Representation Theory in Arbitrary Characteristic

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Table of contents :
§1. Semisimple modules and algebras
§2. Decompositions of Algebras
§3. Group-algebras
§4. Projective KG-modules
§5. The number of simple and indecomposable KG modules
§6. Clifford Theory
§7. Brauer's permutation lemma
§8. Symmetric Algebras
§9. Modules of solvable, p-solvable and p-constrained groups
§10. p-adic Numbers
§11. Reduction mod p and decomposition numbers
§12. Blocks of defect zero
§13. Examples
§14. The Theorem of Fong and Swan
§15. More on Brauer-Characters and the Cartan-Matrix

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Foreword Introduction Semisimple modules and algebras Decomposition of algebras Group-algebras Projective KG-modules The number of simple and indecomposable KG-modules Clifford-theory Brauer's permutation lemma Symmetric algebras Modules of solvable (p-solvable and p-constrained) groups p-adic numbers Reduction mod p and decomposition numbers Blocks of defect zero Examples The theorem of Fong and Swan

More on Brauer-characters and the Cartan-matrix


These notes contain the material of lectures (slightly revised) at the Villa Madruzzo in Trento, sponsored by the Centro Internazionale per la Ricerca Matematica of the Istituto Trentino di Cultura and organized by B. Huppert and G. Zacher.

5 1 to Q 9 was given from 10.-14. September 1990 by R. Gow (Dublin), B. Huppert, 0.Manz and W. Willems (all Mainz). § 10 to Q 15 followed, again at Trento, from

7.-11. October 1991, presented by R. K n i h (Essen), B. Huppert and W. Willems (Mainz). I thank Frau Gerlach for the typing of the manuscript.

Mainz, September 1992

B. Huppert

Introduction Representation-theory of a finite group G over a field K of characteristic 0 (often assumed to be algebraically closed) is usually called "classical representation-theory". The situation is not changed very much if char K = p is a prime, but p

4 I GI. But if char K = p and p divides

I G I , we enter a much more complex situation, the so called "modular representation-theory". It is distinguished from the classical theory by the fact, that the group-algebra KG has a nontrivial radical. This is closely related to the fact that now we have to distinguish simple and indecomposable modules. As a study of all indecomposable KG-modules is in most cases out of question - there are too many of them - one concentrates on classes of KG-modules which contain only a finite number of isomorphism-types, namely the simple and the indecomposable projective modules. (These modules still contain much information about the group.)

We take some elementary facts from general ring -theory for granted, namely: (1) The theorem of Jordan-Holder about composition series of modules of finite length. (2) The Krull-Schmidt-theorem about unique direct decomposition of modules in

indecomposable ones (Nagao-Tsushima, p. 27). ( 3 ) Schur's lemma: If V is a simple R-module, then HomR(V,V) is a division-algebra. If R is

an algebra over an algebraically closed field K, then even HomR(V,V) = K. In Q 1 we consider semisimple modules and algebras. The direct decomposition of an algebra as a right-module leads us in

9 2 to the study of projective modules.

Theorem 2.11

describes the direct decomposition of the algebra into twosided ideals. We apply these ideas in § 3 to group-algebras KG. I n particular, we prove some results about the Jacobson-radical of

group-algebras. The structure of projective modules is studied in more detail in Q 4. Here we also show that projective modules are injective, hence the notion "injective" does not appear explicitely in our study of group-algebras. In

5 we determine the number of simple modules

and show that there are infinitely many indecomposable KG-modules if char K = p and the Sylow-p-subgroups of G are not cyclic. In Q 6 we consider the restriction of modules for G to

normal subgroups N of G and the construction of KG-modules from KN-modules. § 7 contains Brauer's permutation lemma and several applications. In § 8 w e study those properties of KG, which follow from the fact that KG is a symmetric algebra. The first part of the lectures is closed by § 9, where several theorems are proved for solvable or p-solvable groups, which do not hold for all groups.

The sections 10 to 1 5 are devoted to the fundamental connection between representation-theory in characteristic 0 and in characteristic p. This connection is established by making a representation in characteristic 0 "integral" and then reducing mod p. For this purpose it is

natural and convenient to work with p-adic numbers and not with algebraic numbers. This guarantees also the vital theorem of Krull-Schmidt (see 11.7). The connection between the Cartanmatrix and the decompositionmatrix (see 11.12) is a powerful tool to study the ideal-structure of the group ring. In


13 we illustrate this by the calculation of the

3 Cartanmatrix for the groups Sq, As, r(2 ), PSL (2,7) and S L (2,s). In § 14 we come back to

p-solvable groups and prove the important lifting-theorem of Fong-Swan. We finish in

5 1 5 by

proving Brauer's result that the determinant of the Cartanmatrix is always a power of the characteristic p.

$ 1 Semisimple modules and algebras


General Assumvtions.

All rings R considered here are algebras of finite dimension over a field K. All R-modules are finitely generated R-modules, hence also are of finite K-dimension.


Definition. a) An R-module M


0 is called simple (irreducible), if 0 and M are the only

R-submodules of M. b) An R-module M is called semisimple, if every submodule N of M has a complementary submodule N ' such that M = N o N '


c) An algebra R is called semisimple, if R, considered as a right-R-module, is semisimple. 1.3

Definition. We define the Jacobson-radical J(R) of R by

J(R) = {r 1 r E R, Mr = 0 for every simple R-module M)


Obviously J(R) is an ideal (two-sided) in R. 1.4

Theorem. a) J(R) is nilpotent.

b) J(R) contains all nilpotent right-ideals of R. c) J(R) is the intersection of all maximal right-ideals of R. Proof: a) As dimK R < w , there exists a composition series R = M >MI> 0

... > M k = O

of R as a right-R-module. As Mi/Mi+l is simple, s o (Mi/Mi+l) J(R) = 0 . This shows MiJ(R)



. Hence

b) Let r be a nilpotent right-ideal and M a simple R-module. Then M r is a submodule of M, hence is 0 or M. But M r = M implies the contradiction M=Mr



Hence M r = 0 and therefore r C J(R). c) If r is a maximal right-ideal of R, then (R/r)J(R) = 0 ,hence J(R) C r . Let A be the intersection of all maximal right-ideals of R and M a simple R-module. Then M

= R/r for some

maximal right-ideal r. As A C t , so (R/r)A = (r

+ A)/r = 0 .

This shows A C J(R). 1.5

Remarks. a) From 1.4 c) follows easily J(R/I) = J(R)/I if I C J(R).

In particular, J(R/J(R)) = 0. b) It can be shown that J(R) is also the intersection of all maximal left ideals of R. Hence there

is a complete right-left-symmetry concerning J(R). 1.6

Lemma. (Nakayama) Let M be an R-module and N a submodule of M such that

M =N

+ MJ(R).

Then M = N.

Proof: We have (M/N)J(R) = (N

+ MJ(R))/N = M/N .

Hence for large k k M/N = (M/N)J(R) = 0 ,

(In 11.9 we shall prove Nakayama's lemma for arbitrary rings.) 1.7

Lemma. If M is a semisimple R-module, then every submodule of M is semisimple.

Proof: Let N '




M. There exists a submodule N" such that M = N ' o N" . Then

N = N f l ( N ' o Nu) = N ' o (N Hence N

n N")

n N" is a complement of N ' in N.

(Dedekind's identity).


Theorem. Equivalent are:

a) M is a semisimple R-module.

k b) M = o Mi with simple R-modules M.. 1 i=l

k c) M =


Mi with simple R-modules M.. 1

i=1 Proof a)

b): As dim

M = M1 o M '


M < m , M contains a simple submodule MI. As M is semisimple,

. By 1.7, also M ' is semisimple. Apply induction on dimK M.

b) a c): This is trivial. c) d a): Let N be a submodule of M. We choose a submodule N ' with N fl N ' = 0 and N ' of maximal dimension. Suppose N

+ N r < M. Then as M =

k i=l

Mi , there exists a simple M j

j N + N ' , hence (N + N ' ) fl M.

= 0. Suppose J n = n ' + m E N n ( N 1 +M.), J w h e r e n E N , n ' E N r , m EM..Then J m=n-n'EM.fl(N+Nf)=0 J and hence

such that M. J

n=nfENnN'=0. This shows N fl (N'

+ M.) = 0. As N ' J

< N'

+ M., this contradicts the choice of N ' . J

Hence M = N o N ' . (Using Zorn's lemma, we can prove theorem 1.8 also for arbitrary modules.)

1 M with semisimple M.. Then M is semisimple. j=1 j J b) If M is semisimple, s o is every epimorphic image of M. 1.9

Corollary. a) Suppose M =

Proof: This follows from 1.8. 1.10

Theorem. Equivalent are:

b) R is a semisimple algebra.

c) Every R-module is semisimple. Proof. a) a b): By 1.4 c), the intersection of all maximal right-ideals of R is 0. As dimK R

Similarly m. = 0 for all j. J Hence n n dimKR r dimK ,o m.= dim m. r dimKR J=I J j=1 K J and therefore

By 1.8, R is semisimple. k b) + c): Let M = C miR be an R-module. As miR is an epimorphic image of R, it is i=l semisimple by 1.9 b). By 1.9 a), then M is semisimple. n

c ) : Now R is semisimple, hence R = o

Mi with simple Mi. Therefore


n J(R) = o MiJ(R) = 0. i=l 1.11

Corollary. Equivalent are:

a) M is a semisimple R-module.

Proof. a) b)

b): is trivial.

a a): As MJ(R)

= 0, M becames an R/J(R)-module by


+ J(R)) = mr

(m E M , r E R ) .

By 1.5 a) J(R/J(R)) = 0. Hence by 1.10, M is a semisimple R/J(R)-module. As the R/J(R)-submodules of M are exactly the R-submodules of M, s o M is also a semisimple

Exercises. 1) Let R = (K), be the ring of all (n,n)-matrices over the field K. Let e.. be the matrix with 1J

entry 1 in position (ij), 0 otherwise. a) R is simple, it has no 2-sided ideals different from 0 and R. n b) R = o e R and e R is a simple R-module. j=1 l j lj 2) Let R={(a..)I a . . E K , a . . = O i f j > i ) 1J 'J IJ by the ring of (n,n)-triangular matrices over the field K.

a) Show that

J(R) = {(a..) 1J

b) Prove R/J(R)


I a..1J E K, a..1J = 0 if j 2 i) .

K o ... o K (as algebra and as module). How many isomorphism-types of

simple R-modules does R have? c) Determine the largest semisimple R-submodule of R. How many types of composition-factors does this have? 3) Let

R = { ( a . . ) ( a . . E K , a i j = O i f j > i , a l l = ...= a nn } 'J 'J be the ring of triangular matrices of type (n,n) over K with constant diagonal. a) Calculate J(R). b) Show that R is indecomposable as a right-R-module.


4) Let Mi(i=l,

... ,k) be submodules of M with semisimple M / M i Then M/ fl Mi is i=l


52 2.1

Decompositions of Algebras

Theorem. a) Let R = PI


... 0 Pn

be a decomposition of R into right-ideals P.. Let J 1 = e + ... + e , where e. E P 1 n J j. Then e.e. = 6.. e. and P. = e.R 1 J

? ) I




b) Suppose conversely, that

1= e


+ ... + en

and e.e. = 6.. e. 1 J

1 J '


Then R=e Ro 1

... o e n R .

Proof: a) We have e. = e e. + ... + e e. , where e.e. E P i . J 1J "J 'J As the decomposition is direct, we obtain e.e. I J = 6i jei . Also R = l R = e R+ 1


...+ e nR = P l o ... o P n ,

e.R C P. implies P. = eiR. 1



b) If r=e r 1 I


e r n n'

then e. r. = e.r is uniquely determined by r. J J J 2.2

Definition. If the Pi in 2.1 are chosen to be indecomposable, we call them the

indecomposable projective modules of R. By the theorem of Krull-Schmidt they are uniquely determined up to isomorphism. (But the sets Pi are in general not uniquely determined.) 2.3

Theorem a) Suppose R = B @...@Bk 1

with 2-sided ideals B. which cannot be written as a direct sum of proper 2-sided ideals. J

a) If

1= f


+ ... + fk , where

f. E B J j'

then f.f = 6.. f. , fi E Z(B) and B. = f.R = R f.1 1J

1 J J




b) If R = I 1 o I 2 is a decomposition into 2-sided ideals I.J of R, then each I.J is a sum of some of the Bi in a). Hence the decomposition in a) is unique up to numbering. c) If M is an indecomposable R-module, there is exactly one i such that

M f . = M and M f. = 0 for j + i. 1 J d) If r is an indecomposable right-ideal of R, then r is contained in exactly one Bi. We call the Bi the blocks of R. Proof: a) As in 2.1 a) we obtain f. f. = 6.. f. . J 1 J l Now r=f r+ 1

...+ fkr -- r f 1 +...+


and f.r , r fi E Bi implies f.r = r fi , hence fi E Z(R). 1


Finally Bi = fiR = R fi follows as in 2.1 a). b) We have


with two-sided ideals B.I.. This forces Bi = BiIl I1 or Bill = 0. J k Hence 1 - R I = o B . I = o Bi . i=l B+~C o



c) As fi E Z(R) , so M fi is a right-R-module. From

k M = o M f .1 i=l

follows the conclusion. d) By c), there exists exactly one fi such that


Lemma. Let e and e ' be idempotents in R.

a) HomR(eR,e ' R) s e ' Re (as K-vectorspace)


b) HomR(eR,eR) is antiisomorphic to eRe as an algebra.

c) eR is indecomposable if and only if 0 and e are the only idempotents in eRe. (In this case we call eRe a local ring.) d) If eR is a simple module, then eRe is a division-algebra. (The converse is not true!) Proof: a) If -

2 a E Hom (eR,elR) , then e a = e a = (ea)e E e ' R e


. The mapping

a -, e n is obviously K-linear. If e a = 0, then for all r E R

(er)a = (ea)r = 0 . If a = e'ae, then a defined by (er)a = aer = ar

is an R-homomorphism of eR into e'R. b)If a , /3 E HomR(eR,eR), we have e ( 4 ) = (ea)B = (e(ea>)B = (eB)(ea)


Hence a + e a is an antisomorphism of HomR(eR,eR) onto eRe


c) Any direct decomposition of eR corresponds to a decomposition e = e l

+ e2 with

idempotents el, e2 in eRe. d) If eR is simple, by Schur's lemma HomR(eR,eR) is a division-algebra. 2.5

Theorem (Lifting idempotents).

Let R be an algebra. We put

R = R/J(R).

a) There exist polynomials pn(n = 0,1, ...) with rational integral coefficients and pn(0) = 0 having the following properties: If

e = eo + J(R) is an idempotent in R,then for sufficiently large n, the element

e = pn(eo) is an idempotent in R with p (e ) + J(R) = E. Also, if a E R and n o e 0a = ae0 (or e 0a = 0 or ae 0 = 0), then ea = ae (or ea = 0 or ae = 0)

b) Suppose

I = ; 1 +...+ en ' where 6. E R and I


= 6 . i . . Then there exist ei E R such that lJ1

1 = e + . . . + e n , e.e. = 6.. e. and e. 1 'J 1 J 1 1

+ J(R) = 6 1. .

C) If eiR is simple, then eiR is indecomposable. Its only maximal submodule is e.J(R). 1 Proof: a) We define the polynomials pn recursively by po(t) = t and 2 3 pn = 3 pnTl - 2 pn-l

for n 2 1 .

Then pn(0) = 0. Also

By induction on n we prove

This is true for n = 0 and follows in general from 2 3 2 pn (e o) - e o = 3pn-1 (eo) - 2pn-1(eo) - eo = 3e0 - 2e: - eo = 0 (mod J(R)) and

2 J(RIzn = 0 , then p (e ) = p (e ) n o n o


If e a = ae 0


' then pn(eo)a = apn(eo)



As pn(0) = 0 , it follows from eoa = 0 that pn(eo)a = 0. b) We construct idempotents el, ..., e in a). Suppose el,

..., em-l

+ J(R) = 6..1 1 Let em = a + J(R) and put

and e.


for 1 s m c n by induction. For m = 1 ihis is done

E R have been constructed such that e.e. = 6. .e.



Then be. = e.b = 0 for j = 1, J J It follows from a + J(R) = b

..., m-1 .



+ J(R)


By a) we choose an idempotent e

6.6 = 6 6. = 0 for i < m that i m m i

+ J(R) =ern.

m = pn(b) such that em

+ J(R) = em. Also by a)

e.e = e e. = 0 for i < m. ~m m~ Finally we put

-...- en-1 2 T h e n e nei. = e.e ~n = 0 for i < n , e n = e n en = I - e l


and en

+ J(R) = I - el -...-E n-1 = 6n

c) Suppose eiR is simple and eiR = M o M 1 2' Then EiR = eiR/eiJ(R) = M1/MIJ(R) o M2/M2J(R). This implies (for instance) M2 = M2J(R), hence M2 = 0 by 1.6. If M is a maximal submodule of eiR, then (eiR/M)J(R) = 0, hence eiJ(R) C M. Conversely, eiR/eiJ(R) = eiR/(eiR

n J(R)) = (eiR + J(R))/J(R)

= eiR

is a simple R-module, hence e.J(R) is maximal in e.R. 1


2.6 Remark. In the study of integral representNions over rings of p-adic integers, a generalization of 2.5 will be important in Q 11. It can be formulated as follows: Let R be a ring and I a twosided ideal in R with the following properties:

2) We define a metric on R by d(a,b) =

if a - b E

,f I

k+ 1.

Also we require that R is complete with respect to this metric. Then idempotents from R/I can be lifted to R. (In the applications in 11.5 this is guaranteed by using the complete ring of p-adic integers.)

2.7 Definition. An R-module P is called projective, if the following property holds:

If M is any R-module and a an epimorphism of M onto P, then ker a is a direct summand of M. Hence

M = ker a o P ' , where P ' 2.8



Theorem. Let P be an R-module. Equivalent are:

a) P is projective. b) P is a direct summand of a free R-module F. (If P is finitely generated, then F can be chosen to be also finitely generated.) c) Suppose M and N are R-modules. If a is an epimorphism of M onto N and

fi E HomR(P,N),

Proof a)

there exists an y E HomR(P,M) such that ya =



b): Let F be a free R-module and a an epimorphism of F onto P.

Then F = ker a o P ' , where P ' b) 3 c): Now suppose F = P


o P ' , where F is free. Suppose given are M, N, a, P as in c).

Let n be the projection of F onto P with kernel P ' . Let F be free in the generators fi(i E I). As

a is surjective, there exist m. E M such that I

m . a = f.nP I


(i E I) .

Define y' E Horn (F,M) by fiyl = mi and let y be the restriction of y' to P.


For p E P we obtain

c ) : We apply c) for

Hence ya =


. We claim M = Py o ker n. If m = py E

Py fl ker a , then

0 = pya = p. Hence Py fl ker a = 0. If rn E M , then

+ m ( -~ny) , m ( -~ ny)a = m a - (mn)yn = rnn -

m = may where may E Py and

ma)^ = 0 .

This shows M=Pyokera. 2.9

Theorem. a) The indecomposable projective R-modules are (up to isomorphism) the eiR

from 2.5 b), where R/J(R) = R = e l R o ... o enR with simple eiR. b) Let e and e ' be idempotents in R. Then eR


e'R if and only if eR

= E'R.

Proof: a) By 2.8, a projective module P is a direct summand of a free module F. As we assume that P is finitely generated, we can choose F also to be finitely generated. Then F is a direct sum of modules of the type eiR (i= I , ...,n). By the KruIl-Schmidt-theorem, then P is isomorphic to some eiR. b) Let a be an isomorphism of eR onto E'R. We consider

where n and n' are the natural epimorphism. Then ncr is an epimorphism. As eR is projective, there exists

/3 E HomR(eR,elR) with

/3n1 = na. This shows

e ' R/e ' J(R) = (eR)/3n1 = ((eR)/3 + e ' J(R))/e ' J(R) , hence e ' R = (eR)/3+ e l J ( R ) . By 1.6 then e ' R = (eR)/3. If we assume dim e ' R


dim eR, then /3 is injective and

Lemma. Use the notation of 2.9. Then e.Rei it 0 if and only if e.R has a J J composition-factor isomorphic to 6.R. 2.10


Proof: If V < U 5 e.R and U/V J Hence 0 z Uei C e.Rei .


e i R , then 0 z (U/V)ei = (Uei

+ V)/V.


If conversely e.Re. z 0, then by 2.4 HomR(eiR,e.R) z 0. Hence there exists an epimorphism of J J J eiR onto a submodule U z 0 of e.R, and then U has a composition-factor i.K on top J 1 (by 2.5 c)). n Theorem. Suppose R = o e.R with indecomposable e.R as in 2.5. We write 1 1 i=l


if there exists a chain ' - e.R J



such that ei R and e. R have a composition factor in common. Obviously, is an 1 S s+l equivalence relation. If A ... ,A are the equivalence classes of - , then the decomposition k 1' --

R = o B. into indecomposable twosided ideals B. is given by 1 1 i=l

Proof: Suppose e.R and e.R have the common composition factor isR Then by 2.10 1 J e.Re z 0 i. e.Re . I s J S By 2.3 d) there exist block ideals B, B ' , B" in {B1 esR

C B". But then

,..., Bk) such that eiR C B, e.R C B ' and J

0 + eiRes C BB" and 0 + e.Re C B r B " . J sThis implies B = B" = B r . Hence eiR and e.R are in the same block ideal. If J eiR e.R, this argument also shows that eiR and e.R are in the same block ideal. If we put J J


then R = o Ci and each Ci is contained in a block ideal B i t i= 1


We show that Ci is a twosided ideal, hence is a block ideal: Suppose that eiR C C. and J esR 5 Cm, where j + m . Then e.R and e R have no composition factor in common. In parti1


cular, esR has no composition factor isomorphic to ZiR, hence esRei = 0 by 2.10. This shows C C. = 0 for m z j. Thus m J RC. = ( o C )C. = C. C . C C. . J , m J J J - J Therefore C. is a twosided ideal in R. Then by 2.3 b) C. is the sum of some block ideals, so is J J a block ideal. 2.12

Theorem. Let R=e Ro 1

... o enR

be a semisimple algebra with simple e i R Then the blocks of R are the sums of the eiR of the same isomorphism type (often called homogeneous components). Proof: This follows from 2.11. 2.13

Theorem (Wedderburn). Let R be a simple algebra (which means that 0 and R are the

only twosided ideals of R). a) Then R, as an R-right-module, is a direct sum of n isomorphic simple modules, R-Vo

... o V s a y . n

b)R is anti-isomorphic to the matrix ring (D) over the division-ring D = HomR(V,V), where n V is the only simple R-module.

Proof: a) As R is simple, this follows from 2.12. b) By 2.4 b), HomR(R,R) is anti-isomorphic to R. Write R = V c ... c V = { (vl,

... ,vn) I

vj E V } .

If or E HomR(R,R) then

where or.. E HomR(V,V) = D. Jl

One easily checks that or 2.14


(or..) is an algebra-isomorphism of HomR(R,R) onto (D),. '1

Theorem. Let R be a semisimple algebra over an algebraically closed field K. Then k R = o (K),. i=l I

(direct sum as algebra).

The simple R-modules are (up to isomorphism) the V. (i=l, ...,k), where I

dimKVi = ni, and (K),. acts on V. naturally and (K), (j + i) annihilates Vi. I I j Proof: By 2.3 R = B1 @ . . . @ B k' where the Bi are the blocks of R. By 2.12, the Bi are the homogeneous components of the semisimple right R-module R. By 2.13, we obtain that Bi is antiisomorphic to (D.) I

n.1 for some

skew-field Di. As K is algebraically closed, Di = K and Bi is isomorphic to (K),,, for (K),. is 1


antiisomorphic to itself (by transposition of the matrices). Exercises.

5) Let

R = { (aij)

I a.. E K , 'J

ail = O i f j > i }

be the ring of all (n,n) triangular matrices over the field K. Let e.. be the matrix with entry 1 'J in position (i,j), 0 otherwise.

n a)

and eiiR is indecomposable. R = o e..R 1I i=l


e..R = o K e.. 11 j=l 1J ' e..J(R) =



K e.. 'J

and K eil is the only simple submodule of eiiR. c)

R has only one block.


Determine the composition-factors of e..R.


R has (up to isomorphism) n simple factor-modules, but only one simple submodule.


At first we have to decide which group-algebras are semisimple. 3.1

m Lemma. Let U be a subgroup of G with coset-decomposition G = U Uti and let V be a i =1

KG-module. If a E HomKU(VU,VU), then the mapping

lies in HomKG(V,V) and

p defined by

P is independant of the choice of the transval {ti I i = 1,...,m).

(Here VU denotes V, considered as a KU-module.) m

Prooff: We have G =

U tilu

. Suppose g E G

-1 -1 and gti = t i , ui with ui E U




(1 .) is a permutation. Hence


Theorem. Let V be a KG-module and W a KG-submodule of V. Suppose U r G such

that ( 1 ) C h a r K ) c IG: UI


(2) VU = WU o W '

with a KU-module W '


Then V = W o W" with some KG-module W". In particular, if VU is semisimple, so is V. Proof: Consider the projection -

76 E

HomKU(VU'VU), defined by

( w + w l ) n = w for w E W , w l E W '


Let G = U Uti with m = I G : UI . We consider n',defined by i= 1

By lemma 3.1, n' E HomKG(V,V). If w E W, then

vnr = m VZ


(vn1)n' = vn'

(vtjl n)ti E W


i= 1

. Therefore n'

V = im n'

0 ker

is a KG-projection and

n' = W 0 ker n1

with a KG-module ker n' 3.3

Theorem (Maschke). KG is semisimple if and only if Char K

4 I GI.

Proof: a) Suppose Char K 4 I GI. Applying 3.2 with U = E, we see that every KG-submodule of KG has a complement. Hence KG is semisimple.


b) Now suppose Char K

( GI. We consider

C g. gEG Then ig = i = gi for all g E G. Hence i=

i2= IGli=o. As i E Z(KG), so i = KGi is a 2-sided ideal with i2 = K G i2 = O . By 1.4 b), therefore

J(KG) 2 i z 0


Next we consider an extreme case:


3.4 Theorem. Suppose G ( = pn and Char K = p.

a) Up to isomorphism there exists only one simple KG-module, namely K with trivial action of G.

and dim J(KG) = IGI-1.

c) The only simple KG-submodule of KG is K

g. In particular, KG is an indecomposable gEG

d) Every projective KG-module is free. Proof: a) Let V be a simple KG-module and 0 z v E V. Let Ko = GF(p) be the prime-field in K. Then G operates on vKoG. As vKoG is an epimorphic image of KoG (as a


K -vectorspace), I vKoG is a power of p. We decompose vKoG into orbits under G: 0

vK G = W I U 0

... U W r ,

where W1 = (0)and I Wi I is a power of p. Hence there is an orbit W2 = {w) (say) of length 1. Therefore wg = w


0 for all g E G and Kw is a KG-module. This shows V = Kw.

b) This follows immediately from a) and the definition of J(KG) in 1.3 . c) If w =

C a g E KG and wh = w for all h E G, then a = al for all g E G. gEG g g

d) By c), KG is the only indecomposable projective module, and it is free. Hence by the Krull-Schmidt-theorem every projective KG-module is free. 3.5

Theorem (L.E. Dichon). Suppose Char K = p and I GI =

where p

4m. If V is a

projective KG-module, then pa divides dimKV. Proof: Let P be a Sylow-p-subgroup of G. If G = U tip, then i =1

is a free KP-module. Hence every free KG-module is a free W-module. Therefore by 2.8 every projective KG-module is a projective KP-module, hence is a free KP-module by 3.4 d). This implies I PI

1 dimKV.

