CartanÐBrauer Triangle

The corresponding fact in the “variable” W holds as well, of course. The first map therefore induces a well-defined Z-bilinear form

 , K[G]: RK (G) × RK (G) −→ Z

[V ], [W] −→ dimK HomK[G](V, W).

Even though k[G] might not be semisimple any exact sequence 0 → P1 → P → ∼ P2 → 0inMk[G] still satisfies P = P1 ⊕ P2 as a consequence of Lemma 1.6.2.ii. Hence we again have

dimk Homk[G](P, V ) − dimk Homk[G](P1,V)− dimk Homk[G](P2,V)= 0 for any V in Mk[G]. Furthermore, for any P in Mk[G] and any exact sequence 0 → V1 → V → V2 → 0inMk[G] we have, by the definition of projective modules, the exact sequence

0 −→ Homk[G](P, V1) −→ Homk[G](P, V ) −→ Homk[G](P, V2) −→ 0.

Hence once more

dimk Homk[G](P, V ) − dimk Homk[G](P, V1) − dimk Homk[G](P, V2) = 0.

This shows that the second map also induces a well-defined Z-bilinear form

 , k[G]: K0 k[G] × Rk(G) −→ Z

[P ], [V ] −→ dimk Homk[G](P, V ).

If {V1},...,{Vr } are the isomorphism classes of the simple K[G]-modules then [V1],...,[Vr ] is a Z-basis of RK (G) by Proposition 1.7.1.Wehave    dimK EndK[G](Vi) if i = j, [Vi], [Vj ] = K[G] 0ifi = j.

In particular, if K is a splitting field for G then    1ifi = j, [Vi], [Vj ] = K[G] 0ifi = j.

Let {P1},...,{Pt } be the isomorphism classes of finitely generated indecomposable projective k[G]-modules. By Proposition 1.7.4.iii the [P1],...,[Pt ] form a Z-basis of K0(k[G]), and the [P1/ Jac(k[G])P1],...,[Pt / Jac(k[G])Pt ] form a Z-basis of Rk(G) by Proposition 1.7.4.i. We have 2.8 Properties of the CartanÐBrauer Triangle 83

Homk[G] Pi,Pj / Jac k[G] Pj = Homk[G] Pi/ Jac k[G] Pi,Pj / Jac k[G] Pj  [ ] = = Endk[G](Pi/ Jac(k G )Pi ) if i j, {0} if i = j  = = k if i j, {0} if i = j, where the latter identity comes from the fact that the algebraically closed field k is a splitting field for G. Hence     1ifi = j, [Pi], Pj / Jac k[G] Pj = k[G] 0ifi = j.

Exercise 2.8.5 i. If K is a splitting field for G then the map

=∼ RK (G) −→ HomZ RK (G), Z

x −→x,.K[G]

is an isomorphism of abelian groups. ii. The maps

=∼ =∼ K0 k[G] −→ HomZ Rk(G), Z and Rk(G) −→ HomZ K0 k[G] , Z

y −→y,.k[G] z −→ .,zk[G]

are isomorphisms of abelian groups.

Lemma 2.8.6 We have     = y,dG(x) k[G] eG(y), x K[G] for any y ∈ K0(k[G]) and x ∈ RK (G).

Proof It suffices to consider elements of the form y =[P/πRP ] for some finitely generated projective R[G]-module P (see Proposition 2.1.1) and x =[V ] for some finitely generated K[G]-module V .WepickaG-invariant lattice L ⊆ V . The as- serted identity then reads

dimk Homk[G](P /πRP,L/πRL) = dimK HomK[G](K ⊗R P,V). 84 2 The CartanÐBrauer Triangle

We have

Homk[G](P /πRP,L/πRL) = HomR[G](P, L/πRL)

= HomR[G](P, L)/ HomR[G](P, πRL)

= HomR[G](P, L)/πR HomR[G](P, L)

= k ⊗R HomR[G](P, L) (2.8.1) where the second identity comes from the projectivity of P as an R[G]-module. On the other hand

HomK[G](K ⊗R P,V)= HomR[G](P, V )    = −i HomR[G] P, πR L i≥0  = −i HomR[G] P,πR L i≥0  = −i πR HomR[G](P, L) i≥0

= K ⊗R HomR[G](P, L). (2.8.2)

For the third identity one has to observe that, since P is finitely generated as an −→ = −i R-module, any R-module homomorphism P V i≥0 πR L has to have its −i image inside πR L for some sufficiently large i. Both, P being a direct summand of some R[G]m (Remark 2.1.2) and L by defini- tion are free R-modules. Hence HomR(P, L) is a finitely generated free R-module. The ring R being noetherian the R-submodule HomR[G](P, L) is finitely generated ∼ s as well. Lemma 2.2.1.i then implies that HomR[G](P, L) = R is a free R-module. We now deduce from (2.8.1) and (2.8.2) that

∼ s ∼ s Homk[G](P /πRP,L/πRL) = k and HomK[G](K ⊗R P,V)= K , respectively. 

Theorem 2.8.7 The homomorphism eG : K0(k[G]) −→ RK (G) is injective and its image is a direct summand of RK (G).

Proof Step 1: We assume that R is splitting for G. By Theorem 2.8.3 the map dG : RK (G) −→ Rk(G) is surjective. Since, by Proposition 1.7.1, Rk(G) is a free abelian group we find a homomorphism s : Rk(G) −→ RK (G) such that dG ◦s = id. It follows that

Hom(s, Z) ◦ Hom(dG, Z) = Hom(dG ◦ s,Z) = Hom(id, Z) = id . 2.8 Properties of the CartanÐBrauer Triangle 85

Hence the map

Hom(dG, Z) : HomZ Rk(G), Z −→ HomZ RK (G), Z is injective and

HomZ RK (G), Z = im Hom(dG, Z) ⊕ ker Hom(s, Z) .

But because of Lemma 2.8.6 the map Hom(dG, Z) corresponds under the isomor- phisms in Exercise 2.8.5 to the homomorphism eG. Step 2: For general R we use, as described at the beginning of this section a larger (0,p)-ring R for k which contains R and is splitting for G.LetK denote the field of fractions of R .ItP is a finitely generated projective R[G]-module then

R ⊗R P = R [G]⊗R[G] P is a finitely generated projective R [G]-module such that

R ⊗R P /πR R ⊗R P = R /πR R ⊗R P = (R/πRR) ⊗R P

= P/πRP

(recall that k = R/πRR = R /πR R ). This shows that the diagram

κ RK (G) K0(R[G]) ρ

[V ]−→[K ⊗K V ] [P ]−→[R ⊗RP ] K0(k[G])

ρ κ RK (G) K0(R [G]) is commutative. Hence

eG Tr K0(k[G]) RK (G) Cl(G, K)

⊆ [V ]−→[K ⊗K V ] eG

RK (G) Cl(G, K ) Tr is commutative. The oblique arrow is injective by Step 1 and so then is the upper left horizontal arrow. The two right horizontal arrows are injective by Corollary 2.3.7.i. This implies that the middle vertical arrow is injective and therefore induces an injective homomorphism

RK (G)/ im(eG) −→ RK (G)/ im(eG). 86 2 The CartanÐBrauer Triangle

By Step 1 the target RK (G)/ im(eG) is isomorphic to a direct summand of the free abelian group RK (G). It follows that RK (G)/ im(eG) is isomorphic to a subgroup in a finitely generated free abelian group and hence is a free abelian group by the elementary divisor theorem. We conclude that ∼ RK (G) = im(eG) ⊕ RK (G)/ im(eG). 

We choose R to be splitting for G, and we fix the Z-bases of the three involved Grothendieck groups as described before Exercise 2.8.5.LetE, D, and C denote the matrices which describe the homomorphisms eG, dG, and cG, respectively, with respect to these bases. We, of course, have

DE = C.

Lemma 2.8.6 says that D is the transpose of E. It follows that the quadratic Cartan matrix C of k[G] is symmetric. Chapter 3 The Brauer Character

As in the last chapter we fix an algebraically closed field k of characteristic p>0, and we let G be a finite group. We also fix a (0,p)-ring R for k which is splitting for G.LetmR = RπR denote the maximal ideal and K the field of fractions of R.

3.1 Definitions

As in the semisimple case we let Cl(G, k) denote the k-vector space of all k-valued class functions on G, i.e. functions on G which are constant on conjugacy classes. For any V ∈ Mk[G] we introduce its k-character

χV : G −→ k g −→ tr(g; V),

α β which is a function in Cl(G, k).Let0→ V1 −→ V −→ V2 → 0 be an exact sequence g in Mk[G].Fori = 1, 2weletAi be the matrix of Vi −→ Vi with respect to some k-basis e(i),...,e(i). We put e := α(e(1)) for 1 ≤ j ≤ d and we choose e ∈ V , 1 di j j 1 j ≤ + = (2) −→g for d1

Since the trace is the sum of the diagonal entries of the respective matrix we see that

= + ∈ χV (g) χV1 (g) χV2 (g) for any g G.

It follows that

P. Schneider, Modular Representation Theory of Finite Groups, 87 DOI 10.1007/978-1-4471-4832-6_3, © Springer-Verlag London 2013 88 3 The Brauer Character

Tr: Rk(G) −→ Cl(G, k)

[V ]−→χV is a well-defined homomorphism.  Proposition 3.1.1 The k-characters χV ∈ Cl(G, k), for {V }∈k[G], are k-linearly independent.  ˆ Proof Let k[G]={{V1},...,{Vr }} = A where A := k[G]/ Jac(k[G]). Since k[G] is artinian A is semisimple (cf. Proposition 1.1.2.vi). By Wedderburn’s structure theory of semisimple k-algebras we have r = A EndDi (Vi) i=1 where Di := Endk[G](Vi). Since k is algebraically closed we, in fact, have Di = k. For any 1 ≤ i ≤ r we pick an element αi ∈ Endk(Vi) with tr(αi) = 1. By the above product decomposition of A we then find elements ai ∈ k[G],for1≤ i ≤ r, such that 

ai αi if i = j, (Vj −→ Vj ) = 0ifi = j.  : [ ]−→ If we extend each χVj by k-linearity to a k-linear form χVj k G k then  1ifi = j, χV (ai) = tr(ai; Vj ) = j 0ifi = j.  r = ∈ We now suppose that j=1 cj χVj 0inCl(G, k) with cj k. Then also r  = [ ] = r  = j=1 cj χVj 0inHomk(k G ,k) and hence 0 j=1 cj χVj (ai) ci for any 1 ≤ i ≤ r. 

Since, by Proposition 1.7.1, the vectors 1 ⊗[V ],for{V }∈k[G], form a basis of the k-vector space k ⊗Z Rk(G) it follows that

Tr: k ⊗Z Rk(G) −→ Cl(G, k)

c ⊗[V ]−→cχV is an injective k-linear map. But the canonical map Rk(G) −→ k ⊗Z Rk(G) is not in- jective. Therefore the k-character χV does not determine the isomorphism class {V }. This is the reason that in the present setting the k-characters are of limited use.

Definition An element g ∈ G is called p-regular, resp. p-unipotent, if the order of g is prime to p, resp. is a power of p.

Lemma 3.1.2 For any g ∈ G there exist uniquely determined elements greg and guni in G such that 3.1 Definitions 89

− greg is p-regular, guni is p-unipotent, and − g = gregguni = gunigreg; moreover, greg and guni are powers of g.

Proof Let psm with p m be the order of g. We choose integers a and b such that s aps bm ap + bm = 1, and we define greg := g and guni := g . Then

aps +bm g = g = gregguni = gunigreg.

s m = aps m = p = bmps = Furthermore, greg g 1 and guni g 1; hence greg is p-regular and guni is p-unipotent. Let g = gr gu = gugr be another decomposition with p-regular s gr and p-unipotent gu. One checks that gr and gu have the order m and p , respec- s = aps = 1−bm = 1−bm ap = =  tively. Then greg g g gr gu gr and hence also guni gu.

Let e denote the exponent of G and let e = eps with p e. We consider any finitely generated k[G]-module V and any element g ∈ G.Letζ1(g, V ), . . . , ζd (g, V ), where d := dimk V , be all eigenvalues of the k-linear endomorphism g V −→ V . We list any eigenvalue as many times as its multiplicity as a zero of the g characteristic polynomial of V −→ V prescribes. We have

χV (g) = ζ1(g, V ) +···+ζd (g, V ).

The following are easy facts:

1. The sequence (ζ1(g,V),...,ζd (g, V )) depends up to its ordering only on the isomorphism class {V }. g 2. If ζ is an eigenvalue of V −→ V then ζ j , for any j ≥ 0, is an eigenvalue of j g order(g) V −−→ V . It follows that ζi(g, V ) = 1 for any 1 ≤ i ≤ d. In particular,  each ζi(g, V ) is an eth root of unity. Is gp-regular then the ζi(g, V ) are e th roots of unity. 3. If 0 → V1 → V → V2 → 0 is an exact sequence in Mk[G] then the se- quence (ζ1(g,V),...,ζd (g, V )) is, up to a reordering, the union of the two

sequences (ζ1(g, V1),...,ζd1 (g, V1)) and (ζ1(g, V2),...,ζd2 (g, V2)) (where di := dimk Vi ).

Lemma 3.1.3 For any finitely generated k[G]-module V and any g ∈ G the se- quences (ζ1(g, V ), . . . , ζd (g, V )) and (ζ1(greg,V),...,ζd (greg,V)) coincide up to a reordering; in particular, we have

χV (g) = χV (greg).

Proof Since the order of greg is prime to p the vector space

V = V1 ⊕···⊕Vt 90 3 The Brauer Character decomposes into the different eigenspaces Vj for the linear endomorphism greg V −−→ V . The elements gregguni = gunigreg commute. Hence guni respects the eigenspaces of greg,i.e.guni(Vj ) = Vj for any 1 ≤ j ≤ t. The cyclic group guni is a p-group. By Proposition 2.2.7 the only simple k[ guni ]-module is the trivial −−→guni module. This implies that 1 is the only eigenvalue of V V . We therefore find 1 ∗ a basis of Vj with respect to which the matrix of guni|Vj is of the form .The 01 ζ 0 j matrix of greg|Vj is the diagonal matrix where ζj is the corresponding 0 ζj ζ ∗ j eigenvalue. The matrix of g|Vj then is . It follows that g|Vj has a single 0 ζj eigenvalue which coincides with the eigenvalue of greg|Vj . 

The subset

Greg := {g ∈ G : g is p-regular} consists of full conjugacy classes of G. We therefore may introduce the k-vector space Cl(Greg,k) of k-valued class functions on Greg. There is the obvious map Cl(G, k) −→ Cl(Greg,k) which sends a function on G to its restriction to Greg. Proposition 3.1.1 and Lemma 3.1.3 together imply that the composed map

Tr Trreg : k ⊗Z Rk(G) −→ Cl(G, k) −→ Cl(Greg,k) still is injective. We will see later on that Trreg in fact is an isomorphism.

Remark 3.1.4 Let ξ ∈ K× be any root of unity; then ξ ∈ R×.

= j ∈ × ∈ Z m = ≥ Proof We have ξ aπR with a R and j .Ifξ 1 with m 1 then = m jm jm ∈ × = 1 a πR which implies πR R . It follows that j 0 and consequently that ξ = a ∈ R×. 

× × Let μe (K) and μe (k) denote the subgroup of K and k , respectively, of all   e th roots of unity. Both groups are cyclic of order e , μe (K) since R is splitting  for G, and μe (k) since k is algebraically closed of characteristic prime to e . Since × μe (K) ⊆ R by Remark 3.1.4 the homomorphism

μe (K) −→ μe (k)

ξ −→ ξ + mR is well-defined.

=∼ Lemma 3.1.5 The map μe (K) −→ μe (k) is an isomorphism.

e Proof The elements in μe (K) are precisely the roots of the polynomial X − 1 ∈ K[X].Ifξ1 = ξ2 in μe (K) were mapped to the same element in μe (k) then the 3.2 Properties 91

 same polynomial Xe − 1 but viewed in k[X] would have a zero of multiplicity >1. But this polynomial is separable since p e. Hence the map is injective and then also bijective. 

We denote the inverse of the isomorphism in Lemma 3.1.5 by

μe (k) −→ μe (K) ⊆ R ξ −→[ξ].

The element [ξ]∈R is called the Teichmüller representative of ξ. The isomorphism in Lemma 3.1.5 allows us to introduce, for any finitely gener- ated k[G]-module V (with d := dimk V ), the K-valued class function

βV : Greg −→ K

g −→ ζ1(g, V ) +···+ ζd (g, V ) on Greg. It is called the Brauer character of V . By construction we have

βV (g) ≡ χV (g) mod mR for any g ∈ Greg.

The first fact in the list before Lemma 3.1.3 implies that βV only depends on the isomorphism class {V }, whereas the last fact implies that

= + βV βV1 βV2 for any exact sequence 0 → V1 → V → V2 → 0inMk[G]. Therefore, if we let Cl(Greg,K)denote the K-vector space of all K-valued class functions on Greg then

TrB : Rk(G) −→ Cl(Greg,K)

[V ]−→βV is a well-defined homomorphism.

3.2 Properties  Lemma 3.2.1 The Brauer characters βV ∈ Cl(Greg,K), for {V }∈k[G], are K- linearly independent.  = ∈ Proof Let {V }∈k[G] cV βV 0 with cV K. By multiplying the cV by a high enough power of πR we may assume that all cV lie in R. Then

(cV mod mR)χV = 0, {V } 92 3 The Brauer Character and Proposition 3.1.1 implies that all cV must lie in mR. We therefore may write cV = πRdV with dV ∈ R and obtain

0 = cV βV = πR dV βV , hence dV βV = 0. {V } {V } {V }

∈ ∈ 2 Applying Proposition 3.1.1 again givesdV mR and cV mV . Proceeding induc- ∈ i ={ } { }∈ [ ]  tively in this way we deduce that cV i≥0 mR 0 for any V k G .

Proposition 3.2.2 The diagram

dG RK (G) Rk(G)

Tr TrB res Cl(G, K) Cl(Greg,K), where “res” denotes the map of restricting functions from G to Greg, is commutative.

Proof Let V be any finitely generated K[G]-module. We choose a lattice L ⊆ V which is G-invariant and put W := L/πRL. Then dG([V ]) =[W], and our assertion becomes the statement that

βW (g) = χV (g) for any g ∈ Greg holds true. Let e1,...,ed be a K-basis of V such that L = Re1 + ··· + Red . Then e¯1 := e1 + πRL,...,e¯d := ed + πRL is a k-basis of W .Wefixg ∈ Greg, g and we form the matrices Ag and Ag of the linear endomorphisms V −→ V and g W −→ W , respectively, with respect to these bases. Obviously Ag is obtained from Ag by reducing its entries modulo mR.Letζ1(g, V ), . . . , ζd (g, V ) ∈ K, resp. ζ1(g, W), . . . , ζd (g, W) ∈ k, be the zeros, counted with multiplicity, of the characteristic polynomial of Ag, resp. Ag.Wehave

χV (g) = tr(Ag) = ζi(g, V ) and χW (g) = tr(Ag) = ζi(g, W). i i

Clearly, the set {ζi(g, V )}i is mapped by reduction modulo mR to the set {ζi(g, W)}i . Lemma 3.1.5 then implies that, up to a reordering, we have

ζi(g, V ) = ζi(g, W) .

We conclude that

χV (g) = ζi(g, V ) = ζi(g, W) = βW (g).  i i 3.2 Properties 93

Corollary 3.2.3 The homomorphism

=∼ TrB : K ⊗Z Rk(G) −−→ Cl(Greg,K)

c ⊗[V ]−→cβV is an isomorphism.

Proof From Proposition 3.2.2 we obtain the commutative diagram

id ⊗dG K ⊗Z RK (G) K ⊗Z Rk(G)

Tr TrB res Cl(G, K) Cl(Greg,K) of K-linear maps. The left vertical map is an isomorphism by Corollary 2.3.7.ii. The map “res” clearly is surjective: For example, ϕ(g)˜ := ϕ(greg) is a preimage of ϕ ∈ Cl(Greg,K). Hence TrB is surjective. Its injectivity follows from Lemma 3.2.1. 

Corollary 3.2.4 The number of isomorphism classes of simple k[G]-modules coin- cides with the number of conjugacy classes of p-regular elements in G.

Proof The two numbers whose equality is asserted are the dimensions of the two K-vector spaces in the isomorphism of Corollary 3.2.3. 

=∼ Corollary 3.2.5 The homomorphism Trreg : k ⊗Z Rk(G) −→ Cl(Greg,k) is an iso- morphism.

Proof We know already that this k-linear map is injective. But by Corollary 3.2.4 the two k-vector spaces have the same finite dimension. Hence the map is bijective. 

We also may deduce a more conceptual formula for Brauer characters.

Corollary 3.2.6 For any finitely generated k[G]-module V and any g ∈ Greg we have

= −1 [ ] βV (g) Tr d g V (g).

Proof By construction the element βV (g) only depends on V as a k[ g ]-module where g ⊆G is the cyclic subgroup generated by g. Since the order of g is prime to p the decomposition homomorphism d g is an isomorphism by Corollary 2.2.6. The assertion therefore follows from Proposition 3.2.2 applied to the group g .  94 3 The Brauer Character

Lemma 3.2.7 Let M be a finitely generated projective R[G]-module and put V := K ⊗R M; we then have

χV (g) = 0 for any g ∈ G \ Greg.

Proof We fix an element g ∈ G \ Greg. It generates a cyclic subgroup g ⊆G whose order is divisible by p. The trace of g on V only depends on M viewed as an R[G]-module. We have R[G] =∼ R[ g ][G: g ]. Hence M, being isomorphic to a direct summand of some free R[G]-module, also is isomorphic to a direct summand of a free R[ g ]-module. We see that M also is projective as an R[ g ]-module (cf. Proposition 1.6.4). This observation reduces us (by replacing G by g ) to the case of a cyclic group G generated by an element whose order is divisible by p. We write G = C × P as the direct product of a cyclic group C of order prime to p and a cyclic p-group P = {1}. First of all we note that by repeating the above observation for the subgroup P ⊆ G we obtain that M is projective also as an R[P ]- module. Let g = gCgP with gC ∈ C and gP ∈ P . We decompose

V = V1 ⊕···⊕Vr into its isotypic components Vi as an R[C]-module. Since K is a splitting field for any subgroup of G the Vi are precisely the different eigenspaces of the linear gC endomorphism V −−→ V .Letζi ∈ K be the eigenvalue of gC on Vi . Each Vi ,of course, is a K[G]-submodule of V . In particular, we have

r χV (g) = ζi tr(gP ; Vi). i=1

It therefore suffices to show that tr(gP ; Vi) = 0 for any 1 ≤ i ≤ r. The element r 1 − ε := ζ j gj ∈ K[C] i |C| i C j=1 is the idempotent such that Vi = εiV . But since p |G| and because of Remark 3.1.4 we see that εi ∈ R[C]. It follows that we have the decomposition as an R[G]-module

M = M1 ⊕···⊕Mr with Mi := M ∩ Vi.

Each Mi , being a direct summand of the projective R[P ]-module M, is a finitely generated projective R[P ]-module, and Vi = K ⊗R Mi . This further reduces us (by replacing M by Mi and g by gP ) to the case where G is a cyclic p-group with generator g = 1. We know from Propositions 1.7.4 and 2.2.7 that in this situation the (up to isomorphism) only finitely generated indecomposable projective R[G]-module is R[G]. Hence M =∼ R[G]m 3.2 Properties 95 for some m ≥ 0, and

tr(g; V)= m tr g; K[G] .

Since g = 1wehavegh = h for any h ∈ G.UsingtheK-basis {h}h∈G of K[G] g we therefore see that the corresponding matrix of K[G] −→ K[G] has all diagonal entries equal to zero. Hence  tr g; K[G] = 0.

The above lemma can be rephrased by saying that there is a unique homomor- phism

Trproj : K0 k[G] −→ Cl(Greg,K) such that the diagram

eG K0(k[G]) RK (G)

Trproj Tr

ext0 Cl(Greg,K) Cl(G, K), where “ext0” denotes the homomorphism which extends a function on Greg to G by setting it equal to zero on G \ Greg, is commutative.