If in particular the trivial KG-module K is a direct summand of KG, then by 3.5 pa divides dimKK = 1, hence p

4 I GI.


Remark. Let K be an algebraically closed field of characteristic p and let P be an


indecomposable projective KG-module. Suppose GI = a) If G is psolvable, then

with p I/ m.

$T dimKP, (see 9.6).

b) The statement in a) is for p > 2 in general not true. c) Let PI be the indecomposable projective module with trivial head. If G has a p-complement, then dim PI = pa (see 4.5 c)). If p = 2, then

za T dim PI

for all groups G

(see 7.14). But there is no correspnding result for p > 2. If G = PSL(2,7), then for p = 3 the projective module PI has dimension 9 (see 13.8). 3.7

Theorem. For any KG-module V we define ker V = {g


g E G, vg = v for all v E V ) .

Obviously ker V a G. a) If Char K = o, then



V simple b) If Char K = p, then fl

V simple

ker V = 0 (G) P

is the largest normal p-subgroup of G. Proof: a) If g E f l ker V, then by 3.3, g - 1 E J(KG) = 0 . V simple b) Let V be a simple KG-module. Then V contains a simple KO (G)-module. By 3.4 a) P therefore W = { v J v E V , v g = v forall g E O ( G ) ) * O . P B u t i f w E W , g E O (G)andhEG,then P

So W is a KG-submodule of V. As V is simple, so W = V. This shows

" v simple ker . Suppose conversely that g E

ker V, hence g fl V simple

- 1 E J(KG). As J(KG) is nilpotent

(1.4 a)),

for sufficiently large pS we obtain


n ker V s 0 (G) V simple P


We now describe a way to construct KG-modules from modules for a subgroup H of G.

m 3.8

Definition. Suppose H


G and G = U H ti. Let W he a KH-module. We form the i= 1

induced module

WG becomes a KG-module by (w

ti)g = w o t.g = wh. o t . ,





if t.g = h.I t i ' E H t i , . Hence application of g permutes the K-subspaces W I


t i in the same

way as the cosets H t. are permuted. I


Theorem (Nakayama's reciprocity theorem). Suppose H


G and let W be a KH-module

and V a KG-module. We have the following isomorphisms as K-spaces: a)

G HomKG(V,W )



where VH is the restriction of V to H. b)





Proof: Let G = U H ti i=1 a)

and t l = 1 .

G Let n E HomKG(V,W ). If v E V, then VCY =

1 i=l

v n .1 o t .I

with ai E HomK(V,W). If h E H, then

Comparing the component for i=l, we obtain (vh)al o 1 = v a o h = (val)h o 1 . 1 Hence al E HomKH(VH,W). From m (va)t:l = J



v a . o t.tyl = v a . o 1 + ... 1



= (vt.-1)a = 2 (vt.-1) a . o t. = (vt.- 1) a o 1 + ... J i=l J 1 1 J 1 we conclude

-1 v a . = (vt. ) a J J 1' Hence a is uniquely determinded by al and a -I a1 is a K-linear monomorphism. If conversely al E HomKH(VH,W) is given, we define a by

If g E G and tig = hi ti , then (va)g =

-1 .I (vti ) a l I= 1

o t.g =



G Hence a E HomKG(V,W ) and therefore a


(vt.1 )alhi o t 1. ~

al is bijective.

p E HomKG(W G,V), we define PI E HomK(W,V) Obviously P1 E Homm(W,VH). Also

b) If


o ti)P = ((w o ])ti)@ = ((w o 1)P) ti = wplti

Hence #I1 determines defined by

by (w o 1)j3 = wPl.

uniquely. Conversely if


P1 E HomKH(W,VH),

the mapping j3

H lies in HomKG(W ,V), as a calculation similar to a) shows. Hence




dules of M isomorhic to V is by 1.9 a semisimple submodule


of M. If a E HomKG(V,M), then V a


/3 /31 is bijective. -I

Remark. Let V be a simple KG-module and M any KG-module. The sum of all submo-

S. Hence

k HomKG(V,M) = HomKG(V,S) =





In particular, dim HomKG(V,M) = k dimKHomKG(V,V). If moreover K is algebraically K closed, by Schur's lemma we obtain dim HomKG(V,M) = k K


We call k the multiplicity of V as a submodule of M. Similarly, dim HomKG(M,V) = l dimKHomKG(V,V), if M/N K


V e ... o V



where N is the intersection of all submodules T of M with M/T = V. (M/N is the largest semisimple factormodule of M with only composition factors isomorphic to V.) 3.11

Theorem. If V is a KG-module, we define H(V) = V/V J(KG)

(head of V)

S(V) = AnV J(KG)

(socle of V).

and By 1.11, H(V) is the largest semisimple factormodule of V and S(V) the largest semisimple submodule of V. Let K be algebraically closed. Suppose H


G, V is a simple KG-module and W a simple KH-module. The

G multiplicity of V in H(W ) is equal to the multiplicity of W in S(VH). The multiplicity of V

G in S(W ) is equal to the multiplicity of W in H(VH). Proof: By 3.9 we have -

dimK HomKG(V,W G) = dimK HomKH(VH,W)


This implies by 3.10 our assertion. 3.12

Theorem. Let K be algebraically closed.

a) The multiplicity of a simple KG-module V in KG/J(KG) and in S(KG) is dimKV.

In particular KG/J(KG)

s S(KG).

b) If P is an indecomposable projective KG-module, then S(P) is simple.

(We shall see in 5 8 that even S(P) = P/PJ(KG).) Proof: a) KG is induced from the trivial module K for the subgroup E = (1). If V is a simple KG-module, then the multiplicity of V in H(KG) resp. S(KG) is by 3.11 equal to the multiplicity of

K in S(VE) resp. K in H(VE) , and both are equal to dimKV. This shows KG/J(KG)


b) We have KG = o P. with indecomposable projective modules Pi. As I

Pi/Pi J(KG) is simple, by a) also S(P.) must he simple. I


Theorem. Suppose K is algebraically closed and KG semisimple. Then KG=



V simple

In particular



( d i m K ~ ) L. V simple Proof: Now KG = S(KG). By 3.12, the multiplicity of any simple KG-module



in KG

is dimKV. 3.14

Theorem (Clifford). Let V be a simple KG-module and N a G. Then VN is a

semisimple KN-module.

Proof: Let W be a simple KN-submodule of VN. Then also Wg for any g E G is a simple KN-module, for if x E N, then -1 W g x = W gxg g = W g .



Wg is a KG-submodule z 0 of V. So by simplicity of V we obtain


v= 1

Wg. gEG Hence by 1.9 a) VN is a semisimple KN-module. Theorem 3.14 is only the first of many more detailed statements. We come back to these problems in Q 6. To obtain more information about J(KG), we first consider a lemma.


Lemma. Suppose N 4 G and Char K

4 I GIN 1. If W is a simple KN-module, then W G

is a semisimple KG-module. ti Proof: By 3.2 it suffies to show that (W )N is semisimple. But if G = U Nt, then -




= o


t€T If w E W and y E N, then

G Hence W o t is a KN-module, which obviously is simple. Therefore (W )N is semisimple. 3.16

a G. Theorem. Suppose N -

a) J(KN) C J(KG)


b) (Villamayor, Green, Stonehewer) If Char K

4 I G/N I

, then

J(KG) = J(KN)KG and hence dimKJ(KG) = dimKJ(KN) . I GIN I


c) Suppose Char K = p. Suppose also that G has a normal Sylow-p-subgroup P and G = U Pt, t€T

where T is a transversal of P in G. Then J(KG) = J(KP)KG = < (y-l)t

I y E P, t E T >




Proof: a) If V is a simple KG-module, then by 3.14 we have -

with simple KN-modules W.. Hence VJ(KN) = o W.J(KN) = 0 . J j=1 J This shows J(KN) C J(KG). b) Suppose G = U



Suppose a =

Nt, hence KG = o KN tET

. t.

at t E J(KG), where at E KN. Let W be a simple KN-module. By 3.15, W

G is

t€T a semisimple KG-module. Thus W


a = 0, hence for all w E W

This forces w at = 0, so W at = 0 for every simple KN-module W. This shows at E J(KN), so

c) By b) and 3.4 b) now J(KG) = J(KP)KG = < (y-l)t

I y E P, t E T >.

A special case of induced modules are the permutation modules, which we consider in

example 3.17. 3.17

Examples. a) Suppose H


m G and G = U H ti. Consider K as the trivial KH-module i= 1

m and put K~ = o K(l rn ti). i=l If g E G and H tig = H t i , , then tig = h t for some hi E H and i i' (1 o ti)g = 1 o t.g = 1 63 hitiI = 1 h.1 @ i' t = 1 @ it ' I


If we put v.I = 1 o t.,I then KG = a K v . 1 i= 1 G and v.g = v if H tig = H t . , . We call K the permutation-module for G:H. Obviously I i' I

Kw with w =

1 i= 1



are KG-submodules of K ~ where , wg = w for all g E G, vg - v E A for all v E V. Hence G KwcK /AcK is the trivial KG-module. If Char K b) Let G = S and H = S . Then K 4 3

m = I G:HI, then w $ A,i!hence K~ = Kw



G IS . the "natural" permutation-module for G of degree 4.

Suppose at first Char K + 2. Then by a) K


= Kw o A.

If A is not simple, it has a I-dimensional sub- or factormodule U. As U corresponds to a homomorphism of S4 into K',

U belongs to the trivial or the signum representation.

Now by 3.9 G HomKG(K ,U) = HomKH(K,UH) =




= U~

0 otherwise. (Observe that for G = S4 then UH corresponds to the signum representation of H = S3 if U belongs to the signum representation of S4). Hence



HomKG(Kw o A U ) = HomKG(K,LJ) o HomKG(A,U)


This shows HomKG(A,U) = 0. Similarly HomKG(U,A) = 0. Hence A is simple. Now assume Char K = 2 and G = S4. Then we have by a) 0 C Kw C A C KG and dimKA/Kw = 2. If A/Kw is not simple, it has only the composition factors K,K for now K is G the only KG-module of dimension 1. Hence K has only the composition-factor K, with multiplicity 4. But if g E G and o(g) = 3, then K


is a semisimple K< g >-module, so g would

act trivially on K ~But . this is not true, hence A/Kw is simple of dimension 2. Write A/Kw = V. By the same argument VH is a simple KH-module. Hence G G HomKG(K ,V) = HomKH(K,VH) = 0 and HomKG(V,K ) = Hom

KH(VH,K) = 0 .

Therefore V is not contained in H(KG) and in S(KG). Hence the only composition-series of KG is 0 C s(K')

C K ~ J ( K GC) KG , where

S(KG) r KG/K GJ(KG) = K G G K J(KG)/S(K ) V is simple.



Already the group-algebra KS shows a remarkable variety of structure, depending on 3 the characteristic of the field K 3.18

Example. Let G = be the symmetric group ST where a 2 = b 3 = 1, aba = b-1

a) Suppose that Char z 3 and K contains a primitive third root


of unity. By Maschke's

theorem, KG is a semisimple module for < b >. More explicitely: We put

Then e.b = J e . . Using 1 + E + E~ = 0, we see easily J J 1 = e + e + e2 , e.e. = 6..e. 0 1 '1 1JI Hence putting R = KG, we obtain R = eOR o e l R o e 2 R . Also e.R = Ke. o Ke.a I




and the actions of a and b on e R are described by the matrices 1

(Observe e ab = e aba 1 1

. a = e 1b-la

= c-le a.) As



E-l, we see easily that elR is a simple

R-module. The same is true of e2R. Case 1: Assume Char K


2,3. Then R is semisimple. We already know three simple

R-modules, namely K


gEG Hence by 2.14

g, K

sgn g gEG

. gand e l R .

R = (K) o (K) o (K)2 . This shows elR e e2 R. Case 2: Assume Char K = 2 and K algebraically closed. Then the signum-representation is the trivial representation. AF now J(R)


0 and we know already simple modules of dimensions 1

and 2, we obtain R/J(R)


(K) o (K)2 and dim J(R) = 1 .

On eoR =Keg o Keoa we have the representation given by

Hence eoR is indecomposable and contains the submodule


b) Now we suppose Char K = 3. By 3.15 c), dim J(R) = 4. We put e l = 1 (l+a) , e2 = 1 (l-a) . Then 1 = e1

+ e2,

e.e = Dijei . I j

Therefore R = e l R o e2R , where dim eiR = 3 by 3.5. As J(R) = e l J(R) o e2J(R) has dimension 4, so dim eiR/eiJ(R) = 1 and eiR is indecomposable. We put

w- = (1-a)(l+b+bL) =


sgn g.g .

gEG As b-1, bL-1 E J(R) (see 3.6 b), so

Obviously, w+g = w + and w-g = sgn g wfor all g E G. We put

we have


Therefore elR/elJ(R) is a module for the trivial representation. Furthermore

Hence w

+ = sl(l-b) E elJ(R) 2

and the module

belongs to the signum-representation. Similar statements hold for e 2R.

so elR and e2R belong by 2.10 to the same block. Hence KS3 has only one block if Char K = 3. 3.19 Example. Now suppose that G = A4 is the alternating group of degree 4. Then

G = (al,a2,b) where a 2l = a 22 = b 3 = 1 , a l a 2 = a 2 a l , b - 1a 1b = a 2 , b- 1a b = a a 2 1 2 ' Suppose that K contains a primitive third root of unity if Char K + 3. a) Suppose Char K z 2. In 3.17 b) we constructed a simple KS4-module V of dimension 3. If


would be not simple, it would be by 3.14 a direct sum of modules of dimension 1. Then A4

is simple. As A ' would act trivially on V, which by construction of V is not true. Hence V 4 A4

I GIG' I = 3, there are 1 or 3 simple KG-modules of dimension 1 if Char K = 3 resp. Char K z 3. Hence if Char K + 2,3, we obtain R = KG


(K) o (K) o (K) o (K)3


(For the modules of dimension 1 we need that K contains primitive third roots of unity.)

If Char K = 3, then we have only one simple module of dimension I and one of dimension 3. As J(KG) + 0,so 11 r dim KG/J(KG) =

1 nL 1

(ni the dimensions of the simple KG-modules). This forces

R = KG/J(KG) E (K) o (K)3 and

Lifting idempotens, we get R = eoR o elR o e2R o e3R


As 3 divides dim eiR (see 3.9, so dim eiR = 3 (i=O ,..., 3). Therefore e.R 1

= eiR is simple for


i=1,2,3. As eoR has at the top the 1-dimensional module K, and as there is now (Char K = 3)


only one 1-dimensional and no 2-dimensional simple module, eoR has only the composition


factors K,K,K.


b) Now let Char K = 2 and let


be a primitive third root of unity in K. Put

1 -2j2 e. = 3 (1+E-J b + ~b ) (j=0,1,2) . J Then as in 3.18 a) we obtain KG = eOKG o elKG r e2KG. As 4 divides the dimension of every projective KG-module by 3.5, the e.KG are


indecomposable. We analyze eOKG:

As eob = eo7 we obtain e KG = (eo,e a 0

eOa2 eOala2) (as K-space).

By 3.16 c) follows



eoJ(KG) = eo (h (1-al) , bl(l-a2) , bl(l-ala2)) = (eo(l-al) , eo(l-a2) ea(l-ala2))


As Char K = 2, also

e (1-al)(l-a 1a 2) = eo(l-a2)(1-a 1a 2). e 0(1-al)(l-a2) = eO(l+a1+a 2+a 1a 2) - 0 This shows 2 eOJ(KG) = KeO(l+al+a2+ala2) = S(eOKG) . e0(1-al)b = eob-eOba2 = eO(l-a2) , e (1-al) eo(l-a2)b = e b - e ba a e -e a a 0 0 1 2 = 0 0 1 2 = 0 2 Hence b operates on eOJ(KG)/eOJ(KG) by the matrix hence can be diagonalized. Finally we obtain

+ e0(1-a2) + eo(l+a 1+a 2+a 1a 2) .

(7 I)

, which has eigenvalues



- 1,

Using e.b = $e.b we obtain a corresponding result for e.KG (i=1,2), namely J J J el KG

and similarly for e2KG. 3.20 Example. Let G = S4 and K algebraically closed with Char K dimensions of the simple KG-modules. As G/V


2,3. We want to find the

S3 (V the Klein-subgroup), by 3.18 we know

that KG has simple modules of dimensions 1,1,2. By 3.17 b) we also have a simple module V of dimension 3. Hence

where the mi are the dimensions of the missing simple modules. But 9 = 1mf

and mi > 1


has only the solution m: = 9. Hence one module is missing, it is also of dimension 3. This

module is obtained by multiplying the matrices of the representation on V by the signum. Wedderburn's theorem says KG = (K) @ (K) @



(KI3 @ (Q3 .

(This is true for any K with char K z 2,3, K not necessarily algebraically closed.) 3.21 Remark. If N a G and char

4 ( N 1,


is a central idempotent in KG. If 1 N I > 1, then KG hence has more than one block.

If Char K = p and G is p-solvable, there is a converse by W. Gaschiitz and P. Fong: G has only one block if and only if 0 ,(G) = E (see 9.14).


Exercises 6 ) Let G = (gl

a) KG s K[yl

,..., gd) be an elementary abelian p-group with IGI = pd. Suppose Char K = p. ,..., yd]/i,

where i is the ideal generated by

yp ,..., yd.P

b) Show that dim J(KG)/J(KG)~= d. 7) Let G be a p-group. The minimal number d of generators of G is given by


d = p (Burnside). Let G = (gl d

,..., gd). Suppose again Char K = p.


hence dim J(KG)IJ(KG)~5 d .

b) Put

G = G/q!(G).

Using the natural epimorphism of KG onto K c and exercise 6 show that = d~ .~ dim J(KG)/J(KG)~a dim J ( K ~ / J ( K

(S. Jennings has given a complete, but rather complicated description of dim J(KG)'/J(KG)


in terms of the structure of the p-group G; see Blackburn-Huppert 11, p. 259). 8) Suppose K is algebraically closed, A

that dimKM



G and A abelian. If M is a simple KG-module, show

I G:A I .

9) Let G = (g) be a cyclic group of order I GI =

contains a primitive m-th root of unity

a . a) eiSP = r'e.I


with p

4m. Suppose Char K = p and K



b ) l = e o + ...+ e


ande.e.=d..e.. ' J 'J'

c) dimKeiKG = pa and eiKG is an idecomposable projective KG-module. d) KG has m blocks. e) Show J(KG$ = ( g r n - 1 $ ~ ~ and hence a J(KG)P

a 3 J(KG)P =


10) Let G = GL(n,q) and V = V(n,q) the natural G-module. Show: If n > 1 and (n,q)

* (2,2),

then V is not a projective KG-module. 11) Let P be a Sylow-p-subgroup of G and K with Char K = p the trivial module for KP. If V G G is any simple KG-module, then V is contained in H(K ) and in S(K ). 12) Let G = SL(2,3), K algebraically closed, Char K z 2,3. a) G permutes the 4 1-dimensional subspaces of the Zdimensional vectorspace over GF(3). The kernel of this permutation-representation is (-E) and GI(-E)


b) Simple KG-modules with -E in the kernel have dimensions 1,1,1,3.





If -E is not in the kernel of a simple KG-module V, then dim V is even.

d) KG = (K) I




6)(KI3 @




(KI2 @ (KI2


13) Let Char K = p. Equivalent are: a)


= pn.

b) KG is an indecomposable KG-module.

c) KG/J(KG) is a simple KG-module. 14) Suppose that V is a KG-module, H s G and Char K

4 1 G : HI. I f VH is a projective

KH-module, then V is a projective KG-module. 15) Let G = GL(n,p) with n > 1 and let V = V(n,p) be the natural KG-module

(where Char K = p). a) If (n,p) z (2,2), then V is not projective. b) V(2,2) is a projective module for GL(2,2).

9 4 Projective KG-modules We collect informations from 9 2. 4.1

Notations. Let

n K G = o e. KG i= 1 1 with indecomposable projective eiKG. Then eiKG/eiJ(KG) is simple, and the isomorphismtype of eiKG is determined by eiKG/eiJ(KG). The multiplicity of eiKG/eiJ(KG) in KG/J(KG) is by 3.12 a) equal to dimKeiKG/eiJ(KG), if K is algebraically closed. We denote by MI ,..., Mk the isomorphism-types of simple KG-modules. If M.1 = eiKG/eiJ(KG), we denote eiKG by P(Mi). Then if K is algebraically closed, we can state k KG E o dimKMi P(Mi) . i=l We keep the notations of 4.1. 4.2

Theorem. Let M be any KG-module and k

Then we define the projective cover P(M) of M by

a) There exists an epimorphism a of P(M) onto M with ker a r P(M)J(KG). b) Let P be any projective KG-module and y an epimorphism of P onto M. Then P


P(M) o P'

Proof. a) Let n be the natural epimorphism of P(M) onto k


o ni(P(Mi)/P(Mi)J(KG)) r o niMi E M/MJ(KG) i=l

with ker n = P(M)J(KG).


We consider

As P(M) is projective, there exists a such a@ = n. Hence

Nakayamas Lemma (1.6) implies M = P(M)a

+ MJ(KG) = P(M)a .

Hence a is surjective. If v E Ker a, then O=va/3=vn, therefore v E P(M)J(KG). b) Let P be projective and y an epimorphism of P onto M. We consider

where 6 a = y, and 6 exists as P is projective.

If v E P(M), then there exists w E P with va=wy=w6a. Hence v - w6 E Ker a




Using again 1.6, we conclude P(M) = Pb

+ P(M)J(KG) = P 6 .

As P(M) is projective, this implies

P = ker b o P" = ker 6 o P(M)



Definition. Let V and W be KG-modules. Then V

W becomes a KG-module by

( v o w ) g = v g o w g for v E V , w E W , g E G . (This procedure does not work for general algebras in place of KG!)


Theorem. a) Let F be a free and V an arbitrary KG-module. Then F o

KG-module in rank F


V is a free

dimKV free generators. V is projective.

b) If P is a projective and V an arbitrary KG-module, then P o Proof. a) It obviously suffices to show that KG o -

V is free, for the tensor-product is distri-

butive. Let V be the K-vectorspace V with trivial action of G (vg = v for all v E Vo , g E G). 0 Then KG o V0 is a free KG-module in dimKV free generators. We define a K-linear mapping


of KG o

V onto KG o

V0 by

(gov)t=govg Obviously



for g E G , v E V .

is bijective. For g,h E G and v E V we have

( ( g o v ) h ) r = (gh o v h ) r = gh o(vh)(gh)-' = gh o v g Hence KG o



KG o


= ( ( g o v)t)h

Vo as KG-modules.

b) If P is projective and F = P o P ' free, then by a) (PoKV)o(P1 o K V ) z F o K V is free, hence P o 4.5

V is projective.

Theorem. Suppose H



a) If W is a projective KH-module, then b) Suppose Char K = p and p

wG is a projective KG-module.

4 1 H I . Let KH resp. KG be the trivial module for H resp. G.

Then ( K ~ is)projective ~ and ( K ~ ) ' dim P(KG)



P(K G) o P ' . In particular



c) Suppose that H is a p-complement of G, hence I G:H 1 = pa and p 1 I HI. Then P(KG)


(KHIG is the permutation-module for G:H. (By Phillip Hallf s theorem all

p-solvable groups have p-complements.)

Proof. a) Let F be a free KH-module such that F -









W e W ' . Then










is KG-free, for KHoKHKG=KG. Hence WG is a projective KG-module.

.&I H 1,

b) As p

so every KH-module is projective. Hence by a) ( K ~ is)KG-projective. ~ If a is

the epimorphism of KG onto KG, given by



a is an epimorphism of

G (KH) = KH o KH KG onto KH o KH KG

= KG. By 4.2 b) this implies G (KH)

= P(KG) o P' .

G a c) Now dimK(KH) = I G:H I = p .

By Dickson' s theorem 3.5, pa divides dim P(KG), hence by b) ( K ~ = )P(K3. ~ 4.6

Theorem. If M is a simple KG-module, then P(M) is isomorphic to a direct summand of

P(KG) o


Proof. By 4.4 b) P(KG) o -

M is projective. If n is the epimorphism of P(KG) onto KG, then

n o L is an epimorphism of P(KG) o

M onto KG Q

M = M. Apply 4.2 b).

Now we can prove some crude estimates for dim J(KG). 4.7

Theorem. Suppose K is algebraically closed of characteristic p and


with p ) r m .

a) dimKJ(KG)


pa -1.

b) dimKJ(KG) a I GI -

&A. G

If in particular G has a p-complement, then dim J(KG)




e. P

Proof. a) We have n KG = o Pi i=l with PI = P(KG) (say). Then

But P1/PIJ(KG) s KG and hence by 3.5

b) By 4.1 we have



.s dimKP(KG)




This implies k




( GI = dim KG = o dimKMi dimKP(Mi) 5 dim P(KG)



But from k KG/J(KG)


o dimKMi




(see 3.13)




Remark. Equality in 4.7 is extremely rare:

a) dimKJ(KG) = pa-lholds if and only if either G is a p-group (ct 3.4 b)) or if G is a


Frobenius-group with Frobenius-kernel F and G/FJ = pa (cf. 7.8 and Feit, page 260). b) (Brockhaus, J. Algebra 95).

dim J(KG) = ( G I -

holds if and only if G has a normal Sylow-psubgroup. dim P(KG)


The proof uses the classification of finite simple groups. (For p=2, the classification can be avoided using duality arguments.) c) It is still an open question whether dim J(KG)




E l pa


This condition is equivalent to ( d i m ~ ) ~ s q = I G l ~ . . P

M simple

Remark (P. Fong). Let K be algebraically closed and Char K = p. Let G be a p-solvable b group and pa I GI. If M is a simple KG-module, then dimKM divides [ G I . If dimKM = p m 4.9


and p

4m, then a dimKP(M) = p m In particular, P(KG) o

(see 9.6)


M is indecomposable if and only if p

4dimK M. This holds

for all simple KG-modules and G arbitrary if and only if G has a normal Sylow-p-subgroup (cf. 3.7 b), 7.12). But in general, dimKM does not divide I GI. The simple group PSL(2,7) of order 168 has in characteristic 7 simple modules of the dimensions 1,3,5,7 (see 5.11).


For general rings, besides the projective modules the injective modules play an important role. For group-algebras over fields projective and injective modules coincide. We use duality theory of finite-dimensional vectorspaces to prove this. 4.10



Definition. Let V be a KG-module. We put V = HomK(V,K). Then V becames a

KG-module by (ng)(v) = n(vg-') for n E v*,v E V and g E G. Observe that -1 - 1 (4g1g2))(v) = 4 v g 2 g1 ) = ("gl)(vg;l) = ((ag1)g2)(v) . * 4.11 Theorem. If P is a projective KG-module, so is P . Proof. As HornK(- ,K) is additive, it suffices to show -


We define the Let { a g E K ) be the K-basis of HomK(KG,K), defined by a (h) = 6 g g gh '


linear mapping h from KG onto (KG) by gh = a . For gl,g2, x E G we have g ((g1g2)h)(x) = a (4 = a 81g2 gl which shows (glg2)h = (glh)g2

= ( a g )(x) = ((g1h)g2)(x) gl



Hence A is a KG-isomorphism. Theorem. Let P be a projective and V any KG-module. If a is a monomorphism of P


into V, then P a is a direct summand of V. (P is "injective".) Proof. From the exact sequence O---+P%V we obtain by dualization



where a is easily seen also to be a KG-homomorphism. (Recall that a is defined by

(A a



)(p) = h(pa) for h E V , p E P .)




By theorem 4.1 1, also P is projective, hence V = Kern a o W for some KG-module W. Dualizing again (and using duality-theory of vectorspaces), we obtain V = (Ker a*)' s W' , where (Ker a*)' = {v

1 h(v) = 0

if h E Ker a * } .


If h a = 0, then for all p E P


0 = ( h a )(p) = h(pa)


Hence (Ker a*)'



From r dim (Ker a*)' dim P a = dim P = dim P* = dim ~ * / ~a e* = we deduce P a = (Ker a*)' , hence V = P a o W'



Theorem (Bryant-Kov5cs)

a) Suppose that for every 1 it g E G there exists a KG-module V on which g does not act as a g scalar. Then o V contains a submodule isomorphic to KG. lzgEG g b) Let V be a faithful G-module. Then IGI-1 0

j =I

v o ... o



has a direct summand isomorphic to P for any indecomposable projective module P. In particular every simple KG-module is isomorphic to a submodule of some V o ... o V Proof. a) A KG-module W contains a submodule isomorphic to K G if there exists w that wf(wgI l*gEG). For then wh@(wgI h * g E G ) , so {wg


g E G ) is linearly independent over K.

Now we prove: Let U,V be KG-modules and R,S subsets of G such that u @ ( u r I r E R ) , V $ ( V S1 s E S ) . Then

By assumption, there exist a E Hom (U,U) and /3 E HomK(V,V) such that K u a = U, (ur)a = 0 if r E R , v/3 = v, (vs)/3 = 0 if s E S . If t E R U S, then

((u o v)t)(a o /3) = (ut)a o (vt)fi = 0 . Therefore U ~ V @ ( ( U ~I ~V E) R~ U S ) .

E W such

By assumption, there exists v E V such that v @ (v g ) for 1 z g E G. g g g g We consider


b) Let V1 be the trivial module. As V is faithful, so every 1 ;t g E G does not act as a scalar on V1 o V . By a) /GI-1 (V1 o V) = o n. (V o ... o V) (for some n.) 1# ~ E G j=1 j J


contains a submodule W isomorphic to KG. By 4.12, W is even a direct summand. By the Krull-Schmidt-theorem, then the indecomposable P is isomorphic to a direct summand of some

Theorem 4.13 has been used in the theory of formations of solvable groups (Huppert I, p. 705-71 1). 4.14

Lemma. Let K be algebraically closed and eKG with eL = e an indecomposable

projective KG-module with simple head M = eKG/eJ(KG). Let V be any KG-module. Then dim HomKG(eKG,V) = dimKVe is the multiplicity of M as a composition-factor of V. Proof. Obvoulsy HomKG(eKG,V) =- Ve (as K-vectorspace) by

a + e a = e 2 a = (ea)e

for a E HomKG(eKG,V). Let W be a maximal KG-submodule of V. Then dim Ve = dim We

+ dim Ve/We

By induction, dim We is the multiplicity of M in We. Also

~e/W= e Ve/(Ve Hence as H(eKG)

n W) = (Ve + W)/W = (V/W)e

-. M, by the remark above dimKG(M,M) =1 if VIW = M

dim(V/W)e = dim HomKG(eKG,V/W) =


Theorem. Let K be algebraically closed and MI, M2 be simple KG-modules. Then

4.15 M1 o

M2 is cyclic, which means there is a w E M1 e

M2 such that

M10KM2=wKG. Proof. Suppose dim M2 r dim MI. We consider M1 e -

P(M2). As this is projective by 4.4,

we obtain

If ni


dim Mi, then MI o

P(M2) is a direct summand of


o dim Mi P(Mi) = KG i= 1 hence is cyclic. As MI o

(4.1) ,

M2 is an epimorphic image of MI o

P(M2), so M1 o

M2 is

cyclic. Now n.I = dim HomKG(Ml o P(M2) , Mi)


, M I o Mi) (see exercise 15 b)) * = multiplicity of M2 in MI o Mi (by 4.14) * dim(M1 o Mi) = dim HomKG(P(M2)

s dim M.






16) Define the concept of an injective envelope of a KG-module and prove the dual of 4.2. 17) Let V. (i=1,2,3) be KG-modules. 1

a) HomK(V1,V2) becames a KG-module by (.g)(v,) for a E HomK(V1,V2)

= (4vlg-l))&

. And then

* = V1 Q




as KG-modules. What corresponds to HomKG(V1,V2) in V1


V2 ?

b) Show HomK(V1 o V2V3)

* = Hom K(V2' V 1 Q V3)

as KG-modules and

18) Let M1 and M2 be simple KG-modules and K algebraically closed. a) The multiplicity of the trivial KG-module K in the head of M1 o



M2 is


b) The multiplicity of K in the socle of M1 o

M2 is





otherwise .


c) Define h E HomK(Ml o M1,K) by


Then A is a KG-epimorphism of M1 o M1 onto the trivial KG-module K.


d) Let {ml ,..., mn) , {al ,..., a n ) be dual bases of M1 and MI, which means a.(m.) = 6... Put 1 J 1J

n w=

C i=l

m.oa.. 1


Show wg = w. e) The trivial KG-module K is a direct summand of M o M* if and bnly if 1'K 1 Char K dim MI


19) Let M be a KG-module. If M is simple (indecomposable), so is M


9 5 The number of simple and indecomposable KG modules


Definition. For a K-algebra A, we define the center of A by Z(A) = {z E A


za = az for all a E A)

and the commutator of A by [A,A] = spanK {ab - ba ( a,b E A)


While Z(A) is a subalgebra, [A,A] only is a K-subspace of A. Let A = A1 s ...s Am with 2-sided ideals Ai. Since A.A. = 0 for i z j, it is immediate that 1 J Z(A) = Z(Al) o ... o Z(Am) and [&A] = [A1,Al] e ... s [A,,Am]


5.2 Example. Wedderburn's Theorem (2.14) reduces the study of semi-simple K-algebras over an algebraically closed field t o the study of matrix algebras A = (K), Elementary linear algebra yields Z(A) = K -

[:I . ] '


and [&A] = Kernel of tho trace map.

In particular, dimKZ(A) = 1 = dimKA/[A,A]. 5.3 Notations: Let KG be a group algebra and let char K = p



a) We denote by R1 ,..., R the conjugacy classes of G. Fix an element gi E Ai in each class h


g € KG (i=l,...,h). Furthermore, let Al ,..., and define the class sums 3. = 1 I gai

those classes which consist of p'-elements. (Clearly k = h if and only if p = 0 or p KG is semi-simple, by Maschke's Theorem 3.3.) b) For convenience, we also set T = [KG,KG] and S = T + J(KG). Observe that S/J(KG) = [KG/J(KG), KG/J(KG)] .

5.4 Theorem. Let KG be a group algebra. a)

{zl,... ,Rh) is a K-basis of YKG).

+ T, ... ,gh + T) is a K-basis of KGIT. Moreover: If g E A., then g + T = g. + T.

b) {gl




h) be

4 I GI, i.e.

Proof: a) Clearly,

- are linearly independent. If g zl, ... ,Ah

ffi E Z(KG). Conversely, if a =


E G, then g-lffig =

Ri and thus

ax x E Z(KG), then


for all g E G and thus ax = a - 1 for all g,x E G. Therefore, the coefficients of a are constant gxg on the conjugacy classes of G and a is a linear combination of R1, ... ,Ah. b) Let x and y he G-conjugate, say y = gxg-l. Then

n - y = x - gxg-l = g-l(gx) - ( g ~ ) ~E- Tl and x+t = y+T . Thus gl+T,

... ,gh+T do span KG/T. In order to establish the linear independence, we define

If follows that p(xy - yx) = q?i(x(yx)x-l -yx) = 0 for all x,y E G and consequently qi(T) = 0 h


(i=l, ...,h). If

h ai(gi+T) = 0 (with n i E K), then

i= 1


oigi E T and


h a. = q?. ( J




aigi) = 0 for all j


5.5 Corollary: Let KG be semi-simple (i.e. char K = 0 or char K

4 I GI) and K algebraically

closed. Then the number of isomorphism types of simple KG-modules equals the number h of conj ugacy classes of G. Proof: By Wedderburn's Theorem (2.14), KG = (K), -



where m is the number of

(K), m

simple KG-modules. It follows from 5.2, that dimKZ(KG) = m = dimKKG/T. It now depends on the taste of the reader whether he wants to apply 5.4 a) or 5.4 b) in order to finish the proof. It should however be clear that the approach via 5.4 a) cannot be adapted easily to the case where char K

I I GI. The problem is that the center of KG/J(KG) is a subspace of a quotient

and hard to study. The following beautiful arguments, which are due to R. Brauer, are inspired by the approach via 5.4 b). We first characterize S = T

+ J(KG) in a different manner and prepare this by the following

lemma. 5.6 Lemma. Let KG be a group algebra and assume that char K = p > 0.

a) If a,b E KG, then n n n (a+blP = aP + bP (mod T) for all n E M. b) TP

c T. --.c

c. E {a,b), be one of the P' 1 2P-2 summands of (a+blP different from aP and bP. Then cyclic rearrangements of the factors

Proof: (1) We first prove assertion a) for n = 1. Let c = c l -

c. of c give rise to p summands in (a+b)P. It thus suffices to show that the sum of all cyclic 1

rearrangements of c lies in T. Now c1c2. - c + c2c3 P Therefore

c = 2 - c c - . .cp + ((c2- - -cP)cl-c1(c2- - -cJ) P l 1 2



2-12 (mod T).

ai(aibi-biai) E T, where ai E K and ai, bi E KG-

(2) We next establish assertion b). Let t = 1

By (I), we obtain tP = &. (a.(a.b. 1 1 1 - b.a.)lP I I = & ap(aibi - biailP (mod T) 1



In order to show tP E T, it thus suffices to assume that t = ab-ba (a,b E KG). Again by (1) it follows that tP r (ablP +

= (ablP - (ba)P = a(ba. - -ab) - (ba. - .ab)a = 0 (mod T).

(3) We finally show assertion a) by induction on n. The induction hypothesis implies that n-1 n-1 n-1 (a+blP = aP + bP + t for some t E T But then (1) yields


n n (a+blP = aP




+ bP + tP = aP + bP

(mod T) ,

since tP E T, by (2). 5.7 Proposition. Let KG be a group algebra and assume that K is algebraically closed and char K = p > 0. Then 1

S := T

+ J(KG) = {a E KG I aP

E T for some i r 0)


Proof. We set 1

I aP

So = {a E KG (1) We show that S


for some i

+ 0)

C So . 1

Let a,b E So and pick i and j, i r j say, such that ap' E T and bPi E T. It follows from 5.6 b) 1

that bP E T, and 5.6 a) implies 1


(a+blP = aP


+ bP



0 (mod T)

Thus a+b E S and So is closed under addition. 0

L So, because each element of J(KG) is nilpotent (1.3). Since trivially and since So is closed under addition, S = T+J(KG) L S 0'

Obviously J(KG) T C Sg'

(2) We conversely show that So C S. We set

= KG/J(KG) and use Wedderburn's Theorem (2.14) to decompose 1

= Al o...o Am into matrix rings A. = (K)


n j

. We fix a E So and

i a 0 such that aP E T. We

also decompose




a = a o...o 1 rn


a.J E. A.J ' . a.J E K G . I


Since A.A = 0 (i ik), it follows that iP = iry m..m 6; J k (T + J(KG))/J(KG) = S/J(KG) = [ Ri=,m 1. Therefore

i and hence iP E [A.,A.] for all j=l, ...,m. j J J


. Recall that

By 5.2 [A..A.] is the kernel of the trace map on A. = (K) . in particular, the trace of the J J J n.J ' matrix



is equal to 0. Since the trace is the sum of the eigenvalues and since char K = p, it I


follows that (tr(a.))P = tr(i? ) = 0. Therefore tr(a.) = 0, a . E [A. A.] and J J J J J'J finally means a E S, which was to be shown.

a E [ RTf,m].This

The next Theorem is the natural generalization of 5.4 b). 5.8 Theorem (R. Brauer). Let KG be a group algebra over an algebraically closed field'K of characteristic p > 0. Recall that g l ,..., g were chosen as representatives of the p'-classes



,..., ffk of

G. Then {gl

+ S ,..., gk+ S) is a K-basis of KGIS.

Proof (1). For g E G, let g = ux be its unique factorization into an element u of p-power order I

and a p'-element x such that ux = xu. If up = 1, then I



(g-x)P = gp - xp = 0 and 5.7 yields g-x E S. By 5.4 b), we also have x-g. E T


a p'-element. Consequently, g

S for some j E (1 ,..., k), since x is

+ S = g. + S and the elements gl + S ,..., gk + S do span KG/S.

J (2) To prove their linear independence, suppose that k


a . g . E S (a.E K) . J J J k 1 By 5.7, there exists i 2 0 such that ( 1 C Z . ~ E. )T,~ and 5.6 a) yields i= 1 J J k i i k I 1 apgp = ~ . g . I) 0~ (mod T) . j = 1 J .I j=1 J J i=1


Let m be the pf-part of I G I. We choose integers a and b such that ap


+ bm = 1. Since m is a



multiple of the order of each g. Cj=l,...,k), it follows that g = (g? )a. Conseque'ntly, the gP as J J J j well are representatives of the pf-classes 4il ,..., Rk of G, possibly in a different order. 1

By 5.4 b), they are linearly independent modulo T. So a? = 0, that is a. = 0 G = l , ...,k). J J

5.9 Theorem (R. Brauer). Let K G be a group algebra over an algebraically closed field K of characteristic p > 0. Then the number of isomorphism types of simple KG-modules equals the number k of conjugacy classes of G consisting of p'-elements. Proof. Set KC = KG/J(KG). -

By Wedderburn's Theorem (2.14),

~ s ( K o...o(K), ) ~ , 1 m where m is the number of simple

m-and KG-modules. Since S = [ m,m],it follows from

5.2 that dimKRi7/s = m. On the other hand, 5.8 implies that dimKRe/S = dimKKG/S = k. 5.10 Remark (Alperin's weight conjecture). Let K be algebraically closed of characteristic p > 0. We consider pairs (Q,V) where Q is a p-subgroup of G and V is a simple

H = NG(Q)-module. Next we put (Q,V)

- (Q',V1)

relation. Let [(Q,V)]


- defines an equivalence denote the equivalence class containing (Q,V). Since Q a H and Q acts

if there exists g E G such that Q' = Q~ and V '

Vg. Clearly

trivially on V we may regard V as an KH/Q-module. Now Alperin's weight conjecture may be stated as follows: The number of simple KG-modules is equal to


{ [Q,V)]


Q a p-subgroup of G, V a simple H = NG(Q)-module which is projective as a KH/Q-module }


This means that the number of simple modules can be computed p-locally. In the meantime, the conjecture has been proved for several classes of groups, including p-solvable groups. If the underlying field K is not assumed to be algebraically closed, the question about the number of simple KG-modules becomes more difficult. The answer, which is not as explicit as above, was given by Berman (see [HB; VII, 3.1 I]). The techniques how to extend K to a

sufficiently large field L have not been studied here and we thus will not give a proof of Berman's Theorem. An immediate - but rather heavy-handed - consequence of 5.9 is that a p-group P over an

algebraically closed field K of characteristic p has the trivial module K as its only simple module (cf. 3.4). Other solvable groups can only be treated when so-called Clifford techniques (96) are available. We thus postpone this and study here the group SL(2,p) in characteristic p as an application of 5.9.

5.11 Example. Let G = SL(2,p) and K an algebraically closed field of characteristic p. a) Inside the polynomial ring K[x,y], we consider the K-subspaces Vm consisting of the i m-i ( i=O,...,m ) is a K-basis of Vm homogeneous polynomials of degree m (m r 0). Thus {x y a b and dimKVm = m+l. For A = ( c d ) E SY2,p), we put (xiym-i)A = (ax + b ~ ) ~ ( + c xdy)m-i . It is easy to check that this turns Vm into a KG-module. We show that Vm is simple for m = O,l, ...,p- 1. To do so, let W be a non-zero submodule of Vm and pick n


j m-j a.xy E W , wherea z O and n s m . n i=o J


For 0 r t E GF(p), we consider the matrices S(t) =


:) E G. Then

. n sit) = 2 a.(x+tyyym-J= fj(x,y)J E w j=o J j =O n


for suitable polymomials f. E K[x,y]. Observe that fo = f and f = a y J n n K-vector space,




. t-'(f.s(t)) E W for all i E Z .

Using the fact that p-1 . t=l

we obtain

-1 i f p-1 dividesi 0 otherwise


. Also, since W is a

Since 0 s n s m s p-1, this sum equals - fn if n c p -1, and - fo- fn if n = m = p-1. m As fo = f E W and fn = anym (an L 0), we conclude y E W in any case.


We now play a similar game with the matrices T(t) = (:

E G. Namely

p-1 -i m. p-1 . & t (y T(t)) = & t - ' ( t x + ~ )E~W for all i E Z t=l t=l and this is equal to p-1






. m

m ( T ) ~ J ~ ~ -1 J ( ~I =) j ~ m-j y p-& j=O j =0 t=l







j=o p-1 1 j-i

(m) x iy m-j J

for 0 r i srnc p-1 and O c i < p-1 = m

yp-l for m = p-I, i = O


i m-i E W for all Recall that for m = p-1 we have already seen that yP-I E W. It follows that x y i=O,l, ...,m. Thus W = V


and V


is simple for m = 0,...,p-1.

b) In fact the Vo ,..., V

are all the simple KG-modules. This follows from Brauer's P-1 Theorem 5.9, because G has exactly p conjugacy classes consisting of p'-elements. Representatives for these are 1 0 (O

-10 0 1 ( , s) , where s


for p > 2

for p = 2. (see [HB; VII 3.10 a)]). c) If p z 2, then Z(G) =

((-A -7))and G/Z(G) = PSL(2,p). Now i m-i


i m-i -x y

if m is even if m is odd


Thus Z(G) acts trivially exactly on the modules Vg, Vt, V4 ,..., Vp3

, Vp-l. Consequently,

this is a complete set of isomorphism types of simple K PSL(2,p)-modules. 5.12 Remarks: Let K be an algebraically closed field of characteristic p > 0. 4


a) Consider G = SL(2,7). Then GI = 2 -3.7, and by 5.10, there exists a simple KG-module

V with dimKV = 5. Thus


4 [GI .

b) The phenomenon of a) does not occur for psolvable groups G. This is an immediate

consequence of the Fong-Swan-Theorem (see 14.2). In 9.3 however we give an easier proof for the fact that if G is solvable, then dimKV


for all simple KG-modules V.

Theorem 5.9 tells us that the number of simple KG-modules is finite. On the other hand, the number of indecomposable KG-modules is usually infinite. The situation is described by the following beautiful theorem of D.G. Higman:

5.13 Theorem. Let K be a field of characteristic p. Suppose P E Syl G. P a) If P is not cyclic, there exist indecomposable KG-modules of arbitrary large K-dimension. In particular, there exist infinitely many isomorphism types of indecomposable KG-modules.


b) If P is cyclic, there exist at most GI non-isomorphic indecomposable KG-modules.

To prove theorem 5.13, we need several preparations. 5.14 Lemma. Suppose U


G and let T be a transversal of U in G such that 1 E T.

a) If V is a KU-module, then

V o l = { v o l l vEV) G is a direct summand of (V )" with V b) Suppose Char K




V o 1 as KU-module.

4 I G:U I. If W is a KG-module, then W is isomorphic to a direct summand

Proof. a) We have -

vG = v o l o v t , where

Vt= Vot. IztET

But V' is a KU-module, for tu $ U for every u E U if 1 + t E T. G b) By a), WU is a direct summand of ((WU) )U. Hence by 3.2, W is isomorphic to a direct G summand of (WU) . 5.15 Theorem. Let V be a KG-module. Then V is directly indecomposable if and only if S = HomKG(V,V) is a local ring, which means that S/J(S) is a skew-field.

Proof. Direct decompositions V = V o V 2 as KG-module correspond to the projections 1 L n = n E HomKG(V,V), where

+ v ) = v (v. E V . ) . 1 2 1 J J Hence V is indecomposable if and only if S = HomKG(V,V) contains only the trivial 4 v

idempotents 0 and 1. As idempotents can be lifted from S/J(S) to S by 2.5, we have only to decide which semisimple algebras

k @

i= 1

(Di)n. (Di skew-fields) I

have only trivial idempotents. This obviously enforces k = 1 = n 1' 5.16 Example. Let G = ( g l ) x (g2) be elementary abelian of type (p,p) and char K = p. If E is the identity-matrix and N any matrix of type (n,n), we define a representation D of G of degree 2n by

We only have to observe that E O p E 0 E O E O p E O (E = (pE = ( O E ) . (N = (O and

To show that D is indecomposable for certain N, we have to show by 5.15 that the matrices commuting with D(gl) and D(g2) form a local ring. If

commutes with D(gl) and D(g2), by simple matrix-calculation we obtain B = 0, A = D and .AN = NA. If we choose now

then AN = NA implies

If V is a KG-module for D, then HomKG(V,V) corresponds to all matrices

The mapping of this matrix onto a l is obviously an algebra-homomorphism of HomKG(V,V) onto K with a nilpotent kernel. Hence HomKG(V,V) is a local ring and by 5.15, V is an indecomposable KG-module.

5.17 Theorem. Let G be a p-group and char K = p. a) If G is not cyclic, there are indecomposable KG-modules of every even dimension.


b) If G = ( g ) is cyclic and GI = pa, then there are (up to isomorphism) exactly I GI

indecomposable KG-modules, namely

with dim V. = i 1

Vi = KG/(~a (i = 1,...,p ).


Proof. a) If G is not cyclic, then G has a factorgroup of type (p,p). Apply 5.16. b) By 3.4, KG has only one maximal submodule, namely J(KG) = (g-1)KG. Then also KG/(~-I)'KG has only one maximal submodule, therefore is indecomposable. Let conversely V be an indecomposable KG-module. Then the canonical Jordan-form of g on V is given by

a This forces N~ = 0, hence dim V 5 pa. As we have described all possible matrices for g, this gives exactly one indecomposable KG-module Vi with dim Vi = i for 1 6 i



5.18 Proof of 5.13. a) Suppose P E Syl G and P not cyclic. By 5.17 a), there exists an indecomposable KP-module P G If V of dimension 2k (k=1,2,...). By 5.14 a), Vk is isomorphic to a direct summand of (Vk)p.


with indecomposable KG-modules W., then by the Krull-Schmidt-theorem Vk is isomorphic to J a direct summand of some (W ) . But then jp dim W. r dim Vk = 2k . J b) Now let P be cyclic. By 5.17 b), there are exactly I P I indecomposable KP-modules Vi, where dim V. = i (i=l,..., /PI). We show that every indecomposable KG-module W is 1

isomorphic to a direct summand of some vG, where j a dim W. j

As p

4 I G:P 1, by 5.14 b) W is isomorphic to a direct summand of (Wp) G. If

with indecomposable KP-modules Ui, by the Krull-Schmidt-theorem W is isomorphic to a G direct summand of some U i , where dim U.



dim W. The number of non-isomorphic direct

summands of U G of dimension at least dim Ui is at most I G:P I as dim UG = IG:PI dim Ui.


Hence there are at most G:P

I I P I = 1 GI

indecomposable KG-modules.

5 6 Clifford Theory 6.1 Definition. Let N a G and W a KN-module.

a) For g E G, we consider the KN-module WogLWoKNKG. -1

Observe that (w o g)n = wng


W o g is called the g-conjugate module of W. b) The subgroup

I(W) = {g E G ( W

= W o g as KN-modules)

is called the inertia group of W. Since W o n = W


1 = W for n E N, we see that N



As promised in 3.13, we study in more detail how a simple KG-module restricts to a normal a G. subgroup N -

6.2 Theorem (Clifford). Let N a G and V a simple KG-module.

1 Wg and Wg a W o g is simple. gEG In particular, VN is semi-simple and all composition factors of VN have the same dimension.

a) Let W be a simple KN-submodule of VN. Then VN =

b) Let VN = V1 o ... o Vt be the decomposition of VN into homogeneous components Vi. Then right-multiplication by the elements of G transitively permutes the Vi. In particular, all V. have the same composition length and we may write 1



-. Wi o ... o Wi = eW.1

for a simple KN-module Wi and e E DI, being independent of i .

I Vig = Vi) = I(Wi) I G : I(Wi) I = t .

c) We have {g E G In particular, d) V



V. o a(,i)


(i = 1,.-.,t) .

In particular, Vi is a simple KI(Wi)-module.

for i = I, ...,t.

e) Let W be as in a) and {rl

,..., rt} be a coset representative of I(W) t

in G. Then


Proof. a) It is trivial that Wg = W o g and the rest has been proved in 3.13. b) For g E G, Vig again is a direct sum of pairwise isomorphic simple modules, so Vig C V. J for some j. But then Vi = ( v i g ) g l v . ~ - ' V for some k. This implies i = k and V. = V.g.