Proposition 3.2.8 =∼ i. The map Trproj : K ⊗Z K0(k[G]) −→ Cl(Greg,K)is an isomorphism. eG ii. im(K0(k[G]) −−→ RK (G)) ={x ∈ RK (G) : Tr(x)|G \ Greg = 0}.

Proof i. We have the commutative diagram

id ⊗eG K ⊗Z K0(k[G]) K ⊗Z RK (G)

Trproj Tr

ext0 Cl(Greg,K) Cl(G, K).

The maps id ⊗eG and Tr are injective by Theorem 2.8.7 and Corollary 2.3.7.ii, re- spectively. Hence Trproj is injective. Using Proposition 1.7.4 and Corollary 3.2.3 we obtain

dimK K ⊗Z K0 k[G] = dimK K ⊗Z Rk(G) = dimK Cl(Greg,K).

It follows that Trproj is bijective. 96 3 The Brauer Character

ii. As a consequence of Lemma 3.2.7 the left-hand side im(eG) of the asserted identity is contained in the right-hand side. According to Theorem 2.8.7 there exists a subgroup Z ⊆ RK (G) such that RK (G) = im(eG) ⊕ Z and a fortiori

K ⊗Z RK (G) = im(id ⊗eG) ⊕ (K ⊗Z Z).

Suppose that z ∈ Z is such that Tr(z)|G \ Greg = 0. By i. we then find an element x ∈ im(id ⊗eG) such that Tr(z) = Tr(x). It follows that

z = x ∈ Z ∩ im(id ⊗eG) ⊆ (K ⊗Z Z) ∩ im(id ⊗eG) ={0}.

Hence z = 0, and we see that any x ∈ RK (G) such that Tr(x)|G \ Greg = 0mustbe contained in im(eG).  Chapter 4 Green’s Theory of Indecomposable Modules

4.1 Relatively Projective Modules

Let f : A −→ B be any ring homomorphism. Any B-module M may be considered, via restriction of scalars, as an A-module. Any homomorphism α : M −→ N of B-modules automatically is a homomorphism of A-modules as well.

Definition A B-module P is called relatively projective (with respect to f )iffor any pair of B-module homomorphisms

P

γ β M N for which there is an A-module homomorphism α0 : P −→ M such that β ◦ α0 = γ there also exists a B-module homomorphism α : P −→ M satisfying β ◦ α = γ .

We start with some very simple observations.

Remark

1. The existence of α0 in the above definition implies that the image of β contains the image of γ . Hence we may replace N by im(β). This means that in the above definition it suffices to consider pairs (β, γ ) with surjective β. 2. Any projective B-module is relatively projective. 3. If P is relatively projective and is projective as an A-module then P is a projec- tive B-module. 4. Every B-module is relatively projective with respect to the identity homomor- phism idB .

Lemma 4.1.1 For any B-module P the following conditions are equivalent:

P. Schneider, Modular Representation Theory of Finite Groups, 97 DOI 10.1007/978-1-4471-4832-6_4, © Springer-Verlag London 2013 98 4 Green’s Theory of Indecomposable Modules i. P is relatively projective. ii. For any (surjective) B-module homomorphism β : M −→ P for which there is an A-module homomorphism σ0 : P −→ M such that β ◦ σ0 = idP there also exists a B-module homomorphism σ : P −→ M satisfying β ◦ σ = idP .

Proof i. =⇒ ii. We apply the definition to the pair

P

idP β M P. ii. =⇒ i. Let P

γ β M N be any pair of B-module homomorphisms such that there is an A-module homo- morphism α0 : P −→ M with β ◦ α0 = γ . By introducing the B-module M := (x, y) ∈ M ⊕ P : β(x) = γ(y) we obtain the commutative diagram

pr2 M P

pr1 γ β M N

:= := where pr1((x, y)) x and pr2((x, y)) y. We observe that  σ0: P −→ M y −→ α0(y), y ◦ = is a well-defined A-module homomorphism such that pr2 σ0 idP . There ex- ists therefore, by assumption, a B-module homomorphism σ : P −→ M satisfying ◦ = := ◦ pr2 σ idP . It follows that the B-module homomorphism α pr1 σ satisfies ◦ = ◦ ◦ = ◦ ◦ = ◦ ◦ =  β α (β pr1) σ (γ pr2) σ γ (pr2 σ) γ.

Lemma 4.1.2 For any two B-modules P1 and P2 the direct sum P := P1 ⊕ P2 is relatively projective if and only if P1 and P2 both are relatively projective. 4.1 Relatively Projective Modules 99

Proof Let pr1 and pr2 denote, quite generally, the projection map from a direct sum onto its first and second summand, respectively. Similarly, let i1, resp. i2, denote the inclusion map from the first, resp. second, summand into their direct sum. We first suppose that P is relatively projective. By symmetry it suffices to show that P1 is relatively projective. We are going to use Lemma 4.1.1. Let therefore β1 : M1 −→ P1 be any B-module homomorphism and σ0 : P1 −→ M1 be any A-module ◦ = homomorphism such that β1 σ0 idP1 . Then

⊕ β1 idP2 β: M := M1 ⊕ P2 −−−−−→ P1 ⊕ P2 = P is a B-module homomorphism and

σ ⊕ id : = ⊕ −−−−−0 P→2 ⊕ = σ0 P P1 P2 M1 P2 M is an A-module homomorphism and the two satisfy

◦  = ◦ ⊕ = ⊕ = β σ0 (β1 σ0) idP2 idP1 idP2 idP .

By assumption we find a B-module homomorphism σ  : P −→ M such that  β ◦ σ = idP . We now define the B-module homomorphism σ : P1 −→ M1 as the composite

 i1 σ pr1 P1 −→ P1 ⊕ P2 −→ M1 ⊕ P2 −−→ M1, and we compute

◦ = ◦ ◦  ◦ = ◦ ⊕ ◦  ◦ β1 σ β1 pr1 σ i1 pr1 (β1 idP2 ) σ i1 = ◦ ◦  ◦ = ◦ pr1 β σ i1 pr1 i1 = idP1 .

We now assume, vice versa, that P1 and P2 are relatively projective. Again us- ing Lemma 4.1.1 we let β : M −→ P and σ0 : P −→ M be a B-module and an A-module homomorphism, respectively, such that β ◦ σ0 = idP . This latter relation implies that σ0,j := σ0|Pj can be viewed as an A-module homomorphism

−1 σ0,j : Pj −→ β (Pj ) for j = 1, 2.

−1 −1 We also define the B-module homomorphism βj := β|β (Pj ) : β (Pj ) −→ Pj . We obviously have ◦ = = βj σ0,j idPj for j 1, 2. −1 By assumption we therefore find B-module homomorphisms σj : Pj −→ β (Pj ) ◦ = = such that βj σj idPj for j 1, 2. The B-module homomorphism 100 4 Green’s Theory of Indecomposable Modules

σ : P = P1 ⊕ P2 −→ M

y1 + y2 −→ σ1(y1) + σ2(y2) then satisfies β ◦ σ(y1 + y2) = β σ1(y1) + σ2(y2) = β ◦ σ1(y1) + β ◦ σ2(y2)

= β1 ◦ σ1(y1) + β2 ◦ σ2(y2)

= y1 + y2 for any yj ∈ Pj . 

Lemma 4.1.3 For any A-module L0 the B-module B ⊗A L0 is relatively projective.

Proof Consider any pair

B ⊗A L0

γ β M N of B-module homomorphisms together with an A-module homomorphism α0 : B ⊗A L0 −→ M such that β ⊗ α0 = γ . Then

α: B ⊗A L0 −→ M

b ⊗ y0 −→ bα0(1 ⊗ y0) is a B-module homomorphism satisfying β ◦ α = γ since β ◦ α(b ⊗ y0) = β bα0(1 ⊗ y0)

= b(β ◦ α0)(1 ⊗ y0) = bγ (1 ⊗ y0)

= γ(b⊗ y0). 

Proposition 4.1.4 For any B-module P the following conditions are equivalent: i. P is relatively projective; ii. P is isomorphic to a direct summand of B ⊗A P ; iii. P is isomorphic to a direct summand of B ⊗A L0 for some A-module L0.

Proof For the implication from i. to ii. we observe that we have the B-module ho- momorphism

β: B ⊗A P −→ P b ⊗ y −→ by 4.1 Relatively Projective Modules 101 as well as the A-module homomorphism

σ0: P −→ B ⊗A P y −→ 1 ⊗ y which satisfy β ◦ σ0 = idP . Hence by Lemma 4.1.1 there exists a B-module ho- momorphism σ : P −→ B ⊗A P such that β ◦ σ = idP . Then B ⊗A P = im(σ ) ⊕ σ ker(β), and P −→ im(σ ) is an isomorphism of B-modules. The implication from ii. to iii. is trivial. That iii. implies i. follows from Lem- mas 4.1.2 and 4.1.3. 

We are primarily interested in the case of an inclusion homomorphism R[H ]⊆R[G], where R is any commutative ring and H is a subgroup of a fi- nite group G.AnR[G]-module which is relatively projective with respective to this inclusion will be called relative R[H]-projective.

Example If R = k is a field then a k[G]-module is relatively k[{1}]-projective if and only if it is projective.

We recall the useful fact that R[G] is free as an R[H ]-module with basis any set of representatives of the cosets of H in G.

Lemma 4.1.5 An R[G]-module is projective if and only if it is relatively R[H ]- projective and it is projective as an R[H]-module.

Proof By our initial Remark it remains to show that any projective R[G]-module P also is projective as an R[H]-module. But P is a direct summand of a free [ ] =∼ [ ] [ ] R G -module F i∈I R G by Proposition 1.6.4. Since R G is free as an R[H ]-module F also is free as an R[H]-module. Hence the reverse implication in Proposition 1.6.4 says that P is projective as an R[H ]-module. 

As we have discussed in Sect. 2.3 the R[G]-module R[G]⊗R[H ] L0, for any [ ] G R H -module L0, is naturally isomorphic to the induced module IndH (L0). Hence we may restate Proposition 4.1.4 as follows.

Proposition 4.1.6 For any R[G]-module P the following conditions are equiva- lent: i. P is relatively R[H]-projective; G ii. P is isomorphic to a direct summand of IndH (P ); G [ ] iii. P is isomorphic to a direct summand of IndH (L0) for some R H -module L0.

Next we give a useful technical criterion. 102 4 Green’s Theory of Indecomposable Modules

Lemma 4.1.7 Let {g1,...,gm}⊆G be a set of representatives for the left cosets of H in G; aleftR[G]-module P is relatively R[H ]-projective if and only if there exists an R[H ]-module endomorphism ψ : P −→ P such that

m −1 = giψgi idP . i=1

Proof We introduce the following notation. For any g ∈ G there are uniquely de- termined elements h1(g), . . . , hm(g) ∈ H and a uniquely determined permutation π(g,.) of {1,...,m} such that

ggi = gπ(g,i)hi(g) for any 1 ≤ i ≤ m.

Step 1:Letα0 : M −→ N be any R[H]-module homomorphism between any two R[G]-modules. We claim that

α := N (α0): M −→ N m −→ −1 x giα0 gi x i=1 is an R[G]-module homomorphism. Let g ∈ G. We compute

m m −1 = −1 −1 = −1 α g x giα0 gi g x giα0 (ggi) x i=1 i=1 m m = −1 = −1 −1 giα0 gπ(g,i)hi(g) x giα0 hi(g) gπ(g,i)x i=1 i=1 m m = −1 −1 = −1 −1 gihi(g) α0 gπ(g,i)x g gπ(g,i)α0 gπ(g,i)x i=1 i=1 = g−1α(x) for any x ∈ M which proves the claim. We also observe that, for any R[G]-module homomorphisms β : M −→ M and γ : N −→ N ,wehave

γ ◦ N (α0) ◦ β = N (γ ◦ α0 ◦ β). (4.1.1)

Step 2: Let us temporarily say that an R[H ]-module endomorphism ψ : M −→ M of an R[G]-module M is good if N (ψ) = idM holds true. The assertion of the present lemma then says that an R[G]-module is relatively R[H ]-projective if and only if it has a good endomorphism. In this step we establish that any 4.1 Relatively Projective Modules 103

R[G]-module P with a good endomorphism ψ is relatively R[H ]-projective. Let

P

γ β M N be a pair of R[G]-module homomorphisms and α0 : P −→ M be an R[H ]-module homomorphism such that β ◦ α0 = γ .Using(4.1.1) we see that α := N (α0 ◦ ψ) satisfies

β ◦ α = β ◦ N (α0 ◦ ψ)= N (β ◦ α0 ◦ ψ)= N (γ ◦ ψ)= γ ◦ N (ψ) = idP .

Step 3: If the direct sum M = M1 ⊕ M2 of R[G]-modules has a good en- domorphism ψ then each summand Mj has a good endomorphism as well. Let ij prj Mj −→ M −−→ Mj be the inclusion and the projection map, respectively. Then := ◦ ◦ ψj prj ψ ij satisfies N = N ◦ ◦ = ◦ N ◦ = ◦ = (ψj ) (prj ψ ij ) prj (ψ) ij prj ij idMj where, for the second equality, we again have used (4.1.1). Step 4: In order to show that any relatively R[H]-projective R[G]-module P has a good endomorphism it remains, by Step 3 and Proposition 4.1.4, to see that any R[G]-module of the form R[G]⊗R[H ] L0,forsomeR[H ]-module L0, has a good endomorphism. We have

R[G]⊗R[H ] L0 = g1 ⊗ L0 +···+gm ⊗ L0 and, assuming that g1 ∈ H , we define the map

ψ: R[G]⊗R[H ] L0 −→ R[G]⊗R[H ] L0 m gi ⊗ xi −→ g1 ⊗ x1. i=1

Let h ∈ H and observe that the permutation π(h,.)has the property that π(h,1) = 1. We compute m m m ψ h gi ⊗ xi = ψ hgi ⊗ xi = ψ gπ(h,i)hi(h) ⊗ xi i=1 i=1 i=1 m = ψ gπ(h,i) ⊗ hi(h)xi = gπ(h,1) ⊗ h1x1 i=1 104 4 Green’s Theory of Indecomposable Modules

= gπ(h,1)h1(h) ⊗ x1 = hg1 ⊗ x1 m = hψ gi ⊗ xi . i=1

This shows that ψ is an R[H]-module endomorphism. Moreover, using that −1 = = π(gj ,i) 1 if and only if i j, we obtain

m m −1 ⊗ gj ψ gj gi xi j=1 i=1 m m = −1 ⊗ gj ψ gj gi xi j=1 i=1 m m −1 = gj ψ g −1 hi g ⊗ xi π(gj ,i) j j=1 i=1 m m −1 = gj ψ g −1 ⊗ hi g xi π(gj ,i) j j=1 i=1 m m −1 −1 = gj g −1 ⊗ hj g xj = gj g −1 hj g ⊗ xj π(gj ,j) j π(gj ,j) j j=1 j=1 m m = −1 ⊗ = ⊗ gj gj gj xj gj xj . j=1 j=1

Hence ψ is good. 

If k is a field of characteristic p>0 and the order of the group G is prime to p then we know the group ring k[G] to be semisimple. Hence all k[G]-modules are projective (cf. Remark 1.7.3). This fact generalizes as follows to the relative situation.

Proposition 4.1.8 If the integer [G : H] is invertible in R then any R[G]-module M is relatively R[H]-projective.

1  Proof The endomorphism [G:H ] idM of M is good. 4.2 Vertices and Sources 105

4.2 Vertices and Sources

For any R[G]-module M we introduce the set

V(M) := set of subgroups H ⊆ G such that M is relatively R[H]-projective.

For trivial reasons G lies in V(M).

Lemma 4.2.1 The set V(M) is closed under conjugation, i.e. for any H ∈ V(M) and g ∈ G we have gHg−1 ∈ V(M).

Proof Let

M

γ β L N

−1 be any pair of R[G]-module homomorphisms and α0 : M −→ L be an R[gHg ]- −1 module homomorphism such that β ◦ α0 = γ . We consider the map α1 := g α0g : M −→ L.Forh ∈ H we have −1 −1 −1 −1 −1 α1(hy) = g α0(ghy) = g α0 ghg gy = g ghg α0(gy) = hα1(y) for any y ∈ M. This shows that α1 is an R[H]-module homomorphism. It satisfies −1 −1 β ◦ α1(y) = β g α0(gy) = g β ◦ α0(gy) = g−1γ(gy)= γ g−1gy = γ(y) for any y ∈ M. Since, by assumption, M is relatively R[H ]-projective there exists an R[G]-module homomorphism α : M −→ L such that β ◦ α = γ . 

We let V0(M) ⊆ V(M) denote the subset of those subgroups H ∈ V(M) which are minimal with respect to inclusion.

Exercise 4.2.2

i. V0(M) is nonempty. ii. V0(M) is closed under conjugation. iii. V(M) is the set of all subgroups of G which contain some subgroup in V0(M).

Definition A subgroup H ∈ V0(M) is called a vertex of M. 106 4 Green’s Theory of Indecomposable Modules

Lemma 4.2.3 Let p be a fixed prime number and suppose that all prime numbers = p are invertible in R (e.g., let R = k be a field of characteristic p or let R be a (0,p)-ring for such a field k); then all vertices are p-groups.

Proof Let H1 ∈ V0(M) for some R[G]-module M and let H0 ⊆ H1 be a p-Sylow subgroup of H1. We claim that M is relatively R[H0]-projective which, by the min- imality of H1, implies that H0 = H1 is a p-group. Let

M

γ β L N be a pair of R[G]-module homomorphisms together with an R[H0]-module ho- momorphism α0 : M −→ L such that β ◦ α0 = γ . Since [H1 : H0] is invertible in R it follows from Proposition 4.1.8 that M viewed as an R[H1]-module is relatively R[H0]-projective. Hence there exists an R[H1]-module homomorphism α1 : M −→ L such that β ◦ α1 = γ .ButH1 ∈ V(M). So there further must exist an R[G]-module homomorphism α : M −→ L satisfying β ◦ α = γ . 

In order to investigate vertices more closely we need a general property of induc- tion. To formulate it we first generalize some notation from Sect. 2.5.LetH ⊆ G be a subgroup and g ∈ G be an element. For any R[gHg−1]-module M, corre- −1 sponding to the ring homomorphism π : R[gHg ]−→EndR(M), we introduce the R[H ]-module g∗M defined by the composite ring homomorphism −1 π R[H]−→R gHg −→ EndR(M) h −→ ghg−1.

Proposition 4.2.4 (Mackey) Let H0,H1 ⊆ G be two subgroups and fix a set {g1,...,gm}⊆G of representatives for the double cosets in H0\G/H1. For any R[H1]-module L we have an isomorphism of R[H0]-modules ∗ ∗ G ∼ H0 −1 H0 −1 Ind (L) = Ind − g L ⊕···⊕Ind − g L . H1 ∩ 1 1 ∩ 1 m H0 g1H1g1 H0 gmH1gm

Proof As an H0-set through left multiplication and simultaneously as a (right) H1- set through right multiplication G decomposes disjointly into

G = H0g1H1 ∪···∪H0gmH1.

Therefore the induced module IndG (L),viewedasanR[H ]-module, decomposes H1 0 into the direct sum of R[H0]-modules

IndG (L) = IndH0g1H1 (L) ⊕···⊕IndH0gmH1 (L) H1 H1 H1 4.2 Vertices and Sources 107 where, for any g ∈ G, we put   IndH0gH1 (L) := φ ∈ IndG (L) : φ g = 0 for any g ∈/ H gH . H1 H1 0 1

We note that IndH0gH1 (L) is the R[H ]-module of all functions φ : H gH → L H1 0 0 1 = −1 ∈ ∈ such that φ(h0gh1) h1 φ(h0g) for any h0 H0 and h1 H1. It remains to check that the map

∗ H0 −1 −→ H0gH1 Ind −1 g L Ind (L) H0∩gH1g H1 −→ := −1 φ φ (h0gh1) h1 φ(h0), where h0 ∈ H0 and h1 ∈ H1, is an isomorphism of R[H0]-modules. In order to ˜ ˜ ˜ check that φ is well defined, let h0gh1 = h0gh1 with hi ∈ Hi . Then = ˜ ˜ −1 −1 ˜ −1 −1 ∈ ∩ −1 h0 h0 gh1h1 g with gh1h1 g H0 gH1g and hence = ˜ ˜ −1 −1 = ˜ −1 −1 ˜ φ(h0) φ h0 gh1h1 g h1h1 φ(h0) which implies

= −1 = ˜−1 ˜ = ˜ ˜ φ (h0gh1) h1 φ(h0) h1 φ(h0) φ (h0gh1).

It is clear that the map φ : H gH −→ L lies in IndH0gH1 (L).Forh ∈ H we 0 1 H1 0 compute h = −1h = −1 −1 φ (h0gh1) h1 φ(h0) h1 φ h h0 −1 = φ h h0gh1 h = φ (h0gh1).

This shows that φ −→ φ is an R[H0]-module homomorphism. It is visibly injec- tive. To establish its surjectivity we let ψ ∈ IndH0gH1 (L) and we define the map H1

φ: H0 −→ L

h0 −→ ψ(h0g).

−1 For h ∈ H0 ∩ gH1g we compute −1 −1 −1 φ(h0h) = ψ(h0hg) = ψ h0gg hg = g hg ψ(h0g) −1 −1 = g h gφ(h0) 108 4 Green’s Theory of Indecomposable Modules

∈ H0 −1 ∗ which means that φ Ind −1 ((g ) L). We obviously have H0∩gH1g

= −1 = −1 = φ (h0gh1) h1 φ(h0) h1 ψ(h0g) ψ(h0gh1), i.e. φ = ψ. 

Proposition 4.2.5 Suppose that R is noetherian and complete and that R/Jac(R) is artinian. For any finitely generated and indecomposable R[G]-module M its set of vertices V0(M) consists of a single conjugacy class of subgroups.

Proof We know from Exercise 4.2.2.ii that V0(M) is a union of conjugacy classes. In the following we show that any two H0,H1 ∈ V0(M) are conjugate. By Propo- sition 4.1.6 the R[G]-module M is isomorphic to a direct summand of IndG (M) H0 as well as of IndG (M). The latter implies that IndG (M) is isomorphic to a di- H1 H0 rect summand of IndG (IndG (M)). Together with the former we obtain that M is H0 H1 isomorphic to a direct summand of IndG (IndG (M)). At this point we induce the H0 H1 R[H0]-module decomposition

m ∗ G ∼ H0 −1 Ind (M) = Ind − g M H1 ∩ 1 i H0 gi H1gi i=1 from Mackey’s Proposition 4.2.4 to G and get, by transitivity of induction, that

m G G ∼ G −1 ∗ Ind Ind (M) = Ind − g M H0 H1 ∩ 1 i H0 gi H1g1 i=1 as R[G]-modules. Our assumptions on R guarantee the validity of the KrullÐ RemakÐSchmidt Theorem 1.4.7 for R[G], i.e. the “unicity” of the decompo- sition into indecomposable modules of the finitely generated R[G]-modules G G G −1 ∗ Ind (Ind (M)) and Ind − ((g ) M) . It follows that any indecompos- H0 H1 ∩ 1 i H0 gi H1gi able direct summand of IndG (IndG (M)), for example our M, must be a direct H0 H0 G −1 ∗ summand of at least one of the Ind ∩ −1 ((gi ) M). This implies, by Propo- H0 gi H1gi −1 sition 4.1.6, that M is relatively R[H0 ∩ gH1g ]-projective for some g ∈ G. −1 Hence H0 ∩ gH1g ∈ V(M). Since V(M) is closed under conjugation accord- −1 ingtoLemma4.2.1 we also have g H0g ∩ H1 ∈ V(M). Since H0 and H1 are both −1 minimal in V(M) we must have H0 = gH1g . 

Definition Suppose that M is finitely generated and indecomposable, and let V ∈ V0(M). Any finitely generated indecomposable R[V ]-module Q such that M G is isomorphic to a direct summand of IndV (Q) is called a V -source of M. A source of M is a V -source for some V ∈ V0(M). 4.2 Vertices and Sources 109

−1 We remind the reader that NG(H ) ={g ∈ G : gHg = H } denotes the normal- izer of the subgroup H of G. Since G is finite any of the inclusions gHg−1 ⊆ H or −1 gHg ⊇ H already implies g ∈ NG(H ).