Let w.l.0.g. {V

J k J Vm} be the G-orbit generated by V1 (m r t). Then V1 o ... o Vm is a


KG-submodule of V and the simplicity of V implies m = t. c) By b), we have

d),e) By b) and c), it is no loss to assume that i = 1, W = W and Vi = Vlri (i=l, ...,t). Thus, 1



each v E V can be written as v =

i= 1

We define

v.r. with unquely determined vi E V1. I I

a E HomKG(V,V1 o KT(W) KG) by

Obviously, a is surjective. Since


dim V = t.dimKVl = G:T(W)I dimKVl = dim ( V e K 1 KT(W)K G ) , a is an isomorphism, and d) holds.

To see e), note that V. = V r implies Wr. = I T r I li I l i


W. I



The assertion thus follows from b). 6.3 Example: An analogous assertion unfortunately does not hold for indecomposable modules: Let G = (g) x ( h ) be elementary abelian of order p2, char K = p and V a 3-dimensional K-vector space with K-basis {vl,

" "Y vlg=vl, v h = v

v ). IVe define a KG-module structure on

v2' 3

v2g=v2, v h = v 2 , v g = v l + v 3 , v 3 h = v + v 2 3 2 3'

An easy calculation with matrices yields

EndKG(V) =


a 0 0 0 a 0 b c a


a,b,c € K


and thus 0 and 1 are the only idempotents of EndKG(V). Therefore, V is an indecomposable KG-module.

- G. Then v, = ( u ) o (v,w)

We consider N = ( h )


and a similar argument as above shows that (v,w) is an indecomposable KN-module. From $ 4 one might expect that indecomposable projective modules behave better. Indeed, 6.4 Theorem (Nakayama). Let N Q G and V a projective indecomposable KG-module. Then


where e E PI, W is an indecomposable KN-module and ti € G. Proof. Since V is projective indecomposable, there exists a KG-module X such that V o X = KG. We decompose

into projective indecomposable KN-modules Wi. Since

it follows that V o X = W

G o ... o WG .

By the Krull-Schmidt Theorem, we may assume that V is a direct summand of W:


set W = W1. Let T be a transversal of N in G. Then WG = o W o t tET

with KN-modules W o t


Since W is an indecomposable KN-module, all conjugate modules W o t as well are indecomposable. We write

with indecomposable, painvise non-isomorphic KN-modules Vi and mi E N. Since VN is a

G direct summand of (W )N , it again follows from the Krull-Schmidt-Theorem that there exist ti E T satisfying Vi a W o t. (i=l, ...,k). In particular, given j E { I , ...,k) there exists g E G such I

that V. .I

= Vlg. Since V is a KG-module, we have

with non-isomorphic indecomposable KN-modules Vig. Comparing the multiplicity of V. in


both decompositions, it follows that m. = m This completes the proof. J 1' We now conversely start from modules for a normal subgroup N

a G.

6.5 Theorem. Let N a G and V be an indecomposable KS-module. Set I = I(V), and


vI = v1 s . . . e V m into indecomposable KI-modules V.. I a) Then V? is an indecomposable KG-module. b) If V. is even simple, then VG. and V are simple as *-ell. 1

c) V?


= vG if and only if Vi V.. E

j J Proof. Let T be a transversal of N in I. Since h' a I, -

VI = o V tET But I = I(V) implies that V




with KS-modules V s t


V o t for all t E T, and by the Krull-Schmidt Theorem we obtain

e ni V for some ni E Ol (i=l .....m) . N Let now R be a transversal for I in G, and fix r E R. I t follows as above that

(V.) I

Vi o r e n.I (V o r)

(as KN-modules).


a) Suppose now that VG = W o X with KG-modules W We first observe that

o l+rER


Vi o r is a KI-module. Since

0 and X.

and since V. o 1 G V. is an indecomposable KT-module, we may assume by the Krull-Schmidt 1


theorem that Vi is a direct summand of WI. Set WI = Vi o W ' for a KT-module W '


If we restrict further down to N. we obtain

But W was chosen as a KG-module. Hence, for r E R , WN = W N r = ( W B ~=)n i~( V o r ) o ( W k o r ) . Recall that it follows from the choice of R that V o r

# V o r'

for r + r'. Consequently, the

multiplicity of the indecomposable KN-modules V o r (r E R) in W N is at least n..I By (*), G therefore (V . ) I


is isomorphic to a direct summand of WN.

In particular, dimKW a dimKvY = dim W


+ dimKX, forcing X to be 0.

b) Suppose now that V. is simple (for some i) and choose a simple factor module W of V




By Nakayama-reciprocity (3.8),

and the simplicity of Vi implies that Vi is isomorphic to a submodule of WI. Consequently, WN contains a submodule isomorphic to (V.) I N


niV. By Clifford's Theorem (6.2), WN is

semi-simple and hence the indecomposable KN-module V is simple. This proves the second assertion of b). Since W is a KG-module, we again use (*) to obtain that

G is a direct summand of WN. In particular, dim K W + dimK V? and V 1. s W is simple. G : = vG.Thus assume V . = vG but Vi $ V.. c) It is clearly trivial that Vi = V implies V J j 1 j7 J Recall first that

and analogously for j.

By our assumption, Vi must be isomorphic to a direct summand of V ' , and therefore (Vi)N is j isomorphic to a direct summand of (V') . Now j N (v!) J



(Vj o r)N =

I t r ER


and (Vi)N = ni V. By Krull-Schmidt follows V

(n-(Y 8 r)Is J

n \'

e r for some 1 z r E R, which is a

contradiction since r $ I. The prof is complete. Starting from an indecomposable module V of a normal subgroup N a G, the step between

I(V) and G is fairly well-understood, by Theorem 6.5. The step between N and I(V) however is much more delicate. In the remainder of this section we shall concentrate on simple modules V and shall mostly assume that K is algebraically closed. 6.6 Theorem. Suppose N a G and G K cyclls. L r K h: algehra~callyclosed and V a

KN-module with I(V) = G. a) If V is simple, there exists a (necessar~lyslmpls) KG module W such that W N = V. b) If V is indecomposable, char K = p and p

4 j G S I. there exlsts a (necessarily

indecomposable) KG-module W such that W s = Proof. Let n = I GIN I. By -

assumption. G = \



h t c ~ rwrnr h E G. Let p : N -, AutK(V) denote

h the representation of N on V. S ~ n c ep and p arc equnalenr, there exists a E AutK(V) such that -1 h a p(x)a = p(x ) for all x f S .

Hence as hn E N , a-"p(x)a


= p(x

This shows I


p ( h n ) n n t HomKN(V,V) .

hn ) = p(hn)-ldx)dhn)

a) Now suppose that V is a simple KN-module. Then by Schur's lemma (as K is algebraically closed) p(hn)a-" = a1 for some 0 z a E K Choose b E K such that bn = a and put


fi = b a . Then

p(xh) = p-lp(x)fi and p(hn) =

pn .

Now we define a : G + AutK(V) by o(hlx) = plp(x) for x E N , 0 5 i < n If j is any integer and j = nq



with 0 r i < n, then

u(dx) = u(hi hnqx) = /3ip(hnqx) = flip(hn)qp(x) = pitnqp(x) = $p(x)


Hence if x, y E N, we obtain j+kxh k u(h'xhky) = u(h y) = &+kp(xhk)p(y) +k -k = B P(x)pkP(Y) = Bip(x)Bkp(y) = u(fjX)u(hkY).


Hence a is a representation of G on V with uN = p. b) As V is indecomposable and K algebraically closed, we have Homm(V,V)/J(HomKN(V,V))


K, hence

Homm(V,V) = K1 o J(Homm(V,V))


Therefore by (*) p(hn)cc-" = a1


where a E K" and y E J(HomKN(V,V)). Again we choose b E K such that bn = a and put

p = ba. Then still p(xh) = B-lp(x)B and


p(hn)p-n = 1 + 6

where 6 is a nilpotent element in HomKN(V,V). Hence there exists an integer m such that m dP = 0. As p(hn) commutes with 1 + 6, it also commutes with


by (**). Hence




Now put as in a)


and show that a is multiplicative. The assumption p (GI = p

4 1 GIN 1

in 6.6 b) is necessary: Let G = (g) and N = (gP) with

2 . Then the indecomposable representation of N of degree 2 cannot be extended to a

(necessarily faithful) representation of G. 6.7 Remarks. Let N Q G and V a simple KN-module. a) If there exists a KG-module W with WN t V, then V is said to be extendible and W is called an extension of V. It should be clear from the proof of 6.6, that even in the very special case "GIN cyclic and K algebraically closed" we have not explicitly constructed an extension. b) If

IN I and I G/N I are coprime and I(V) = G, then V is extendible to G. We do not give a

proof of this (very useful) result and refer instead to lsaacs (Joum. Algebra 68, 1981). This is even true for any field K. c ) Suppose that V is extendible to the KG-module W. By 6.2 e), it follows that



= WN .-. e o V @ r.1' where t = I G:I(V) I. Therefore, I(V) = G. -

i= 1 J ) The condition 1(V) = G, while necessary by c), is not sufficient to extend V to a KG-module W:

Let G = Q8 or D8 and N = Z(G). Let K be any field of char K


2 and consider the l-dimen-

iional KN-module V on which N acts non-trivially. Clearly, I(V) = G holds. If there exists W

such that WN


V, then dimKW = 1 and thus G ' s ker W. Consequently, V

-. WN e K, a

contradiction. We next give another criterion for the extendability of a simple KN-module V. For example it applies if ~ G = )1 and N is an abelian minimal normal subgroup of G. 6.8 Proposition. Suppose that N a G and G/N splits over N, i.e. there exists H s G such that

G = NH and N f? H = 1. If V is a 1-dimensional KN-module with I(V) = G, then V is extendible to G. Proof. Let p : N + AutK(V) be the representation afforded by the KN-module V. Since V = V o g for all g E G, them exists n E AutK(V) such that p(xg) = d l p ( x ) n for all a N. Since x E N, g E G. But dimKV = 1, and thus p(xg) = p ( x ) Let C = ker V -

p(cg) = p(c) = 1 for all c E C, g E G, it follows that C a G. Moreover,


p([x,g]) = p ( x ) - ' p ( x ~ = 1 for all x E N g E G implies that N/C


Z(G/C). Since by assumption G splits over N, we have G/C = N/C x HC/C.

For g E G, write gC = xC. hC with uniquely determined xC E N/C, hC E HC/C. Define a : G + AutK(V) by

4 g ) = p(x).

Observe that a is well-defined, is a representation of G and satisfies aN= p.

Let V be a simple A-module for a K-algebra A so that D := End (V) is a skew-field. A Jacobson's Density Lemma states that given D-linear independent elements vl ,..., vn E V and arbitrary w1 ,..., w E V, then there exists a E A with v.a = w. (i=l, ...,n). For an elementary 1 I n proof, cf. [Hu, V, 4.21. We shall use this in the proof of the next result. 6.9 Theorem. Let N a G, K algebraically closed and V a simple KN-module. Suppose that V

is extendible to the KG-module W. a) If X is a simple K(G/N)-module, then W o KX is a simple KG-module (with module structure given as in 4.3). b) If X,Y are simple K(G/N)-modules and W o KX


W o KY, then X

= Y.

Proof. a) Let U be a non-zero KG-submodule of W e KX, and 0 * u E U. We may then write n



w.ox. 1 i=1 1 for a K-basis {wl, ... , wn) of W and suitable elements xi E X; without loss of generality, x 1 z 0.

By hypothesis, WN



V is a simple KN-module, and since K is algebraically closed,

Endm(V) = K Consequently, the elements w ,..., w are as well linear independent over 1 n Endm(V), and Jacobson's Density Lemma implies that given w E W there exists a E KN such that wla = w and wia = 0 (i > 1). Setting a =


ahh (ah E K), and using xh = x for all


x E X, h E N, we obtain (wi o xi)a =


ah(wi o xi)h =


Thus ua = w o xl and W o xl



a (w.h e x.h) =



a (w.h o x.) = w.a e x. . h l I 1 1

L U. Hence for all g E G,

W o xlg = W g o x l g = ( W o x l ) g C u g = U . Since x z 0, we have x KG = X, and therefore W s X = U. This proves the simplicity of 1 1 K

w o Kx. b) Since W o KX = W o Y, it follows that K dim W.dimKX = dimK(W o KX) = dimK(W o Y) = dimKW-dimKY , K K hence dim X = dimKY. We fix K-bases {xl K

,..., xm)

and {yl

respectively. We also fix an isomorphism a E Hom KG(W

,..., ym) of X and Y, KY). Observe that the

KX , W

following rules define endomorphisms a.. E End (W) (i,j=I, ...,m): 1J K m wa.. o y. (w E W, i=l,...,m ) . (woxi)a = J = 1 'J J


For h E N, we have m


(wh)a.. o y. = (wh o x . ) a = (wh e x.h)a = (w e x.)ha 'J



= (w o x.)ah = (





wa.. o y.)h = 1 (wa..)h e y. 'J J j=1 'J J

Comparing coefficients of y., we obtain that J






Thus a . = c i j l W for some c . E K (i,j = 1,...,m) 11 1J


Let (aij(@) and (b..(g)) be matrix-representations of the action of g E G on X and Y with IJ

respect to the bases {xl ,..., xm) and {yl ,...,.)y,


and this is equal to

Comparing the coefficients of yP this implies



~ . ~ ( w g ) a ~=~ (1 g ) c..(wg)b (g) for all w C W , J 1 J ~k

and therefore ---



1 c..b. (g) for all g E G, iJ = I ,...,m j = 1 11 ~k j=1 Consequently, the map y defined by aij(g)cjk =


m xiy =

,Ic..y. (i=l,...,m) J = 1 'J J

. is in HomKG(X,Y). Applying the same procedure to a-1 ~nstead, we see that y is invertible.

Hence X

= Y as KG-modules.

6.10 Lemma. a) Suppose U


G. Let V be a KU-module and W a KG-module. Then

(Wu m KV)


= W m K ~ G(as KG-modules).

b) Let N 4 G and W a KG-module. If we consider K(G/N) in the obvious way as a KG-module, then

Proof. a) We obviously have ( w U o KV)G = ( w U o K ~ o)K



W ~ U e~K(V e KUKG) = WU o KV


as K-vectorspaces by the mapping a defined by ( ( w o v) o t ) a = wt o (v o t) , where v E V, w E W and t is in a transversal T of U in G. If tg = ut' with u E U and t ' E T, then ((w O V )o t ) g ) a = ( ( w o v ) o u t f ) a = ( ( w u e v u ) e t ' ) a = wut' e ( v u o t l ) = wtg o (v o tg) = wtg o (v o t)g = ((w


v) e t)a g .

G. b) This is the special case of a) where V is the trivial KN-module, hence V IS isomorphic to K(G/N). 6.11 Corollary. Let N Q G, K algebraically closed and V a simple KN-module. Suppose that V is extendable to the KG-module W. a) Let U be any simple KG-module which has V as a constituent in UN. Then there exists a simple K(G/N)-module X such that U"W0



b) If W ' is another extension of V to G, then W19woKX for a 1-dimensional K(G/N)-module X. Proof. a) By Nakayama-reciprocity (3.8), 0 + Homm(V,UN)


HomKG(V ,U) , G and the simplicity of U yields that U is a factor module of V . We now apply 6.10 and obtain



that U as well is a composition factor of VG


(W )G






0 = So C S1 C ... C Sm = K(G/N)

be a composition series of the K(G/N)-module K(G/N). Then by 6.9 a), (W


KSi)/(W 0

" W 0 K(Si/Si-l)

is a simple KG-module (i=l,...,m). The Jordan-Holder theorem thus yields that U a W 6x1 KX where X = Si/Si-l for some i. b) Since dimKW1 = dimKW, a) implies dimKX = 1.

We conclude this section with some examples of how to construct simple KG-modules over an algebraically closed field K of characteristic p. If p

4 I GI, then KG n

o (dimKV).V is V simple

semi-simple (3.3, 3.12) and the number of simple KG-modules equals the number of conjugacy classes of G (5.5). Moreover, constructing a simple KG-module V (or its representation, or its character) in this case is essentially the same task as to work over the complex numbers Q: (see 11.13). It is not the aim of this course to treat this so-called "classical" situation and we refer to ([Trento 871). We also take for granted that dimKV

I I GI, provided that V is simple and

p l , IGI. We first investigate groups G which are close to p-groups. 6.12 Example. Let K be an algebraically closed field of characteristic p.

a) Suppose that G has a normal Sylow-p-subgroup P = 0 (G). By 3.6 b), we know that P P s; ker V for each simple KG-module V and V can be considered as a simple K(G/P)-module.


Since p l, G/P 1, the above comments apply in this case. b) Suppose now that G is p-nilpotent with normal p-complement N = 0 ,(G). We decompose P m KN = o Wi into simple KN-modules Wi and again assume that these are known. Then i= 1

with projective KG-modules WG (2.8).

(1) Let W be one of the Wi. The W Let X


G is indecomposable:

0 be an indecomposable direct summand of

Diclcson's theorem ( 3 . 9 , 1 G/N)

wG.Then X as well is projective, and by

1 dimKX. On the other hand, Nakayama's theorem (6.4)

implies the existence of an indecomposable (= simple) KN-module U such that XN is a direct sum of some G-conjugates of U. Since by definition ( w ~ =) t ~~ W T

(T = transversal of N in G) ,


it follows that U is G-conjugate to W. In particular, dim W = dim U divides dimKX K K

Since dimKW


I IN I and (1 GIN I , dimKW) = 1, we obtain dimKWG = I GIN I . dimKW s dim KX s dimKWG ,

and X = W

G is indecomposable and projective.

(2) Let W1

,..., Wr be representatives of the G-conjugacy classes of the KN-modules


,.., Wm (r 5 m). Then Vi := H(W G) = W G/(W G)J(KG)


are representatives of the isomorphism types of simple KG-modules. Moreover,



dimKVi = G:I(Wi) dimKWi, and r equals the number of conjugacy classes of G in N: G By 2.5 c), H(W ) is simple for i=l, ...,m,. Since

G each simple KG-module is isomorphic to H(W ) for some i E (1,...,m). Recall now that by 2.9 b) H(w?) Thus the V1

= H(W G)

,..., Vr are painvise




wG j

Wi, W. are G-conjugate. J


Also, by Nakayama-reciprocity (3.8), dimKHomKN(Wj,(Vi)N) = dimKHom


1 if Wi , W . are G-conjugate J

Thus (Vi)N is the direct sum of the G-conjugates of Wi, each occuring with multiplicity 1. In particular,


dimKVi = G:I(Wi) 1 dimKWi

Finally, by Brauer's theorem 5.9, r equals the number of conjugacy classes of G consisting of p'-elements. These are exactly the G-classes contained in N.

c) Suppose that G is p-nilpotent with abelian normal p-complement N. Then dimKW = 1 for all simple KN-modules W and b) immediately yields: 1) The number of isomorphism types of simple KG-modules equals the number of orbits of

P E Sylp(G) on N. G 2) Each simple KG-module V occurs as H(W ) for some simple (1-dimensional) KN-module W and dimKV = ( G:I(W)


6.13 Example. Let K be algebraically closed of characteristic p. We consider a Frobenius-group G = FH with abelian Frobenius-kernel F and Frobenius complement H. Since ( I F \ , 1HI) = 1, p cannot divide both IF1 and \ H I . I f p


I F [ , then G has a normal

Sylow-p-subgroup, and we refer to 6.12 a). We may thus assume that p 1) Let W


4 I FI, but p I [ H I .

I F be a non-trivial simple (I-dimensional) KF-module, I = I(W) and p the

representation corresponding to W. For f E F and h E I, it follows that P(fh) = p(t3 and therefore, p(thfl ) = 1. Observe that C := ker p a I, and I acts trivially on F/C. Since I/C again is a Frobenius group and C < F, this implies that I = F. We may thus apply 6.5 b) to see that W G is simple. Let

m z I F be another non-trivial simple KF-module. Then WG s mG W , m are G-conjugate . I

Since I = F, the G-conjugacy class of W consists of H I KF-modules. We have thus constructed

1 s

simple KG-modules of the form w G , satisfying F s ker WG.

2) Let k be the pr-class number of H. By Brauer's theorem 5.9, G/F


H has exactly k

different simple K(G/F)-modules Vi. Considered as KG-modules, they satisfy F s ker Vi. Consequently, v. I




for all i, and all simple KN-modules W + IF. 3) In 1) and 2), we have found


different simple KG-modules. The Frobenius


yields that each pf-class of H gives rise to exactly one p ' c l a s s of G. Since p )I I F ] , it follows that G has exactlv

By Brauer's theorem 5.9, the modules described in 1) and 2) are representatives for all isomorphism types of simple KG-modules.

4) Let us in addition assume that H is abelian and therefore cyclic. Then KG has exactly

1% I HI


simple modules of dimension I HI , and simple modules of dimension 1 .

The case of a Frobenius-group with nonabelian Frobenius-kernel will be treated in 7.8, using Brauer's permutation-lemma. Exercises. 17) Construct the simple KG-modules Vi of 6.12 b) using 6.5 and the unproved Remark 6.7 b). 4 G with abelian GIN. Suppose 18) (Roth) Let K be algebraically closed and N -

char K ./,

1 G/N 1.

Let V be a simple KG-module and

where W is a simple KN-module and { g ,...,gm) a transversal of I(W) in g. Then 1

where the U. are simple KG-modules of dimension 1 with N in the kernel. The V o J ~~j simple and pairwise non-isomorphic. In particular,

2 e rm = ( G I N ( . (It follows that the number em of simple components of V


divides I GIN I .)

$ 7 Brauer's permutation lemma

7.1 Lemma. Let A be the (n,n)-matrix over I with entries a.. = ( i j ) = largest common divisor of i and j. 1J Then det A = d l ) where

... d n ) z 0,

is the Euler-function.

i . ~

Proof: We put B = (b..), where ?I

1 if i l j b.. =

Hence B is a triangular matrix with diagonal entries 1, so det B = 1. Let F be the diagonal matrix

The k,l-entry of B'FB is

(Here we use the well-known equation

which is easily proved by a counting argument in the cyclic group of order m.) Hence we have shown

B'FB = A and therefore det A = det F = d l )

... d n ) .


7.2 Lemma (L. Kovacs). Let K be any field, let P and Q be permutation-matrices over K of -1 type (n,n). Suppose there exists a nonsingular matrix T in (K), such that P = T QT. If w.(P) J is the number of cycles of length j in P, we have for all j = 1,2,-

w.(P) = o.(Q) J




Proof: We put w(P) = 2 o.(P). J j We let P and Q operate on a K-vector-space V of dimension n by permutation of the basis. Then obviously k w(P ) = dim {v ( v E V, vpk = v), k -1 for each cycle of P gives rise to a 1-dimensional fixed-space. The assumption P = T Q T k k implies therefore 4 P ) = 4 Q )

for all k.

k If C is a cycle in P of length j, then C has order

and has ( k j ) cycles of length

it's cycle-decomposition. Hence W k ) = 1 (k,j)oj(P). j k k As w(P ) = w(Q ), we obtain

2 (k,j)w.(P) = 2 (k,j)o.(Q) J j

for k = 1,...,n.


By 7.1 this implies w.(P)=o.(Q) J J

f o r j = I ,...,n.

7.3 Notation. Let K be algebraically closed of characteristic p and let V1 ..., Vk be the simple KG-modules (up to isomorphism). We define class-functions Pi(g) = trace Di(g), where D. is the representation of G on Vi. 1

pi (i=l, ...,k) on G by

7.4 Lemma. a) Suppose g = g g , g is the unique decomposition of g E G into its P P ! = ~ PP p-component g and its p f -component g Then Pi(g) = pi(gp. ). P P" b) Let gl, ...,gk be representatives of the conjugacy classes of p f -elements of G. Then pl,


are linearly independent on {gl, ...,gk). In particular, Vi is determined (up to isomorphism) by it's trace-function


c) The (k,k)-matrix caicgj)) is non-singular. Proof: a) As K is algebraically closed, we can assume that the commuting matrices Di(gJ and D.(g ,) are simultaneously in triangular form, say P


Di(gp) = [la..:


I 1 bl O



D.(g ,) = l






m If gP =1, then P

as char K = p. Hence a. = 1 (i=l,...,n). Therefore 1

and hence

+...+ bn = trace D.(g ,) = p.(g 1 P P b) By linearity, we extend pi to KG. p.(g) = b 1


Suppose k 1

c.P.(g.) = 0 1 1 J

for j = I ,...,k.


h , is conjugate to some g.. Suppose gh = $ g = gghwith the p'somponent P J 1 J P h g. of g . Hence by a) J If g E G then g

k Therefore


ciPi = 0 on KG.

i= 1 But


and D. is the projection of KG onto the summand (K),,. We pick an element J J


If a. is an inverse image of in KG, then p.ia ) = b - - ,hence 0 = 1c.P.(a.) = c.. J 1 J '1 1 J J c) is an immediate consequence of b), as the rows (Pi(g,)7


... Pi(gk)) 7

are linearly independent.

7.5 Theorem (Brauer's permutation lemma). Let G and H be finite groups and K an algebraically closed field of characteristic p. Let H operate as a permutation group

Pk (corresponding to G) by

1) on the /3





for h E H,

2) on the p'-classes of G by +(gj) We assume that



h-l $(gj) = @(gj ) for all iJ=l,...,k, and all h E H. a) If P(h) denotes the permutation-matrix of h on {PI, ..., $1 and Q(h) the permutation-matrix of h on {gl, ..., gk), then for all h E H,

P(h)T = T Q(h)

where T = (/3.(g.)) and det T z 0. J b) For every h E H, the number of cycles of length j of h on


Pk) and {g

same. c) The number of orbits of H on {PI, ..., Pk) and {gl ,..., gk) is the same. Proof: a) We define permutation-matrices P(h) = (p..) and Q(h) = (q..) by 1J 1J

1 if/$ Pij =


0 otherwise


Then the (u,v)-entry of P(h)T is

as pui * 0 only for

h Pi = pu. Similarly, the (u,v)-entry of T Q(h) is

1 pu(gj) qjv = pu(&h-I

J Gh G since q. z 0 only for (g . ) = g v . Jv J Hence we obtain for all h E H

P(h)T = T Q(h). By 7.4 c), det T z 0.

gk) is the


b) follows immediately from a) and 7.2. c) It is an elementary fact (Huppert I, V, 20.2) that

is the number of orbits of H on {PI,.,,, Ph), if wl(h) is the number of fixed points of h on {PI,..., Pk). Hence the conclusion follows from b).

7.6 Remark. a) If char K = p, the equation P(h)T = T a h ) does not immediately imply the statement in 7.5 b), it only shows w.(P(h)) = wj(Q(h)) (mod p). J b) One would like very much to conclude from P(h)T = T Q(h) (det T

* 0)

the existence of a permutation-matr~xS such that for

P(h)S = S Q(h)


h f H.