Proposition 4.2.6 Suppose that R is noetherian and complete and that R/Jac(R) is artinian. Let M be a finitely generated indecomposable R[G]-module and let V ∈ V0(M). We then have: i. M has a V -source Q which is a direct summand of M as an R[V ]-module; ii. if Q is a V -source of M then (g−1)∗Q, for any g ∈ G, is a gVg−1-source of M; iii. for any two V -sources Q0 and Q1 of M there exists a g ∈ NG(V ) such that ∼ ∗ Q1 = g Q0.

Proof i. We decompose M,asanR[V ]-module, into a direct sum

M = M1 ⊕···⊕Mr of indecomposable R[V ]-modules Mi . Then r G = G IndV (M) IndV (Mi). i=1 By Proposition 4.1.6 the indecomposable R[G]-module M is isomorphic to a direct G summand of IndV (M). As noted already in the proof of Proposition 4.2.5 the KrullÐ RemakÐSchmidt Theorem 1.4.7 implies that M then has to be isomorphic to a direct G := summand of some IndV (Mi0 ). Put Q Mi0 . ii. That Q is a finitely generated indecomposable R[V ]-module implies that (g−1)∗Q is a finitely generated indecomposable R[gVg−1]-module. The map ∼ G −→= G −1 ∗ IndV (Q) IndgVg−1 g Q φ −→ φ(.g) is an isomorphism of R[G]-modules. Hence if M is isomorphic to a direct summand of the former it also is isomorphic to a direct summand of the latter. iii. By i. we may assume that Q0 is a direct summand of M as an R[V ]-module. But, by assumption, M,asanR[G]-module, and therefore also Q0,asanR[V ]- G G module, is isomorphic to a direct summand of IndV (Q1) as well as of IndV (Q0). G G This implies that Q0 is isomorphic to a direct summand of IndV (IndV (Q1)).Bythe same reasoning with Mackey’s Proposition 4.2.4 as in the proof of Proposition 4.2.5 we deduce that V −1 ∗ Q0 is isomorphic to a direct summand of IndV ∩gVg−1 g Q1 (4.2.1) ∈ G for some g G. Then IndV (Q0) and hence also M are isomorphic to a direct G ((g−1)∗Q ) V M summand of IndV ∩gVg−1 1 . Since is a vertex of we must have 110 4 Green’s Theory of Indecomposable Modules

−1 −1 |V |≤|V ∩ gVg |, hence V ⊆ gVg , and therefore g ∈ NG(V ). By inserting this information into (4.2.1) it follows that Q0 is isomorphic to a direct summand −1 ∗ −1 ∗ of (g ) Q1. But with Q1 also (g ) Q1 is indecomposable. We finally obtain ∼ −1 ∗ Q0 = (g ) Q1. 

Exercise 4.2.7 If Q is a V -source of M then V(Q) = V0(Q) ={V }.

4.3 The Green Correspondence

Throughout this section we assume the commutative ring R to be noetherian and complete and R/Jac(R) to be artinian. We fix a subgroup H ⊆ G. We consider any finitely generated indecomposable R[G]-module M such that H ∈ V(M).Let

M = L1 ⊕···⊕Lr be a decomposition of M as an R[H]-module into indecomposable R[H ]- modules Li . The KrullÐRemakÐSchmidt Theorem 1.4.7 says that the set INDH {M} := {L1},...,{Lr } of isomorphism classes of the R[H]-modules Li only depends on the isomorphism class {M} of the R[G]-module M.

Lemma 4.3.1 Let V ∈ V0(M) and Vi ∈ V0(Li) for 1 ≤ i ≤ r; we then have: ≤ ≤ ∈ ⊆ −1 i. For any 1 i r there is a gi G such that Vi giVgi ; G ≤ ≤ ii. M is isomorphic to a direct summand of IndH (Li) for some 1 i r; G = −1 iii. if M is isomorphic to a direct summand of IndH (Li) then Vi giVgi and, in particular, V0(Li) ⊆ V0(M); iv. if V0(Li) ⊆ V0(M) then M and Li have a common Vi -source. (The assertions i. and iv. do not require the assumption that H ∈ V(M).)

Proof We fix a V -source Q of M. Then M,asanR[G]-module, is isomorphic G to a direct summand of IndV (Q). Using Mackey’s Proposition 4.2.4 we see that L1 ⊕···⊕Lr is isomorphic to a direct summand of H −1 ∗ IndH ∩xVx−1 x Q x∈R where R ⊆ G is a fixed set of representatives for the double cosets in H \G/V .We conclude from the KrullÐRemakÐSchmidt Theorem 1.4.7 that, for any 1 ≤ i ≤ r, there exists an xi ∈ R such that Li is isomorphic to a direct summand of H −1 ∗ ∩ −1 ∈ V Ind ∩ −1 ((xi ) Q). Hence H xiVxi (Li) by Proposition 4.1.6 and, H xi Vxi 4.3 The Green Correspondence 111 since V0(Li) is a single conjugacy class with respect to H by Proposition 4.2.5,we −1 ⊆ ∩ −1 ∈ := −1 have hVih H xiVxi for some h H . We put gi h xi and obtain ⊆ −1 Vi giVgi which proves i. Suppose that Vi ∈ V0(M). Then |Vi|=|V | (cf. Proposition 4.2.5) and hence = −1 = −1 −1 ⊆ Vi giVgi h xiVxi h H. −1 −1 ∗ −1 In particular, xiVxi is a vertex of Li and (xi ) Q is an xiVxi -source of Li . −1 ∗ Using Proposition 4.2.6.ii we deduce that (gi ) Q is a Vi -source of Li .Onthe −1 ∈ V other hand we have, of course, that giVgi 0(M), and Proposition 4.2.6.ii again −1 ∗ implies that (gi ) Q is a Vi -source of M. This shows iv. Since H ∈ V(M) the R[G]-module M, by Proposition 4.1.6, also is isomorphic to a direct summand of r G = G IndH (M) IndH (Li). i=1 Hence ii. follows from the KrullÐRemakÐSchmidt Theorem 1.4.7. Now suppose that [ ] G M,asanR G -module, is isomorphic to a direct summand of IndH (Li) for some ≤ ≤ G 1 i r.LetQi be a Vi -source of Li . Then IndH (Li), and hence M, is isomorphic to a direct summand of IndG (Q ). It follows that V ∈ V(M) and therefore that Vi i i | |≤| | = −1  V Vi . Together with i. we obtain Vi giVgi . This proves iii.

We introduce the set VH := ∈ V : ⊆ 0 (M) V 0(M) V H . VH Obviously the subgroup H acts on 0 (M) by conjugation so that we can speak of VH the H -orbits in 0 (M). We also introduce 0 { } := { }∈ { } : V ⊆ V INDH M Li INDH M 0(Li) 0(M) and 1 { } := { }∈ { } : INDH M Li INDH M M is isomorphic to a direct G summand of IndH (Li) . By Lemma 4.3.1 all three sets are nonempty and 1 { } ⊆ 0 { } INDH M INDH M . Furthermore, using Proposition 4.2.5 we have the obvious map M : 0 { } −→ VH vH INDH M set of H-orbits in 0 (M) {Li}−→V0(Li). 112 4 Green’s Theory of Indecomposable Modules

∈ VH ≤ ≤ ∈ V Lemma 4.3.2 For any V 0 (M) there is a 1 j r such that V 0(Lj ); in this case M and Lj have a common V -source.

Proof By Proposition 4.2.6.i the R[G]-module M has a V -source M0 which is a direct summand of M as an R[V ]-module. Since V ⊆ H the KrullÐRemakÐSchmidt Theorem 1.4.7 implies that M0 is isomorphic to a direct summand of some Lj as R[V ]-modules. This means that we have a decomposition

Lj = X1 ⊕···⊕Xt ∼ into indecomposable R[V ]-modules Xi such that X1 = M0. We note that V is a ver- tex of X1 by Exercise 4.2.7.LetVj be a vertex of Lj . We now apply Lemma 4.3.1.i to this decomposition (i.e. with H,Lj ,Vj ,V instead of G, M, V , V1) and obtain −1 that V ⊆ hVj h for some h ∈ H . But Lemma 4.3.1.i applied to the original decom- −1 position M = L1 ⊕···⊕Lr gives |V |≥|Vj |. Hence V = hVj h which implies V ∈ V0(Lj ). The latter further implies Vj ∈ V0(Lj ) ⊆ V0(M). Therefore M and Lj , by Lemma 4.3.1.iv, have a common Vj -source and hence, by Proposition 4.2.6.ii, also a common V -source. 

M Lemma 4.3.2 implies that the map vH is surjective.

∈ VH Lemma 4.3.3 For any V 0 (M) there exists a finitely generated indecomposable R[H ]-module N such that V ∈ V0(N) and M is isomorphic to a direct summand of G IndH (N).

Proof Let Q be a V -source of M so that M is isomorphic to a direct summand of G = G H IndV (Q) IndH (IndV (Q)). By the usual argument there exists an indecomposable H direct summand N of IndV (Q) such that M is isomorphic to a direct summand of G ∈ V IndH (N). According to Proposition 4.1.6 we have V (N). We choose a vertex     V ∈ V0(N) such that V ⊆ V .LetQ be a V -source of N. Then N is isomorphic H  to a direct summand of IndV  (Q ). Hence M is isomorphic to a direct summand of G H  = G   ∈ V IndH (IndV  (Q )) IndV  (Q ), and consequently V (M) by Proposition 4.1.6. The minimality of V implies V  = V which shows that V is a vertex of N. 

Lemma 4.3.4 Let N be a finitely generated indecomposable R[H ]-module, and let U ∈ V0(N); we then have: G [ ] i. The induced module IndH (N), as an R H -module, is of the form

G =∼ ⊕ ⊕···⊕ IndH (N) N N1 Ns

with indecomposable R[H]-modules Ni for which there are elements gi ∈ G\H ∩ −1 ∈ V such that H giUgi (Ni); 4.3 The Green Correspondence 113

G =  ⊕  [ ]   ii. IndH (N) M M with R G -submodules M and M such that: − M is indecomposable, − N is isomorphic to a direct summand of M,   − U ∈ V0(M ), and M and N have a common U-source; iii. if NG(U) ⊆ H then V0(Ni) = V0(N) for any 1 ≤ i ≤ s and N is not isomorphic to a direct summand of M.

Proof i. Since U ∈ V0(N) we have H =∼ ⊕  IndU (N) N N for some R[H ]-module N . On the other hand, let G  = ⊕ ⊕···⊕ IndH N N0 N1 Ns be a decomposition into indecomposable R[H]-modules Ni . Furthermore, it is easy to see (but can also be deduced from Mackey’s Proposition 4.2.4) that, as R[H ]- modules, we have G  =∼  ⊕  IndH N N N for some R[H ]-module N , and correspondingly that N is isomorphic to a direct summand of IndG (N). The latter implies that N is isomorphic to one of the N .We H ∼ i may assume that N = N0. Combining these facts with Mackey’s Proposition 4.2.4 we obtain   N ⊕ N1 ⊕···⊕Ns ⊕ N ⊕ N =∼ G ⊕ G  IndH (N) IndH N =∼ G H = G IndH IndU (N) IndU (N) =∼ H −1 ∗ IndH ∩gUg−1 g N g∈R = H ⊕ H −1 ∗ IndU (N) IndH ∩gUg−1 g N g∈R,g∈ /H =∼ ⊕  ⊕ H −1 ∗ N N IndH ∩gUg−1 g N g∈R,g∈ /H as R[H ]-modules, where R is a fixed set of representatives for the double cosets in H \G/U such that 1 ∈ R. It is a consequence of the KrullÐRemakÐSchmidt Theo- rem 1.4.7 that we may cancel summands which occur on both sides and still have an isomorphism ⊕···⊕ ⊕  =∼ H −1 ∗ N1 Ns N IndH ∩gUg−1 g N . g∈R,g∈ /H 114 4 Green’s Theory of Indecomposable Modules

Moreover, it follows that Ni ,for1≤ i ≤ s, is isomorphic to a direct summand of H −1 ∗ ∈ ∩ −1 ∈ V Ind ∩ −1 ((gi ) N) for some gi / H . Hence H giUgi (Ni). H gi Ugi ii. Let G =  ⊕···⊕  IndH (N) M1 Mt be a decomposition into indecomposable R[G]-modules. By i. and the KrullÐ RemakÐSchmidt Theorem 1.4.7 there must exist a 1 ≤ l ≤ t such that N is iso-   :=   :=  morphic to a direct summand of Ml . We define M Ml and M i =l Mi .   Lemma 4.3.1.iii/iv (for (M ,N) instead of (M, Li)) implies that V0(N) ⊆ V0(M ) and that M and N have a common U-source. iii. Suppose that NG(U) ⊆ H and V0(Ni) = V0(N) for some 1 ≤ i ≤ s. Then ∩ −1 ∈ V = V H giUgi (Ni) (N) and therefore ⊆ ∩ −1 −1 = ∩ −1 ⊆ −1 U h H giUgi h H hgiU(hgi) hgiU(hgi) for some h ∈ H . We conclude that hgi ∈ NG(U) ⊆ H and hence gi ∈ H which is a contradiction. If N were isomorphic to a direct summand of M then N ⊕ N would G =∼ be isomorphic to a direct summand of IndH (N). Hence, by i., N Ni for some 1 ≤ i ≤ s which contradicts the fact that V0(N) = V0(Ni). 

∈ VH ⊆ Proposition 4.3.5 For any V 0 (M) such that NG(V ) H there is a unique index 1 ≤ j ≤ r such that V ∈ V0(Lj ), and M is isomorphic to a direct summand of G IndH (Lj ).

Proof According to Lemma 4.3.3 we find a finitely generated indecomposable R[H ]-module N such that

− V ∈ V0(N) and − G M is isomorphic to a direct summand of IndH (N). Lemma 4.3.4 tells us that

G =∼ ⊕ ⊕···⊕ IndH (N) N N1 Ns with indecomposable R[H]-modules Ni such that

V0(Ni) = V0(N) for any 1 ≤ i ≤ s.

But by Lemma 4.3.2 there exists a 1 ≤ j ≤ r such that V ∈ V0(Lj ) and hence V0(Lj ) = V0(N). If we now apply the KrullÐRemakÐSchmidt Theorem 1.4.7 to the fact that

L1 ⊕···⊕Lr is isomorphic to a direct summand of N ⊕ N1 ⊕···⊕Ns then we see that, for i = j,wehave

 V0(Li) = V0(Ni ) = V0(N) for some 1 ≤ i ≤ s. 4.3 The Green Correspondence 115

Hence j is unique with the property that V0(Lj ) = V0(N). Since V0(Lj ) = ∼ V0(N) = V0(Ni) for any 1 ≤ i ≤ s we furthermore have Lj = N1,...,Ns and there- ∼ fore necessarily Lj = N. This shows that M is isomorphic to a direct summand of G  IndH (Lj ).

M In terms of the map vH the Proposition 4.3.5 says the following. If the H -orbit V ⊆ VH ⊆ ∈ V 0 (M) has the property that NG(V ) H for one (or any) V then there is { }∈ 0 { } VM { } = V a unique isomorphism class L INDH ( M ) such that H ( L ) , and in fact { }∈ 1 { } L INDH ( M ). We now shift our point of view and fix a subgroup V ⊆ G such that NG(V ) ⊆ H . Then Proposition 4.3.5 gives the existence of a well-defined map

isomorphism classes of finitely isomorphism classes of finitely Γ : generated indecomposable −→ generated indecomposable R[G]-modules with vertex V R[H ]-modules with vertex V where the image {L}=Γ({M}) is characterized by the condition that L is isomor- phic to a direct summand of M.

Theorem 4.3.6 (Green) The map Γ is a bijection. The image {L}=Γ({M}) is characterized by either of the following two conditions: a. L is isomorphic to a direct summand of M; G b. M is isomorphic to a direct summand of IndH (L). Moreover, M and L have a common V -source.

Proof Let L be any finitely generated indecomposable R[H ]-module with vertex V . By Lemma 4.3.4.ii we find a finitely generated indecomposable R[G]-module M with vertex V such that − L is isomorphic to a direct summand of M; − G =  ⊕  [ ] IndH (L) M M as R G -module, and − M and L have a common V -source. In particular, Γ({M}) ={L} which shows the surjectivity of Γ . Let M be any other finitely generated indecomposable R[G]-module with ver- tex V . First we consider the case that M is isomorphic to a direct summand of IndG (L). Suppose that M = ∼ M. Then M must be isomorphic to a direct sum- H  mand of M .LetM = X1 ⊕ ··· ⊕ Xt be a decomposition into indecompos- able R[H ]-modules Xi . Then Lemma 4.3.4 implies that V0(Xi) = V0(L) for any 1 ≤ i ≤ t. But by Lemma 4.3.2 there exists a 1 ≤ i ≤ t such that V ∈ V0(Xi) which is a contradiction. It follows that M =∼ M, hence that L is isomorphic to a direct summand of M, and consequently that Γ({M}) ={L}. This proves that the condi- tion b. characterizes the map Γ . Secondly we consider the case that Γ({M}) ={L}, i.e. that L is isomorphic to a G direct summand of M. Then M is isomorphic to a direct summand of IndH (L) by 116 4 Green’s Theory of Indecomposable Modules

Proposition 4.3.5. Hence we are in the first case and conclude that M =∼ M.This establishes the injectivity of Γ . 

The bijection Γ is called the Green correspondence for (G,V,H). We want to establish a useful additional “rigidity” property of the Green correspondence. For this we first have to extent the concept of relative projectivity to module homomor- phisms.

Definition Let H be a fixed set of subgroups of G. [ ] H =∼ i. An R G -module M is called relatively -projective if M i∈I Mi is iso- morphic to a direct sum of R[G]-modules Mi each of which is relatively R[Hi]-projective for some Hi ∈ H. ii. An R[G]-module homomorphism α : M −→ N is called H-projective if there exist a relatively H-projective R[G]-module X and R[G]-module homomor- phisms α0 : M −→ X and α1 : X −→ N such that α = α1 ◦ α0. In case H ={H } we simply say that α is H -projective.

Exercise 4.3.7

i. An R[G]-module M is relatively R[H]-projective if and only if idM is H -projective. ii. If the R[G]-module homomorphism α : M −→ N is H-projective then γ ◦α◦β, for any R[G]-module homomorphisms β : M −→ M and γ : N −→ N ,is H-projective as well. iii. If α : M −→ N is an H -projective R[G]-module homomorphism between two finitely generated R[G]-modules M and N then the relatively R[H ]-projective R[G]-module X in the above definition can be chosen to be finitely generated as well. (Hint: Replace X by R[G]⊗R[H ] X.)

We keep fixing our subgroups V ⊆ H ⊆ G, and we introduce the family of sub- groups − h := H ∩ gVg 1 : g ∈ G \ H .

Lemma 4.3.8 Let α : M −→ N be a V -projective R[H ]-module homomorphism between two finitely generated R[G]-modules M and N; then there exists a V - projective R[G]-module homomorphism β : M −→ N and an h-projective R[H ]- module homomorphism γ : M −→ N such that α = β + γ .

Proof By assumption we have a commutative diagram

α M N

α0 α1 X 4.3 The Green Correspondence 117 where X is a relatively R[V ]-projective R[H]-module. By Exercise 4.3.7.iii we may assume that X is a finitely generated R[H]-module. The idea of the proof G consists in the attempt to replace X by IndH (X). We first of all observe that, X H [ ] G being isomorphic to a direct summand of IndV (X),theR G -module IndH (X) is G [ ] isomorphic to a direct summand of IndV (X) and hence is relatively R V -projective as well. We will make use of the two Frobenius reciprocities from Sect. 2.3.Let

−→ι G −→π X IndH (X) X be the R[H ]-module homomorphisms given by g−1x if g ∈ H , ι(x)(g) := 0ifg/∈ H and π(φ) := φ(1). ◦ = G = ⊕ [ ] We obviously have π ι idX and therefore IndH (X) im(ι) ker(π) as an R H - module. By applying Lemma 4.3.4.i to the indecomposable direct summands of X we see that ker(π) is relatively h-projective. But we have the commutative diagram

ι◦π−id G G IndH (X) IndH (X)

− pr ⊆ ker(π).

It follows that the R[H]-module homomorphism ι ◦ π − id is h-projective. By the first and second Frobenius reciprocity we have the commutative diagrams

α˜0 α˜1 G G M IndH (X) and IndH (X) N,

π ι α0 α1 X X respectively, where α˜0 and α˜1 are (uniquely determined) R[G]-module homomor- phisms. Then β :=α ˜ 1 ◦˜α0 is a V -projective R[G]-module homomorphism, and γ :=α ˜ 1 ◦ (ι ◦ π − id) ◦˜α0 is a h-projective R[H]-module homomorphism (cf. Ex- ercise 4.3.7.ii). We have

α = α1 ◦ α0 =˜α1 ◦˜α0 + (α1 ◦ α0 −˜α1 ◦˜α0)

= β + (α˜ 1 ◦ ι ◦ π ◦˜α0 −˜α1 ◦˜α0) = β +˜α1 ◦ (ι ◦ π − id) ◦˜α0 = β + γ.  118 4 Green’s Theory of Indecomposable Modules

Proposition 4.3.9 Let M be a finitely generated indecomposable R[G]-module with vertex V such that NG(V ) ⊆ H ⊆ G, and let L be its Green correspondent (i.e. {L}=Γ({M})). For any other finitely generated R[G]-module M we have: i. If M is indecomposable such that L is isomorphic to a direct summand of M then M =∼ M; ii. M is isomorphic to a direct summand of M (as an R[G]-module) if and only if L is isomorphic to a direct summand of M (as an R[H ]-module).

Proof i. By the injectivity of the Green correspondence it suffices to show that V is a vertex of M. The assumption on M guarantees the existence of R[H ]-module   homomorphisms ι : L −→ M and π : M −→ L such that π ◦ ι = idL. Since V is a vertex of L the R[H]-module homomorphism α := ι ◦ π is V -projective. We therefore, by Lemma 4.3.8, may write α = β + γ with a V -projective R[G]-module endomorphism β : M −→ M and an h-projective R[H ]-module endomorphism γ : M −→ M.  Step 1: We first claim that V ∈ V(M ),i.e.thatidM is V -projective. For this we consider the inclusion of rings   EG := EndR[G] M ⊆ EH := EndR[H ] M , and we define

JV := {ψ ∈ EG : ψ is V -projective},

Jh := {ψ ∈ EH : ψ is h-projective}.

Using Lemma 4.1.2 one checks that JV and Jh are additively closed. Moreover, Exercise 4.3.7.ii implies that JV is a two-sided ideal in EG and Jh is a two-sided ideal in EH .Wehaveβ ∈ JV and γ ∈ Jh. Furthermore

n n α = α = (β + γ) ≡ β mod Jh (4.3.1) for any n ∈ N. By Lemma 1.3.5.ii, on the other hand, EH and hence EH /Jh is finitely generated as a (left) EG-module (even as an R-module). Proposition 1.3.6.iii then implies that the EG-module EH /Jh is Jac(EG)-adically complete. This, in particular, means that n Jac(EG) EH + Jh ⊆ Jh. (4.3.2) n∈N  We now suppose that JV = EG. Since M is indecomposable as an R[G]-module the ring EG is local by Proposition 1.3.6 and Proposition 1.4.5. It follows that β ∈ JV ⊆ Jac(EG). We then conclude from (4.3.1) and (4.3.2) that α ∈ Jh. Hence idL = −1 π ◦ α ◦ ι ∈ Jh. We obtain that L is relatively R[H ∩ gVg ]-projective for some g ∈ G \ H . But as we have seen in the proof of Lemma 4.3.4.iii the conclusion that −1 H ∩ gVg ∈ V(L) for some g ∈ G \ H is in contradiction with V ∈ V0(L).We therefore must have JV = EG which is the assertion that idM is V -projective. 4.4 An Example: The Group SL2(Fp) 119

   Step 2: We now show that V ∈ V0(M ).LetV be a vertex of M .ByStep1 we have |V |≤|V |, and it suffices to show that |V |=|V |. But Lemma 4.3.1.i applied to M and V  says that the order of the vertex V of the indecomposable direct summand ι(L) of M is ≤|V |. ii. The direct implication is clear since L is isomorphic to a direct summand  =  ⊕···⊕  of M. For the reverse implication let M M1 Mt be a decomposition into [ ]  indecomposable R G -modules. Then L is isomorphic to a direct summand of Mi ≤ ≤ =∼   for some 1 i t, and the assertion i. implies that M Mi .

4.4 An Example: The Group SL2(Fp)

We fix a prime number p = 2, and we let G := SL2(Fp). There are the following important subgroups 10 a 0 U := : c ∈ F ⊆ B := ∈ G : ad = 1 c 1 p cd of G.

Exercise i. [G : B]=p + 1 and [B : U]=p − 1. ii. U =∼ Z/pZ is a p-Sylow subgroup of G. iii. B = NG(U).

A much more lengthy exercise is the following.

Exercise The group G has exactly p + 4 conjugacy classes which are represented by the elements: − 10 , 10 1. 01 0 −1 − − 2. 11 , 11 , 1 ε , 1 ε (with ε ∈ F× \ F×2 a fixed element), 01 0 −1 01 0 −1 p p a 0 × −1 3. − where a ∈ F \{±1} (up to replacing a by a ), 0 a 1 p − 0 1 X2 − aX + ∈ F [X] 4. 1 a where the polynomial 1 p is irreducible. The elements in 2. have order divisible by p, all the others an order prime to p.

Let k be an algebraically closed field of characteristic p.

Lemma 4.4.1 There are exactly p isomorphism classed of simple k[G]-modules.

Proof Since, by the above exercise, there are p conjugacy classes of p-regular ele- ments in G this follows from Corollary 3.2.4.  120 4 Green’s Theory of Indecomposable Modules

We want to construct explicit models for the simple k[G]-modules. Let k[X, Y ] = ab ∈ be the polynomial ring in two variables X and Y over k. For any g cd G we define the k-algebra homomorphism

g: k[X, Y ]−→k[X, Y ] X −→ aX + cY Y −→ bX + dY.

An easy explicit computation shows that in this way k[X, Y ] becomes a k[G]-mod- ule. We have the decomposition k[X, Y ]= Vn n≥0 where n i n−i Vn := kX Y i=0 is the k[G]-submodule of all polynomials which are homogeneous of total degree n. We obviously have:

− dimk Vn = n + 1; − V0 = k is the trivial k[G]-module; ∼ 2 − The G-action on V1 = kX + kY = k is the restriction to SL2(Fp) of the natural 2 GL2(k)-action on the standard k-vector space k .

We will show that V0,V1,...,Vp−1 are simple k[G]-modules. Since they have dif- ferent k-dimensions they then must be, up to isomorphism, all simple k[G]-modules. The subgroup U is generated by the element u+ := 10 . Similarly the subgroup 11 U − := { 1 b : b ∈ F } u− := 11 01 p is generated by 01 . In the following we fix an n ≥ 0, and we consider in Vn the increasing sequence of vector subspaces

{0}=W0 ⊂ W1 ⊂···⊂Wn ⊂ Wn+1 = Vn defined by i−1 j n−j Wi := kX Y . j=0

Lemma 4.4.2 For 0 ≤ i ≤ n we have:

i. Wi+1 is a k[U]-submodule of Vn; ii. Wi+1/Wi is the trivial k[U]-module; iii. if i

i n−i + + Proof Obviously Wi+1 = kX Y ⊕ Wi .Wehaveu X = X + Y and u Y = Y and hence i i u+ XiY n−i = (X + Y)iY n−i = Xj Y i−j Y n−i j j=0 − i 1 i = XiY n−i + Xj Y n−j j j=0 i n−i ≡ X Y mod Wi.

We now argue by induction with respect to i. The assertions i.Ðiii. hold trivially n true for W1 = kY . We assume that they hold true for Wi . The above congruence immediately implies i. and ii. for Wi+1. For iii. let v ∈ Wi+1 \ Wi be any vector. Then i n−i × v = aX Y + w with a ∈ k and w ∈ Wi. + Using that u w − w ∈ Wi−1 by ii. for Wi , we obtain u+v − v = a u+ XiY n−i − XiY n−i + u+w − w

i−1 n−i+1 ≡ aiX Y mod Wi−1.

Since i

For convenience we insert the following reminder.

Remark Let H be any finite p-group. By Proposition 2.2.7 the trivial k[H ]-module is, up to isomorphism, the only simple k[H]-module. It follows that the socle of any k[H]-module M (cf. Lemma 1.1.5) is equal to

soc(M) ={x ∈ M : hx = x for any h ∈ H }.

Suppose that M = {0}. For any 0 = x ∈ M the submodule k[H ]x is of finite k-dimension and nonzero and therefore, by the Jordan–Hölder Proposition 1.1.2, contains a simple submodule. It follows in particular that soc(M) = {0}.

Lemma 4.4.3 For 0 ≤ n

Proof i. This is a special cases of Lemma 4.3.2.iii. ii. The socle in question is equal + n + to {v ∈ Vn : u v = v}. It contains W1 = kY by Lemma 4.4.2.ii. If u v = v then 122 4 Green’s Theory of Indecomposable Modules k[U]v has k-dimension ≤ 1. On the other hand, any 0 = v ∈ Vn is contained in Wi+1 \ Wi for a unique 0 ≤ i ≤ n. By Lemma 4.4.2.iii we then have k[U]v = Wi+1. + Since Wi+1 has k-dimension i + 1 we see that, if u v = v, then necessarily v ∈ W1. iii. Suppose that Vn = M ⊕ N is the direct sum of two nonzero k[U]-submodules M and N. Then the socle of Vn is the direct sum of the nonzero socles of M and N and therefore has k-dimension at least 2. This contradicts ii. 

Proposition 4.4.4 The k[G]-module Vn, for 0 ≤ n

Proof First of all we observe that − n k U Y = Vn holds true. This is proved by the same reasoning as for Lemma 4.4.3.i with only interchanging the roles of X and Y . Let W ⊆ Vn be any nonzero k[G]-submodule. Its socle as a k[U]-module is nonzero and is contained in the corresponding socle of Vn. Hence Lemma 4.4.3.ii n − n implies that kY ⊆ W . Our initial observation then gives Vn = k[U ]Y ⊆ W . 

The k[U]-module structure of Vn,for0≤ n

k[Z]/Zpk[Z]−→k[U] + Z −→ u − 1.

Because of (u+)i = ((u+ − 1) + 1)i it is surjective. But both sides have the same k-dimension p. Hence it is an isomorphism. If we view Vn, via this isomorphism, as a k[Z]/Zp[Z]-module then we claim that

∼ n+1 Vn = k[Z]/Z k[Z] holds true. This amounts to the statement that ∼ + n+1 Vn = k[U] u − 1 k[U].

By Lemma 4.4.3.i we have the surjective k[U]-module homomorphism

k[U]−→Vn σ −→ σXn.

+ On the other hand, Lemma 4.4.2.ii implies that (u − 1)Wi+1 ⊆ Wi and hence by induction that (u+ − 1)n+1Xn = 0. We deduce that the above homomorphism induces a homomorphism + n+1 k[U] u − 1 k[U]−→Vn 4.4 An Example: The Group SL2(Fp) 123 which has to be an isomorphism since it is surjective and both sides have the same k-dimension n + 1. ∼ Remark 4.4.5 Vp−1 = k[U] as a k[U]-module.

Proof This is the case n = p − 1 of the above discussion. 

This latter fact has an interesting consequence. For this we also need the follow- ing general properties.

Lemma 4.4.6 i. If H is a finite p-group then any finitely generated projective k[H ]-module is free. ii. If H is a finite group and V ⊆ H is a p-Sylow subgroup then the k-dimension of any finitely generated projective k[H]-module is divisible by |V |.

Proof i. It follows from Remark 1.6.9 and Proposition 2.2.7 that k[H ] is a projective cover of the trivial k[H]-module k. Since the trivial module, up to isomorphism, is the only simple k[H]-module it then is a consequence of Proposition 1.7.4.i that k[H], up to isomorphism, is the only finitely generated indecomposable projective k[H]-module. This implies the assertion by Lemma 1.1.6. ii. Let M be a finitely generated projective k[H]-module. By Lemma 4.1.5 it also is projective as an k[V ]-module. The assertion i. therefore says that M =∼ k[V ]m for some m ≥ 0. It follows that dimk M = m|V |. 

Proposition 4.4.7 Among the simple k[G]-modules V0,...,Vp−1 only Vp−1 is a projective k[G]-module.

Proof By Lemma 4.2.3 we have U ∈ V(Vp−1),i.e.Vp−1 is relatively k[U]- projective. But Vp−1 also is projective as a k[U]-module by Remark 4.4.5. Hence it is projective as a k[G]-module. On the other hand, if some Vi with 0 ≤ i

Let M be a finitely generated indecomposable k[G]-module. The vertices of M, by Lemma 4.2.3,arep-groups. It follows that either {1} or U is a vertex of M. We have observed earlier that {1} is a vertex of M if and only if M is a projec- tive k[G]-module. We postpone the investigation of this case. In the other case we may apply the Green correspondence Γ with B = NG(U) to obtain a bijection be- tween the isomorphism classes of finitely generated indecomposable nonprojective k[G]-modules and the isomorphism classes of finitely generated indecomposable nonprojective k[B]-modules.

Example V0,...,Vp−2 of course are indecomposable k[G]-modules which, by Proposition 4.4.7, are nonprojective. As a consequence of Lemma 4.4.3.iii they are indecomposable as k[B]-modules. Hence they are their own Green correspondents. 124 4 Green’s Theory of Indecomposable Modules

In the following we will determine all finitely generated indecomposable k[B]-modules. Another important subgroup of B is the subgroup of diagonal matrices − a 1 0 × =∼ × T := : a ∈ F −−→ F 0 a p p a−1 0 −→ a. 0 a

Since the order p − 1ofT is prime to p the group ring k[T ] is semisimple. As T is abelian and k is algebraically closed all simple k[T ]-modules are one- dimensional. They correspond to the homomorphisms

× χi: T −→ k for 0 ≤ i

Exercise i. The map

pr B −−→ T −1 −1 a 0 −→ a 0 ca 0 a

is a surjective homomorphism of groups with kernel U. ii. B = TU = UT as sets. iii. All simple k[B]-modules are one-dimensional and correspond to the homomor- phisms χi ◦ pr for 0 ≤ i ≤ p − 2. (Argue that U acts trivially on any sim- ple k[B]-module since in any nonzero k[B]-module L the subspace {x ∈ L : u+x = x} is nonzero.)

Let Si ,for0≤ i ≤ p − 2, denote the simple k[B]-module corresponding to χi ◦ pr.

Lemma 4.4.8 For any k[B]-module L we have rad(L) = (u+ − 1)L.

Proof According to the above exercise U acts trivially on any simple k[B]-module + and hence on L/ rad(L). It follows that (u − 1)L ⊆ rad(L). The binomial for- (u+)i − = ((u+ − ) + )i − = i i (u+ − )j (u+ − mula 1 1 1 1 j=1 j 1 implies that [ ]= − [ ] + − 1)k U u∈U (u 1)k U . Since U is normal in B it follows that (u 1)L is a k[B]-submodule and that the U-action on L/(u+ − 1)L is trivial. Hence k[B] acts on L/(u+ − 1)L through its quotient k[T ], and therefore L/(u+ − 1)L is a semisimple k[B]-module. This implies rad(L) ⊆ (u+ − 1)L.  4.4 An Example: The Group SL2(Fp) 125

Let L = {0} be any finitely generated k[B]-module. According to Lemma 4.4.8 we have the sequence of k[B]-submodules + + F 0(L) := L ⊇ F 1(L) := u − 1 L ⊇···⊇F i(L) := u − 1 iL ⊇··· .

Each subquotient F i(L)/F i+1(L) is a semisimple k[B]-module. It follows from Proposition 1.2.1 that there is a smallest m(L) ∈ N such that F m(L)(L) ={0} and F i(L) = F i+1(L) for any 0 ≤ i

This shows that

i i+1 ∼ F (Vp−1)/F (Vp−1) = Wp−i/Wp−i−1 = S2i mod p−1 for 0 ≤ i ≤ p − 1.

1 ∼ In particular, Vp−1/ rad(Vp−1) = Vp−1/F (Vp−1) = S0 is the trivial k[B]-module. Hence Vp−1 is a projective cover of the trivial module. It follows from Proposition 1.7.4.i that there are exactly p − 1isomor- phism classes {Q0},...,{Qp−2} of finitely generated indecomposable projective k[B]-modules which can be numbered in such a way that 1 ∼ Qj / rad(Qj ) = Qj F (Qj ) = Sj for 0 ≤ j

Proof (See Sect. 2.3 for the tensor product of two modules.) Because of Lem- ma 1.6.8 it suffices to show that Vp−1 ⊗k Sj is a projective cover of Sj .But Vp−1 ⊗k Sj obviously is finitely generated. It is indecomposable even as a k[U]- module. Moreover, using Lemma 4.4.8, we obtain + Vp−1 ⊗k Sj rad(Vp−1 ⊗k Sj ) = (Vp−1 ⊗k Sj ) u − 1 (Vp−1 ⊗k Sj ) + = (Vp−1 ⊗k Sj ) u − 1 Vp−1 ⊗k Sj + = Vp−1 u − 1 Vp−1 ⊗k Sj ∼ ∼ = S0 ⊗k Sj = Sj . 126 4 Green’s Theory of Indecomposable Modules

Hence, by Remark 1.6.9, it remains to show that Vp−1 ⊗k Sj is a projective k[B]-module. Let

Vp−1 ⊗k Sj

γ β M N 0 be an exact “test diagram” of k[B]-modules. Let 0 ≤ j 

Vp−1 α˜ γ ⊗id Sj

M ⊗k Sj  N ⊗k Sj  0. β⊗id Sj

Since Vp−1 is projective we find a k[B]-module homomorphism α˜ such that := ˜ ⊗ : ⊗ −→ the completed diagram commutes. But then α α idSj Vp−1 k Sj (M ⊗k Sj  ) ⊗k Sj = M satisfies γ = β ◦ α. 

Using Lemma 4.4.9 and our knowledge about Vp−1 we deduce the following properties of the indecomposable projective k[B]-modules Qj : ∼ − Qj = k[U] as a k[U]-module. i i+1 ∼ i − F (Qj )/F (Qj ) = S2i+j mod p−1 for 0 ≤ i ≤ p − 1, in particular, the F (Qj ) form a composition series of Qj . ∼ − Qj / rad(Qj ) = soc(Qj ). − TheCartanmatrixofk[B] is of size (p − 1) × (p − 1) and has the form ⎛ ⎞ 30202... ⎜ ⎟ ⎜03020...⎟ ⎜ ⎟ ⎜20302...⎟ ⎜ ⎟ ⎜02030...⎟ . ⎜ ⎟ ⎝20203...⎠ ......

In addition, the modules Qj have the following remarkable property.

Definition Let H be a finite group. A finitely generated k[H ]-module is called uniserial if it has a unique composition series.

Exercise In a uniserial module the members of the unique composition series are the only submodules. 4.4 An Example: The Group SL2(Fp) 127

Lemma 4.4.10 Any Qj is a uniserial k[B]-module.

Proof We show the apparently stronger statement that Qj is uniserial as a k[U]- ∼ module. Since Qj = k[U] it suffices to prove that k[U] is uniserial. By our earlier discussion this amounts to showing that k[Z]/Zpk[Z] is uniserial as a module over itself. This is left as an exercise to the reader. 

It follows immediately that all

i Qj,i := Qj /F (Qj ) for 0 ≤ j

Remark 4.4.11 Let H be a finite group and let M be a (finitely generated) ∗ k[H]-module. The k-linear dual M := Homk(M, k) is a (finitely generated) k[H]-module with respect to the H -action defined by

H × M∗ −→ M∗ (h, l) −→ hl(x) := l h−1x .

Suppose that M is finitely generated. Then the map N −→ N ⊥ := {l ∈ M∗ : l|N = 0} is an inclusion reversing bijection between the set of all k[H ]-submodules of M and the set of all k[H]-submodules of M∗ such that

⊥ ∗ ∗ ⊥ ∗ N = (M/N) and M /N = N .

It follows that M is a simple, resp. indecomposable, resp. uniserial, k[H ]-module if and only if M∗ is a simple, resp. indecomposable, resp. uniserial, k[H ]-module.

Lemma 4.4.12 Let H be a finite group; if the finitely generated indecomposable projective k[H ]-modules are uniserial then all finitely generated indecomposable k[H ]-modules are uniserial.

Proof Let M be any finitely generated indecomposable k[H ]-module. We consider the family of all pairs of k[H]-submodules L ⊂ N of M such that N/L is nonzero and uniserial. Such pairs exist: Take, for example, any two consecutive submodules in a composition series of M.WefixapairL ⊂ N in this family for which the length of the factor module N/L is maximal. Inafirststepweclaimthatthereexistsak[H ]-submodule N0 ⊆ N such that N = N0 ⊕L. Since N/Lis uniserial there is a unique k[H ]-submodule L ⊆ L1 ⊂ N such that N/L1 is simple. Let β : P −→ N/L1 be a projective cover of N/L1.By 128 4 Green’s Theory of Indecomposable Modules the projectivity we find a k[H]-module homomorphism α : P −→ N such that the diagram

P α β pr N N/L1 is commutative. We define N0 := im(α).Asβ issurjectivewehaveN0 + L1 = N. α pr If the composite map P −→ N −→ N/L were not surjective its image would be contained in L1/L since this is the unique maximal submodule of N/L. This would imply that N0 ⊆ L1 which contradicts the above. The surjectivity of this composite map says that

N0 + L = N.

On the other hand N0, as a factor module of the indecomposable projective module P (cf. Proposition 1.7.4), is uniserial by assumption. Hence {0}⊂N0 is one of the pairs of submodules in the family under consideration. Because of the surjection N0  N/L and the Jordan–Hölder Proposition 1.1.2 the length of N0 cannot be smaller than the length of N/L. The maximality of the latter therefore implies that =∼ N0 −→ N/L is an isomorphism and hence that

N0 ⊕ L = N.

The above argument, in particular, says that we may assume our pair of submod- ules to be of the form {0}⊂N. In the second step, we pass to the dual module M∗. The Remark 4.4.11 implies that N ⊥ ⊂ M∗ is a pair of k[H ]-submodules of M∗ such that M∗/N ⊥ is a nonzero uniserial module of maximal possible length. Hence the argument of the first step applied to M∗ leads to existence of a k[H ]-submodule M ⊆ M∗ such that M ⊕ N ⊥ = M∗. But with M also M∗ is indecomposable. We conclude that N ⊥ ={0} and consequently that M = N is uniserial. 

Proposition 4.4.13 Any finitely generated indecomposable k[B]-module Q is iso- morphic to some Qj,i.

Proof It follows from Lemmas 4.4.10 and 4.4.12 that Q is uniserial. Let β : Qj −→ Q/F 1(Q) be a projective cover of the simple k[B]-module Q/F 1(Q). We then find a commutative diagram of k[B]-module homomorphisms

Qj α β

Q Q/F 1(Q). pr 4.4 An Example: The Group SL2(Fp) 129

Since β is surjective the image of α cannot be contained in the unique maximal submodule F 1(Q) of Q. It follows that α is surjective and induces an isomorphism ∼ Qj,i = Q for an appropriate i. 

Among the p(p − 1) modules Qj,i exactly the p − 1 modules Qj,p = Qj are projective. Hence using the Green correspondence as discussed above we derive the following result.

Proposition 4.4.14 There are exactly (p − 1)2 isomorphism classes of finitely gen- erated indecomposable nonprojective k[G]-modules; they are the Green correspon- dents of the k[B]-modules Qj,i for 0 ≤ j

∼ Example For 0 ≤ n

Next we will compute the Cartan matrix of k[G]. But first we have to establish a useful general fact about indecomposable projective modules. For any finite group H the group ring k[H] is a so-called Frobenius algebra in the following way. Using the k-linear form

δ1: k[H]−→k ahh −→ a1 h∈H we introduce the k-bilinear form

k[H]×k[H]−→k

(x, y) −→ δ1(xy).

Remark 4.4.15

i. The above bilinear form is symmetric, i.e. δ1(xy) = δ1(yx) for any x,y ∈ k[H ]. ii. The map =∼ ∗ k[H] −−→ k[H] = Homk k[H ],k

x −→ δx(y) := δ1(xy)

is a k-linear isomorphism. = = Proof i. If x h∈H ahh and y h∈H bhh then we compute = = = δ1(xy) ahbh−1 bhah−1 δ1(yx). h∈H h∈H 130 4 Green’s Theory of Indecomposable Modules = = ii. It suffices to show that the map is injective. Let x h∈H ahh such that δx 0. For any h0 ∈ H we obtain = −1 = −1 = = 0 δx h0 δ1 ahhh0 δ1 ahh0 h ah0 h∈H h∈H and hence x = 0. 

Remark 4.4.16 For any finitely generated projective k[H ]-module P we have: i. P ∗ is a projective k[H]-module; ii. P ⊗k M, for any k[H]-module M, is a projective k[H ]-module. iii. let M be any k[H]-module and let L ⊆ N ⊆ M be k[H ]-submodules such that ∼ N/L= P ; then there exists a k[H]-submodule M0 ⊆ M such that ∼ ∼ ∼ M = P ⊕ M0,M0 ∩ N = L, and M0/M0 ∩ N = M/N.

Proof i. By Proposition 1.6.4 the module P is isomorphic to a direct summand of some free module k[H]m. Hence P ∗ is isomorphic to a direct summand of k[H ]∗m. Using Remark 4.4.15.ii we have the k-linear isomorphism

=∼ k[H] −−→ k[H]∗ −→ ahh ahδh−1 . h∈H h∈H The computation = −1 −1 = −1 = h δ(hh)−1 (y) δ1 h h y δh−1 h y δh−1 (y), for any h, h ∈ H and y ∈ k[H], shows that it is, in fact, an isomorphism of k[H ]- modules. ii. Again by Proposition 1.6.4 it suffices to consider the case of a free module P = k[H ]m. It further is enough to treat the case m = 1. But in the proof of Propo- sition 2.3.4 (applied to the subgroup {1}⊆H and W := k, V := M) we have seen that

∼ dimk M k[H]⊗k M = k[H] holds true. iii. (In case N = M the assertion coincides with the basic characterization of projectivity in Lemma 1.6.2.) In a first step we consider the other extreme case where L ={0}. Then N is a finitely generated projective submodule of M. Denoting ι by N −→ M the inclusion map we see, using i., that N ∗ through ι∗ : M∗  N ∗ is a projective factor module of M∗. Lemma 1.6.2 therefore implies the existence of ∗ ∗ ∗ a k[H ]-module homomorphism σ : N −→ M such that ι ◦ σ = idN∗ . Dualizing again and identifying M in the usual way with a submodule of M∗∗ we obtain M = N ⊕ ker(σ ∗|M). 4.4 An Example: The Group SL2(Fp) 131

In the general case we first use the projectivity of N/L to find a submodule L ⊆ N such that N = L ⊕ L and L =∼ N/L=∼ P . Viewing now L as a projective submodule of M we apply the first step to obtain a submodule M0 ⊆ M such that   M = L ⊕ M0. Then N = L ⊕ (M0 ∩ N).  Let k[H ]={{P1},...,{Pt }} be the set of isomorphism classes of finitely gen- erated indecomposable projective k[H]-modules. It follows from Remarks 4.4.11 { }−→{ ∗} [ ] and 4.4.16.i that Pi Pi induces a permutation of the set k H .Re- mark 4.4.11 also implies that the k[H]-modules =∼ ∗ ∗ ∗ soc(Pi) Pi / rad Pi are simple. Using Proposition 1.7.4.i we see that k[H ]= P1/ rad(P1) ,..., Pt / rad(Pt ) = soc(P1) ,..., soc(Pt ) . ∼ Proposition 4.4.17 soc(Pi) = Pi/ rad(Pi) for any 1 ≤ i ≤ t.