This is not true in general. I t 1s equl\alent u ~ t h1


fact that for every L 5 H the number of

common fixed points of {P(h) ( h E L ) and { Qh) 1 h f L) is the same. We study two consequences of 7.5. 7.7 Theorem. Suppose N a G. Let B1,..., Dl,

..., Dk of N over K (Char K = p) and

Bk 05 the traces of the irreducible representations let gl.

classes of N. Then G permutes the sets b 1 =

....-k r be representatives of the pf-conjugacy

{B1,.... &) and A?- = { gNl ,..., g Nh ) . Any g E G has

the same number of fixed points on both of these sets. Proof. G operates on the pl-classcs of N by y -- )g -1

D ~ ( Y=) Di(Yg We put



= trace


DY. Then

= trace ~

-1 -1 ~ ) =( ~ ~ ~ ).g(



E S ) and on the Di by

By 7.5, we obtain that g has the same number of fixed points on A~ and A

~ .

As an application of 7.7 we consider again Frobenius-groups.

7.8 Example. Let as in 6.13 G = FH be a Frobenius-group. We assume char K = p and

PY IFIThen H acts fixed-point-freely on F. By a very elementary argument, H acts also fixed-point-freely on the conjugacy classes of F (see Huppert I, p. 500, theorem 8.9 c).) Hence by 7.7, H acts also fixed-point-freely on the trace-functions pi of F, hence by 7.4 b) on the simple KF-modules Vi(i=2,..., m). (We put V1 = K, the trivial module.) This shows I(Vi) = F if i > 1. Then as in 6.13 VG is simple and V?

= vGif and only if Vi and V are conjugate .i J

under H. As H acts fixed-point-freely on {V2 ,..., V,),

all nontrivial orbits of H have length

I H I, hence there are - l such orbits. In this way we construct m -1 simple KG-modules with



F not in the kernel. If k is the number of p'-classes of H, there are k simple KG-modules with F in the kernel. As in 6.13 we see easily that

is the number of p'-classes in G. Hence we have constructed all simple KG-modules from the simple KF-and KH-modules. Suppose in particular, that G = FH is a Frobenius-group with I HI = pa and hence p J lFI. Then

if we put n. = dim V.. As J J G d i m V . = IHI n J j

u = 2,...,m ) ,

the sum of the squares of the degrees of simple KG-modules is

This shows

dim J(KG) = pa - 1 . (Compare 4.8). We prepare a second application of 7.5. 7.9 Lemma. Let V be a KG-module, {VI,..., vn) a basis of V and . .

vig =



aij(g)vj J=1

If (a ,,..., a-} is the basis of V* = Hom.,(V.KI. &Sad bv a.(v.i = d... then the action of G - L I1 -


on V is described by


p is the trace-function

on V and

on V-.then 6-dg)= B ( ~ - ' ) .

Proof: By definition 4.10 we have -


This shows n aig =


The trace





a..(g )a..

1x1 J'


f i ' of g on V ' is hence given by

p is the trace-function on V

7.10 Theorem (R. Brauer). The number of self-dual simple KG-modules is equal to the

number of pl-classes gG such that g G = (g-1)G .

Proof: By 7. 4 b) and 7.9, we have Vi

* = Vi

if and only if

for j = 1,...,k, where the g. are representatives of the p'-classes of G. To apply 7.5, we take a 3 group H = ( h ) of order 2 and let H operate on the Pi by P h. = P *. , on the p'-classes by 1


we can apply 7.5. 7.11 Theorem (P. Fong). Suppose K is algebraically closed of characteristic 2 and V is a self-dual simple KG-module. If V is not the trivial KG-module K, then V carries a symplectic, non-singular G-invariant form. In particular, dim V is even. Proof: Suppose at first that V = Kv is 1-dimensional, hence vg = a(g)v with a(g) E K'.


If V = V , we obtain by 7.9 -1 -1 a(g) = a(g ) = a(g) .

As char K = 2, this implies a(g) = 1 for all g E G, hence V is the trivial KG-module.

Suppose dim V r 2. At first we construct a bilinear form (.;)

on V by

(v1'v2) = (hv1)(v2)' where h is a KG-isomorphism of V onto V 0 = (v 1,v2) - (hvl)(v2)


. If

for all v2 E V,

then hvl = 0, hence vl = 0. Therefore (., - ) is non-singular. For g E G we have

Hence (.;)

is G-invariant. We put q(v) = (v,v). Then

q(vl + v2) = (vl + v2' v1 + v2) = (vl,vl) + (v1*v2) + (v2'v1) = 9(v1)




q(v2) + [v1'v21'

where the obviously G-invariant bilinear form


is defined by

[v1'v21 = (v1'v2) + (v2'v1).

As char K = 2, so

0 = 4 q(v) = q(2v) = q(v) for all v E V. Therefore


+ q(v) +

[v,v] = [v.v]

is symplectic.

[.;I is not identically 0. Then R = {v I v E V, [v,w] = 0 for all u f \')

Case 1: Suppose that

is a proper G-invariant subspace of \'. .%s \' is simple, so R = 0. Hence


is non-singular

and symplectic. Case 2: Now suppose that [ ., .] is identically 0 on V, hence

+ v2) = q(vl) + g(v,)-

for all v - E V. J If q is identically zero on V, then (.;) is syrnplectic. Suppose q(vl)




0 for some vl E V.

Then by (*),

U = {v


v E V, q(v) = 0)

is a G-invariant subspace of V. If v E V, there exists a E K such that q(avl

+ V) = a 2 q(vl) + q(v) = 0.

(We only need here that K is perfect). Hence dim U = dim V-1 > 0, contradicting the simplicity of V. 7.12 Theorem. Suppose Char K = 2 and 2 l, dim C' for every simple KG-module. Then G has a normal Sylow-2-subgroup. Proof: (Okuyama). By 3.6 b), the maximal normal 2-subgroup 02(G) of G is in the kernel of every simple KG-module. Hence we can assume that 0 (G) = E, then we have to 2 show 2 4 I G I .

By 7.11 and our assumption, the trivial KG-module K is the only simple KG-module which is self-dual. Therefore by 7.10, the only self-inverse 2'-class of G is the class (1). -1 Hence if g + 1 is a 2'-element, then g is not conjugate to g . Suppose 2

1 I G I . Let

i j be any involutions of G. Then < i,j > is a dihedral-group

and 1

(ij) = ji = (ij)



Hence the order of ij is a power of 2. In particular, for any g E G we obtain that < i,ig > is a 2-group. By an elementary lemma of R. Baer this implies i E OZ(G), a contradiction.(See Blackburn-Huppert 11, p. 500; a more direct proof of Baer's lemma is in M. Suzuki, Group Theory 11, p.195-196.)

7.13 Remark. a) Theorem 7.12 is still true if we replace 2 by any odd prime. But the proof of this important fact needs at present the classification of simple groups and detailed results about the modular representations of Chevalley-groups (G. Michler, Journal of Algebra 104 (1986), 220-230.) For ocld primes, there is no analogue of 7.1 1 and no substitute for the fact that two involutions generate a dihedral group. (Only B a e r ' s lemma works for odd p as well.) b) The converse of theorem 7.12 (and now for all primes p) is much easier: If G has a normal Sylow-p-subgroup P, then by 3.6 b), P is in the kernel of any simple KG-module V. Hence V is a simple module for the pr-group G/P, and by Huppert, Endliche Gruppen I, V, 12.11 or 11.13 of these notes follows dim V

I I G/P 1, so p 4 dim V.

7.14 Theorem. Suppose I GI = 2"m with 2 & m. Let K he an ;~lgebraicallyclosed field with +

Char K = 2 and let P1 he the principil indecnmpos;ible projective module. Then 2a


dim PI.

Proof: Let P. (i = 1,...,k) be the projective cover of the simple KG-module M., where M is the I 1 1 trivial module. By lemma 7.15 we have


P i = P(Mi) Hence



= P(M.). 1

KG = P1 e

* dim M.1 . P.1 o


Mi+ M i


If Mi = M*i


M1 , then 2 dim


1 dim Pi (3.5), so

dim M i by 7.11. Hence as 2a

o * dim Mi Pi M.a M . I


is divisible by 2"".

* dim Mi (Pi o P i ) .


Mi= M .

The same is obviously true for -


* dimMi(PiaPi).

o Mi+ M

As 2'

T dim KG

this implies

za T dim PI.



7.15 Lemma. If Mi is a simple KG-module. then Pf51



P(\li) .

Proof. If

then by duality-theory



Now P(Mi) is projective (4.1 I), indecomp~ssble(,exercise 19) \vith socle M i . T h e n by 8.8 we obtain P(Mi)




P(M i).

tj 8 Symmetric Algebras

8.1 Definition. Let K be a field. A finite dimensional K-algebra A is called a symmetric


algebra if there exists a h E A = HomK(A,K) which satisfies the following two conditions. (1) h does not contain any non-zero right ideal in its kernel.

(2) h(ab) = @a) for all a,b E A. 8.2 Examples. a) KG is a symmetric algebra:


a g E KG we put h(a) = al. Obviously, h E A gEG g contained in ker A. If a = & a g E I, then gEG g For a =


. Let I be a right ideal of A

0 = h(ah-l) = ah for all h E G .

Hence a = 0, as required. Finally,

for all a,b E KG. Thus KG is a symmetric algebra. b) If A is a symmetric algebra and e an idempotent of A, then EndA(eA) is a symmetric algebra: Since EndA(eA) is anti-isomorphic to eAe (see 2.4b)), it is sufficient to show that eAe is a symmetric algebra. Let h be the linear function which makes A into a symmetric algebra and let A' be the restriction of h to eAe. It remains only to show that ker A' does not contain a non-zero right ideal of eAe. Suppose hl(xeAe) = 0 for some x E eAe. Then x = exe and

A' (xeAe) = h(xeAe) = h(exeA) = h(xA) , proving x = 0. 8.3 Theorem. Let A be a finite dimensional K-algebra. Then A is a symmetric algebra if and only if there exists a nun-singular symmetric K-bilinear form (,):AxA+K which is associative, i.e.

(ab,c) = (a,bc) for all a,b,c E A

. 8

Proof. Suppose that A is a symmetric algebra w.r.t. (a,b) := h(ab)

h E A . For a,b E A we put


Then (,) is K-bilinear, associative and symmetric. Moreover, if (a,b) = 0 for all b E A, then ker h contains the right ideal aA. Thus by assumption on ?,. a = 0 and the form (,) is


non-singular. Conversely, given the bilinear form (,) we define h E A by h(a):= ( a , l ) for a E A . If 0 = h(aA) = ( a 4 1 ) = (a,A)


then a = 0, since (,) is non-singular. Thus ker i. contains no non-zero right ideal of A. Finally, the symmetry of (,) implies q a b ) = (ab.1) = (a.b) = \b.a) = jba,l) = Wba) for all a,b, E A.

8.4 Definition. Let A be a ring. For any subset i of A we put R(i)={a L(i)= { a

I a E A , i a = O } and 1 aEA,ai=O).

Clearly, R(i) is a right ideal and L(i) is a left ideal of .-t 8.5 Theorem. Let A be a symmetric algebra u-.r.t. to the bilinear form (,). Let r denote a right ideal and [ a left ideal of A. a)

L(r) = { a R([) = { a

I a E A , (a,r) = 0 for all r E r ) and I a E A , (l,a) = 0 for all I E L ) .

+ dimKL(r) = dimKA

and dimK[ + dimKR(I) = d i m K A .


dim r


R(L(r)) = r and L(R(I)) = I .


The map r -+ L(r) is a duality of the lattice of right ideals of A onto the lattice of left


ideals of A, that is


for all right ideals t i of A. Similarly I -+ R(Q is a duality of the lattice of left ideals of A onto the lattice of right ideals of A. Proof. a) Put 'r = { a


a E A , (a,r) = 0 for all r E r )

and let x E 'r. Then, for all a E A and all r E r , we have 0 = (x,ra) = (xr,a)


Since (,) is non-singular, there follows xr = 0 for all r E r. Thus $ C L(r). Conversely, if x E L(r), then 0 = (xr,l ) = (x,r) for all r E r ,

proving L(r) = 'r. The proof for left ideals is similar. b) Since (,) is non-degenerate, we obtain by a well-known result on bilinear forms dim A = dim r K K

+ dim&

= dimKr

+ dim&(r) .

For left ideals the assertion follows similarly. c) Obviously r

R(L(r)) and by b), dim K R(L(r)) = dim K A - dim KL(r) = dimKr ,

proving equality. Similarly, L(R([)) = I. d) According to c), the map r


L(r) is bijective map from the set of right ideals of A onto the

set of left ideals of A. The assertions L(rl R(Ll

+ C2) = R (Il)

+ r2)

= L(rl) fl L(r2) and

fl R(r2) are obvious. Furthermore by c)

We again omit the proof for left ideals which is similar. 8.6 Definition. For a finite dimensional algebra A let Sr(A), respectively S1(A) denote the

socle of A regarded as a right, respectively left module. By 3.11, we have Sr(A) = L(J(A)). Using I.%), we also have SI(A) = R(J(A)).

8.7 Theorem. Let A be a symmetric algebra. a) L(1) = R(1) for every 2-sided ideal I of A. b) L(J(A)) = R(J(A)) = Sr(A) = S1(A). c) If e is a primitive idempotent of A, then the socle S(eA) of eA is simple and

S(eA) = eSr(A). (For group-algebras see 3.12 b).) Proof. a) 8.5 a) and the symmetry of the bilinear form yield L(1) = { a I a € A , (a,x) = 0 for all x E I ) = { a ( a E A ,

(x,a) = 0 for all x E I ) = R(1).

b) This follows immediately from a) and 8.6. c) Let r be a simple submodule of eA. Clearly X( 1-e) 2 4 r ) . By a), we have


c S,(A)

= S1(A) = RIJcAI)

and by 8.5,


Let - denote the canonical epimorphism from A onto X = NJ(A). Now observe that AI[A(I-~)+ J(A))] as left A-modules and EndA@) Since


- -

& End&=)

ais simple, & is a division algebra. Hcnic

(antiisomorphism by 2.4 b). is also simple, since A is semisimple

as left-A-module. (This depends on 1.5 b)!) Thus u-eobtain

L(r) = A(1-e)

+ J(A) .

By part a) of the Theorem and 8.5 follows

r = R(L(r)) = R(A(1-e)

+ J(A)) = RiAt I-el) ? R(J(A)) = e A n Sr(A) = eSr(A) .

This proves that S(eA) is simple and equal to eS ( A ) r

Let A be a finite dimensional algebra and let P be a projective indecomposable A-module. Then the head H(P) = P/PJ(A) of P is simple by 2.9. If in addition A is a symmetric algebra, then by 8.7 the socle S(P) of P is also simple. Thus we may ask: Is there a connection between head and socle? Indeed there is one as the next result shows. 8.8 Theorem. Let P be a projective indecomposable module of a symmetric algebra A. Then

Proof. We may assume that P = eA with a primitive idempotent e of A. By 8.7, S(eA) is simple and S(eA) = eS (A). Obviously, r HomA(eA, eSr(A)) = eSr(A)e (as K-spaces) by the map a -, e a = e 2 a = (ea)e for a E Hom A (eA, e Sr (A)). Thus it is sufficent to show that eSr(A)e z 0. Suppose eSr(A)e = 0. Let h be the linear form which makes A into a symmetric algebra. Then 0 = h(eSr(A)e) = h(eSr(A))


By assumption on h follows e S (A) = 0, a contradiction. r 8.9 Remark. A finite dimensional K-algebra is called a Frobenius algebra if there exists a


h E A = HomK(A,K) which does not contain any non-zero right ideal in its kernel. Thus symmetric algebras are in particular Frobenius algebras. a) A finite dimensional K-algebra A is a Frobenius algebra if and only if there exists a non-singular associative K-bilinear form (,) : A x A -, K. (The proof is similar to the proof of 8.3.) b) All the assertions of 8.5 are true for a Frobenius algebra by the same proof. c) If A is a Frobenius algebra and P is a projective indecomposable A-module, then the socle S(P) of P is simple. But there are Frobenius algebras for which S(P) f H(P). 8.10 Definition. Let A be a finite dimensional K-algebra and let P1 ,..., Pk be a complete set of representatives of the isomorphism classes of projective indecomposable A-modules. For i = I ,...,k we put Vi = Pi/PiJ(A). By 2.9, V1 ,..., Vk is a complete set of representatives of the

isomorphism classes of simple A-modules. Let c.. denote the multiplicity of V. as a 'J J composition factor of Pi. (This is well defined by the Jordan-Hblder Theorem.) The integers c.. 'J are called the Cartan invariants of A and the k x k matrix

c = (c..) ,

'J the Cartan matrix of A. Note that the Cartan matrix is only defined up to a permutation of columns and rows. 8.11 Remarks. Let C be the Cartan matrix of k a) C is the identity matrix if and only if A is semisimple. b) If Pi and V. belong to different blocks. then c - = 0. by 2.10 and 2.11. Thus the Cartan J IJ matrix decomposes into a block diagonal matrix. c) Let 1 = e


+ ... + en be a decomp~sitionof

1 into mutually orthogonal and primitive

idempotents ei. Suppose the numbering is chosen such that Pi = eiA for i = I , ...,k. If K is algebraically closed, then

c.. = dim Hom ir A. elAl = dim e.Ae. '1 K X J K I J' This follows from 4.14 replacing KG by A. The prcr~fis exactly the same. 8.12 Theorem. Let K be algebraically closed and k t A be a symmetric K-algebra. Then the Cartan matrix of A is a symmetric matrix Proof. Let 1 = e l -

+ ... + en be a decomposition of 1 into mutually orthogonal and primitive

idempotents ei. Then n n n A = o e.A = o Ae. = e e.&. 1 1 i=l i=l i.~=l J as K-vector spaces. Let (,) denote the bilinear form associated with the symmetric algebra A. For j


r and all a, a ' E A, we have

(, e a t e ) = ( a ' e ) = 0 . 1 J r S I J r ' S Using the symmetry of (,), we obtain for i z s and all a E A (, e Ja t es ) ~= ( era l es , = 0. I l j = ( e ra t ese l. , ae.) J


With 8. 11 c) follows c.. = dim 1J

= c.. e.Ae. = dimKA - dim ( e . ~ e . ) 5l dim A 1 dim erAes = dimKe.Ae. K 1 J K l J (r,~)*O,i) J 1 Jl

Similarly, c.. J'


c.. , so c.. = c... 1J 'J J'

8.13 Example. Let K be an algebraically closed field of characteristic p and let G be the symmetric group S on 4 letters. In the following section we compute the Cartan matrix 4 C = (c..) for KG in case p = 2 and p = 3. 1J

Case p = 2. By 3.6, the Klein four group is contained in the kernel of every simple KG-module. Thus the simple KG-modules are obtained by inflation from the simple KS3-modules. So, by 3.17, the trivial module K


and a Zdimensional module, say V, are the only simple KG-modules.

Let PI = PG(KG) and P2 = PG(V) be the projective covers of KG and V respectively. By 4.5 c), PI is the permutation module induced from a 2-complement of G. Hence dimKP1 = 8. Since K is algebraically closed, we have KG

= P1 o P2 o P2 (see 2.5). Thus the dimension of

P is also 8. Since KG is a symmetric algebra, c.. = c.. by 8.12. Now follows 1J J' 2 8 = d i m P -- c 11 + c12dim V = c l l

+ 2c 1 2

and 8 = dimKP2 = c21 In particular, c12 is even. Note that c l l Next we claim c12


1. Otherwise c l

+ c~~ dim V = c 1 2






= 8 and G induces on P1 a group of uni-triangular

matrices, hence a 2-group. Therefore A4 is in the kernel of PI. But this is not true as P1 is a faithful permutation-module for G. Thus c12


1 and the relations 8 = c l 1 + 2 c I 2 , 8 = c 1 2 + 2 ~ 2 2 , c I 2 = 0 ( m o d 2 ) , c1121



ell = 4 , c 1 2 =

2 , cZ2 = 3 .


Note that det C = 8 = (char K) 3. Case p = 3. Since G has four 3'-classes, KG has four simple modules V1,V2 Vy V4 (see 5.9). We may suppose that V1 = KG and V is the module corresponding to the signum representation. By 2 G = KG o A, where H = S and A is a simple KG-module. Put V3 = A. Next we 3.17 b), (KH) 3 claim that V4 E V3 o Vz We may suppose H = ((1.2.3). (1,2)). Let g = (1,2) and let Di be


the representation on V i It is easy to see that o D+gj = I . Furthermore


tr(D3 o D2)(g) = t r D3(gl-tr D,tg) = -tr D3(g) = -1 , and therefore V3



=# V3 o V2. Since V,- IS o m d~mens~onal. V3 e V2 is simple. This proves

4 -= V3 o V2. Now let P.I = PG( V i ) be the pmjectix-e cover of Vi. By 4.5 c), dimKPl = 3.

According to 8.8, the module P I has the f o l l o u i n structure

with a I-dimensional module X. If X


V , then S1 opsrates on P1 as a group of unitriangular 1

matrices, hence has a non-trivial 3-factor group, a contradiction. Hence X = V2.


To determine the structure of P 2 , note that P, by 1.6 is a direct summand of P1 o V

- r PI 8

result of Dickson ( 3 4 , 3 divides dimKPz Thus P, structure of P2 we obtain

. By the

2 V and for the composition 2

Now, 4

24 = dim KG =


(dimKVi)(dimKPi) = 3

+ 3 + 3 -dimKP3 + 3 - dimKP4.

i=l r3

+ 3 + 3.dimKV3 + 3.dimKV4=


This forces P3 = V3 and P4 = V4. Thus for the Cartan matrix we obtain 2 1 0 0 1 2 0 0

In particular, KG has 3 blocks. Furthermore, det C = 3 = char K. 8.14 Remark. In 8.13, the determinant of the Cartan matrix was always a power of the characteristic of the underlying field. By a result of R. Brauer, this is true for any finite group and any algebraically closed field of prime characteristic p (see 15.15). Exercises 22) If A is a Frobenius (symmetric) algebra, then the full matrix algebra (A),



F~robenius 1

(symmetric) algebra. 23) A finite dimensional semisimple algebra over an algebraically closed field is a symmetric algebra. (Remark: One can drop the assumption on the field, but this involves an argument on division algebras which we have not prepared so far.) 24) Let A be a finite dimensional K-algebra. Suppose Sr(A) = r l o ... o r t with pairwise non-isomorphic simple modules r.. If t = 1 or I KI = 1


then A is a Frobenius algebra.

25) Let V be a finite dimensional K-vector space and let A = AV be the exterior algebra of V. Then A is a Frobenius algebra. Only in case 2 )r dim V or char K = 2, A is a symmetric algebra.


26) Let A be a symmetric K-algebra w.r.t. to h E A

. Let M be a simple A-module with

character f3. Furthermore, let {al ,..., an) and {bl ,..., bn) be dual K-bases for 4 i.e.

then a) @(a)= h(za) for all a E A

27) Let the notation be as in 26. Furthermore. let e be a primitive idempotent of A such that


a= eA/eJ(A). Then the right multiplication with z on eA is an A-epimorphism from eA

onto S(eA). 28) Use the duality * defined in 4.10 to prove that the scxle of an indecomposable projective


KG-module is simple.

9 9 Modules of solvable, p-solvable and p-constrained groups


9.1 Lemma. Let K be algebraically closed and let N be a normal subgroup of G with I GIN a prime. If V is a simple KG-module, then one of the following two cases occurs. VN is simple

(i) or



uGwith a simple KN-module U.

Proof. Let U be a simple submodule of VN. If IG(U) -

= N, then the second case occurs by 6.5.

Thus we may assume that I (U) = G and by 6.6 there exists a simple KG-module, say W, with G G WN z U. Since HomKG(V,U ) z Hom KN (VN ,U) z 0 (see 3.9), V is a composition factor of G U . By 6.10, U ~ = W ~ K G / N . Hence V


W o X with a 1-dimensional K G/N-module X. This proves

VN=(WoX)NeWN=U. 9.2 Theorem (R. Swan) Let K be algebraically closed and let G/N be solvable. If V is a simple KG-module, then the number of composition factors of VN divides I GIN


Proof. We argue by induction on I GIN I . Let M be a maximal normal subgroup of G containing N. By 9.1, we have VM =


v1 0 ... 0 vs

where s = 1 or s = G/M I and the V. are G-conjugate simple KM-modules. 1

Let t denote the number of composition factors of V

l I N. Since the V.J are all G-conjugate, t is

. By the induction hypothesis, t divides the number of composition factors of every V j l ~ M/N Thus the number of composition factors of VN, which is equal to st divides




9.3 Corollary. Let K be algebraically closed and let G be solvable. If A is an abelian normal subgroup of G and V is a simple KG-module, then dim V ( I GIAI. In particular, dim V


Proof. By Clifford's theorem, VIA = V1 e ... e Vs with s~mpleKA-modules Vi. Since K is algebraically closed and A is abelian, dim Vi = 1. Thus the assertion follows by 9.2.

9.4 Remarks. a) 9.2 and 9.3 hold true under the weaker hypothesis that K is algebraically closed of prime characteristic p and G is a psol\able group. This is a consequence of the famous Fong-Swan theorem whose proof 1s given In 5 I4 (see 14.1). b) Note that 9.3 is false in general. For instana. PSUL?) has by 5.10 a simple module of dimension 5 in characteristic 7. But 5 4 I PSU2.7, : = 2 ) - 3 - 7 . To investigate the projective covers of s~mpler n d u k s u s need Green's indecomposability theorem. It plays a role at m a n places


modular representation theory. We shall state it only.

For a proof see [Hu - BI, Chap. VI1, $161.

9.5 Green's Indecomposability Theorem. Let K be algebraically closed of characteristic

p > 0. Let N be a subnormal subgroup of G u ~ t hI G:S I a power of p. If V is an indecomposable KN-module, then the induced rndule VG is indecornposable.

9.6 Theorem (P. Fong). Let K be algebraically c l a d of characteristic p > 0 and let G be a

solvable group. Let V be a simple KG-module and let PG(V) denote its projective cover. Then


dimK PG(V) = GI (dim V) P K P' Proof. We argue by induction on I GI . Let N be a maximal normal subgroup of G and let U be a submodule of VN. At first we claim that PG(V) ( PN(U)G :


0 L Homm(U,VN)

G HomKG(U ,V), there is a KG-epimorphism from UG onto V,


G hence from PG(U ) onto V. Thus PG(V) G by 4.5 a). Furthermore PN(U) has PG(U G)



G G PG(U ) by 4.2 b). But PN(U) is KG-projective

uG as a factor module. Hence again by 4.2 b)

P ~ ( u ) and ~ , the claim is proved.

Now proving the theorem we have to distinguish two cases.