∗ ∼ Proof For any 1 ≤ i ≤ t there is a unique index 1 ≤ i ≤ t such that soc(Pi) = ∗ Pi∗ / rad(Pi∗ ). We have to show that i = i holds true. For this we fix a decomposi- tion

k[H]=N1 ⊕···⊕Ns into indecomposable (projective) submodules. Defining Mi := Nj for any 1 ≤ i ≤ t ∼ Nj =Pi we obtain a decomposition

k[H]=M1 ⊕···⊕Mt .

We know from Corollary 1.7.5 that Mi = {0} (and hence soc(Mi) = {0}) for any 1 ≤ i ≤ t. We also see that soc k[H] = soc(M1) ⊕···⊕soc(Mt ) is the isotypic decomposition of the semisimple k[H ]-module soc(k[H ]) with soc(Mi) being {soc(Pi)}-isotypic. Let 1 = e1 +···+et with ei ∈ Mi . By Proposition 1.5.1 the e1,...,et are pair- wise orthogonal idempotents in k[H] such that Mi = k[H ]ei . We claim that ∗ Mi soc(Mj ) ={0} whenever i = j .

Let x ∈ soc(Mj ) such that Mix = {0}. Since Mix is contained in the {soc(Pj )}- ·x isotypic module soc(Mj ) it follows that Mix and a fortiori Mi (through Mi −→ 132 4 Green’s Theory of Indecomposable Modules ∼ Mix) have a factor module isomorphic to soc(Pj ) = Pj ∗ / rad(Pj ∗ ).ButMi/ rad(Mi) ∗ by construction is {Pi/ rad(Pi)}-isotypic. It follows that necessarily i = j .This shows our claim and implies that ej ∗ x = 1 − ei x = x for any x ∈ soc(Mj ). i =j ∗

Suppose that j ∗ = j. We then obtain, using Remark 4.4.15.i, that

δ1(x) = δ1(ej ∗ x) = δ1(xej ∗ ) = δ1(xej ej ∗ ) = δ1(0) = 0 and therefore that {0}=δ1 k[H]x = δ1 xk[H] = δx k[H ] for any x ∈ soc(Mj ). By Remark 4.4.15.ii this implies soc(Mj ) ={0} which is a contradiction.  = ∪ = ∪ := 01 Exercise (Bruhat Decomposition) G B BwU B UwB with w −10 . Lemma 4.4.18 i. The k[G]-module Vp is uniserial of length 2, and there is an exact sequence of k[G]-modules α β 0 −→ V1 −→ Vp −→ Vp−2 −→ 0. p p p−1 ii. The socle of Vp as a k[U]-module is equal to kY ⊕ k(X − XY ).

Proof We define the k-linear map β : Vp −→ Vp−2 by − − − − β XiY p i := iXi 1Y p 1 i for 0 ≤ i ≤ p.

In order to see that β is a k[G]-module homomorphism we have to check that β(gv) = gβ(v) holds true for any g ∈ G and v ∈ Vp. By additivity it suffices to consider the vectors v = XiY p−i for 0 ≤ i ≤ p. Moreover, as a consequence of the Bruhat decomposition it also suffices to consider the group elements g = u+,w and g ∈ T . All these cases are easy one line computations. Obviously β is surjective with ker(β) = kXp + kYp. On the other hand we have the injective ring homomorphism α : k[X, Y ]−→k[X, Y ] defined by

α(X) := Xp,α(Y):= Y p, and α|k = id . = ab ∈ For g cd G we compute α(gX) = α(aX + cY) = aXp + cY p = (aX + cY)p = (gX)p = g Xp = gα(X) 4.4 An Example: The Group SL2(Fp) 133 and similarly α(gY) = gα(Y). This shows that α is an endomorphism of k[G]- =∼ modules. It obviously restricts to an isomorphism α : V1 −→ ker(β). Hence we have the exact sequence in i. Since V1 and Vp−2 are simple k[G]-modules the length of Vp is 2. Next we prove the assertion ii. The socle of Vp (as a k[U]-module) contains the image under α of the socle of V1 and is mapped by β into the socle of Vp−2.Using Lemma 4.4.3.ii we deduce that p p−2 kY ⊆ soc(Vp) and β soc(Vp) ⊆ kY .

In particular, soc(Vp) is at most two-dimensional. It also follows that −1 p−2 p p p−1 soc(Vp) ⊆ β kY = kY ⊕ kX ⊕ kXY . We have u+Y p = Y p,u+Xp = Xp + Y p,u+ XY p−1 = XY p−1 + Y p,

+ p p−1 p p−1 p p−1 hence u (X − XY ) = X − XY and therefore X − XY ∈ soc(Vp). p It remains to show that Vp is uniserial. Suppose that it is not. Then Vp = (kX + p ∼ kY ) ⊕ N for some k[G]-submodule N = Vp−2. We are going to use the filtration p p kY = W1 ⊂ W2 ⊂···⊂Wp ⊂ Wp+1 = Vp with Vp = Wp ⊕ kX introduced before Lemma 4.4.2. The identity p p p−1 p kY ⊕ k X − XY = soc(Vp) = kY ⊕ soc(N)

p implies that the socle of N contains a vector of the form u0 + X with u0 ∈ Wp.Let p p v = u + aX with u ∈ Wp and a ∈ k be any vector in N. Then v − a(u0 + X ) = u − au0 ∈ Wp ∩ N.Ifu − au0 = 0 then this element, by Lemma 4.4.2.iii, gen- p erates, as a k[U]-module, Wi for some 1 ≤ i ≤ p. It would follow that kY = p W1 ⊆ Wi ⊆ N which is a contradiction. Hence u = au0, v = a(u0 + X ), and p N ⊆ k(u0 + X ).ButVp−2 has k-dimension at least 2, and we have arrived at a contradiction again. 

Lemma 4.4.19 i. For any n ≥ 1 there is an exact sequence of k[G]-modules

γ μ 0 −→ Vn−1 −→ V1 ⊗k Vn −→ Vn+1 −→ 0. ∼ ii. For 1 ≤ n ≤ p − 2 we have V1 ⊗k Vn = Vn−1 ⊕ Vn+1.

Proof i. Since G acts on k[X, Y ] by ring automorphisms the multiplication

μ: k[X, Y ]⊗k k[X, Y ]−→k[X, Y ]

f1 ⊗ f2 −→ f1f2 134 4 Green’s Theory of Indecomposable Modules is a homomorphism of k[G]-modules. It restricts to a surjective map

μ: V1 ⊗k Vn −→ Vn+1.

On the other hand we have the k-linear map

γ : Vn−1 −→ V1 ⊗k Vn v −→ X ⊗ Yv− Y ⊗ Xv. = ab ∈ For g cd G we compute

gγ(v) = g(X ⊗ Yv− Y ⊗ Xv) = gX ⊗ gYgv − gY ⊗ gXgv = (aX + cY) ⊗ (bX + dY)gv − (bX + dY)⊗ (aX + cY)gv = abX ⊗ Xgv + adX ⊗ Ygv+ cbY ⊗ Xgv + cdY ⊗ Ygv − (baX ⊗ Xgv + bcX ⊗ Ygv+ daY ⊗ Xgv + dcY ⊗ Ygv) = (ad − bc)X ⊗ Ygv− (ad − bc)Y ⊗ Xgv = X ⊗ Ygv− Y ⊗ Xgv = γ(gv).

Hence γ is a k[G]-module homomorphism. We obviously have im(γ ) ⊆ ker(μ). Since V1 ⊗k Vn = X ⊗ Vn ⊕ Y ⊗ Vn it also is clear that γ is injective. But

dimk Vn−1 = n = 2(n + 1) − (n + 2)

= dimk V1 ⊗k Vn − dimk Vn+1

= dimk ker(μ).

It follows that im(γ ) = ker(μ) which establishes the exact sequence in i. ii. In the given range of n the modules Vn−1 and Vn+1 are simple and noniso- morphic. As a consequence of i. the assertion ii. therefore is equivalent to V1 ⊗k Vn having a k[G]-submodule which is isomorphic to Vn+1, resp. to the nonvanishing of Homk[G](Vn+1,V1 ⊗k Vn).Forn = p − 2 this is clear, again by i., from the projec- tivity of Vp−1. We now argue by descending induction with respect to n. Suppose ⊗ =∼ ⊕ ∗ that V1 k Vn+1 Vn Vn+2 holds true. We observe that with V1 also V1 is a sim- ple k[G]-module by Remark 4.4.11. But since V1, up to isomorphism, is the only [ ] ∗ =∼ simple two-dimensional k G -module we must have V1 V1. We also recall from linear algebra that the map

∼ ⊗ −−=→ ⊗ ∗ Homk[G](Vn+1,V1 k Vn) Homk[G] Vn+1 k V1 ,Vn −→ ⊗ −→ ⊗ A v l (l idVn ) A(v) is bijective. It follows that 4.4 An Example: The Group SL2(Fp) 135 ⊗ =∼ ⊗ ∗ Homk[G](Vn+1,V1 k Vn) Homk[G] Vn+1 k V1 ,Vn ∼ = Homk[G](Vn+1 ⊗k V1,Vn) ∼ = Homk[G](Vn ⊕ Vn+2,Vn) = {0}. 

For the remainder of this section let {P0},...,{Pp−1} be the isomorphism classes of finitely generated indecomposable projective k[G]-modules numbered in such a way that ∼ Pn/ rad(Pn) = Vn for 0 ≤ n

Proof With Vp−1 also the k[G]-module V1 ⊗k Vp−1 is projective by Proposi- tion 4.4.7 and Remark 4.4.16.ii. Let

V1 ⊗k Vp−1 = L1 ⊕···⊕Ls be a decomposition into indecomposable projective k[G]-modules Li . By Lemma 4.4.19.i we have an injective homomorphism of k[G]-modules γ : γ Vp−2 −→ V1 ⊗k Vp−1. Obviously, not all of the composed maps Vp−2 −→ V1 ⊗k pr Vp−1 −→ Li can be equal to the zero map. Since Vp−2 is simple we therefore find a 1 ≤ i ≤ s together with an injective k[G]-module homomorphism Vp−2 −→ Li .But according to Proposition 4.4.17 the only indecomposable projective k[G]-module ∼ whose socle is isomorphic to Vp−2 is, up to isomorphism, Pp−2. Hence Pp−2 = Li , i.e. Pp−2 is isomorphic to a direct summand of V1 ⊗k Vp−1. To prove our assertion it remains to check that Pp−2 and V1 ⊗k Vp−1 have the same k-dimension or rather that dimk Pp−2 ≥ 2p. As already recalled from Proposition 4.4.17 we have ∼ ∼ Pp−2/ rad(Pp−2) = soc(Pp−2) = Vp−2.

Since Pp−2 is indecomposable and Vp−2 is simple and nonprojective (cf. Propo- sition 4.4.7) we must have soc(Pp−2) ⊆ rad(Pp−2) (cf. Remark 4.4.22.i below). It follows that dimk Pp−2 ≥ 2dimk Vp−2 = 2p − 2. On the other hand we know from Lemma 4.4.6.ii that p divides dimk Pp−2.Asp = 2 we conclude that dimk Pp−2 ≥ 2p. 

Corollary 4.4.21 Pp−2 is a uniserial k[G]-module of length 3 such that [Pp−2]= 2[Vp−2]+[V1] in R(k[G]). 136 4 Green’s Theory of Indecomposable Modules

Proof Since we know Vp−2 to be isomorphic to the socle of the indecomposable k[G]-module Pp−2 by Proposition 4.4.17 it follows from Proposition 4.4.20 and Lemma 4.4.19.i that [Pp−2]=[Vp−2]+[Vp] and that Pp−2 is uniserial if Vp is uniserial. It remains to invoke Lemma 4.4.18.i. 

We point out the following general facts which are used repeatedly.

Remark 4.4.22 Let H be a finite group, and let M be a finitely generated projective k[H]-module; we then have: i. If M is indecomposable but not simple then soc(M) ⊆ rad(M); if moreover, rad(M)/ soc(M) is simple then M is uniserial; ii. suppose that M has a factor module of the form N1 ⊕···⊕Nr where the Ni are simple k[H ]-modules; let Li  Ni be a projective cover of Ni ; then L1 ⊕···⊕ Lr is isomorphic to a direct summand of M.

Proof i. By Proposition 4.4.17 we have the isomorphic simple modules soc(M) =∼ M/rad(M).Ifsoc(M) ⊆ rad(M) it would follow that the projection map soc(M) =∼ −→ M/rad(M) is an isomorphism. Hence M = soc(M) ⊕ rad(M) which is a con- tradiction. Since M is indecomposable soc(M) is the unique minimal nonzero and rad(M) the unique maximal proper submodule of M. The additional assumption on M therefore guarantees that there are no other nonzero proper submodules. ii. Let

M = M1 ⊕···⊕Ms be a decomposition into indecomposable projective k[H ]-modules Mi . Then N1 ⊕···⊕Nr is a factor module of the semisimple k[H ]-module M/rad(M) = M1/ rad(M1) ⊕···⊕Ms/ rad(Ms). The Jordan–Hölder Proposition 1.1.2 now im- plies the existence of an injective map σ :{1,...,r} →{1,...,s} such that ∼ Ni = Mσ(i)/ rad(Mσ(i)) ∼ and hence Li = Mσ(i) for any 1 ≤ i ≤ r. 

Proposition 4.4.23 For p>3 we have: ∼ i. V ⊗ P − = P − ⊕ V − ⊕ V − ; 1 k p ∼2 p 3 p 1 p 1 ii. V1 ⊗k Pn = Pn−1 ⊕ Pn+1 for any 2 ≤ n ≤ p − 3; iii. [P ]=2[V ]+[V − − ]+[V − − ] in R(k[G]) for any 1 ≤ n ≤ p − 3; n ∼n p 1 n p 3 n iv. V1 ⊗k P1 = P0 ⊕ P2 ⊕ Vp−1; v. P0 is a uniserial k[G]-module of length 3 such that [P0]=2[V0]+[Vp−3]. ∼ If p = 3 then V1 ⊗k P1 = P0 ⊕ V2 ⊕ V2 ⊕ V2 and [P0]=3[V0].

Proof We know from Corollary 4.4.21 that ∼ ∼ Pp−2/ rad(Pp−2) = Vp−2, rad(Pp−2)/ soc(Pp−2) = V1, ∼ soc(Pp−2) = Vp−2. 4.4 An Example: The Group SL2(Fp) 137

Hence V1 ⊗k Pp−2 has submodules N ⊃ L such that ∼ ∼ V1 ⊗k Pp−2/N = V1 ⊗k Vp−2 = Vp−3 ⊕ Vp−1, ∼ ∼ N/L= V1 ⊗k V1 = V0 ⊕ V2, ∼ ∼ L = V1 ⊗k Vp−2 = Vp−3 ⊕ Vp−1, where the second column of isomorphisms comes from Lemma 4.4.19.ii. Since Vp−1 is projective we may apply Remark 4.4.16.iii iteratively to obtain that V1 ⊗k Pp−2 has a factor module which is isomorphic to Vp−3 ⊕ Vp−1 ⊕ Vp−1, resp. V0 ⊕ V2 ⊕ V2 ⊕ V2 if p = 3. Using Remark 4.4.22.i we then see that the module Pp−3 ⊕ Vp−1 ⊕ Vp−1, resp. P0 ⊕ V2 ⊕ V2 ⊕ V2 if p = 3, is isomorphic to a direct summand of V1 ⊗k Pp−2. It remains to compare k-dimensions. We have dimk V1 ⊗k Pp−2 = 4p. Hence we have to show that dimk Pp−3 ≥ 2p if p>3, resp. dimk P0 ≥ 3ifp = 3. From Propositions 4.4.17 and 4.4.7 we know (cf. Re- mark 4.4.22.i) that

dimk Pp−3 ≥ 2dimk Vp−3 = 2p − 4 and from Lemma 4.4.6.ii that p divides dimk Pp−3. Both together obviously imply the asserted inequalities. This establish the assertion i. as well as the first half of the case p = 3. We also obtain

[Pp−3]=[V1 ⊗k Pp−2]−2[Vp−1]

=[Vp−3]+[Vp−1]+[V0]+[V2]+[Vp−3]+[Vp−1]−2[Vp−1]

= 2[Vp−3]+[V2]+[V0] (4.4.1) if p>3, resp.

[P0]=[V1 ⊗k P1]−3[V2]=3[V0] if p = 3. This is the case n = p − 3 of assertion iii. and the second half of the case p = 3. Next we establish the case n = p − 3 of assertion ii. (in particular p ≥ 5). Using (4.4.1) and Lemma 4.4.19.ii we obtain

[V1 ⊗k Pp−3]=2[V1 ⊗k Vp−3]+[V1 ⊗k V0]+[V1 ⊗k V2]

= 2[Vp−4]+2[Vp−2]+[V1]+[V1]+[V3]

= 2[Vp−2]+2[Vp−4]+[V3]+2[V1]. ∼ On the other hand V1 ⊗k Pp−3 has the factor module V1 ⊗k Vp−3 = Vp−4 ⊕ Vp−2. Hence Remark 4.4.22.ii says that Pp−4 ⊕ Pp−2 is isomorphic to a direct summand of V1 ⊗k Pp−3. By Corollary 4.4.21 the summand Pp−2 contributes 2[Vp−2]+[V1] to the above class, whereas the summand Pp−4 contributes at least 138 4 Green’s Theory of Indecomposable Modules

[Pp−4/ rad(Pp−4)]+[soc(Pp−4)]=2[Vp−4]. The difference is [V1]+[V3] which cannot come from other indecomposable projective summands Pm of V1 ⊗k Pp−3 since each of those would contribute another 2[Vm]. It follows that ∼ V1 ⊗k Pp−3 = Pp−4 ⊕ Pp−2 and [Pp−4]=2[Vp−4]+[V3]+[V1].

This allows us to establish ii. and iii. by descending induction. Suppose that

[Pn]=2[Vn]+[Vp−1−n]+[Vp−3−n] and

[Pn+1]=2[Vn+1]+[Vp−2−n]+[Vp−4−n] hold true for some 2 ≤ n ≤ p − 4 (the case n = p − 4 having been settled above). Using Lemma 4.4.19.ii we obtain on the one hand that

[V1 ⊗k Pn]=2[V1 ⊗k Vn]+[V1 ⊗k Vp−1−n]+[V1 ⊗k Vp−3−n]

= 2[Vn−1]+2[Vn+1]+[Vp−2−n]+[Vp−n]+[Vp−4−n]+[Vp−2−n]

= 2[Vn+1]+2[Vn−1]+[Vp−n]+2[Vp−2−n]+[Vp−4−n]. ∼ On the other hand V1 ⊗k Pn has the factor module V1 ⊗k Vn = Vn−1 ⊕ Vn+1 and hence, by Remark 4.4.22.ii, a direct summand isomorphic to Pn−1 ⊕ Pn+1.This summand contributes to the above class at least 2[Vn−1]+2[Vn+1]+[Vp−2−n]+ [Vp−4−n]. Again the difference [Vp−n]+[Vp−2−n] cannot arise from other inde- composable direct summands of V1 ⊗k Pn. It therefore follows that ∼ V1 ⊗k Pn = Pn−1 ⊕ Pn+1 and [Pn−1]=2[Vn−1]+[Vp−n]+[Vp−2−n].

It remains to prove the assertions iv. and v. Using iii. for n = 1 and Lem- ma 4.4.19.ii we have

[V1 ⊗k P1]=2[V1 ⊗k V1]+[V1 ⊗k Vp−2]+[V1 ⊗k Vp−4]

= 2[V0]+2[V2]+[Vp−3]+[Vp−1]+[Vp−5]+[Vp−3]

=[Vp−1]+2[Vp−3]+[Vp−5]+2[V2]+2[V0].

In particular, V1 ⊗k P1 has a subquotient isomorphic to the projective module ∼ Vp−1. It also has a factor module isomorphic to V1 ⊗k V1 = V0 ⊕ V2.UsingRe- marks 4.4.16.iii and 4.4.22.ii we see first that V1 ⊗k P1 has a factor module iso- morphic to V0 ⊕ V2 ⊕ Vp−1 and then that it has a direct summand isomorphic to P0 ⊕ P2 ⊕ Vp−1. By iii. for n = 2 the latter contributes to the above class at least 2[V0]+2[V2]+[Vp−3]+[Vp−5]+[Vp−1]. For a third time we argue that the difference [Vp−3] cannot come from another indecomposable direct summand of 4.4 An Example: The Group SL2(Fp) 139

V1 ⊗k P1 so that we must have ∼ V1 ⊗k P1 = P0 ⊕ P2 ⊕ Vp−1 and [P0]=2[V0]+[Vp−3].

It finally follows from Remark 4.4.22.i that P0 is uniserial (for all p ≥ 3). 

Remark 1

i. dimk P0 = dimk Pp−1 = p. ii. dimk Pn = 2p for 1 ≤ n ≤ p − 2. iii. A closer inspection of the above proof shows that for any of the modules P = P1,...,Pp−3 the subquotient rad(P )/ soc(P ) is semisimple of length 2 so that P is not uniserial.

Exercise V1 ⊗k P0 = P1.

We deduce from Proposition 4.4.7, Corollary 4.4.21, and Proposition 4.4.23 that the Cartan matrix of k[G], which has size p × p, is of the form ⎛ ⎞ 300 ⎝030⎠ for p = 3 001 and ⎛ ⎞ 210 ⎜ ⎟ ⎜ 2 . 1 ⎟ ⎜ ⎟ ⎜ 2 . 1 ⎟ ⎜ ⎟ ⎜ ...⎟ ⎜ ⎟ ⎜ ... ⎟ ⎜ ⎟ ⎜ 21. ⎟ ⎜ ⎟ ⎜ 31 ⎟ ⎜ ⎟ ⎜ 13 ⎟ for p>3. ⎜ ⎟ ⎜ . 12 ⎟ ⎜ ⎟ ⎜ .. . ⎟ ⎜ ⎟ ⎜ .. . ⎟ ⎜ ⎟ ⎜ .. . ⎟ ⎜ ⎟ ⎜11 2 ⎟ ⎝ 12⎠ 01

If we reorder the simple modules Vn into the sequence

V0,V(p−1)−2,V2,V(p−1)−4,V4,...,V(p−1)−1,V1,V(p−1)−3,V3,...,Vp−1 140 4 Green’s Theory of Indecomposable Modules and correspondingly the Pn then the Cartan matrix becomes block diagonal, with 3 blocks, of the form ⎛ ⎞ 21 ⎜ ⎟ ⎜12 1 ⎟ ⎜ ⎟ ⎜ 12 ⎟ ⎜ ⎟ ⎜ ...... ⎟ ⎜ . . . ⎟ ⎜ ⎟ ⎜ 21 ⎟ ⎜ ⎟ ⎜ 130 ⎟ ⎜ ⎟ ⎜ 0 21 ⎟ . ⎜ ⎟ ⎜ 12 1 ⎟ ⎜ ⎟ ⎜ 12 ⎟ ⎜ ⎟ ⎜ . . . ⎟ ⎜ ...... ⎟ ⎜ ⎟ ⎜ 21 ⎟ ⎝ 130⎠ 0 1

This reflects the fact, as we will see later on, that k[G] has exactly three different blocks.

4.5 Green’s Indecomposability Theorem

We fix a prime number p and an algebraically closed field of characteristic p.Be- fore we come to the subject of this section we recall two facts about the matrix algebras Mn×n(k).

Lemma 4.5.1 Any automorphism T : Mn×n(k) −→ Mn×n(k) of the k-algebra Mn×n(k), for n ≥ 1, is inner, i.e. there exists a matrix T0 ∈ GLn(k) such that = −1 ∈ T(A) T0AT0 for any A Mn×n(k).

Proof The algebra Mn×n(k) is simple and semisimple. Its, up to isomorphism, unique simple module is kn with the natural action. We now use T to define a new module structure on kn by

n n Mn×n(k) × k −→ k (A, v) −→ T(A)v which we denote by T ∗kn. Since any submodule of T ∗kn also is a submodule of kn we obtain that T ∗kn is simple as well. Hence there must exist a module isomorphism ∼ n = ∗ n k −→ T k . This means that we find a matrix T0 ∈ GLn(k) such that n T0Av = T(A)T0v for any v ∈ k and A ∈ Mn×n(k).

It follows that T0A = T(A)T0 for any A ∈ Mn×n(k).  4.5 Green’s Indecomposability Theorem 141

Lemma 4.5.2 Let T be an automorphism of the k-algebra Mp×p(k) with the prop- erty that

T(ei) = ei+1 for 1 ≤ i ≤ p − 1 and T(ep) = e1, where ei ∈ Mp×p(k) denotes the diagonal matrix with a 1 for the ith entry of the diagonal and zeros elsewhere. Then the subalgebra Q := A ∈ Mp×p(k) : T(A)= A is local with Q/ Jac(Q) = k.

Proof According to Lemma 4.5.1 we find a matrix T0 ∈ GLp(k) such that T(A)= −1 ∈ T0AT0 for any A Mp×p(k). In particular

T0ei = ei+1T0 for 1 ≤ i ≤ p − 1 and T0ep = e1T0.

This forces T0 to be of the form ⎛ ⎞ 0 ··· ··· ··· tp ⎜ ⎟ ⎜ . ⎟ ⎜t1 0 . ⎟ ⎜ ⎟ × ⎜ . .. . ⎟ with t ,...,t ∈ k , ⎜ .t2 . . ⎟ 1 p ⎜ ⎟ ⎝ . . . ⎠ . .. 0 . 0 ··· ··· tp−1 0 which is conjugate to ⎛ ⎞ 0 ··· ··· ··· tp ⎜ ⎟ ⎜ . ⎟ ⎜10 . ⎟ ⎜ ⎟ ⎜ . .. . ⎟ p = · ··· · ⎜ . 1 . . ⎟ where t t1 tp. ⎜ ⎟ ⎝ . . . ⎠ . .. 0 . 0 ··· ··· 10

The minimal polynomial of this latter matrix is Xp − tp = (X − t)p. It follows that the Jordan normal form of T0 is ⎛ ⎞ t 0 ⎜ ⎟ ⎜1 t ⎟ ⎜ ⎟ ⎜ .. .. ⎟ ⎜ . . ⎟ , ⎜ . ⎟ ⎝ .. t ⎠ 01t 142 4 Green’s Theory of Indecomposable Modules and hence that the algebra Q is isomorphic to the algebra ⎧ ⎛ ⎞ ⎛ ⎞⎫ ⎪ t t ⎪ ⎪ ⎜ ⎟ ⎜ ⎟⎪ ⎨ ⎜1 t ⎟ ⎜1 t ⎟⎬  := ∈ : ⎜ ·· ⎟ = ⎜ ·· ⎟ Q ⎪A Mp×p(k) ⎜ ⎟ A A ⎜ ⎟⎪ . ⎪ ⎝ · ⎠ ⎝ · ⎠⎪ ⎩⎪ t t ⎭⎪ 1 t 1 t

We leave it to the reader to check that: − Q is the subalgebra of all matrices of the form ⎛ ⎞ a 00 ⎜ ⎟ ⎜∗ a ⎟ ⎜ ⎟ ⎜ · ⎟ . ⎝ a 0⎠ ∗∗a − ⎧⎛ ⎞⎫ ⎪ 00⎪ ⎪⎜ ⎟⎪ ⎨⎜∗ 0 ⎟⎬ := ⎜ · ⎟ I ⎪⎜ ⎟⎪ ⎪⎝ ⎠⎪ ⎩⎪ 0 ⎭⎪ ∗∗0 is a nilpotent two-sided ideal in Q. − Q/I = k. In particular, Q is local with I = Jac(Q) (cf. Proposition 1.2.1.v and Proposi- tion 1.4.1). 

We now let R be a noetherian, complete, local commutative ring such that R/Jac(R) = k (e.g., R = k or R a (0,p)-ring for k), and we let G be a finite group. We fix a normal subgroup N ⊆ G as well as a finitely generated indecomposable R[N]-module L. In this situation Mackey’s Proposition 4.2.4 simplifies as follows. Let {g1,...,gm}⊆G be a set of representatives for the cosets in N\G/N = G/N. [ ] ⊗ ⊆ [ ]⊗ = G We identify L with the R N -submodule 1 L R G R[N] L IndN (L). Then G = ⊕···⊕ =∼ −1 ∗ IndN (L) g1L gmL and giL gi L (4.5.1) as R[N]-modules (cf. Remark 2.5.2.i). With L any giL is an indecomposable R[N]-module (cf. Remark 2.5.2). Hence (4.5.1) is, in fact, a decomposition of G [ ] IndN (L) into indecomposable R N -modules. In slight generalization of the dis- cussion at the beginning of Sect. 2.5 we have that ∗ ∼ IG(L) := g ∈ G : g L = L as R[N]-modules is a subgroup of G which contains N. 4.5 Green’s Indecomposability Theorem 143

= [ ] G Lemma 4.5.3 If IG(L) N then the R G -module IndN (L) is indecomposable.

[ ] G = ⊕ Proof Suppose that the R G -module IndN (L) M1 M2 is the direct sum of two submodules. By the KrullÐRemakÐSchmidt Theorem 1.4.7 one of this summands, say M1, has a direct summand M0,asanR[N]-module, which is isomorphic to L. =∼ −1 ∗ =∼ −1 ∗ =∼ Then giL (gi ) L (gi ) M0 giM0 is (isomorphic to) a direct summand of giM1 = M1. Using again Theorem 1.4.7 we see that in any decomposition of M1 into indecomposable R[N]-modules all the giL must occur up to isomorphism. On the other hand our assumption means that the R[N]-modules giL are pairwise G nonisomorphic, so that any giL occurs in IndN (L) with multiplicity one. This shows that necessarily M2 ={0}. 

Remark 4.5.4 EndR[N](L) is a local ring with EndR[N](L)/ Jac EndR[N](L) = k.

Proof The first half of the assertion is a consequence of Propositions 1.3.6 and 1.4.5. From Lemma 1.3.5.ii/iii we know that EndR[N](L) is finitely generated as an R- module and that Jac(R) EndR[N](L) ⊆ Jac EndR[N](L) .

It follows that the skew field D := EndR[N](L)/ Jac(EndR[N](L)) is a finite- dimensional vector space over R/Jac(R) = k. Since k is algebraically closed we must have D = k. 

Lemma 4.5.5 Suppose that [G : N]=p and that IG(L) = G; then the ring EG := G = EndR[G](IndN (L)) is local with EG/ Jac(EG) k.

Proof We have the inclusions of rings = G ⊆ := G EG EndR[G] IndN (L) EN EndR[N] IndN (L) ⊆ := G E EndR IndN (L) .

Fixing an element h ∈ G such that hN generates the cyclic group G/N we note that G the action of h on IndN (L) is an R-linear automorphism and therefore defines a unit θ ∈ E×. We obviously have

EG ={α ∈ EN : αθ = θα}.

We claim that the ring automorphism

Θ: E −→ E α −→ θαθ−1 144 4 Green’s Theory of Indecomposable Modules satisfies Θ(EN ) = EN and therefore restricts to a ring automorphism Θ : EN −→ EN .Letα ∈ EN , which means that α ∈ E satisfies α(gx) = gα(x) ∈ ∈ G for any g N and x IndN (L). Then θαθ−1(gx) = hα h−1gx = hα h−1ghh−1x = hh−1ghα h−1x = ghα h−1x = g θαθ−1 (x)

∈ ∈ G −1 ∈ −1 ∈ for any g N and x IndN (L). Hence θαθ EN , and similarly θ αθ EN . This proves our claim, and we deduce that EG = α ∈ EN : Θ(α) = α .

An immediate consequence of this identity is the fact that

× = ∩ × EG EG EN .

Using Proposition 1.2.1.iv it follows that ∩ = ∈ : + ⊆ × EG Jac(EN ) α EG 1 EN α EN ⊆ ∈ : + ⊆ × ∩ α EG 1 EGα EN EG = ∈ : + ⊆ × α EG 1 EGα EG

= Jac(EG).

p−1 In order to compute EN we note that {1,h,...,h } is a set of representatives for the cosets in G/N. The decomposition (4.5.1) becomes

G = ⊕ ⊕···⊕ p−1 IndN (L) L hL h L.

∼ −1 ∗ ∼ Furthermore, by our assumption that IG(L) = G we have hL = (h ) L = L and hence G =∼ ⊕···⊕ IndN (L) L L as R[N]-modules with p summands L on the right-hand side. We fix such an iso- morphism. It induces a ring isomorphism

∼ −−→= 0 0 := EN Mp×p EN with EN EndR[N](L) ⎛ ⎞ α11 ··· α1p ⎜ ⎟ −→ . . α ⎝ . . ⎠ αp1 ··· αpp 4.5 Green’s Indecomposability Theorem 145

∈ G ∈ ⊕···⊕ where, if x IndN (L) corresponds to (x1,...,xp) L L, then α(x) corre- sponds to p p α1j (xj ), . . . , αpj (xj ) . j=1 j=1

The automorphism Θ of EN then corresponds to an automorphism, which we de- 0 note by T ,ofMp×p(EN ). On the other hand, using Lemma 1.2.5 and Remark 4.5.4, we have ∼ −→= 0 0 = EN / Jac(EN ) Mp×p EN / Jac EN Mp×p(k). As any ring automorphism, T respects the Jacobson radical and therefore induces a ring automorphism T of Mp×p(k). Introducing the subring Q := A ∈ Mp×p(k) : T(A) = A we obtain the commutative diagram of injective ring homomorphisms

EG/EG ∩ Jac(EN ) Q

=∼ EN / Jac(EN ) Mp×p(k).

We remark that Jac(R)EN ⊆ Jac(EN ) by Lemma 1.3.5.ii/iii. It follows that the above diagram in fact is a diagram of R/Jac(R) = k-algebras. Next we observe that the automorphism Θ has the following property. Let αi ∈ EN ,for1≤ i ≤ p, be the endomorphism such that = − j id if j i 1, αi|h L = 0 otherwise.

Since θ(hj L) = hj+1L for 0 ≤ j

−1 −1 θαiθ = αi+1 for 1 ≤ i

Θ(αi ) = αi+1 for 1 ≤ i

Theorem 4.5.6 (Green) Let N ⊆ G be a normal subgroup such that the index [G : N] is a power of p; for any finitely generated indecomposable R[N]-module L [ ] G the R G -module IndN (L) is indecomposable.

Proof Since G/N is a p-group we find a sequence of normal subgroups

N = N0 ⊂ N1 ⊂···⊂Nl = G in G such that [Ni : Ni−1]=p for any 1 ≤ i ≤ l. By induction we therefore may as- sume that [G : N]=p.IfIG(L) = N the assertion follows from Lemma 4.5.3. = := G Suppose therefore that IG(L) G. Then the ring EG EndR[G](IndN (L)) is local with EG/ Jac(EG) = k by Lemma 4.5.5. It also is complete by Proposi- tion 1.3.6.iii. Hence Proposition 1.5.11 says that 1 is the only idempotent in EG. G On the other hand, the projection of IndN (L) onto any nonzero direct summand as [ ] G an R G -module is obviously an idempotent in EG. It follows that IndN (L) must be indecomposable.  Chapter 5 Blocks

Throughout this chapter we fix a prime number p, an algebraically closed field k of characteristic p, as well as a finite group G.LetE := E(G) := {e1,...,er } be the set of all primitive idempotents in the center Z(k[G]) of the group ring k[G]. We know from Proposition 1.5.5.iii/iv that the ei are pairwise orthogonal and satisfy e1 +···+er = 1. We recall that the ei -block of k[G] consists of all k[G]-modules M such that eiM = M. An arbitrary k[G]-module M decomposes uniquely and naturally into submodules

M = e1M ⊕···⊕er M where eiM belongs to the ei -block. In particular, we have: Ð If a module M belongs to a block then any submodule and any factor module of M belongs to the same block. Ð Any indecomposable module belongs to a unique block. By Proposition 1.5.3 the block decomposition

k[G]=k[G]e1 ⊕···⊕k[G]er of k[G] is a decomposition into two-sided ideals. By Corollary 1.5.4 it is the finest such decomposition in the sense that no k[G]ei can be written as the direct sum of two nonzero two-sided ideals.

5.1 Blocks and Simple Modules

We define an equivalence relation on the set k[G] as follows. For {P }, {Q}∈k[G]  ∼ we let {P }∼{Q} if there exists a sequence {P0},...,{Ps} in k[G] such that P0 = P , ∼ Ps = Q, and

Homk[G](Pi−1,Pi) = {0} or Homk[G](Pi,Pi−1) = {0}

P. Schneider, Modular Representation Theory of Finite Groups, 147 DOI 10.1007/978-1-4471-4832-6_5, © Springer-Verlag London 2013 148 5Blocks for any 1 ≤ i ≤ s. We immediately observe that, if {P } and {Q} lie in different equivalence classes, then

Homk[G](P, Q) = Homk[G](Q, P ) ={0}. Let

k[G]=Q1 ⊕···⊕Qm be a decomposition into indecomposable (projective) submodules. For any equiva- lence class C ⊆ k[G] we define  PC := Qi,

{Qi }∈C and we obtain the decomposition  k[G]= PC. C⊆k[G]

Lemma 5.1.1 For any equivalence class C ⊆ k[G] there exists a unique eC ∈ E such that PC = k[G]eC; the map ∼ set of equivalence classes in k[G] −→ E

C −→ eC is bijective.

Proof First let {P }, {Q}∈k[G] such that there exists a nonzero k[G]-module ho- momorphism f : P −→ Q. Suppose that P and Q belong to the e- and e -block, re- spectively. Then P/ker(f ) belongs to the e-block and im(f ) to the e -block. Since P/ker(f ) =∼ im(f ) = {0} we must have e = e . This easily implies that for any equivalence class C there exists a unique eC ∈ E such that PC belongs to the eC- block. Secondly, as already pointed out we have Homk[G](PC,PC ) ={0} for any two equivalence classes C = C . This implies that any k[G]-module endomorphism f : k[G]−→k[G] satisfies f(PC) ⊆ PC for any C. Since multiplication from the right by any element in k[G] is such an endomorphism it follows that each PC is a two- sided ideal of k[G]. Hence, for any e ∈ E,  k[G]e = ePC C⊆k[G] is a decomposition into two-sided ideals. It follows that there is a unique equivalence class C(e) such that ePC(e) = {0}; then, in fact, k[G]e = ePC(e). Since PC(e) belongs to the eC(e)-block we must have e = eC(e) and k[G]eC(e) = PC(e). On the other hand, given any equivalence class C we have PC = {0} by Corollary 1.7.5. Hence PC = eCPC implies C(eC) = C.  5.1 Blocks and Simple Modules 149

 Remark 5.1.2 For any {P }, {Q}∈k[G] we have Homk[G](P, Q) = {0} if and only if the simple module P/rad(P ) is isomorphic to a subquotient in some composition series of Q.

Proof First we suppose that there exists a nonzero k[G]-module homomorphism f : P −→ Q. The kernel of f then must be contained in the unique maximal submodule =∼ rad(P ) of P . Hence P/rad(P ) −→ f(P)/f(rad(P )). f Vice versa, let us suppose that there are submodules L ⊂ N ⊆ Q such that N/L=∼ P/rad(P ). We then have a surjective k[G]-module homomorphism P −→ N/L and therefore, by the projectivity of P , a commutative diagram of k[G]- module homomorphisms

P f

N N/L, pr where f cannot be the zero map. It follows that Homk[G](P, Q) contains the f ⊆ nonzero composite P −→ N −→ Q. 

Proposition 5.1.3 Let C ⊆ k[G] be an equivalence class; for any simple k[G]- module V the following conditions are equivalent:

i. V belongs to the eC-block; ii. V is isomorphic to P/rad(P ) for some {P }∈C; iii. V is isomorphic to a subquotient of Q for some {Q}∈C.

Proof i. =⇒ ii. Let V belong to the eC-block. Picking a nonzero vector v ∈ V we obtain the surjective k[G]-module homomorphism

k[G]−→V x −→ xv.

It restricts to a surjective k[G]-module homomorphism k[G]eC −→ eCV = V . Lemma 5.1.1 says that k[G]eC = PC is a direct sum of modules P such that {P }∈C. Hence V must be isomorphic to a factor module of one of these P . ii. =⇒ iii. This is trivial. iii. =⇒ i. By assumption V belongs to the same block as some Q with {Q}∈C. But Lemma 5.1.1 says that Q, and hence V , belongs to the eC-block. 

Corollary 5.1.4 If the simple k[G]-module V is projective then it is, up to isomor- phism, the only simple module in its block. 150 5Blocks

Proof We consider any {Q}∈k[G] different from {V }. By Remark 4.4.16.iii the projective module V cannot be isomorphic to a subquotient of Q. Hence

Homk[G](Q, V ) = Homk[G](V, Q) ={0}. It follows that the equivalence class of {V } consists of {V } alone. 

Let us consider again the example of the group G = SL2(Fp) for p>2. The simple k[G]-modules V0,V1,...,Vp−1, as constructed in Proposition 4.4.4,aredis- tinguished by their k-dimension dimk Vn = n + 1. Let Pn −→ Vn be a projective cover. In Proposition 4.4.7, Corollary 4.4.21, and Proposition 4.4.23.iii/v we have determined the simple subquotients of each Pn.ByusingRemark5.1.2 in order to translate the existence of certain subquotients into the existence of certain nonzero k[G]-module homomorphisms between the Pn we deduce the following facts:

Ð All simple subquotients of Pn for n even, resp. odd, are odd-, resp. even-, di- mensional. This implies that Homk[G](Pn,Pm) ={0} whenever n and m have different parity.

Ð Vp−1 is projective. Ð There exist nonzero k[G]-module homomorphisms:

Pp−3 P0

Pp−5 P2

. P4

..

P2 .

Pp−3 5.2 Central Characters 151

and

Pp−4 P1

Pp−6 P3

. P5

..

P1 .

Pp−2

Together they imply that the equivalence classes in k[G] are       {P0}, {P2},...,{Pp−3} , {P1}, {P3},...,{Pp−2} , and {Vp−1} .

We conclude that SL2(Fp) has three blocks. More precisely, we have

E ={eeven,eodd,eproj} such that a simple module V lies in the

eeven-block ⇐⇒ dimk V is even,

eodd-block ⇐⇒ dimk V is odd and = p, ∼ eproj-block ⇐⇒ V = Vp−1.

5.2 Central Characters

We have the decomposition      Z k[G] = Ze with Ze := Z k[G] e e∈E of the center as a direct product of the rings Ze (with unit element e). Since E is the set of all primitive idempotents in Z(k[G]) the rings Ze do not contain any other 152 5Blocks idempotent besides their unit elements. Therefore, by Proposition 1.5.11, each Ze is a local ring. Moreover, the skew fields Ze/ Jac(Ze) are finite-dimensional over the algebraically closed field k. Hence the maps

ιe: k −→ Ze/ Jac(Ze)

a −→ ae + Jac(Ze) are isomorphisms. This allows us to introduce the k-algebra homomorphisms

  −1 pr pr ιe χe: Z k[G] −→ Ze −→ Ze/ Jac(Ze) −−→ k, which are called the central characters of k[G].

Exercise The simple Z(k[G])-modules all are one-dimensional and correspond bi- jectively to the characters χe.

Proposition 5.2.1 Let e ∈ E, and let V beasimplek[G]-module; then V belongs to the e-block if and only if zv = χe(z)v for any z ∈ Z(k[G]) and v ∈ V .

Proof The equation ev = χe(e)v = 1v = v immediately implies eV = V and there- fore that V belongs to the e-block. Since k is algebraically closed Schur’s lemma implies that Endk[G](V ) = k idV . This means that the homomorphism Z(k[G]) −→ Endk[G](V ) induced by the action of Z(k[G]) on V can be viewed as a k-algebra homomorphism χ : Z(k[G]) −→ k such that zv = χ(z)v for any z ∈ Z(k[G]) and v ∈ V . The Jacobson radical [ ] = Jac(Z(k G )) e Jac(Ze), of course, lies in the kernel of χ. Suppose that V be- longs to the e-block. Then ev = v and e v = 0fore = e ∈ E and any v ∈ V .It follows that χ(e)= 1 and χ(e ) = 0 for any e = e, and hence that χ = χe. 

Let O(G) denote the set of conjugacy classes of G. For any conjugacy class O ⊆ G we define the element

Oˆ := g ∈ k[G]. g∈O

We recall that {Oˆ : O ∈ O(G)} is a k-basis of the center Z(k[G]). The multiplication in Z(k[G]) is determined by the equations

ˆ ˆ ˆ O1O2 = μ(O1, O2; O)O with μ(O1, O2; O) ∈ k. O∈O(G)

(Of course, the μ(O1, O2; O) lie in the prime field Fp.) 5.3 Defect Groups 153

5.3 Defect Groups

The group G × G acts on k[G] by

(G × G) × k[G]−→k[G]   (g, h), x −→ gxh−1.

In this way k[G] becomes a k[G × G]-module. The two-sided ideals in the ring k[G] coincide with the k[G × G]-submodules of k[G]. We see that the block de- composition  k[G]= k[G]e e∈E is a decomposition of the k[G × G]-module k[G] into indecomposable submodules. For any e ∈ E we therefore may consider the set V0(k[G]e) of vertices of the in- decomposable k[G × G]-module k[G]e. Of course, these vertices are subgroups of G × G. But we have the following result. To formulate it we need the “diagonal” group homomorphism

δ: G −→ G × G g −→ (g, g).

Proposition 5.3.1 The k[G × G]-module k[G] is relatively k[δ(G)]-projective.

Proof Because of (gxh−1, 1)δ(G) = (gx, h)δ(G) the map

∼ G −→ (G × G)/δ(G) x −→ (x, 1)δ(G) is an isomorphism of G × G-sets. By Lemma 2.4.6.i it induces an isomorphism

∼ [ ] −→= G×G k G Indδ(G) (k) of k[G × G]-modules. The assertion therefore follows from Proposition 4.1.6. 

Corollary 5.3.2 For any e ∈ E the indecomposable k[G × G]-module k[G]e has a vertex of the form δ(H) for some subgroup H ⊆ G; if H ⊆ G is another subgroup such that δ(H ) is a vertex of k[G]e then H and H are conjugate in G.

Proof By Proposition 5.3.1 and Lemma 4.1.2 the k[G × G]-module k[G]e, being a direct summand of k[G], is relatively k[δ(G)]-projective. This proves the first half of the assertion. By Proposition 4.2.5 there exists an element (g, h) ∈ G × G such that δ(H ) = (g, h)δ(H )(g, h)−1. It follows that H = gHg−1.  154 5Blocks

Definition Let e ∈ E; the subgroups D ⊆ G such that δ(D) ∈ V0(k[G]e) are called the defect groups of the e-block.

Defect groups exist and are p-subgroups of G by Lemma 4.2.3. The defect groups of a single block form a conjugacy class of subgroups of G.

Lemma 5.3.3 Let e ∈ E, and let D be a defect group of the e-block; then k[G]e as a k[δ(G)]-module is relatively k[δ(D)]-projective.

Proof Let

k[G]e = L1 ⊕···⊕Ls be a decomposition of the indecomposable k[G × G]-module k[G]e into indecom- posable k[δ(G)]-modules. By Lemma 4.1.2 it suffices to show that δ(D) ∈ V(Li) for any 1 ≤ i ≤ s. According to Lemma 4.3.1.i we find elements (gi,hi) ∈ G × G such that −1 (gi,hi)δ(D)(gi,hi) ∩ δ(G) ∈ V(Li). −1 But the equation (gi,hi)(d, d)(gi,hi) = (g, g) with d ∈ D and g ∈ G implies −1 = −1 gidgi hidhi and therefore

−1 −1 (gi,hi)δ(d)(gi,hi) = δ(gi)δ(d)δ(gi) .