Case 1: Suppose G/N = q



Suppose at first that V = uG, hence IG(U) = N. By the induction hypothesis then dim PN(U) = I N / (dim U)p, = IGlp(dim w p , . P Hence G dim PN(U) = 1 GI (dim V) P P" As N = IG(U) = IG(PN(U)), the module pN(ulG is indecomposable by 6.5.

The remark above then shows that PG(V)




and we are done.

By 9.1, we may now assume that VN = U. Since p

4 I G/N I we have J(KG) = J(KN) KG = KG J(KN), by 3.16. Clearly,

PG(V)/PG(V) J(KN) is the largest semisimple KN-factor module of P G ( q N . But


pG(v)/pG(V J ( W ) I N = PG(V)pG(V)KG J ( W ) N = PG(V)pG(V) J(KG) I N

VN = U.

This shows that PG(V)N is indecomposable. Since U is an epimorphic image of P G ( q N we obtain P (U) N


P G ( V N Thus by the induction hypothesis, dim PG(V) = dim PN(U) = I N I (dim U)p, = I Gp 1 (dim P



Case 2: Suppose 1 G/N = p. ~ By Green's indecomposability theorem 9.5, the module P ~ ( u is) indecomposable. Hence by the remark above PG(V) in both cases (dim V)



G G PN(U) . Now either VN = U or V = U by 9.1. Hence

, = (dim U) , and by the induction hypothesis we obtain P

G dim PG(V) = dim PN(U) = p 1 N 1 (dim U) - G p(dim V)p,. P P' -

I 1

9.7 Corollary. Let K be algebraically closed of prime characteristic p and let G be a solvable group. If V is a simple KG-module with I GI


I dim V then V is projective.

Proof. By 9.3 and the hypothesis d i m V = IGI ( d i m 9 P P" By 9.6, dim PG(V) = I GI (dim V) , . Hence PG(V) = V. P P 9.8. Corollary. Let K be algebraically closed of pnme characteristic p. Let G be solvable with p-complement H. If V is a simple KG-module with p x d i m V, then VIH is simple. G Proof. Let KG denote the trivial module. By 4.5 c), PG(KG) = (KH) . Furthermore 4.6 yields PG(V) ( PG(KG) e V. Since and


dim PG(V) = I GI (dim V) = I G 1 p(dirn V) P P dim PG(KG) 0 V = ( GI dim V, we haw P PG(V) = PG(KG) s V.

Hence by 6.10

by 9.6,

P G ( V )G s \ ' . l K H * \ . i H ) G = ( ~ ( H ) G andVIHissimple.

9.9 Remarks. a) The statements in 9.6 - 9.8 are still uue for psolvable groups. b) Landrock and Michler have pointed out, that the fim Janko group J1 of order 23 . 3 . 5 - 7.11 19 has two simple modules of dimension 73 - 7 in characteristic 2, both of which

are not projective. c) Sometimes the following statement, also due to P. Fong, is quite useful. Let K be algebraically closed of characteristic p > 0 and let G he a psolvable group with p-complement

H. Then for every simple KG-module V there exists a simple KH-module W such that PG(V) = wG. (For the trivial module V, this is just 4.5 c). Feit, Chap. X, theorem 3.4.)

9.10 Lemma. Let N be a normal p-subgroup of G and let K be any field of characteristic p > 0. If a : KG + K GIN is the canonical epimorphism g 4 gN, then Ker a = J(KN)KG is nilpotent.

Proof. Note that by 3.4 -

Thus KG J(KN) C Ker a. Comparing dimensions yields equality. Since J(KN) KG = KG J(KN), the kernel of a is nilpotent.

9.11 Theorem. Let K be a field of characteristic p > 0. Let N be a normal p-subgroup of G

with CG(N) C N. Then KG is an indecomposable algebra, i.e. KG consists only of the principal p-block which is denote by Bo(G) throughout his lecture. Proof. We may assume that N > 1. Let f be any block idempotent. First we claim a decomposition (i)

f = fo

+ fl

with fo E Z(KCG(N)) and fl E J(Z(KG)):

For x E G let ffx denote the G-conjugacy class of x and x =


y. Now, for the

yE.R x decomposition in (i), it is enough to show that

;is nilpotent

for any x f CG(N). (Note that


then generates a nilpotent Ideal in Z(KG)). Let x !$ CG(N) Clearly, N acts by conjugation on

Rx and all the orbits have a length divisible by p. Thus, if a : KG -I KG/N is the canonical epimorphism, then a(;) = 0 and (ii)

;is nilpotent by 9.10.

For n large enough we obtain by (i)

n n f = fP = fp o


+ @1

n = fp E Z(KCG(N)).

Now observe that CG(N) is a p-group by hypothesis. Hence 1 is the only non-zero idempotent in Z(KCG(N)) by 3.4. This proves f = 1.

9.12 Lemma. Let K be a field of characteristic p > 0 and let H = 0 ,(G). Put P 1 h. Then = hEH a) f is an central idempotent in KG. b) fKG

= KG/H (as algebras and KG-modules).

c) Bo(G>

I fKG.

d) Op,(G) acts trivially on BJG).

Proof. Obviously

f2 = f E Z(KG) and KG = K G

5 ( 1 -nKGis

a 2-sided decomposition of KG.

l c Bo(G) is a direct summand of K G . Since f a c t s like the identity on the trlvlal r n ~ ~ u 1(G.

Hence c) and d) follow. Let a : KG -4 KGM be the canon~calrp~morphlua-Ckarly (I-f)KG Ker a fl fKG = 0. Thus we obta~n tKG

9.13 Definition. A group G



Imn = KG H


Ker a and

proving b).

called p-c~~-rnstnincj ~f

CG(Opt,p(G).0 . ( G Ir P

T Op


Note that p-solvable groups are p- constrained (reHuppen Endliche Gruppen, Chap. VI, 6.5).

9.14 Theorem (Cossey, Fong, Gaschiitz) Let K be a field of characteristic p > 0. If G is p-constrained, then the following are equivalent. a) KG is an indecomposable algebra. b) 0 ,(G) = 1. P Proof. a) =, b): Suppose H -

:= 0 , (G)


* 1. Put f =

Then KG = K G o (1-f)KG yields a non-trivial decomposition. b) 3 a): Since 0 ,(G) = 1, C ( 0 (G)) C 0 (G) and the assertion follows by 9.11. P G P P

9.15 Remarks a) (M. Harris, G. Michler) Let G be a finite non abelian simple group and let K be a field of characteristic p. Then KG is an indecomposable algebra if and only if p = 2 and G is one of the Mathieu groups M22, M24. b) Suppose the characteristic p of K is odd. Then for arbitrary G, the group algebra KG is an indecomposable algebra if and only if G is p-constrained with 0 ,(G) = 1. A consequence of this fact is the converse of 9.14 for p odd. P

9.16 Corollary. Let K be a field of prime characteristic p. If G is p-constrained, then

is the blockidempotent of the principal p-block. Proof. By 9.12, K G s KG10 ,(G). P

,(G) is p-constrained with 0 ,(GI0 (G)) = 1. P P P Thus K G is an indecomposable algebra (see 9.14) and fKG si. Bo(G) by 9.12 c).

Note that GI0

9.17 Theorem. Let K be a field of prime characteristic p. Let G be p-constrained and let V be a simple KG-module. Then the following are equivalent: a) V belongs to the principal p-block. b) Op,(G) acts trivially on B. (G) acts trivially on V. c, Opr,p 1

Proof: a) =, b): Since f = -


g is the block idempotent of the principal block,

we have V = Vf. Hence 0 ,(G) acts trivially on V.



c): V may be regarded as a module for GI0


,(G). By 3.7, 0 (GI0 ,(G)) acts then P


trivially on V and the statement in c) follows. c) =$ a): By the assumption 0 ,(G) acts trivially on V. Thus by 9.16, the block idempotent of P the principal p-block acts trivially on V, which means that V belongs to the principal p-block.

9.18 Lemma. Let K be a field of characteristic p > 0 and let P be a projective module. Then a) Ker P is a normal p'-subgroup of G. b)

n Ker V where the intersection runs through the composition factors of P is


p-nilpotent. Q G. Moreover PI Proof: a) Let H = Ker P. Clearly H module for H is projective. This implies p 4 1 H 1.

= KH


... o KH. Thus the trivial

b) Put A = fl Ker V, the intersection over all composition factors of P. Then A a G and






G. Obviously A/H acts faithfully on P. S i m A acts trivially on each composition

factor of P,

A/H is a p-group. This shows that A

is pnilpotent.

9.19 Theorem (R. Brauer) Let K be a field of characteristic p > 0. Then 0 (G) = P',P


Ker V

where the intersection runs through the set of simple KG-modules belonging to the principal p-Block Bo(G). Proof: By 9.12, 0 ,(G) Ker V for every simple KG-module V belonging to Bo(G). Hence P 0 (G) L Ker V by 3.7. Thus 0 ( G ) " Ker \' and equality holds because of 9.18. P',P P




9.20 Remark. The following result, due to Pahlinp. strengthens 9.19 in an obvious way. Let

K be a field of characteristic p > 0 and let V be a simple KG-module. If 6 is the set of 7

composition factors of PG(V)J(KG)/PG(V)J(KG)-, then


w s

Ker W = O p t,p(Ker V).

9.21 Definition. Let K = IF be the prime field of characteristic p > 0. A chief factor V = L/H P of a group G is called a p-chief factor, if V is a p-group. If V has a complement in G/H, we call V complemented. Let

5 (G) resp. @(G) denote the set of p-chief factors resp.

P P complemented p-chief factors of G in a given chief series of G.

Note that each p-chief factor may be regarded as a simple F G-module. P

Furthermore the isomorphism types (including multiplicities) in

5 (G) do not depend on the

P choosen chief series of G. (Actually this is also true for @(G), but we do not use this fact.) P 9.22 Lemma. Let H = a) Opt ,p(G)

n Ker V VESp(G)

" H.

b) If P E Syl (H), then CG(P) C H. P Proof. a) Let V = M/N -

E 5 (G). Then


oP ,(G)N/N n v = 1, hence 0 ,(G) C CG(V). As V is simple, by 3.7 then also Op, ,p(G) C CG(V) P b) Suppose H < HCG(P). By the Frattini argument we have HCG(P) HNG(P) = G. Let E/H be a chief-factor of G with E/H s HCG(P)/H. Then E centralizes all p-chief factors of G above H. But p-chief factors below H are covered by P, hence are centralized by HCG(P). Therefore E centralizes all p-chief factors of G, which implies E


H, a contradiction.

9.23 Theorem (W. Willems) If K = IF , then the following statements are equivalent: P a) G is p-constrained. b) Each simple KG-module belonging to Bo(G) is a composition factor of a suitable tensor product of p-chief f.actors.

Proof: -


M :=

o V VGP(G)

and H :=

f l Ker V. "Gp(G)

a) 3 b).By 9.22 a), we have O p t,p(G) C H. We claim equality. If 0

(G) C H, we may find a p'-element h E H \ 0





Since h acts faithfully on 0 (G)/O ,(G) (note that G is p-constrained) and trivially on all P1,P P p-chief factors, we obtain h E 0 (G), a contradiction. P'>P

(G) and M is a faithful K G/H-module. By the Bryant-Kovacs theorem (see Thus H = 0 P',P 4.13), the regular K G/H-module regarded as a KG-module is a direct summand of )G/H)-1

o j=1

n. (M o .. . o M) for suitable integers n. (j-times) J


Since 0 (G) = H acts trivially on all simple modules in Bo(G) (see 9.17), these are P'?P contained in K G/H and the statement in b) follows. I



+ a). By 9.19,

and by the assumption in b),


n Ker V vGYP(G)


r\ Ker V = 0 (G). VEBo(G ) P'*P

Hence H = 0 (G) and H = 0 ,(G)P with P E SyI (H). P',P P P Now by 9.22, CG(H/O ,(G)) 5 CG(P) r H and G is p-constrained. P The rest of this section is devoted to the proof of theorem 9.24. T o state it, let K = F and for


a simple KG-module V, let m(V) denote the number of complemented p-chief factors

isomorphic to V in a given chief series of G. I

With this notation we have

9.24 Theorem a) If V E g ( G ) , then V occurs as a component of

P P~(K~)J(KG)/P~(KG)J(K with G ) ~multiplicity m(V).

b) (W. Gaschiitz) Let G be p-solvable. If V is a composition factor of

P ~ ( K ~ ) J ( K C ) I P , ( K ~ ) J ( K G=) ~X then V E $(GI and its multiplicity in X is exactly m(V).

We prove 9.24 through several Lemmas. The approach presented here is quite elementary of ring theoretical nature, due to Okuyama and Tsushima.

T o be brief in the following, let I(G) denote the augmention ideal of KG and J = J(KG). Let e be an idempotent such that eKG = PG(KG). Without reference we shall use quite often the fact eJ =eI(G).

9.25 Lemma. Let H a G. Then there is a KG-epimorphism

where G acts by conjugation on I(H)/I(H)



Proof. Define /3 : I(H) -+ eI(H)KG/eJI(H) by (h-1)P = e(h-1)

+ eJI(H)

(h E H).

For h E H and g E G observe the following facts


e(h- 116 = e(h- 1)g + e(1-6)(hg-1)

(i i)

e(1-6)(hg-1) E eJI(H)


eg(1-h) = e(l-h) + e(g-l)(l-h)

The first two facts imply that /3 is a KG-homomorphism. Since I(H) KG = KG I(H), (iii) shows that (3 is an epimorphism. Thus we obtain a since I ( H ) C ~ Ker


9.26 Lemma. Let V be an elementary abelian normal p-subgroup of G. 2 2 Then V n I(V)/I(V) as KG-modules via the map v -I (v-1) + I(V)

Proof: Let V = (vl, ...,v n ) be elementary abelian of order p


. If vg, = v dil(g) ... v nd i n k )

n (g E G, 0 r dij(g) a p-1), then g operates on the K-space V = o Kvi (V written additively) by i=l

For v,w E V we have (v-l)(w-1) = (vw-1) - (v-1) - (w-1) = 0 mod 1 ( v 2 . Hence vw - 1 = (v-1)

+ (w-1)

mod l(V)



The action of G on the K-basis {vi-1 + 1 ( v 2

(v.-l)g = vg - 1 = I

Hence V





I i = 1. .... n )

of 1 ( v ) b ( q 2 is therefore

dilW dink) n 2 - 1 1 1 d (EX v.-1) (mod I(V) ). v1 ...v n J= 1 J J



9.27 Lemma. Let V be a minimal normal psuwoup

of G which is complemented. 2 a) I f f is any primitive idrrnp,tmt of KG, then fl(V)KG fJ .

b) (eI(V)KG

+ eJ2)/eJ2 = \'

Proof: Write G = T V with T -


G and T

- \' = 1

Obviously I(V) = J(KV) and I(V)KG = Ir \'KT = KT 11k3 = KG I(V). Furthermore by 3.16, I(V)KT C J(KG) = J. Hence KG = K T o I(V)KT and J = Jo


It\')KT where Jo = J(KT). Since I(V)KT is a 2-sided

nilpotent ideal in KG, we may assume that e.f 5 hT b) lifting idempotemts (replace J by I(V)KT in 2.5 and note 2.9). Furthermore, we have


J 2 = [ I ( V ) ~ K T+ I(V)Jo

+ Jolt



VI] i


where the last sum is direct since Jo C KT and


[ I ( V ) ~ K T+ I(v)J~,+ J ( ~ ( V ) ] I ~ V ) K T .

Multiply the equation (i) by f. This yields

C KT and where the last sum is again direct since fJ2 0~ ~ ( v ) ~ +K fl(V)Jo T

a) Suppose

+ fJ$(V) C KT I ( V ) ~ K T+ KT I(V)KT + KT I(V) C I(V) KT.

~ ~ ( v ) K= G ~ - ~ ( v ) KcTf ~ ~ .

Then R(V) KT = R(V) KT


. ~ ~

= fI(V) KT f l ( [ f l ( ~ KT )~ = [ ~ I ( v ) ~ K+ T fl(V)Jo

since [ ]

2 + fI(V)JO + fJ,I(V)] e fJo)

+ fJoI(V)]

o ( f ~ fli fI(v) KT)


Observe that fl? f l fI(V)KT C KT

n I(V)KT = 0,

hence n(v) KT = O ( V ) ~ K T+ O ( V ) J ~+ EJJ(V). Multiplying this from the left by 0 + t E SI(KTf) yields ~ I ( V ) K T= ~ I ( V ) ~ K+T~I(v)J(, since tJo = 0. (see 8.7%)). Hence by the Nakayama Lemma (1.6),


2 2 ~KTJ(KV)= KT I(V) = ~I(V)KT= ~ I ( V ) ~ K=T KT I(V) = KT J(KV)

and again by the Nakayama Lemma tKT I(V) = 0, a contradiction, since tl(V) + 0. Hence fl(V)KG




b) Consider the composition of KG-maps r)

which is an epimorphism. 2 Since the right module has an epimorphism onto (el(V)KG + eJ )/eJ2

* 0 and V is simple, the

assertion follows.

9.28 Lemma. Let V be a minimal normal psubgruup of G . I f eI(V)KG C e ~then ~ V, is complemented. Proof: Throughout the proof we freely use again the fact that 1(V) C J (3.16). Now let L be a KG-submodule of eJ such that

Put T = {g

I g E G, e(g-1) E L ) . \Vtt

c l a ~ mthat T is a complement of V.

For g, h E T, we have e(gh-1) = e(g-l)h

+ e(h-1).

Suppose V C T. Then (eI(V)KG

+ eJL),eJ'

Hence T


G. 7


-, 2 L cl-. hence eI(V)KG C eJ , a contradiction.

Thus V C T and it is enough to show that G = \' - T Obviously eI(V)

+ eJl(V) C eI(V)KG = eKG Ir\.'r

For v E V, we have e(v-l)g = evg - eg = egvg - eg = eg(vg-1) = eig-l)(vg-1) E eI(G)I(V)

Hence It follows that

+ e(vg-1).

+ el(W 2 eJI(W + eI(V).

+ eJI(v) = eI(V)KG. 2 eJ = L + eI(V)KG = L + eI(V), since eJI(V) C eJ



Thus, for g E G, we have e(g-1) = y Since the isomorphism V


+ ew

with y E L and w E I(V).

I ( v ) / I ( v ) ~of 9.26 is given by v


("-1) + I ( v ) ~we may write

+ u with u E I(V)2 . e(g-v) = e(g-1) + e(1-v) = y + ew - ew + eu E L, w = (v-1)


since eu E el(V)2 C eJ 2 C L. Therefore e(p-'-1) E L, which means gv

- 1 E T.

Thus g E TV and the proof is complete.

Finally we need 9.29 Lemma. Let G be a p-solvable group and let L 5 H, both normal in G. If HX, is of p'-order, then el(H)KG = eI(L)KG. Proof: Put G = G/L. There is a natural epimorphism from I(H)KG/I(L)KG onto I(R)KG. Comparing dimensions shows that it is an isomorphism. Thus we may assume that L = 1. Let U be a p-complement in G. u , since eKG = K:

For e we may take e =

= PG(KU).

uEU Now H C U implies eI(H) = 0. 9.30 Proof of 9.24: Choose a chief series 1 = G < G i C ~ = - V ~ ~ S . ~ p. I S If V1, V2 are simple and belong to Bo(G), then each composition fac- la 11f \' 1 s V2 belongs to Bo(G). (This is not true for A5 = SL(2,4) and p = 2.)


Let G be solvable and K algebraically ch=d '-1 characteristic p > 0. The following statements are equivalent. a) The Sylow p-subgroup of G is normal. b) p )i dim V for all simple KG-modules \ c) PG(V) = PG(KG) o V for all simple KG-rr.kdulcs V. d) Let H be a p-complement of G. Then \' H


LS srmple

for all simple KG-modules V.

Let G be solvable and K algebraically c l ~ odi characteristic p. The following statements are equivalent a) G is of p-length 1. b) p

I dim V for all simple KG-mcdules \' f Bc{G).

$10 p-adic Numbers

Modular representation theory is - alas - a "p-local theory". Therefore it is only natural to work with p-adic fields instead of algebraic number fields. We give in this section a report on p-adic numbers, without proofs. For details we refer to Cassels, Frohlich: Algebraic number theory (Acad. Press 1967), Serre: Corps locaux (Hermann 1962). 10.1 The rational p-adic field Q

P' Let Q be the field of rational numbers and p a prime. We define a valuation v on v(pi

i)= i

Q by

if a,b,i t I , p &, ah.

We also put v(0) = m. Then v has the following properties:


v(ab) = v(a)




+ v(b);

Min {v(a), v(b))


If we put

11 a-b 11 = 2-v(a-b) then (3)

I(. (1 is a metric on Q, and from (2) follows the "strong triangle inequality" Ila+bll


Max {








Now we can imitate the process of completion by Cauchy-sequences by using the metric The resulting set is again a field, the field every element a

10.2 Algebraic -

Q of rational p-adic numbers. We remark that P

+ 0 of Q can be uniquely written as P

where 0 5 a i < p and a.


0. (This series does converge in the sense of

extensions of


11 - 1) .)


of finite degree (K:Q ) = n. It is a non-trivial fact that the P P valuation v can be uniquely extended to a valuation of K by

Let K be an extension of

11- 11.

1 I

for a E K. Moreover K is complete with respect to v. We put

Then I is easily seen to be an integral-domain, the ring of integers of K. The only maximal ideal of I is p= {a



If we take x E p with ~ ( x minimal, ) then p = rr I is a principal ideal. As p E p, so Char I/P = p. Indeed, R = I/# is a finite field. We put 1 lip1 = pf . {v(a)

I O + a E K } = -e1H

and (K:Q ) = n = e f . P The ring I is a principal ideal ring, its only ideals arc 1 3t

I (i=O,l, ...) and 0.

(This fact makes p-adic numbers for our p u r p x supenor to algebraic numbers.) We remark I n fl U = Up. Hence if n E l and p

n. lhcn n

E I n a n d therefore n-'

E I.

We call K a field of p-adic numbers.

10.3 Definiton. Let G be a finite group and K a pdti field with integers I and K = I/p. We call (K,I,K) a p-modular system for G, if the follou ~ n gconditions are fulfilled: (1)

K is a splitting field for every subgroup U


G. u-hich means

for some m. (depending on U). 1


K is a splitting-field for every U

for some n.. J


G. T h ~ hmeans thy definition) that

10.4 Remark. Without proof we state that a p-modular system of G can be obtained in the --

following way: (E), where E is a primitive I GI -th root of unity. That K is a splitting-field P G follows from a famous theorem by R. Brauer (see Huppert I, p. 591 or Isaacs,

Let K = Q for every U


p. 161). That also R is a splitting-field for every U s G is a less deep result of R. Brauer (Blackburn-Huppert 11, p. 32). This is due to the fact that in characteristic p an absolutely irreducible representation of G can already be realized in the field of its character-values,


hence in the field of I G -th roots of unity over GF(p). As there are no primitive p-th roots of unity in characteristic p, the I GI ,-th roots of unity suffice. P It is much easier to see, that some finite algebraic extension of



supplies a p-modular system

for G, avoiding the use of Brauer's theorem. For most applications this suffices.

$11 Reduction mod p and decomposition numbers.

11.1 Definition. Let (K,I,R) be a p-modular system for G. Let V be a KG-module of finite K-dimension. a) We call an IG-submodule M of V an IG-lattice of V, if MK = V and M is a finitely generated I-module. As I is a principal ideal domain and M is torsion-free, M is a free I-module and rankIM = dimKV. If

then n V = o m. K. j=1


As M is an IG-module, we have for e\.ery g E G

with a. (g) E I. Hence D, defined by ~k D(g) = ("k(6)) is an integral representation of G over I. b) If M is an IG-lattice of V, we form the KG-module M = M/Mx. Hence with the notation of a) we get

where Z = m. t Mn. Also J J

where a ( g ) = a. (g) jk ~k

+ n 1. Hence D, defined by

is a representation of G over K.

11.2 Theorem. Let (K,I,K) be a p-modular system for G. If V is a KG-module, there exists an IG-lattice of V. Proof. Let {v -

v n ) be a K-basis of V. Put

Then M is an IG-module, finitely generated and torsionfree as an I-module. As I is a principal ideal ring, so M is a free I-module. (Here it pays to use p-adic numbers and not algebraic numbers.) Let n M = o m. I. j=1 J Obviously MK = V and { m

m, ) is a K-basis of V

11.3 Example. Let G = ( g ) be the group of order 2 and (K,I,K) a 2-modular system for G. Let -

be the KG-module with v1g=v2, v g = v 1' 2 We put W1

= v1

+ v2 '


= v1 - v2 .

As Char K = 0, so {w 1 , w 2) is also a K-basis of V, and w l g = w1 ' w2g = -

W2 '


M - v I o v 2 1 and M - w l o w I 1- 1 2- I 2 are IG-lattices of V. On M the group G operates trivially, but not so on M 2 1' Hence M l 8 M2 . But the composition-factors of M are independent of the choice of the IG-lattice M:

11.4 Theorem (R. Brauer). Let (K,I,R) be a pmodular system for G and V a KG-module. Let MI and M2 be IG-lattices of V. Then M I and

a2have the same composition factors as

KG-modules (including multiplicities). Proof. - Suppose n M - o m. I. 2-j=1 J

As m. E V = MIK, there exists an s E D1 b ( 0 ) suc-h J Hence M ns C MI. As the multiplication by 2



r, an


lC-isomorphism of M2 onto M 2 n , we

may assume that

for some 0 < t E IN. Now I = I f rr

is an Artinian ring (of length

I+ 1


1 r.



is Artinian, being a finitely generaled i - r n d u k . Thcn SfI M~IT'" is a finitely generated ic-module, hence is also Artinian. Therctllrr r k r h > r r r n of Jordan-Hiilder holds for M ~ I M ~ ~ + ~ . We consider the diagram

We have the following IG-isomorphisms:

+ M2)/MIn = M21(M1n fl M2).




M1/M2 = Mln/M2n




(by multiplication with n).

The Jordan-Hiilder theorem for i ~ - m o d u l e simplies from (2) and (3) that M1/(M1 n + M2) and (MI" f' M2)IMZn

have the same composition-factors. Finally, (4) and (1) show that M I / M l n and

M2/M2n have the same composition-factors,

which was to be proved.

Until now we used the fact that 1 is a principal ideal ring. In the next steps the completeness of K (and I) will be crucial.

11.5 Theorem. Let K be a p-adic field, I its ring of integers and

the maximal ideal of I. Let

n R = o a.1 j=1 J be an I-algebra, which is a free I-module. a) R = R/Rn

becomes a K-algebra by

+ Rn) (a + In) = ra + Rx (rl + Rn) $ (r + R;r) = r I 2 (r


I, +


for r,rl,r2 E R and a E I. b) The mapping

is a ring-epimorphism of R onto I? uirh k z m l R.T. c) R is complete under the topolo:!



d) Suppose

T = e1 +...+ C n with



' e.J = b'J. . 5.Then thsrc exrit dcrnpxents e.J E R with


I = e +...+ e 1


s s I


= J . e.