It follows that

−1 −1 (gi,hi)δ(D)(gi,hi) ∩ δ(G) ⊆ δ(gi)δ(D)δ(gi) and hence that δ(D) ∈ V(Li) (cf. Exercise 4.2.2.ii and Lemma 4.2.1). 

Proposition 5.3.4 Let e ∈ E, and let D be a defect group of the e-block; any k[G]- module M belonging to the e-block is relatively k[D]-projective.

Proof We denote by (k[G]e)ad the k-vector space k[G]e viewed as a k[G]-module =∼ through the group isomorphism G −→ δ(G). This means that G acts on (k[G]e)ad δ by     G × k[G]e ad −→ k[G]e ad (g, x) −→ gxg−1.

As a consequence of Lemma 5.3.3 the module (k[G]e)ad is relatively k[D]- projective. We therefore find, by Proposition 4.1.6,ak[D]-module L such that [ ] ad G (k G e) is isomorphic to a direct summand of IndD(L). On the other hand we 5.3 Defect Groups 155 consider the k-linear maps   α ad β M −→ k[G]e ⊗k M −→ M v −→ e ⊗ v x ⊗ v −→ xv.

Because of

α(gv) = e ⊗ gv = geg−1 ⊗ gv = g(e ⊗ v) = gα(v) and     β g(x ⊗ v) = β gxg−1 ⊗ gv = gxg−1gv = gxv = gβ(x ⊗ v) both maps are k[G]-module homomorphisms. The composite map satisfies βα(v) = ev = v and hence is the identity map. It follows that M is isomorphic to a direct ad summand of (k[G]e) ⊗k M. Together we obtain that M is isomorphic to a direct G ⊗ summand of IndD(L) k M. But, as we have used before in the proof of Proposi- tion 2.3.4, there are isomorphisms of k[G]-modules   G ⊗ =∼ [ ]⊗ ⊗ =∼ [ ]⊗ ⊗ =∼ G ⊗ IndD(L) k M k G k[D] L k M k G k[D] (L k M) IndD(L k M). Therefore, Proposition 4.1.6 implies that M is relatively k[D]-projective. 

The last result says that a defect group of an e-block contains a vertex of any finitely generated indecomposable module in this block. Later on (Proposition 5.4.7) we will see that the defect group occurs among these vertices. Hence defect groups can be characterized as being the largest such vertices. At this point we start again from a different end. For any x ∈ G we let CG(x) := {g ∈ G : gx = xg} denote the centralizer of x.LetO ⊆ G be a conjugacy class.

Definition The p-Sylow subgroups of the centralizers CG(x) for x ∈ O are called the defect groups of the conjugacy class O.

For any p-subgroup P ⊆ G we define   ˆ IP := {kO : P contains a defect group of O} ⊆ Z k[G] .

Exercise 5.3.5 i. Any two defect groups of O are conjugate in G. ii. If P is a p-Sylow subgroup then IP = Z(k[G]). iii. If P ⊆ P then IP ⊆ IP . iv. IP only depends on the conjugacy class of P .

Lemma 5.3.6 Let O1, O2, and O in O(G) be such that μ(O1, O2; O) = 0; if the p-subgroup P ⊆ G centralizes an element of O then it also centralizes elements of O1 and O2. 156 5Blocks

Proof Suppose that P centralizes x ∈ O. We define   X := (y1,y2) ∈ O1 × O2 : y1y2 = x as a P -set through

P × X −→ X     −1 −1 g,(y1,y2) −→ gy1g ,gy2g .

One checks that

|X|=μ(O1, O2; O) = 0ink.

On the other hand let X = X1 ∪···∪Xm be the decomposition of X into its P - orbits. Since P is a p-group the |Xi| all are powers of p. Hence either |Xi|=0in k or |Xi|=1. We see that we must have at least one P -orbit Y ={(y1,y2)}⊆X consisting of one point. This means that P centralizes y1 ∈ O1 and y2 ∈ O2. 

Proposition 5.3.7 Let P1,P2 ⊆ G be p-subgroups; then

I I ⊆ I −1 . P1 P2 P1∩gP2g g∈G

Proof For i = 1, 2letOi ∈ O(G) such that Pi contains a defect subgroup Di of Oi . We have to show that

ˆ ˆ ˆ O O = μ(O , O ; O)O ∈ I −1 . 1 2 1 2 P1∩gP2g O∈O(G) g∈G

Obviously we only need to consider any O ∈ O(G) such that μ(O1, O2; O) = 0. We pick a defect subgroup D of O. By definition D centralizes an element of O. Lemma 5.3.6 says that D also centralizes elements of O1 and O2. We therefore find elements gi ∈ G,fori = 1, 2, such that ⊆ −1 ∩ −1 ⊆ −1 ∩ −1 D g1D1g1 g2D2g2 g1P1g1 g2P2g2 . := −1 Setting g g1 g2 and using Exercise 5.3.5.iv it follows that Oˆ ⊆ = k I −1∩ −1 IP ∩gP g−1 .  g1P1g1 g2P2g2 1 2

Corollary 5.3.8 IP , for any p-subgroup P ⊆ G, is an ideal in Z(k[G]).

Proof Let P ⊆ G be a fixed p-Sylow subgroup. Using Exercise 5.3.5.ii/iii we de- duce from Proposition 5.3.7 that   [ ] = ⊆ ⊆ IP Z k G IP IP IP ∩gP g−1 IP . g∈G  5.3 Defect Groups 157

∈ P ∈ Remark 5.3.9 Let e E, and let be a set of p-subgroups of G;ife P ∈P IP then IQe = Ze for some Q ∈ P.

Proof We have

e = ee ∈ IP e ⊆ Ze. P ∈P

As noted at the beginning of Sect. 5.2 the ring Ze is local. By Corollary 5.3.8 each IP e is an ideal in Ze. Since e/∈ Jac(Ze) we must have IQe ⊆ Jac(Ze) for at least one Q ∈ P.ButIQe then contains a unit so that necessarily IQe = Ze. 

For any subgroup H ⊆ G we have in k[G] the subring   K[G]ad(H ) := x ∈ k[G]:hx = xh for any h ∈ H .

We note that k[G]ad(G) = Z(k[G]). Moreover, there is the k-linear “trace map”   ad(H ) trH : k[G] −→ Z k[G]

x −→ gxg−1. g∈G/H

It satisfies

trH (yx) = trH (xy) = y trH (x) for any x ∈ k[G]ad(H ) and y ∈ Z(k[G]).

Lemma 5.3.10 IP ⊆ im(trP ) for any p-subgroup P ⊆ G. ˆ Proof It suffices to show that O lies in the image of trP for any conjugacy class O ∈ O(G) such that Q := P ∩ CG(x) is a p-Sylow subgroup of CG(x) for some x ∈ O. Since [CG(x) : Q] is prime to p and hence invertible in k we may define the element 1 y := hxh−1 ∈ k[G]ad(P ). [C (x) : Q] G h∈P/Q We compute

1 − − 1 − tr (y) = g hxh 1 g 1 = gxg 1 P [C (x) : Q] [C (x) : Q] g∈G/P G h∈P/Q G g∈G/Q

= gxg−1 = Oˆ .  g∈G/CG(x)

Proposition 5.3.11 Let P ⊆ G be a p-subgroup, and let e ∈ IP be an idempotent; for any k[G]-module M the k[G]-submodule eM is relatively k[P ]-projective. 158 5Blocks

Proof According to Lemma 5.3.10 there exists an element y ∈ k[G]ad(P ) such that e = trP (y). Since y commutes with the elements in P the map ψ: eM −→ eM v −→ yv is a k[P ]-module homomorphism. Let {g1,...,gm} be a set of representatives for m −1 = the cosets in G/P . Then i=1 giygi e, which translates into the identity m −1 = giψgi ideM . i=1 Our assertion therefore follows from Lemma 4.1.7. 

Corollary 5.3.12 Let e ∈ E, and let P ⊆ G be a p-subgroup; if e ∈ IP then P contains a defect subgroup of the e-block.

Proof Byassumptionwehave s ˆ e = aiOi i=1 with ai ∈ k and such that O1,...,Os are all conjugacy classes such that P ∩CG(xi), for some xi ∈ Oi ,isap-Sylow subgroup of CG(xi). We define a central idempotent ε in k[G × G] as follows. Obviously

O(G × G) = O(G) × O(G).

For O ∈ O(G) we let O−1 := {g−1 : g ∈ O}∈O(G). We put s     := O O−1 ∧ ∈ [ × ] ε aiaj i, j Z k G G . i,j=1

2 With e also ε is nonzero. To compute ε it is more convenient to write e = g∈G agg. We note that Ð ag = 0ifg/∈ O1 ∪···∪ Os , and 2 = = Ð that e e implies g g =g ag1 ag2 ag. 1 2 = −1 Using the former we have ε g,h∈G agah(g, h ). The computation        2 = −1 −1 ε ag1 ah1 g1,h1 ag2 ah2 g2,h2 g1,h1∈G g2,h2∈G

  = −1 ag1 ag2 ah1 ah2 g,h g,h∈G g1g2=g,h1h2=h 5.4 The Brauer Correspondence 159

  = −1 ag1 ag2 ah1 ah2 g,h = g,h∈G g1g2 g h1h2=h   −1 = agah g,h g,h∈G = ε now shows that ε indeed is an idempotent. For any pair (i, j) the group        × ∩ −1 = ∩ × ∩ −1 (P P) CG×G xi,xj P CG(xi) P CG xj     = P ∩ CG(xi) × P ∩ CG(xj )

−1 = × is a p-Sylow subgroup of CG×G((xi,xj )) CG(xi) CG(xj ). It follows that   ε ∈ IP ×P ⊆ Z k[G × G] .

Hence we may apply Proposition 5.3.11 to P × P ⊆ G × G and ε and obtain that the k[G × G]-module εk[G] is relatively k[P × P ]-projective. But applying ε = a a (x, y−1) to any v ∈ k[G] gives i,j i j x∈Oi y∈Oj

ˆ ˆ εv = aiaj xvy = aiaj OivOj

i,j x∈Oi y∈Oj i,j

ˆ ˆ = aiOi v aj Oj i j = eve and shows that

εk[G]=ek[G]e = k[G]e.

It follows that P × P contains a subgroup of the form (g, h)δ(D)(g, h)−1 with g,h ∈ G and D a defect subgroup of the e-block. We deduce that P contains the defect subgroup gDg−1. 

Later on (Theorem 5.5.8.i) we will establish the converse of Corollary 5.3.12. Hence the defect groups of the e-block can also be characterized as being the small- est p-subgroups D ⊆ G such that e ∈ ID.

5.4 The Brauer Correspondence

Let e ∈ E,letD be a defect group of the e-block, and put N := NG(D). Obviously N × N contains the normalizer NG×G(δ(D)). Hence, by Green’s Theorem 4.3.6, 160 5Blocks the indecomposable k[G × G]-module k[G]e has a Green correspondent which is the, up to isomorphism, unique indecomposable direct summand with vertex δ(D) of k[G]e as a k[N × N]-module. Can we identify this Green correspondent? We first establish the following auxiliary but general result.

Lemma 5.4.1 Let P be any finite p-group and Q ⊆ P be any subgroup; the k[P ]- P module IndQ(k) is indecomposable with vertex Q.

Proof Using the second Frobenius reciprocity in Sect. 2.3 we obtain   P = = Homk[P ] k,IndQ(k) Homk[Q](k, k) k id .

Since the trivial module k is the only simple k[P ]-module by Proposition 2.2.7 this [ ] P implies that the socle of the k P -module IndQ(k) is one-dimensional. But by the Jordan–Hölder Proposition 1.1.2 any nonzero k[P ]-module has a nonzero socle. It P follows that IndQ(k) must be indecomposable. In particular, by Proposition 4.1.6, P P IndQ(k) has a vertex V contained in Q. So, again by Proposition 4.1.6,IndQ(k) is isomorphic to a direct summand of

  m     P P =∼ P V −1 ∗ IndV IndQ(k) IndV Ind ∩ −1 gi k V gi Qgi i=1 m = P Ind ∩ −1 (k). V gi Qgi i=1

Here {g1,...,gm}⊆G is a set of representatives for the double cosets in V \G/Q, and the decomposition comes from Mackey’s Proposition 4.2.4. By what we have shown already each summand is indecomposable. Hence the KrullÐRemakÐSchmidt P =∼ P ≤ ≤ Theorem 1.4.7 implies that IndQ(k) Ind ∩ −1 (k) for some 1 i m.Com- V gi Qgi paring k-dimensions we deduce that |Q|≤|V | and consequently that Q = V . 

Remark 5.4.2 In passing we observe that any p-subgroup Q ⊆ G occurs as a vertex of some finitely generated indecomposable k[G]-module. Let P ⊆ G be a p-Sylow subgroup containing Q. Lemma 5.4.1 says that the indecomposable k[P ]-module P G IndQ(k) has the vertex Q. It then follows from Lemma 4.3.4 that IndQ(k) has an indecomposable direct summand with vertex Q. Since the trivial k[G]-module k is a G [ : ] direct summand of IndP (k) (as a p G P ) we, in particular, see that the p-Sylow subgroups of G are the vertices of k. We further deduce, using Proposition 5.3.4, that the p-Sylow subgroups of G also are the defect groups of the block to which the trivial module k belongs.

We also need a few technical facts about the k[G × G]-module k[G] when we view it as a k[H ×H ]-module for some subgroup H ⊆ G. For any y ∈ G the double coset HyH ⊆ G is an H × H -orbit in G for the action 5.4 The Brauer Correspondence 161

(H × H)× G −→ G   −→ −1 (h1,h2), g h1gh2 , and so k[HyH] is a k[H × H]-submodule of k[G]. We remind the reader that the centralizer of a subgroup Q ⊆ G is the subgroup CG(Q) ={g ∈ G : gh = hg for any h ∈ G}.

Lemma 5.4.3 × i. k[HyH] =∼ IndH H (k) as k[H × H ]-modules. (1,y)−1δ(H∩yHy−1)(1,y) ii. If H is a p-group then k[HyH] is an indecomposable k[H × H ]-module with vertex (1,y)−1δ(H ∩ yHy−1)(1,y). iii. Suppose that y/∈ H and that CG(Q) ⊆ H for some p-subgroup Q ⊆ H ; then no indecomposable direct summand of the k[H × H ]-module k[HyH] has a vertex containing δ(Q).

Proof i. We observe that     ∈ × : = = ∈ × : −1 = (h1,h2) H H (h1,h2)y y (h1,h2) H H h1yh2 y   −1 = (h1,h2) ∈ H × H : h2 = y h1y    = h, y−1hy : h ∈ H,y−1hy ∈ H   − − = (1,y) 1(h, h)(1,y): h ∈ H ∩ yHy 1   − − = (1,y) 1δ H ∩ yHy 1 (1,y).

Hence, by Remark 2.4.2,themap   − −  H × H/(1,y) 1δ H ∩ yHy 1 (1,y)−→ HyH ··· −→ −1 (h1,h2) h1yh2 is an isomorphism of H × H -sets. The assertion now follows from Lemma 2.4.6.i. ii. This follows from i. and Lemma 5.4.1. iii. By Proposition 4.1.6 the assertion i. implies that each indecomposable sum- mand in question has a vertex contained in (1,y)−1δ(H ∩ yHy−1)(1,y). Suppose therefore that this latter group contains an H × H -conjugate of δ(Q). We then find elements h1,h2 ∈ H such that   −1 −1 −1 −1 (h1,h2)δ(Q)(h1,h2) ⊆ (1,y) δ H ∩ yHy (1,y)⊆ (1,y) δ(G)(1,y), hence −1 (h1,yh2)δ(Q)(h1,yh2) ⊆ δ(G), and therefore −1 = −1 ∈ h1gh1 yh2g(yh2) for any g Q. 162 5Blocks

−1 ∈ ⊆ ∈ It follows that h1 yh2 CG(Q) H . But this implies y H which is a contradic- tion. 

For any p-subgroup D ⊆ G we put

ED(G) := {e ∈ E : D is a defect subgroup of the e-block}.

Theorem 5.4.4 (Brauer’s First Main Theorem) For any p-subgroup D ⊆ G and any subgroup H ⊆ G such that NG(D) ⊆ H we have a bijection ∼ B: ED(G) −→ ED(H ) such that the k[H × H]-module k[H]B(e) is the Green correspondent of the k[G × G]-module k[G]e.

Proof First of all we point out that, for any two idempotents e = e in E(G),the k[G]-modules k[G]e and k[G]e belong to two different blocks and therefore can- not be isomorphic. A fortiori they cannot be isomorphic as k[G × G]-modules. A corresponding statement holds, of course, for the k[H × H ]-modules k[H ]f with f ∈ E(H). Let e ∈ ED(G). Since H × H contains NG×G(δ(D)) the Green correspondent Γ(k[G]e) of the indecomposable k[G × G]-module k[G]e exists, is an indecom- posable direct summand of k[G]e as a k[H × H]-module, and has the vertex δ(D) (see Theorem 4.3.6). But we have the decomposition  k[G]= k[HyH] (5.4.1) y∈H \G/H as a k[H × H ]-module. The KrullÐRemakÐSchmidt Theorem 1.4.7 implies that Γ(k[G]e) is isomorphic to a direct summand of k[HyH] for some y ∈ G.By Lemma 5.4.3.iii we must have y ∈ H . Hence Γ(k[G]e) is isomorphic to a direct summand of the k[H × H]-module k[H]. It follows that   Γ k[G]e =∼ k[H]f for some f ∈ ED(H ). Our initial observation applied to H says that B(e) := f is uniquely determined.

For e = e in ED(G) the k[G × G]-modules k[G]e and k[G]e are noniso- morphic. By the injectivity of the Green correspondence the k[H × H ]-modules Γ(k[G]e) and Γ(k[G]e ) are nonisomorphic as well. Hence B(e) = B(e ).This shows that B is injective. For the surjectivity of B let f ∈ ED(H ). The decomposition (5.4.1) shows that k[H]f ,asak[H × H]-module, is isomorphic to a direct summand of k[G] and hence, by the KrullÐRemakÐSchmidt Theorem 1.4.7, to a direct summand of k[G]e for some e ∈ E(G). It follows from Proposition 4.3.9 that necessarily k[H ]f =∼ Γ(k[G]e) and e ∈ ED(G). In particular, f = B(e).  5.4 The Brauer Correspondence 163

The bijection B in Theorem 5.4.4 is a particular case of the Brauer correspon- dence whose existence comes from the following result.

Lemma 5.4.5 Let f ∈ E(H) such that CG(D) ⊆ H for one (or equivalently any) defect group D of the f -block; we then have: i. There is a unique β(f) := e ∈ E(G) such that k[H ]f is isomorphic, as a k[H × H ]-module, to a direct summand of k[G]e; ii. any defect group of the f -block is contained in a defect group of the e-block.

Proof i. The argument for the existence of e was already given at the end of the proof of Theorem 5.4.4. Suppose that k[H]f also is isomorphic to a direct sum- mand of k[G]e for a second idempotent e =e ∈ E(G). Then k[H ]f ⊕ k[H ]f is [ ]= [ ]˜ [ ] isomorphic to a direct summand of k G e˜∈E(G) k G e. Since k H f , but not k[H]f ⊕ k[H ]f is isomorphic to a direct summand of k[H ] the decomposition (5.4.1) shows that k[H]f must be isomorphic to a direct summand of k[HyH] for some y ∈ G \ H . But this is impossible according to Lemma 5.4.3.iii (applied with Q = D). ii. Let D˜ be a defect group of the e-block. By Lemma 4.3.1.i applied to k[G]e =∼ k[H]f ⊕ ··· and H × H ⊆ G × G we find a γ = (g1,g2) ∈ G × G such that ⊆ ˜ −1 ⊆ ˜ −1  δ(D) γδ(D)γ . Then D g1Dg1 .

For any p-subgroup D ⊆ G we put

E≥D(G) := {e ∈ E : D is contained in a defect group of the e-block}.

We fix a subgroup H ⊆ G and a p-subgroup D ⊆ H , and we suppose that CG(D) ⊆

H . Any other p-subgroup D ⊆ D ⊆ H then satisfies CG(D ) ⊆ CG(D) ⊆ H as well. Therefore, according to Lemma 5.4.5, we have the well-defined map

β: E≥D(H ) −→ E≥D(G).

Theorem 5.4.6 Let f ∈ E(H), and suppose that there is a finitely generated inde- composable k[H ]-module N belonging to the f -block which has a vertex V that satisfies CG(V ) ⊆ H ; we then have: i. CG(D) ⊆ H for any defect group D of the f -block; in particular, β(f) is de- fined; ii. if a finitely generated indecomposable k[G]-module M has a direct summand, as a k[H ]-module, isomorphic to N then M belongs to the β(f)-block.

Proof By Proposition 5.3.4 we may assume that V ⊆ D. Then CG(D) ⊆ CG(V ) ⊆ H which proves i. Let e ∈ E(G) such that M belongs to the e-block. Suppose that e = β(f). We deduce from (5.4.1), by multiplying by f , the decomposition  fk[G]=k[H]f ⊕ fX with X := k[HyH]. y∈H \G/H,y∈ /H 164 5Blocks

Since f commutes with the elements in H this is a decomposition of k[H × H ]- modules. Similarly we have the decomposition

fk[G]=fk[G]e ⊕ fk[G](1 − e) as a k[H × H ]-module. So fk[G]e is a direct summand of k[H ]f ⊕ fX.Byas- sumption the indecomposable k[H × H]-module k[H ]f is not isomorphic to a di- rect summand of k[G]e = fk[G]e ⊕ (1 − f)k[G]e. Hence it is not isomorphic to a direct summand of fk[G]e. It therefore follows from the KrullÐRemakÐSchmidt Theorem 1.4.7 that fk[G ]e must be isomorphic to a direct summand of fX and hence of X = fX ⊕ (1 − f)X as a k[H × H]-module. Using Lemma 5.4.3.iii we conclude that no indecomposable direct summand Y of the k[H × H ]-module fk[G]e has a vertex containing δ(V). We consider any indecomposable direct sum- mand Z of Y as a k[δ(H)]-module. Lemma 4.3.1.i applied to Y = Z ⊕··· and δ(H) ⊆ H × H implies that any vertex of Z is contained in some vertex of Y (as a k[H × H ]-module). It follows that no vertex of Z contains δ(V). Using the notation introduced in the proof of Proposition 5.3.4 we conclude that no indecomposable direct summand of the k[H]-module (f k[G]e)ad has a vertex containing V .Ifwe analyze the arguments in the proof of Proposition 5.3.4 then, in the present context, they give the following facts: ad Ð fM= ef M is isomorphic to a direct summand of (f k[G]e) ⊗k fM. ÐIfZ is an indecomposable direct summand of (f k[G]e)ad with vertex U then Z ⊗k fM is relatively k[U]-projective. This implies that the vertices of the inde- ad composable direct summands of (f k[G]e) ⊗k fM are contained in vertices of the indecomposable direct summands of (f k[G]e)ad. We deduce that the indecomposable direct summands of fM have no vertices con- taining V .ButN = fN is a direct summand of M and hence of fM and it does have vertex V . This is a contradiction. We therefore must have e = β(f). 

Proposition 5.4.7 Let e ∈ E(G) and let D be a defect group of the e-block; then there exists a finitely generated indecomposable k[G]-module M belonging to the e-block which has the vertex D.

Proof Let H = NG(D) and let X be any simple k[H ]-module in the B(e)-block of k[H ]. We claim that X has the vertex D.ButD is a normal p-subgroup of H . Hence as a k[D]-module X is a direct sum

X = k ⊕···⊕k of trivial k[D]-modules k (cf. Proposition 2.2.7 and Theorem 2.5.3.i). On the other hand the B(e)-block has the defect group D so that, by Proposition 5.3.4, X is relatively k[D]-projective. We therefore may apply Lemma 4.3.1 to D ⊆ H and the above decomposition of X and obtain that any vertex of the trivial k[D]-module also is a vertex of the k[H]-module X. The former, by Lemma 5.4.1, are equal to D. Hence X has the vertex D. Green’s Theorem 4.3.6 now tells us that X is the 5.5 Brauer Homomorphisms 165

Green correspondent of some finitely generated indecomposable k[G]-module M with vertex D. Moreover, Theorem 5.4.6 implies that M belongs to the β(B(e)) = e- block. 