.. .


("lifting idempotents"). Pr0of.a) and b) are trivial. c) follows simply irllm the ha that the topology on I induces a topology on the product space R, under u hich R eLlmplsts. 7

d) As in 2.5, we lift at first one idempotent ;= ;i r L m R to R. Suppose we have already found r. E R G = l , ...,m) such that I

i = r. + R n , r. - r. J

We put (as in 2.5)



RJ-'. r2 - r J




Then r m + l - 'm --3 r m - 2 r m


m m = 3 r m - 2 r m - r m i O ( m o d R n ).

In particular r

m+ 1





Hence (rl,r 2...) is a Cauchy-sequence in R, which by c) converges to a limit e. Then

and 2 e 2 - e = l i m ( r -r.)=O. j+cc



Finally we construct the idempotents el, ...,en in R exactly as in 2.5 b). We need the Krull-Schmidt-theorem for IG-modules. We prepare the proof by

11.6 Lemma. Let I be a p-adic ring and R an I-algebra, which as an I-module is free and finitely generated. a) R x C_ J(R), where J(R) is defined as the intersection of all maximal right-ideals of R. b) Suppose 0 and 1 are the only idempotents in R. Then R is a local ring, which means that R/J(R) is a skew-field. c) Let 1 be a p-adic ring and M an IG-module, which is finitely generated and free as an I-module. If M is an indecomposable IG-module, then HomIG(M,M) is a local ring. Proof. a) We show R n C r for every maximal right-ideal r of R, hence R x J(R): -

If not, we obtain R=Rx+r, hence l = r 1x + r 2 with r l E R , r 2 E r .

Now the completeness of R shows

(Observe that this series converges!) But then

a contradiction. b) Now let


= R/R

e be an idempotent in R/J(R). Then e can be lified to an idempotent 6 in

n (by 2.5), and i can be lifted to an ijcmpxrnr c in R by 11.5. As R has only the

idempotents 0 and 1, the same is true for

R J ~ R IBur . then by

Wedderburn's theorem

is a skew-field. c) If k




rn l 1

then HomIG(M,M) L HomI(X1.\l r = I l i n As I is a principal ideal ring, so HomIG(\ t a a finltcly generated free I-module.

As M is indecomposable, 0 and 1 art: the on!! ljrrmpotents in HomIG(M,M).

Hence by b) HomIG(M,M) is a local ring. 11.7 Theorem. Let I be


p-adic ring and X! a n IG-m~lifule.finitely generated and free as an

n I-module. Then M = o M. with indecompsabls IC-modules M.. The isomorphism types of j=1 J J the M . are uniquely determined. J Proof. - M is a Noetherian I-module, hence a Stxtherlan IG-module. Therefore there exists a n decomposition M = o M. with indecompc>sahltt 31 . By 11.6, Hom (M M.) is a local ring. j=1 J 1 IG j' J

Hence our assertion follows from the Azumaya version of the Krull-Schmidt-theorem. (cf. Anderson-Fuller, Rings and Categories of Modules, p. 144 or Lambeck, Lectures on Rings and Modules, p. 78). A direct proof can be given as follows:

Suppose M = M 1 @ ... o M ~ o N , where each Mi is isomorphic to an indecomposable IG-module U and U is not a direct summand of N. We put H = HomIG(U,M) and E = HomIG(U,U)


Obviously EH C H, so H is a left-E-module. We put

Y = {q


q E H , q HomIG(M,U) C J(E)


This makes sense as H HomIG(M,U) L E


Obviously Y is an E-submodule of H and J(E)H

Y. Hence H = H N is a vector-space for the

skew-field E = E/J(E). If we show that dimg H = r, then r is uniquely determined. Let pi be an IG-isomorphism of U onto Mi. We claim that E-basis of H. Let q E H and for u E U

+ ... + uq r + U S

uq = uq, with


uqi E Mi and USE N Then

-1 q . p . = E. for some 1 1





We claim that s E Y, which means t HomIG(M,U)

C J(E) .

Suppose not. Then there is a y E HomIG(M,U) such that ty = 1u


Ei , ... , pr

is an

But then N = U t o Ker y with U t = U, a contradiction. r

Hence zE Y and





p i (mod Y)



which shows


This means

Chose p. E HomIG(M,U) with p. p. = l U , ul p, = 0 for j J J J Then we obtain

This shows that {


6. = 6 . p . p . E J(E), hence b = 0 . J J J J J ... , pr } is an E-basis for HY.


11.8 Remark The Krull-Schmidt-theorem is \ e n n r t l ) true for UG-modules. In exercise 33 we show that it never holds if G is not a p-group.

If G is a p-group, there exists an epimorphrsm of ZG onto the ring U[E] of integers in the cyclotomic field Q(E),where


= 1

* E.




i t LS known

that U[E] is not a principal

ideal ring. If a,b are non-principal ideals In B [ E ] . rhcn a o b = U [ ~ ]~



(see Curtis-Reiner, Representation Theory o i F ~ n l t cGroups and Associate Algebras, p. 150). This shows that the Krull-Schmidt-theorem is n~rttrue for Z [ E ] , hence also not for ilG. Hence the Krull-Schmidt-theorem can h d d at most for pgroups with p


19. (See also

Curtis-Reiner, Methods of Representation Theon. 11. p. 3 3 . ) In 1.6 we proved Nakayama's lemma f o r algebras uith a nilpotent radical. Now we need a more general version:

11.9 Lemma. Let R be a ring and J(R) its Jacobson-radical, defined as the intersection of all maximal right-ideals of R. a) If r E J(R), then there exists s E R with (1-r)s = 1. b) If M is a finitely generated R-module with M = MJ(R), then M = 0. c) (Nakayama) If M is a finitely generated R-module and N a submodule of M with

M =N

+ MJ(R),

then M = N.

d) If M is a simple R-module, then MJ(R) = 0. Proof. a) Suppose (1-r)R C R. Then there exists a maximal right ideal -

r of R such that

(I-r)R 5 r. But then 1 = (1-r)

+ r E r + J(R) = r C R,

a contradiction. Hence (1-r)R = R and therefore (1-r)s = 1 for some s E R. k b) Suppose M =


miR + 0 with minimal k > 0. Then

i= 1 ml =

,Im.r. with ~ = 1J J

r. E J(R). J

By a), there exists s such that (1-r,)s = 1. Then

a contradiction. Hence M = 0. c) We apply b) to the finitely generated R-module M/N. Then

(M/N)J(R) = (MJ(R)

+ N)/N

= M/N


Hence M/N = 0. d) By b) M > MJ(R). As M is simple, so MJ(R) = 0. 11.10 Theorem. a) Let M and P be IG-modules, finitely generated as IG-modules and free as I-modules. Suppose that P is a projective IG-module. If M M is a projective IG-module.

= P, then M = P. In particular, also

b) Let el, e2 be idempotents in IG. We put


e. = e. 1



+ IGn.

el KG s 2 KG, then elIG e e21G.

c) Let M be a finitely generated IG-module, free as an I-module. If M is a projective KG-module, then M is a projective IG-module. d) Suppose p ); I G I . If M is an IG-module, free as an I-module, then M is a projective IG-module. Proof. a) We consider the diagram -


x is an isomorphism and v,- is the epimorphism defined for m E M by m v2 = m

+ Mrrf

I\ \Ix=


We define v for P similarly. We then define /3 h?- /? = v 1 1 so is

0. As P is projective and

v2 is onto. there


a As vl and a are epimorphisms,

an a E HomIG(P, M) such that

P = a v2.

We show that a is an isomorphism: (1) Suppose O

+ u E Ker a.Write u = u ' n I u-hers u ' 5 Prr. Then t

O = u a = ( u ' a)x. As M is a torsion-free I-module, this implies u'm = 0. But then

O = u'nv 2-







* 0.

This is a contradiction as 5 is bijective.

(2)av2 = (3 is an epimorphism, so M=Pa+Mn. But I G x c J(IG) (see 11.6 a)). As M is a finitely generated module, Nakayama's lemma 11.9 implies M = P a .

b) This follows from a) with elIG = P, e21G = M. c) M is isomorphic to a direct sum of projective modules

7 R G G=l,...,r). By 11.5 d) there

exist idempotents e. in IG such that J e. + 1Gx = F J j' hence


e. KG = e.IGle.IGn = 6.iG J J J J r If we put P = o e.1G , then P s M , hence by a) P = M. I i=l d) If p 1 I G I , then every KG-module is projective. Hence M is projective by c). We now fix our notations. 11.11 Definition. a) Let (K,I,K) be a p-modular system for G. Let h = h(G) be the class-number of G and let V1, ...,Vh be the simple KG-modules (up to isomorphism). Let k = k(G) be the number of p'-classes of G and let W1, ...,Wk be the simple KG-modules (see 5.5 and 5.8). By 11.2, there exists an IG-lattice Mi of Vi. Let d.. be the multiplicity of W. as a 1J J composition factor of Mi = Mi/Mix. (By 11.4 this is independent of the choice of the lattice Mi !) The matrix D = (d..) of type (h,k) is called the decomposition matrix for G in ?)

characteristic p. b) Suppose



?KG i=l

with indecomposable KG-modules

5 KG. By 11.5 d) we lift this decomposition to

n IG = o e. IG. 1 i=l with orthogonal idempotents ei. Here e.IG is determined (up to isomorphism) by the 1


KG (see 11.10 b)). This is by 2.9 b) determined by its head

Now choose eiIG (i=l,...,k) such that

5 K G / 5 J(RG).

We keep this notation. 11.12 Theorem (R. Brauer).

a) The multiplicity of V. as a composition-factor (and direst summand) of eiKG is d... J J' b) If C is the Cartan-matrix for KG, then C = D ' D . w-hkh means

(This again implies the symmetry c.. = c.. in 8.12- 1 'J 1' Proof a) Suppose h

Then by 3.9 a.. = dim HomKG(eiKG.\. r = drmK Vj e.I' J' K 1 Hence a.. = dim V c = rank I J' K 1 1 1




= drmR 9.c = dimR HornKG& KG, M.) J 1 J

and this is by 4.14 the multiplicity of the head H:a i

7 KG as a composition factor of M..J

This shows a.. = d... J' Jl b) By a) we have h

Let Mr as before be an IG-lattice of Vr. Then h

is an IG-lattice of e. KG. The multiplicity of Wi J Hence the multiplicity of Wi in N. is J


7 KG !5 J(KG) in Mr is by definition dri.

But also e.IG is an IG-lattice of e.KG. Hence by 11.4, the multiplicity of Wi in 6.IG = KG J J J J is also

But this multiplicity is c . . by definition 8.10 of the Cartan-matrix. 11 Now we can prove that for p'-groups the classical and the p-modular representation-theory "are the same". 11.13 Theorem. Suppose p ./,


and let (K,I,K) be a p-modular system for G. Let V1, ...,Vh

be the simple KG-modules (up to isomorphism) and let Mi be an IG-lattice of Vi. Then MI, ..., Mh are the simple KG-modules. In particular, dim

R M.I = dim K V.1 divides I GI.

Proof. As p ,/. I G I , KG is semisimple and hence the indecomposable projective KG-modules are simple. By 11.12 follows

E = C = D'D. As G has no elements of an order divisible by p, the numbers h(G) and k(G) are the same, say h. Hence

Hence for every i exists exactly one i ' such that d., . = 1,d.. = O 1 ,1 J1

forj # i t .

If i + j, then h o=c..= 1 d.d.=d.,.d.,.=dirj. 1' r=l r l r j I I I J Hence i ' + j '. Therefore i -+ i ' is a bijective mapping on (1, obtain M.



W. (i=l,...,h). 1

..., h).

Renumbering the Wi, we

1 1

11.14 Remark. Every simple KG-module W. appears as a wrnposition-factor in some M.. J 1 m f . If not, then d . = 0 for all i = I ,...h. But then by 11.12 1J

a contradiction.

Exercises 35) Let 52 be a finite G-set on which G operates b>-permutation. We form the UG-module V(Q) = o Uei, iER where G acts by e.g = e. I 1g'

a) If a E HomnG(V(R), V(R)) and

then a.. = a. for all i,j and a l l g f G. 'J lg,jg b) If G is transitive on R, then L'cQ) I!. a n ~n&xkxnptsablr UG-module. (Look at the trace of a projection rn Horn r


36) (A. Dress) Let GI =


n I=




ZGt L-lQ?. \ - t Q ) )

\ r ~ t hd n r l x r prmzs p a . > 0, r r 2. Take Pi E Syl G, let I' I Pi


a) If a is defined on V(Qi) by (Pig) cx = 1 then a E HomuG(M,?Z).


(i=l .....r ~ .

b) Let


with bi E I. If we define


p E HomnG(U,M).

c) Show that M=KeraoimfisKeraon.

As U

V(Ri), hence the Krull-Schmidt-theorem ist not true for W-modules.

9 1 2 Blocks of defect zero.

Throughout this section let

(K,I,R), be a p-modular system for the groups in question. Let R

denote a commutative ring (R may stand for I or K). 12.1 Definition. Let M,N be RG-modules. Let H be a subgroup of G. -

G For a E HomRH(M,N) we define the trace T H ( a ) by


where {gl,


G ... ,gs} is a right transversal of H in G. Sote that the definition of TH(a) is


independent of the chosen transversal and T H ( a ] E HomRG(M,N). (Proof exactly as in 3.1.) 12.2 Theorem. Let M be an RG-module. u-hich b p j e c t i v e and finitely generated over R. Then the following are equivalent: a) M is a projective RG-module.


b) There exists a E EndR(M) such that T l ( a \= rdsl .

Proof: a) =, b): Define y E EndR(RG) b! gY=

[ 01

for g = 1 otherwise

G Then T (y) = idRG This fact generalizes immediately to a free RG-mcdule F. Since M is projective we have F = M


M ' w lth a free RG-module F.

Let n denote the projection of F onto M with kernel 31'. As n is RG-linear, it follows

G G T I ( ~ I M n ) =T 1 ( ~ I M ) x =id M ' b)

+ a): Let N be an RG-module and let y : S


M be an epimorphism. We have to find a

p E HomRG(M,N) such that py = id M ' Since M is R-projective by assumption, there is a x f HomR(M,N) such that ny = idM '

G G T 1 an)^ = T 1 G = T ( a ) = idM


with T G l ( a n ) E HomRG(M,N)

12.3 Lemma. --

(y is RG-linear)


Let M be an IG-lattice. If


q =ml + Mn, ... ,ms = ms + M n )

is a R-basis of M, then ml,

... ,ms is an I-basis of M.

Proof: By assumption and 11.6 a), we have -

Now Nakayama's Lemma 11.9 yields

If we had a linear dependence of the mi over I, there would be a relation

with ai E I and not all ai in In. But then


would be a contradiction. Hence M = o m.1 . i=l 1 12.4 Lemma. Let V be a simple KG-module with IG-lattice M. Let a E EndIG(M) and let tr denote the usual trace of linear maps. Then a) v(rg M)


v(tr a )


b) Equality holds in a) if and only if a is an automorphism of M Proof: a -

a = a id

E EndIG(M) may be considered as an element of EndKG(V) = K . idV. Hence

v with a E K. The action of a on a free I-basis of M forces a E I and we obtain

Moreover equality holds if and only if a is a unit in I. 12.5 Theorem. Let V be a simple KG-module with IG-lattice M. Suppose I GI I P



Then a) M is a projective IG-module


b) M is a projective simple KG-module. Proof: Choose 0 + iii = m + Mn. By DefineaEEndR(M)by

12.3 let m = m



and m a = O I

... ,ms be an I-basis of M. forir2.

Now by 12.4 v(rgM) Since




G v(trTl(a))=v(lG!)tr(trcz)=v(IGI).

dim V , u e also hare

v(l G I ) 5 v(rg M ) . hence equality holds. In particular, T G (a) is an automorph~smof .\I by 12.4 b). Now 1

and by 12.2 M is projective, and thrrrhrc );I G M=MT1(a)LmIGC\1.





In particular M = iii KG for any 0 -L iii f Sl . This proves the simplicity of M. 12.6 Remark. Let V,W be nonisomorphic simple KG-modules with lattices M,N resp. If I GI P divides both the dimensions of V and W, then 51 f

s. This follows from 12.5 since projective

KG-modules are uniquely liftable by 11.lo. By the methods we have worked out so far we can improve 9.3 for Char K = 0 (see also 2.1 in Clifford Theory and Applications, Trento 1987) . 12.7 Theorem (Ito) Let A be an abelian normal subgroup of G. If

x E Irrc(G)

then ~ ( 1 1) I G:AI .

Proof: Let V be a KG-module affording X. Since A is abelian, all a E A have a common eigenvector, say v E V. Put M = vIG. Then M is a lattice of V and {vg


g E G) generates M.

By 12.3, we may choose suitable vg which are an I-basis of M. With respect to this basis A acts via diagonal matrices on M. Thus there exists y E EndIA(M) with tr(y) = 1. Now by 12.4, we obtain v(x(1)) = v (rg M)


G v(tr TA(y)) = V( G:A ) + v(tr y) = v( 1 G:A )





Since this holds for all primes p we obtain ~ ( 1 ) G:AI


12.8 Theorem. Let V he a simple KG-module with character X. Then the following conditions are equivalent a) I G l p




b ) ~ ( g ) = O forall g E G with plo(g). c) ~ ( g =) 0 for all non-trivial p-elements g E G


Proof: a) =, b): Let M be a lattice of V and let g E G with pI o(g). By 12.5 M is a projective 1G-module. Hence M is a projective I(g)-module. To prove b) we may assume that M is an indecomposable I(g)-module, hence by the Krull-Schmidt theorem MI I(g). Write g = xy = yx with o(x) = pa and (o(y),p) = 1. Since K is a splitting field for all subgroups of G and p Ifo(y) O(Y> with primitive orthogonal idempotents e . . we have I ( y ) = o eiI 1 i =l Thus

Note that e i l ( x ) is indecomposable, since



ei i ( x ) ( x )

It follows that M s eiI(x) b)

= K ( x ) is indecomposable .

for some i, which yields the assertion.

c): This is trivial.

c) =+a): Let P be a Sylow p-subgroup of G. Then

12.9 Remark (Knorr): Each of the conditions in -

12.9 is equivalent to

~ ( g =) 0 for all g E G with o(g) = p. The proof depends on Brauer's first main theorem which we d o not prove here. 12.10 Definition. Let B be a block of KG, i.e. a t-sided indecomposable direct summand of KG (see 2.3). B is called of defect zero if B = (K)n as a R-algebra. If we regard B as a KG-right module. then BsnM with a simple projective KG-module .\I and n = d m R X I


Moreover M does not occur as a compxirion factor In a block B ' + B. Remember that by 8.1 1 b) the Cartan-matrix o i RG decomposes into a block diagonal matrix, i.e.

where CB belongs to the block Bi. I

12.1 1 Theorem. Equivalent are: a) Bi is a block of defect 0.

Proof: a) + b):. This follows immediately from the definition Bi being of defect 0. b) a a): Now Bi contains only one simple K G - m d u l e , say Mi. Furthermore Mi is a

projective KG-module. By 2.11, Bi This proves Bi =


(dim Mi)Mi as a KG-right module.

M. as a K-algebra. 1

12.12 Theorem. Let V1,

.. ,Vh denote the simple KG-modules. Suppose pa = I GI p.

a) Let Wi be a simple KG-module belonging to a block Bi of defect 0. There exists exactly one Vr with dri z 0. In this case dri = 1. If Mr is an IG-lattice in VP then Mr = Wi. Also


dim Vr


b) Suppose pa 1 dim Vr. If Mr is a lattice of Vr, then Mr is a simple KG-module belonging to a block Br of defect 0 and dtr = 0 for t z r


proOf: a) By 1 2 1 1 , C B i = (1). Then 11.10 yields

hence there is exactly one r with dri z 0 and for this r even d rl. = 1 . Moreover, if i


j, then o = c..=Cd . d sl sj - drj 'J s

This implies Mr



W i by the definition of the decomposition numbers.

Since Wi is projective, Dickson's theorem (see 3.5) implies IGI



dimKWi=dim V K r'

b) 12.5 yields that Mr is simple and projective. By 2.11, there exists a block Br which contains only Mr. Thus CBr = (1) and Br is of defect zero by 12.1 1. Furthermore





2 dtr

forces dtr = 0 for t z r.

t= 1

12.13 Proposition. Let S be a Sylow p-subgroup of G. If KG has a p-block of defect zero, then S fl

sg= 1 for a suitable g € G. In particular, 0P(G) = 1.


Proof: Let M be a simple KG-module belonging to a p-block of defect zero. Thus MI

e R S for some e € l. As

om^^((^^)^, M) = ~


o m ~ ( 9Q

M is a factor module of the permutation module


e H O ~ ~ ~ ( K ~I , 0,R S )


hence a direct summand since M is

projective. Thus we may write (IQG = M o N

with a R G m a f u l e


Applying Mackey ' s formula, we obtain

= M1 asiS. ~ e . . .



where the sum runs over suitable g E G. As KS is indecomposable and dim

(R sg

yields a gE G with



S : s!:

- S I , the Krull-Schmidt theorem

sgn S = 1.

12.14 Remark. To prove the existence (rtsp. r*.w-crstencu)of a block of defect 0 for a given -

group G is in most cases a rather difficult task

T k reason for this is the fact that some

powerful theorems (for instance, Brauer's fim m r n *orern)

do not work for blocks of defect

zero. 12.15 Remark. Let G be a finite simple groupa) If G is sporadic, then G has p-blocks of &ia-I u u long as p (One easily finds

x E Irrg(G)



x : :sb?



~nspectionof the Atlas.)

b) (Michler, Willems). If G is a group of Lie-r?pc. rhrn G has p-blocks of defect 0 for all primes p. Here the proof depends on DeIigne-Lusaig s character theory. c) It is still an open question whether the a l t e m t l n g g w p s An(n r 5) have always p-blocks of defect 0 for p


5. Using the theory of modular f~~

asymptotically, which means: Given a prime p p-blocks of defect 0 for all n all n





KJyachko proved the existence

5 r k n there exists no(p) E DI such that An has

n (p). (For p = I a d p = 7 . the answer is affirmative for 0


12.16 Example. We determine the simple KG-modules for G = SL(2,2 ) and Char R = 2. a) If E


f A E G, then A is not a scalar multiple of E, since a2 = 1 implies a = 1 for a E GF(2 ).

Therefore there is a basis {v,vA) of the natural G-module V, and with respect to this basis the matrix corresponding to A is

f Hence A is conjugate in GL(2,2 ) to this matrix. As

so A is conjugate in ~ ~ ( 2) ,to2



It is clear by comparing traces that

f 2 are only conjugate for a = b. Hence G has 1 + 2 conjugacy classes. If a = 0, then A = E. b) Let V = V1, V2,

... , Vf be the Galois-conjugates of V, i.e. (a. ) E G acts as ~k

Vi. We put

If A =

[ :] , then the trace of A on W is

c) G has a doubly transitive representation on the 2 f GF(2 ). If again


+ 1 points of the projective line L over

then the number of fixed points of A on L is easily seen to be

a 2= O x +ax+l is irreducible [ 2 othenvi s e Now x = n - 1 E Irr G and ~ ( 1=) 2f . Let M be an IGlattice affording X. Then M is a simple

n(A) =


if if

KG-module by 12.12 b). By 12.8 we obtain X(A) = n(A)

- 1= 0

if o(A) is even. This happens only for a = 0 Hence Z~ is the number of 2'-classes of G. The trace functions of G on the module Ib' from b) and on the simple KG-module M are the same, namely

A? the trace-functions of simple RG-rnidules arc linearly independent by 7.4, we see that W is isomorphic to M, hence is simple. f d) The 2 modules

are simple and painvise nonisornorphic: As factors of the simple module U' the?- certainly are simple. Suppose

Comparison of dimensions shows s = Suppose il !$ {jl, ... , js)

is simple, for il, jl,


. Then

... , js are distinct. But



kri contains the KG-submodule of



symmetric tensors, hence V. o Vi o ... c V. cannot be simple. I '1 1 s Now c) and d) show that the V. r, ... s \'. with 1 s i < I I 1 1

simple KG-modules.


... < iS s; f a r e all the

12.17 Example. We determine the projective indecomposable KG-modules for f G = SL42,2 ) and Char K = 2.

... ,is }

... ,f)

we put VI = V. o ... o V. 1 '1 S (V - K is the trivial module). By 12.16, { VI I I C N } describes a full set of simple

I = { il,



N = (1,


KG-modules. a)




for all I :


Hence tr A = a I I

+ aZ2 = tr A-l. This proves Vi

simple modules. For a general I = { i


v I --( V .l1



0...@Vi) S



V: since the traces uniquely determine

,is ) C N we finally obtain * * v . o . . . o v . EV. o . . . o v . = v 1 I' '1 's '1 S

b) The composition factors of Vi o Vi are K K V i + l

where Vf+l = V1 :

Let trV,(g) denote the trace of g E G on V.. If trV(g) = a, then 1


Since the trace functions of simple KG-modules are linearly independent over fi we see that Vi+l appears in V. o Vi with odd multiplicity, every other simple KG-module with even 1

multiplicity. As dim Vi o Vi = 4, this shows that Vi o Vi has the composition factors

vi+pR, R. c)

VN 0 VN = P(VI) V



N o VN = VN o P(K) o ...

where P(VI) denotes the projective cover of V




First note that V N is projective by 12.16, hence VS e V, also by 4.4. As *

HomKG(VS. Vs Is VI) z HomKG(VN,VN) = K , a) P(VI) is a direct summand of VN o V N I by 3.2. So i t remains to show that the projective module VN is a composition factor of VN e Vs. But this follows from b) since Horn KG - (V N a VN I,V1)



... O Vf) @I (V1



... O Vf) P (V 1 8 \.,)

has a section isomorphic to V1 o ... o Vf =



VN =

--- d V f @ Vf)


P(V,) = \'$ a \'\




I *2

\'?; s \'\ .

Using c) we obtain

I GI = 1(dim VI) (dim PI \.I






Hence equality holds ever\\\ hcrc .inJ J


4 I ZI



(f-111)) - 2f

Then P(K) = (KI Z)

, the t r ~ v ~ Z-rnodulr al KZ I \ prta.cir,\e Hence K- G Z G Horn- ( ( K J Z ) , k) a Hornu KG




e) Let Z be the c y c l ~ csubgroup ~ ) Gt o r tv&r 2 Since 2



P(K) is a direct summand of (K 1 Z)



15 projective,

by 4.5.

K;;IZ ) -= K '


f f By d), dim P(K) = 2 (2 -1) . Thus P t K ) =




Exercises 37) Use 12.17 b) and d) to compute the Camn rn'l!n.r of .A5 = SL(2,4) for p = 2. 38) Compute the Cartan-matrix of SL(3.8) for p = 2 39) Let p


3 and q be primes with p ( q-1. Thrn 'p

r i q P ) of order yP(qP- 1) of the fomm

qP- I. k t G be the normal subgroup of

a) There exist Sylow-p-subgroups S and S of G with S1 f' S2 = E. 1 2 p- 1 , hence none of them is divisible by p2. b) The character-degrees of G are 1, p, P Remark. a) By taking a direct product of k copies of the group G in exercise 39), we obtain a group H with the following properties: 1) p

2k- I GI, but the maximal p-power in character-degrees of G is p k .


2) H has Sy16w-p-subgroups TI, T2 with T1 II T2 = E. b) Espuelas and Navarro have shown: Let G be of odd order

with p

b a ) b2 . 0 (G) = E. Then there exists x E Irr G with p I ~ ( 1 and P with p m, there exists c) If G is solvable, 0 (G) = E and (GI = P b p I ~ ( 1 and ) b a (T. WolQ.


(Conjecture: b) is true for every solvable G.)

4 m and

x E Irr G with

$13 Examples

13.1 Lemma. a) Let A be a matrix of finite order rn in the matrix-ring (I), --

over a p-adic ring I.


A = E (mod n I), then m is a power of p = Char b) Let



be a primitive m-th root of unity in the p a d i c field K, where Char R

Then E =


+ n I is a primitive m-th root of unity




c) Let (K,I,K) be a p-modular system for G. Icr V be a KG-module and M an IG-lattice in V. If the trivial module is the only KG-compcaition-factor of M, then G K e r V is a p-group. Proof a) Suppose q

p is a prime dividing r n \Vc put B = A ~ ' ~Then . Bq = E and


B=E+nSc with s > 0 and C E (I),, not all coefficients rn C dlvisible by n. Hence

This implies rr2S

1 q rrS C, which


not true ss q 5 I x fl H = Up.

b) This follows by application of a ) I 1. We obtain Certainly w = dim K 5

(mod 3 ) , (mod 3)


As there exists a 3'-subgroup S of G of order 8, we know dimRP(W1)


(G:S I = 21.

From (3) follows either dS1 = 0, d61 = 1 or d51 = Z d61 = 0. But in the second case (1) and

(2) imply 5 = dS5 w and 8 = d 65 w . contradicting w > 1. Hence d 51 = 0, d61 = 1 and by = d65 = 1.

(1) and (2) w = 7,


Hence the missing simple KG-module is W5 = Finally,

8 = ~ ~ $=11 )+ 7 d65 shows d65 = 1. Hence

Also now all simple KG-modules are red&-ti.urs of IG-lattices. (Observe det C = 3.) Char


= 7: As G


PSL(2,7), we obtaln fnrm 5.10 all the simple KG-modules

W1, W2' W3, W of dimensions 1,3,5.7. 4

s e ~mmcdiately


3 ~ 3 WZs (by 13.1,





(by 12.5).

(Observe that we do not yet know the Brauer-cParactzr of W3 !) We have

As G has a 7-complement (isomorphic ro Sir. uc ha\c by 1.5 C) 7 = dim-P(W ) = 1



+ 6 dll


This implies d41 = 1, d61 = 0. Then 6

= ( 1~) = 1 + 3 d 4 2 + 5 d J j 4




Finally 8=

x6(1) = 3 d62 + 5 d63

implies d62 = d63 = 1 .

Hence 1 0 0 0 0 1 0 0 D=

1 0 1 0 0 0 0 1 0 1 1 0

2 0 1 0 1 1 2 0 0 0 0 1

Now the simple KC-module W3 is not the reduction of an IC-lattice. (Observe det C = 7.) 13.9 Example. Let G = SL(2,S). The character-table is given by P


allZ =

The characters


and bli2 =

x 1 up to x5 come from G/(g2)

theory A, p. 235.)

= A5. (See L. Dornhoff, Group representation


K = 2: As 02(G) = (g2) is by 3.6 in the kernel of every simpf KG-module, we know

all simple KG-modules from 13.6, and we even know the first 5 rows of the decomposition-matrix D. We have simple KC-modules W.(i = 1, ... , 4) of dimensions 1,4,2,2. I

It is clear that M are simple of dimension 2- As 617

we obtain

= W4 (possibie exchanging x6 and x7).

= W3 ,

The irreducible Brauer-characters of C are tiem x,.

x4, x6, x7. For 2'-elements g we have

This shows

(Observe det C = 26 ; see exercise 12.) Char K = 3: Now on



(i = 6, ... ,9) g, is rtpccscoted by -E, hence cannot contain any

composition-factor Wi ( i = 1,2,3,4) of




we obtain by 13.1 two further simple KG-mcdulcs

% and

of dimension 2. By 12.5

simple of dimension 6. Hence we have found all simple RGrnodules. From

for 3'-elements we obtain the missing decomposition numbers dq. Hence

3 is

and C =

(Oserve det C = 3L.)


Char K = 5: By 5.10, the dimensions of the simple KG-modules Wi are 1,3,5,2,4, where W1, W2' W3 are modules for


As. Obviously


is the simple module W4

Separating the modules with g2 in the kernel and using knowledge of As we obtain



If dg4 = 3, then x9(g) = 3 @g) on 5'-elements g. As this is not true on gq, we obtain dg4 = dg5 = 1. From dimK P(W ) - 4 dg5 + 6 = 0 (mod 5) 5 we obtain dg5 = 1 , dg4 = 0.


-2 1 0 1 3 0 0 0 1

2 . (Observe det C = 5 .) 3 1 1 2

For every characteristic 2,3,5 now e v e n simple KG-module is the reduction of some IG-lattice. 13.10 Remarks. a) The determinant of the Cartan-matrix G 1s always a power of p, hence is never 0. A consequence of det C

* 0 is that tuo project~vrmodules with the same composition-factors

are isomorphic. (See 15.15.) b) For the solvable groups S4 and ~(1'1 u r s 3 u in 13.5 and 13.7 that for every characteristic p all the simple KG-modules are obtained by reduction of some lattices. We remarked already that this is true for all p-sol\dblc g r o u p

1 9 1-11. This property does not hold for As with p = 2

(13.6) and PSL(2,7) with p = 7 I 13 8 I. But SL(2,S) (13.9). Call a character


dtws hold for all primes in the non-solvable group

ot G pratronal. if there exists a number n, prime to p, such

) in the cyclotc~-niciicld Q . Then the following holds: If for every that all values ~ ( g are n characteristic p every KG-module can be I~ticrlt o an IG-module with p-rational character, then G is solvable. ( G. HiR, J. of Algebra 9-1 I 19R5). 1 1 1 (1987).)

(The values a


and b


in 13.9 are n~jt5-nt~onal!)

Exercises. The proof of 13.1 works also for matrices over I , even better: 40) (Minkowski) Let G be a finite group of m;ttrices of type (n,n) over U. We define

a E Hom(G, GL(n,p)) by

(a..) a = (a.. + p I ) J' 1J for (a,.) E G. 1J

a) Ker a = E if p > 2. b) If p = 2, then Ker a is an elementary abelian 2-group. (If g E Ker a and o(g) = 4, then we obtain an equation i = 1 + 2b, where i is a primitive 4-th root of unity and b an algebraic integer.) 41) Using the character-table of Aq, compute the decomposition- and Cartan-matrices of A4 for Char K = 2,3. 42) Using the character-table of SL(2,3) (Trento 1987, p.52), compute the decomposition- and Cartan-matrices of SL(2,3) for Char K = 2,3. (There are simple KG-modules of dimensions 1,1,1 for Char K = 2; simple KG-modules of dimensions 1,2,3 for Char K = 3.) 43) Using the character-table of GL(2,3) (Trento 1937, p.56), compute decomposition- and Cartan-matrices of GL(2,3) for Char K = 2 and Char K = 3.) (Char

R = 2: Simple KG-modules of dimensions 1,2; det

5 C=2 ;

2 Char R = 3: Simple KG-modules of dimensions 1,1,2,2,3,3; det C = 3 .) 44) Let p be a prime. We consider the group G = T ( ~ of~ all ) semilinear mappings i x + axP


over L = G F ( ~ ~Let ) . (K,I,K) be a p-modular system for G. Then G has the normal subgroups

T = { (x:b)


and X S = { (ax+b) I a, b E L, a

a) Show that T = 0 (G), 0 ,(G) = E and






hence I GIG'

I = p(p-1).

b) The character-group of GIG' is of the form (a)x

( p ) , where o(a) = p

Obviously, KG has p-1 simple modules of dimension 1 with trace functions

and o(p) = p-1. 1

(i = 0,

... , p-2).

c) G/T has t = pP-l-l simple modules of dimension p over K. Their reductions mod p are simple and distinct. d) C has over K exactly p simple modules W I . ... , W of dimension pP-l. All simple P KG-modules have been listed. e) Let M. be an IG-lattice in W.. Then T contains every simple KG-module with multiJ J J plicity 1.

Q The decomposition matrix of G is



[: 1 :

is of type (p,l). Therefore

Calculation shows det C = p2p-1 . 45) Let K be a field of characteristic p > 0 and let N be a normal p-subgroup of G. Put J = J ( m ) and let G act by conjugation on the powers J'. Furthermore for a simple KCIN-module V let P resp. P denote its projective cover as a KG- resp. KG/N-module. Prove a) P = P/PJ.


b) P/PJ a J ~ / J ~ +G'


c) dimKP = N I dimKP d) If N


Z(G), then CG = IN CGIN , where C


and GIN.

. G resp. CGIN are the Cartan-matrices w.r.t. C

$14 The Theorem of Fong and Swan.

The purpose of this section is the proof of the following important theorem: 14.1 Theorem (Fong, Swan). Let G be a psolvable group and (K,I,K) a p-modular system for G with algebraically closed K. Let V be a simple KG-module. Then there exists a simple KG-module V0 such that for every IG-lattice M of Vo we have

fi = V.

An immediate consequence is

14.2 Corollary. Let G be p-solvable and R a splitting-field for G of characteristic p. If V is a simple KG-module, then dimR V divides I G I . (We remind the reader that the simple group PSL(2,7) has in characteristic 7 a simple module of dimension 5; see 5.11.) We show at first that a p-modular system with algebraically closed 14.3 Lemma. There exists for G a p m t d u l a r system (&I$) R is algebraically closed. Proof. We cannot take K itself -


K always exists.

with the additional property that

bc algebraically closed, for then the valuation on K could not

be discrete. Hence we have to be a little bit more careful. Let Q be the (complete) rational padic field and let E~ be a primitive root of unity in P the algebraic closure of Q . We put P E~ 1 all n with p n) . KO = Qp("


As K contains a primitive I GI -th root of unity, it is a splitting field for all subgroups of G. 0 Adjoining an n-th root of unity with p n defines a non-ramified extension of Q P' ). Let K be the completion of KO. Then the Hence KO is a non-ramified extension of Q ( E P IGI of valuation on K is still discrete. The residue-field R of K contains all primitive n-th roots


unity for p &, n (see 13.1 b)), hence K is the algebraic closure of GF(p). (For details about p-adic fields see again J.P. Serre, Corps Locaux.)

14.4 Lemma. Suppose N 4 G. Let L be a splitting-field for G and N. Let V be a simple LG-module such that

with a simple LN-module W. Suppose that there exists an LG-module U such that UN Then there exists a simple LGIN-module X such that V



= U o X.

Proof. By Nakayama-reciprocity (3.8) we have 0


G HornLN (V ,W)=HomLG(V,W ) .

By 6.10 we have

wG = (UN )G = U oLLG/N. If X is a simple LG/N-module, then U oLX is simple by 6.9 a). As V is isomorphic to a composition factor of U o LG/N, we obtain V L

= U oLX for some simple LG/N-module X.

Remark: In general, the representation of N on W will not allow an extension to an ordinary representation of G, but it always has an extension which is a projective representation PI of G. If D denotes the representation of G on V, we obtain D(g) = P (g) o P (g), where P2 is a 1 2 projective representation of G/N. This is a generalization of 14.4. (See Huppert I, p. 567.) 14.5 Proof of theorem 14.1 (by J.P. Serre). Let (G,V) be a counter-example with smallest possible dimR V. We can assume that G operates faithfully on V, hence by 3.6 b) 0 (G) = E. P We put N = 0 ,(G). As G is p-solvable, so N > E. P (1) VN is a homogeneous m - m o d u l e : Suppose VN = HI O

... O Hs

with homogeneous components H and s > 1. If T is the stabilizer of HI, then by 6.2 V z HG I , j where H is considered as a T-module, necessarily simple (as V is simple). As 1 dim - H < dim - V, there exists an I T-lattice M1 such that K 1 K


HI. Then

is an IG-lattice in M oIK and obviously

This is impossible as (G,V) is a counter-example. Hence

v N =- W e ... s w with a simple m - m o d u l e W.

(2) dimR W > 1 : Suppose dimR W = 1. As G is faithful on V, so N is represented on V by matrices of type aE, hence N 0


Z(G). But then ( G ) = N x 0 (G) = S = O , ( G ) .




As G is p-solvable, this implies G = 0 .(G).Then by 11.13 we do not have a P counter-example. Hence dimR W > 1. (3) There exists a projecti\r I S - r n ~ d u l eW0 such that WO -? W. In addition p

Hornl N(WO'WO) = 1

As p

4rankIWO and


4 I N I , by 11.13 therc c x ~ s t san I S-lattice Wo with


= W and Wo 4 K is a

simple KN-module. By 11.10 dl. \VO is a pojective I N-module. Also by 11.13 rankI Wo = dimR \\' divides IS I , hence p

4rankI Wo .

As in the proof of 12.1 we see that u a = u-a for a E Hom I N (W0,W0) and w E Wg, where a E I. (4) Let p denote the representation of S on WO. For every g E G there exists an

I-automorphism cx of \YO with drt a = 1 and g I: - 1 p(x)cxg ior a11 x E s : (*) p(xg) = cxg W o ... o W with a simple Rh'-module W, we have W As V N = -

= wg for all g E G.

is an epimorphism of H onto G with kernel

So a = a E and a n I N l = 1, where p

4n. Therefore S is a pl-group and H is still p-solvable. (It

is crucial for our proof that this increase of I GI is allowed, induction is by dimR V !) ( 6 ) For x,y E N we have

p(y1-l P(X) P(Y) = p(xY) , hence p(y) E ?Ly as p(y) 1


= 1. Therefore r with

x t = (x,p( 4 ) is a monomorphism of N into H. N t is normal in H for (xt)(ga) = (x,p(x))(g'a) = (xg, a-'p(x)a) = (xg, p(xg)) = (xg1-c We define a homomorphism


of H into AutI Wo by

(g,a)n2 = a:


Then Wo becomes an I H-module by u(g,a) = u((g,a)n2) = u a for u E WO . For (x,p(x)) E NT we have U(X,P(~)) = u p(x) . Hence

(w-) -W. 0 Nt-

The KG-module V becomes a m - m o d u l e by v(g7a>= v((g,a)nl) = vg For x E N we have v(xt) = v(x,p(x)) =



Hence by (1) VNt By 14.4 we obtain V




... o W .

%o 17; Y with a KH-module Y. As by (2)

dim - V = dim K K



R Y > dimR Y ,

the pair (H,Y) is not a counter-example. As H is p-solvable, there exists an I-free I H-module Yo with

5 e Y. Then

The p'-group S = Ker n operates trivially on V. Hence by 13.1 it also operates trivially on 1 the I H-module Vo = Wo


I Yo. Therefore only G = HIS operates on Vo and (G,V) is not a

counter-example. 14.6 Remark. Isaacs gave the following improvement of 14.1: The IG-lattice M can even been chosen with p-rational character. If p > 2, then this character is uniquely determined. (Isaacs, Pac. J. Math. 53 (1974), 171-188.) 14.7 Theorem. Suppose G is p-solvable and W a simple KG-module such that


dimRW. Then W is a projective KG-module. (For solvable groups this is 9.7.)

Proof. By 14.1 we can lift W to an IG-module Wo. As I GI I P

nos W are projective.

dimKWo, so by 12.5 Wo and

$ 15 More on Brauer-Characters and the Cartan-Matrix

As a motivation for the results to come, we begin with a very special case.

15.1 Example. Let G have a normal Sylow-p-subgroup P. By the Schur-Zassenhaus theorem (Huppert I, p.126), then G = PH is a semidirect product. If "inflate"



to a 2- E Irr G with P



Irr H, we can

ker 2- by ?(y h) = 4 h ) for y E P, h E H

or induce







By 3.6 b), P = 0 (G) is contained in the kernel of every simple KG-module W, so W P is a simple m - m o d u l e . Since H is a p'-group, this provides by 11.13 a bijection between the simple KH-modules and the simple KG-modules; namely an I H-lattice M in the simple



KH-module V goes to Fi = M. If W is a simple m - m o d u l e , then it is projective as p

I HI, therefore wb

is a

projective KG-module by 4.5. If T is a simple m - m o d u l e , then HornKG (W Therefore t.E




,?) = Hornm

(W,T) =


is the indecomposable projective KG-module with head

?. If xi E Irr G


Irr H, then by 11.12 a) the decomposition number is

Hence we can calculate the decomposition-matrix

D and the Cartan-matrix

character table of G. In fact,

If h E H and u E P, then hU E H if and only if u E Cp(h). Therefore CG(h) = Cp(h) CH(h) and h


f' H = h


. Hence

C from. the

G ICG(h)l (h) =


T x€hGn


G This shows ( t )H = n t , where n is the permutation character of H on the H-set P.

Now let h. (i = 1,...,k = h(H)) be representatives for the conjugacy class of H. For 1

each i j we put

The orthogonality relations for H show that U = (u..) is unitary. We also have 'J

If we define the diagonal matrix Co by Co = (hii I Cp(hi) I), then this means

Hence k det C = det Co = Il Cp(hi) i= 1



is a power of p and IPI s d e t C s [ P I h(H) .


Furthermore, det C = I PI if and only if C (h.)] = 1 P '


for all hi


fixed-point-freely on P. While det C = P I h(H) if and only if G = P

1, hence if H operates x


The question naturally arises which of these assertions remain true in general, i.e. if the Sylow-p-subgroup is not necessarily normal.

We start by showing that C is an invertible matrix. For that purpose let si (i = 1,..., k)


be representatives for the pl-conjugacy classes of G and let ri = I CG(si) 'I2. As before, we denote the irreducible Brauer character of the simple KG-module Wi by


Pi (i = 1,..., k).


be the character of the projective I G-lattice Ui, uniquely determined by the fact that Oi

has the head Wi . We recall that

by 11.12 a). Let Z be the (k,k)-matrix over K with entries z... '= 'J

15.2 Lemma. --

B.(s.) r. J

2' C Z = E.

Proof. By 11.12 b) and the ordinary orthogonality relations we have k (2' c Z).. = 1 z . c z . ml mn nj



This is the assertion.

15.3 Notation. Let S denote the set of all p'-elements of G. For class functions --

a,/3 from

S to K, we define

This is an inner product on the K-vectorspace of class functions on S. In particular, we form yij = (/3i,/3j)' and

T = (y..). 'J

15.4 Proposition.


= Z Z ' = C-I

is the inverse of C. In particular, det C is a positive

integer. Proof. From 15.2 follows -

c = ( 2 ' ) - 1 z-1 = ( z z ' ) - l , hence the second equation. Also

= (p. P.)' = yij. 1' J Hence C" = r. As C is a regular matrix over


H, its determinant is a non-zero integer. Since

r = det Z det is positive, so is det C.


Corollary. a) Let




be projective KG-modules which have the same

composition factors, including multiplicities. Then P = P ' . In particular, the Brauer-characters determine the isomorphism-type of projective RG-modules. b) (R. Swan). Let M and M ' be projective IG-modules. Then M


M ' if and only if

M o K E M f @ K. 1 I Proof. a) Suppose k P = o x. P. and P ' = o y. P. I I i=l i=l I 1

are the decompositions of P and P '

into indecomposable projective KG-modules Pi. The

multiplicity of W. in P resp. P' is then J k k .& x . c . . = 1 yI. c .~j (j=l, ...,k). i=l 'J i=l


As C is invertible, this implies x. = y. (i=l,...,k), hence P 1

b) The Brauer-character

M o I K






of M is the character of M o I K, restricted to p'-elements. Hence if

I K, then the projective KG-modules M and M ' have the same

Brauer-character. By a) this implies M


M ' , hence M

M ' by 11.10 a).


Corollary. a) ( $ i , P . ) ' = 6 . . . J 1J b) {(Ii I i=l,..,k) and {Pi i=l, ...,k) are K-bases for the space of class functions from S to K.


Proof. - a) By 11.12 we have h



b) Suppose (on S). Then by a)



(.Ia. (I., 6.)' = a. 1=1






Hence the (Ii are linearly independent. The same argument works with the role of the (Ii and


interchanged. Since there are exactly k p'-classes in G, the assertion follows.

(The linear independence of the


has already been proved in 13.4 a).)

15.7 Remark. a) To obtain deeper results, we need to know that certain "natural" class functions on G are generalized characters, i.e. H-linear combinations of the X.1 (i=l ,...,h). Brauer's theorem gives a very powerful criterion for this. The essence of it is that a class function a on G with values in K is a generalized character of G, if (and only if) for every nilpotent subgroup H of G the restriction aHof a to H is a generalized character of H. (In fact, not all nilpotent subgroups of G are needed, but only those, which have at most one non-cyclic Sylow subgroup; these groups are called elementary.) For two of the many proofs we refer to Huppert I. p. 586-591 or Isaacs, p. 126-131. b) Let n be a set of primes and n ' the complement of n in the set of all primes. Given g E G, there is a unique factorisation g = g n g n ' =g,t where g n is a n-element and g n' p'-part s are



a n'-element. All the elements of G with a fixed




Here, by definition, G is the set of all p-elements in G. (One should not confuse 1 GI and P P Gp 1 ; 1 G 1 is the order of every Sy low-p-subgroup of G, hence is a power of p, while in


general ( G ( is not even a power of p. A relation between (GI and ( G ( is given in P P P 15.10.)

15.8 -

Definition. Let cx be a class function from S to K. We define class functions ai

(i=1,2,3) from G to K by

alid =


IGlpcx(g) a2(6) =


= 1 P otherwise;

if g

and a3(g) =

I CG(g)pl "(g) 0



P otherwise.

Obviously, the a. are class functions on G. 1

Except for 15.10, we will use only al and a2.

15.9 Proposition. Let cx be a class function from S to K and suppose that aH is a generalized character of


for every pf-subgroup H


G. Then

ai (i=1,2,3) is a

generalized character of G. Proof. -

a) For n l and rr


we use Brauer's theorem from 15.7 a). Let F = H x P be a

nilpotent subgroup of G, where H is a p'-group and P a p-group. Then


are obviously generalized characters of F.

b) We consider a3. If

x E Irr G, then -





II CG(s)p l a(s) ji(~)




u s )=




x,(us) = ( a l '

sES uEC (s) G P

-FrsES uECG(s)P

stands for ( x ~ ) AS ~ . by a) al and





are generalized characters of G, s o E I.

(a,, x)G = (a1. Therefore also a3 is a generalized character of G.

15.10 Remark. Before applying 15.9 to the study of the Cartan- and decomposition-matrix, we note two curious consequences: Let a = l G be the trivial character of G and P E Syl G. Then P

Hence / P I = IGI

divides IG P P Similarly, with the same a = 1 G

/ GI




I GP , I .


(It can be proved more generally that

15.1 1 Lemma. a) Let -



I Gn(

/j be a Brauer-character of

for any set n of primes.)

G. Then


is a character for every

p ' -subgroup H of G.

p is a Brauer-character of G, there exists a generalized character q of G such that qS = 13. (If G is p-solvable, by $14 7 can be choosen as an ordinary character of G .)

b) If

Proof. a) Let fi be the Brauer-character of the KG-module -

W. Then


is the

Brauer-character of WH. As H is a p'-subgroup, by 11.13 there is an I H-lattice M such that M

WH. By 13.4 c),

; I


is the character of M.

b) By a),


is even a character on every p'-subgroup H of G. Hence by 15.9 q = PI is

a generalized character of G with q(g) = P(g) for every g E S .

15.12 Remark. The "decomposition map" sends every irreducible character --


of G onto


d.. p. . ~ i l ~ = ~ j: ~ We extend it by U-linearity to a map from the generalized characters to the generalized Brauer-characters. Then 15.11 b) is just the statement that this map is surjective.

15.13 Proposition. Let g~ be a generalized character of G which vanishes on all p-singular classes. Then


is a U-linear combination of the $i. And conversely, any $i vanishes on all

p-singular classes. Proof. By 15.6 b) we have k I# =



a. $. (on S) J


for suitable a. E K. By 15.6 a) we obtain J a.J = (v7 Pj)' = (?, (Pj)1IG (since


0 on G S ) .

Hence a. E lZ as I/) (by assumption) and ( p ) (by 15.9 and 15.11) are generalized characters. J j1 In the proof of 12.8 we showed that an indecomposable projective I(g)-module is of l



a the form P = eI(g ), where e is an idempotent with e g E eI. Hence if o(g ) = p > 1, P P' P then

Therefore the matrix for g on P is of the form

and hence trace g = 0 . 15.14 Lemma. -b)

I Gp I 1


a) Suppose P E Syl G. Then Z(P) is a normal Sylow-p-subgroup of CG(P). P

0 (mod PI.

Proof. a) Clearly Z(P) = P

n cG(p) 5 c ~ ( P ) .

Let u be a p-element in CG(P). Then (P,u) is a p-group, hence u E P fl CG(P) = Z(P). the lengths of all orbits are powers of p. The fixed points P' ' are the p'-elements of CG(P). But by a) CG(P) = Z(P) x H with a p'-group H. Hence

b) P acts by conjugation on G

I CG(P)P , ( = I H I

is the number of fixed points of P on G





. Therefore

= ( H I f O (modp).

15.15 Theorem. det C is a power of p and I G / p s d e t C a \GI

k P


Proof. a) By 15.9 and 15.11, for every irreducible Brauer-character -


(Pi)2 are generalized characters of G. Therefore

1 G )Pr

is a matrix over I . Therefore by 15.4 1 k k det(lG1 r ) =[GI d e t r = IGI (detC)P P P k is an integer. This shows that det C is a power of p and det C a GI .

lies in U. Hence


b) If

is the trivial Brauer-character, then


of G, (pi)l and

As p

4 I GP , I

by 15.14 b), so

1 GI

appears in the denominator of yll after cancellation.

But l- = c-' by 15.4. By a well known result from linear algebra, (det C) c-' = adj C is a matrix over

U. Hence in particular det C - yll E 71. This shows

I GI P 5 det C.