Proposition 5.4.8 For any e ∈ E(G) the following conditions are equivalent: i. The e-block has the defect group {1}; ii. any k[G]-module belonging to the e-block is semisimple; iii. the k-algebra k[G]e is semisimple; ∼ iv. k[G]e = Mn×n(k) as k-algebras for some n ≥ 1; v. there is a simple k[G]-module X belonging to the e-block which is projective.

Proof i. =⇒ ii. By Proposition 5.3.4 any module M belonging to the e-block is relatively k[{1}]-projective and hence projective. In particular, M/N is projective for any submodule N ⊆ M;itfollowsthatM =∼ N ⊕ M/N. This implies that M is semisimple (cf. Proposition 1.1.4). ii. =⇒ iii. k[G]e as a k[G]-module belongs to the e-block and hence is semisim- ple. iii. =⇒ iv. Since e is primitive in Z(k[G]) the semisimple k-algebra k[G]e must, in fact, be simple (cf. Corollary 1.5.4). Since k is algebraically closed any finite- dimensional simple k-algebra is a matrix algebra. iv. =⇒ v. The simple module of a matrix algebra is projective. v. =⇒ i. We have seen in Corollary 5.1.4 that X is, up to isomorphism, the only simple module belonging to the e-block. This implies that any finitely gen- erated indecomposable module belonging to the e-block is isomorphic to X and hence has the vertex {1}. Therefore the e-block has the defect group {1} by Propo- sition 5.4.7. 

5.5 Brauer Homomorphisms

Let D ⊆ G be a p-subgroup. There is the obvious k-linear map     s˜: k[G]−→k CG(D) ⊆ k NG(D)

agg −→ agg. g∈G g∈CG(D)

We observe that the centralizer CG(D) is a normal subgroup of the normalizer NG(D). This implies that the intersection O ∩ CG(D), for any conjugacy class O ∈ O(G), on the one hand is contained in CG(D), of course, but on the other hand is a union of full conjugacy classes in O(NG(D)). We conclude that s˜ restricts to a k-linear map 166 5Blocks      s: Z k[G] −→ Z k NG(D)

Oˆ −→ o.ˆ

o∈O(NG(D)),o⊆O∩CG(D)

Lemma 5.5.1 s is a homomorphism of k-algebras.

Proof Clearly s respects the unit element. It remains to show that s is multiplicative. Let O1, O2 ∈ O(G).Wehave   ˆ ˆ O1O2 = |Bg|g with Bg := (g1,g2) ∈ O1 × O2 : g1g2 = g g∈G and hence

ˆ ˆ s(O1O2) = |Bg|g. g∈CG(D) On the other hand

ˆ ˆ s(O1)s(O2) = g1 g2 = |Cg|g

g1∈O1∩CG(D) g2∈O2∩CG(D) g∈CG(D) with   Cg := (g1,g2) ∈ O1 × O2 : g1,g2 ∈ CG(D) and g1g2 = g .

Obviously Cg ⊆ Bg. We will show that, for g ∈ CG(D), the integer |Bg \ Cg| is divisible by p and hence is equal to zero in k. Since D centralizes g we may view Bg as a D-set via the action

D × Bg −→ Bg     −1 −1 h, (g1,g2) −→ hg1h ,hg2h .

The subset Cg consists of exactly all the fixed points of this action. The complement Bg \ Cg therefore is the union of all D-orbits with more than one element. But since D is a p-group any D-orbit has a power of p many elements. 

Literally the same reasoning works for any subgroup CG(D) ⊆ H ⊆ NG(D) and shows that we may view s also as a homomorphism of k-algebras     sD,H := s: Z k[G] −→ Z k[H ] .

It is called the Brauer homomorphism (of G with respect to D and H ). For any e ∈ E(G) its image s(e) either is equal to zero or is an idempotent. We want to establish a criterion for the nonvanishing of s(e). 5.5 Brauer Homomorphisms 167

Lemma 5.5.2 For any O ∈ O(G) we have:

i. O ∩ CG(D) = ∅ if and only if D is contained in a defect group of O; ii. if D is a defect group of O, then O ∩ CG(D) is a single conjugacy class in NG(D); iii. let o ∈ O(NG(D)) such that o ⊆ O; if D is a defect group of o then D also is a defect group of O and o = O ∩ CG(D).

Proof i. First let x ∈ O ∩ CG(D). Then D ⊆ CG(X), and D is contained in a p- Sylow subgroup of CG(x). Conversely, let D be contained in a p-Sylow subgroup of CG(x) for some x ∈ O. Then D centralizes x and hence x ∈ O ∩ CG(D). ii. We assume that D is a p-Sylow subgroup of CG(x) for some x ∈ O. Then x ∈ O ∩ CG(D).Lety ∈ O ∩ CG(D) be any other point and let g ∈ G such that x = gyg−1. Since D centralizes y the conjugate group gDg−1 centralizes x.It −1 follows that D as well as gDg are p-Sylow subgroups of CG(x). Hence we −1 −1 −1 find an h ∈ CG(x) such that hDh = gDg . We see that h g ∈ NG(D) and (h−1g)y(h−1g)−1 = h−1gyg−1h = h−1xh = x. ∈ iii. We now assume that D is a p-Sylow subgroup of CNG(D)(x) for some x o. Then D centralizes x and hence x ∈ O ∩ CG(D). It follows from i. that D is con- tained in a defect group of O and, more precisely, in a p-Sylow subgroup P of CG(x). If we show that D = P then D is a defect group of O and ii. implies that o = O ∩ CG(D). Let us therefore assume that D P . Then, as in any p-group, also D NP (D).Wepickanelementh ∈ NP (D) \ D and let Q denote the p-subgroup generated by D and h. We then have

D Q ⊆ NP (D) ⊆ P ⊆ CG(x), ⊆ from which we deduce that D Q CNG (x).ButD was a p-Sylow subgroup of  CNG (x). This is a contradiction.

Lemma 5.5.3 Let e ∈ E(G) with corresponding central character χe; if D is mini- mal with respect to e ∈ ID then we have: i. s(e) = 0; ii. χe(IQ) ={0} for any proper subgroup Q D; ˆ iii. if χe(O) = 0 for some O ∈ O(G) then D is contained in a defect group of O.

Proof i. Let S ⊆ O(G) denote the subset of all conjugacy classes O such that D contains a defect group DO of O.Byassumptionwehave

ˆ e = aOO with aO ∈ k. O∈S Suppose that

ˆ s(e) = aOs(O) = aO x = 0.

O∈S O∈S x∈O∩CG(D) 168 5Blocks

It follows that, for any O ∈ S,wehaveaO = 0orO ∩ CG(D) =∅. The latter, by Lemma 5.5.2.i, means that D is not contained in a defect group of O. We therefore obtain that

ˆ e = aOO ∈ IDO . O∈S,DOD O∈S,DOD

Remark 5.3.9 then implies that e ∈ IDO for some O ∈ S such that DO D.This contradicts the minimality of D. ii. By Remark 5.3.9 it follows from e ∈ ID that IDe = Ze is a local ring. Then IQe, for any subgroup Q ⊆ D, is an ideal in this local ring. But if Q = D then the minimality property of D says that e/∈ IQe ⊆ IQ. It follows that IQe ⊆ Jac(Ze) and hence that χe(IQ) = χe(IQe) ={0}. iii. Let DO ⊆ G be a defect group of O. Using Proposition 5.3.7 we have

⊆ ⊆ − eIDO IDIDO ID∩gDOg 1 . g∈G ˆ Because of O ∈ IDO the central character χe cannot vanish on the right-hand sum. − Hence, by ii., there must exist a g ∈ G such that D ⊆ gDOg 1. 

Lemma 5.5.4 Let e ∈ E(G) and O ∈ O(G); if the p-subgroup P is normal in G then we have:

i. If e ∈ IP then δ(P) is a vertex of the k[G × G]-module k[G]e; ii. if e ∈ IQ for some p-subgroup Q ⊆ G then P ⊆ Q; ˆ iii. if O ∩ CG(P ) =∅then O ∈ Jac(Z(k[G])).

Proof i. According to the proof of Corollary 5.3.12 the indecomposable k[G × G]- module k[G]e is relatively k[P × P ]-projective. Lemma 4.3.1 then implies that k[G]e as a k[G × G]-module and some indecomposable direct summand of k[G]e as a k[P ×P ]-module have a common vertex. But by Lemma 5.4.3.ii the summands in the decomposition  k[G]= k[PyP] y∈P \G/P are indecomposable k[P × P ]-modules having the vertex   − − (1,y) 1δ P ∩ yPy 1 (1,y).

It follows that k[G]e has a vertex of the form δ(P ∩ yPy−1) for some y ∈ G. Since P isassumedtobenormalinG the latter group is equal to δ(P). ii. We know from Proposition 5.3.11 that any k[G]-module M belonging to the e-block has a vertex contained in Q.LetX be a simple k[G]-module belonging to the e-block (cf. Proposition 1.7.4.i and Corollary 1.7.5 for the existence). By our assumption that P is a normal p-subgroup of G we may argue similarly as in the 5.5 Brauer Homomorphisms 169

first half of the proof of Proposition 5.4.7 to see that P is contained in any vertex of X. iii. Let χf ,forf ∈ E(G), be any central character, and let the p-subgroup ⊆ ∈ ⊆ Df G be minimal with respect to f IDf . From ii. we know that P Df . On the other hand P cannot be contained in a defect group of O by Lemma 5.5.2.i. ˆ Hence Df is not contained in a defect group of O, which implies that χf (O) = 0 by Lemma 5.5.3.iii. We see that Oˆ lies in the kernel of every central character, and we deduce that Oˆ ∈ Jac(Z(k[G])). 

In the following we need to consider the ideals IP for the same p-subgroup P in various rings Z(k[H]) for P ⊆ H ⊆ G. We therefore write IP (k[H ]) in order to refer to the ideal IP in the ring Z(k[H]). We fix a subgroup

DCG(D) ⊆ H ⊆ NG(D) and consider the corresponding Brauer homomorphism     sD,H : Z k[G] −→ Z k[H ] .

Let e ∈ E(G) be such that sD,H (e) = 0 (for example, by Lemma 5.5.3.i, if D is minimal with respect to e ∈ ID(k[G])). We may decompose

sD,H (e) = e1 +···+er uniquely into a sum of primitive idempotents ei ∈ E(H) (cf. Proposition 1.5.5).

Lemma 5.5.5 If e ∈ ID(k[G]) then e1,...,er ∈ ID(k[H ]).

Proof Again let S ⊆ O(G) denote the subset of all conjugacy classes O such that D contains a defect group of O. By assumption e is a linear combination of the Oˆ for O ∈ S. Hence sD,H (e) is a linear combination of oˆ for o ∈ O(H ) such that o ⊆ O for some O ∈ S. Any defect group of such an o is contained in a defect group of O and therefore in gDg−1 for some g ∈ G (cf. Exercise 5.3.5.i). It follows that   +···+ = ∈ [ ] e1 er sD,H (e) IgDg−1∩H k H g∈G and consequently, by Corollary 5.3.8, that   = +···+ ∈ [ ] ≤ ≤ ei ei(e1 er ) IgDg−1∩H k H for any 1 i r. g∈G

Remark 5.3.9 now implies that for any 1 ≤ i ≤ r there is a gi ∈ G such that   ei ∈ I −1∩ k[H] . gi Dgi H 170 5Blocks

Since D is normal in H we may apply Lemma 5.5.4.ii and obtain that D ⊆ −1 ∩ = −1 ∩  giDgi H and hence, in fact, that D giDgi H .

We observe that  r k[H ]sD,H (e) = k[H]f(e1 +···+er ) = k[H ]ei. f ∈E(H) i=1

Since D is normal in H it follows from Lemmas 5.5.5 and 5.5.4.i that δ(D) is a vertex of the indecomposable k[H × H]-module k[H ]ei for any 1 ≤ i ≤ r. On the other hand, G as an H × H -set decomposes into G = H ∪ (G \ H). Correspondingly, k[G] as a k[H × H]-module decomposes into k[G]=k[H ]⊕ k[G \ H ].WeletπH : k[G]−→k[H] denote the associated projection map, which is a k[H × H ]-module homomorphism.

Lemma 5.5.6 πH (e) ∈ sD,H (e) + Jac(Z(k[H])).

= Oˆ ∈ Proof Let e O∈O(G) aO with aO k. Then

πH (e) = aO x O∈O(G) x∈O∩H whereas s (e) = aO( x). It follows that π (e)−s (e) D,H O∈O(G) x∈O∩CG(D) H D,H is of the form

booˆ with bo ∈ k.

o∈O(H ),o∩CG(D)=∅ Lemma 5.5.4.iii (applied to D ⊆ H ) shows that this sum lies in Jac(Z(k[H ])). 

Proposition 5.5.7 The k[H × H]-module k[H]sD,H (e) is isomorphic to a direct summand of k[G]e.

Proof We have the k[H × H]-module homomorphism

α: k[H]sD,H (e) −→ k[G]e ⊗k[H ] k[H ]sD,H (e) v −→ e ⊗ v.

By Lemma 5.5.6 the composite β := (πH ⊗ id) ◦ α ∈ Endk[H ×H ](k[H ]sD,H (e)) satisfies   β(v) = (πH ⊗ id) ◦ α(v) = πH (e)v = sD,H (e) + z v for any v ∈ k[H ]sD,H (e) and some element z ∈ Jac(Z(k[H ])). We note that (ei + z)ei = ei +zei , for any 1 ≤ i ≤ r, is invertible in the local ring Z(k[H ])ei . Therefore 5.5 Brauer Homomorphisms 171 the element   r    −1 × y := (ei + zei) × 0 ∈ Z k[H ] sD,H (e) i=1 f ∈E(H),f=ei is well defined and satisfies   sD,H (e) + z y = sD,H (e).

Then     sD,H (e) + z k[H]sD,H (e) = sD,H (e) + z yk[H ]sD,H (e) = k[H ]sD,H (e) which implies that the map β is surjective. Since k[H ]sD,H (e) is finite-dimensional over k the map β must be bijective. It follows that k[H ]sD,H (e) is isomorphic to a direct summand of k[G]e ⊗k[H ] k[H]sD,H (e). Furthermore the latter is a direct summand of

k[G]e = k[G]e ⊗k[H ] k[H]      = k[G]e ⊗k[H ] k[H]sD,H (e) ⊕ k[G]e ⊗k[H ] k[H ] 1 − sD,H (e) . 

Theorem 5.5.8 For any e ∈ E(G) and any defect group D0 of the e-block we have: ∈ [ ] i. e ID0 (k G ); = ∈ = ii. if H NG(D0) then sD0,H (e) E(H) and B(e) sD0,H (e).

Proof With the notations from before Lemma 5.5.5, where D ⊆ G is a p-subgroup which is minimal with respect to e ∈ ID(k[G]), we know from Lemma 5.5.5 that e1,...,er ∈ ID(k[H ]). Applying Lemma 5.5.4.i to the normal p-subgroup D of H we obtain that δ(D) is a vertex of the indecomposable k[H × H ]-modules k[H ]ei . But by Proposition 5.5.7 these k[H]ei are isomorphic to direct summands of k[G]e. −1 Hence it follows from Lemma 4.3.1 that D ⊆ gD0g for some g ∈ G. We see that e ∈ I ⊆ I −1 = I D gD0g D0 which proves i. In fact Corollary 5.3.12 says that D has to contain some defect subgroup of −1 the e-block. We conclude that D = gD0g and therefore that D0 is minimal with ∈ [ ] respect to e ID0 (k G ) as well. This means that everywhere in the above reasoning we may replace D by D0 (so that, in particular, D0CG(D0) ⊆ H ⊆ NG(D0)). We obtain that D0 also is a defect group of the ei -blocks for 1 ≤ i ≤ r.LetH = NG(D0) so that NG×G(δ(D0)) ⊆ H ×H . By the Green correspondence in Theorem 4.3.6 the indecomposable k[G × G]-module k[G]e with vertex D0 has, up to isomorphism, a unique direct summand, as a k[H × H]-module, with vertex D0. It follows that r = = ∈ 1, which means that sD0,H (e) e1 E(H), and that this summand is isomorphic to k[H ]e1, which means that B(e) = e1. This proves ii. 

Proposition 5.5.9 For any e ∈ E(G) and any defect group D0 of the e-block we have: 172 5Blocks i. If the p-Sylow subgroup Q ⊆ G contains D0 then there exists a g ∈ G such that −1 D0 = Q ∩ gQg ; ii. any normal p-subgroup P ⊆ G is contained in D0.

Proof i. By assumption and Exercise 4.2.2.iii the indecomposable k[G×G]-module k[G]e is relatively k[Q × Q]-projective. Lemma 4.3.2 therefore implies that k[G]e as a k[Q×Q]-module has an indecomposable direct summand X with vertex δ(D0). Applying Lemma 5.4.3.ii to the decomposition  k[G]= k[QyQ] y∈Q\G/Q we see that X also has a vertex of the form (1,y)−1δ(Q ∩ yQy−1)(1,y) for some y ∈ G. It follows that   −1 −1 −1 δ(D0) = (q1,q2)(1,y) δ Q ∩ yQy (1, y)(q1,q2)

−1 for some (q1,q2) ∈ Q × Q. We obtain on the one hand that |D0|=|Q ∩ yQy | and on the other hand that   −1 = −1 −1 −1 ∈ ∩ −1 ∈ q1 dq1 yq2 d yq2 Q yQy for any d D0. := −1 ⊆ ∩ −1 The element g q1yq2 then lies in CG(D0). Hence D0 Q gQg . Moreover      ∩ −1 =  ∩ −1 −1 −1 Q gQg Q q1yq2 Qq2y q1     =  −1 ∩ −1 −1 =  ∩ −1 q1 Qq1 yq2 Qq2y Q yQy

=|D0|.

ii. Since P is contained in any p-Sylow subgroup of G it follows from i. that P ⊆ D0. 

Theorem 5.5.10 Let D ⊆ G be a p-subgroup and fix a subgroup DCG(D) ⊆ H ⊆ NG(D). We then have:

i. E≥D(H ) = E(H); ii. the Brauer correspondence β : E(H) −→ E≥D(G) is characterized in terms of central characters by

χβ(f) = χf ◦ sD,H for any f ∈ E(H); iii. if e ∈ E(G) such that sD,H (e) = 0 then e ∈ im(β) and

sD,H (e) = f ; f ∈β−1(e) iv. ED(G) ⊆ im(β). 5.5 Brauer Homomorphisms 173

Proof i. Since D is normal in H this is immediate from Proposition 5.5.9.ii. ii. For any f ∈ E(H) the composed k-algebra homomorphism χf ◦ sD,H must be a central character of k[G],i.e.χf ◦ sD,H = χe for some e ∈ E(G). Since 1 = χe(e) = χf (sD,H (e)) we have sD,H (e) = 0. Let sD,H (e) = e1 +···+er with ei ∈ E(H). Then 1 = χf (e1) +···+χf (er ) and hence f = ej for some 1 ≤ j ≤ r.By Proposition 5.5.7 the indecomposable k[H × H]-module k[H ]f is isomorphic to a direct summand of k[G]e. It follows that e = β(f). iii. Let sD,H (e) = e1 +···+er with ei ∈ E(H). Again it follows from Propo- sition 5.5.7 that e = β(e1) =···=β(er ). On the other hand, if e = β(f) for some f ∈ E(H) then we have seen in the proof of ii. that necessarily f = ej for some 1 ≤ j ≤ r. iv. If e ∈ ED(G) then D is minimal with respect to e ∈ ID(k[G]) by Corol- lary 5.3.12 and Theorem 5.5.8.i. Hence sD,H (e) = 0 by Lemma 5.5.3.i. 

Proposition 5.5.11 For any e ∈ E(G) and any defect group D0 of the e-block we have: ˆ i. If χe(O) = 0 for some O ∈ O(G) then D0 is contained in a defect group of O; ˆ ii. there is an O ∈ O(G) such that χe(O) = 0 and D0 is a defect group of O.

Proof i. By Corollary 5.3.12 and Theorem 5.5.8.i the defect group D0 is minimal ∈ [ ] with respect to e ID0 (k G ). The assertion therefore follows from Lemma 5.5.3.iii. ∈ [ ] ii. Since e ID0 (k G ) we have

ˆ e = aOO with aO ∈ k O∈S ⊆ O O O where S (G) is the subset of all such that D0 contains a defect group of . = = Oˆ O ∈ Then 1 χe(e) O∈S aOχe( ). Hence there must exist an S such that ˆ χe(O) = 0. This O has a defect group contained in D0 but by i. also one which contains D0. It follows that D0 is a defect group of O.  References

1. Alperin, L.: Local Representation Theory. Cambridge University Press, Cambridge (1986) 2. Craven, D.: The modular representation theory of finite groups. Univ. Birmingham, Thesis (2006) 3. Curtis, C., Reiner, I.: Methods of Representation Theory, Vol. I. Wiley, New York (1981) 4. Dixon, J., Puttaswamaiah, B.: Modular Representations of Finite Groups. Academic Press, New York (1977) 5. Dornhoff, L.: Group Representation Theory, Part B: Modular Representation Theory. Dekker, New York (1972) 6. Feit, W.: The Representation Theory of Finite Groups. North-Holland, Amsterdam (1982) 7. Isaacs, I.M.: Character Theory of Finite Groups. Academic Press, New York (1976) 8. Lam, T.Y.: A First Course in Noncommutative Rings. Springer, Heidelberg (1991) 9. Schneider, P.: Die Theorie des Anstiegs. Course at Münster, 2006/2007. Available at www.math.uni-muenster.de/u/schneider/publ/lectnotes/ 10. Serre, J.P.: Linear Representations of Finite Groups. Springer, Heidelberg (1977)

P. Schneider, Modular Representation Theory of Finite Groups, 175 DOI 10.1007/978-1-4471-4832-6, © Springer-Verlag London 2013 Index

A F Algebra, 7 Fitting, 10 Augmentation, 53 Frobenius reciprocity ideal, 53 first, 56 second, 57 B Block, 20 G Block decomposition, 20 G-orbit, 60 Brauer, 71, 74, 76 G-set, 59 Brauer character, 91 simple, 60 Brauer correspondence, 163 standard, 60 Brauer homomorphism, 166 Green, 115, 146 Brauer’s first main theorem, 162 Green correspondence, 116 Bruhat decomposition, 132 Grothendieck group, 34 Burnside ring, 60 Group ring, 43 C H Cartan homomorphism, 38 H-projective homomorphism, 116 Cartan matrix, 41 Hyper-elementary group, 65 CartanÐBrauer triangle, 50 Central character, 152 Centralizer, 155, 161 I Character, 58 I -adically Clifford, 69 complete, 5 Complete, 8 separated, 5 Composition series, 1 Idempotent, 15 central, 15 D orthogonal, 15 Decomposition homomorphism, 50 primitive, 15 Defect group of a block, 154 J of a conjugacy class, 155 Jacobson radical, 3 Jordan–Hölder, 1 E e-block, 20 K Elementary group, 71 k-character, 87 Essential, 32 KrullÐRemakÐSchmidt, 15

P. Schneider, Modular Representation Theory of Finite Groups, 177 DOI 10.1007/978-1-4471-4832-6, © Springer-Verlag London 2013 178 Index

L p-unipotent, 88 Lattice, 47 Permutation module, 61 invariant, 48 Projective cover, 32 Length, 2 Projective limit, 4 Local ring, 9 R M Radical, 3 Mackey, 106 Ring Module (0,p)-, 43 decomposable, 2 of Witt vectors, 44 free, 29 splitting, 78 indecomposable, 2 induced, 57 S of finite length, 2 Socle, 2 projective, 27 Solomon, 65 relatively H-projective, 116 Source, 108 relatively projective, 97 Splitting, 78 relatively R[H ]-projective, 101 Splitting field, 58 semisimple, 2 trivial, 53 T uniserial, 126 τ -isotypic component, 2 Teichmüller representative, 91 N Nakayama, 4 U Normalizer, 66, 109 Uniserial, 126

P V p-regular, 88 Vertex, 105