157 92 3MB
English Pages 189 Year 1993
Table of contents :
Title
Contents
Foreword
Introduction
§1. Semisimple modules and algebras
§2. Decompositions of Algebras
§3. Groupalgebras
§4. Projective KGmodules
§5. The number of simple and indecomposable KG modules
§6. Clifford Theory
§7. Brauer's permutation lemma
§8. Symmetric Algebras
§9. Modules of solvable, psolvable and pconstrained groups
§10. padic 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 BrauerCharacters and the CartanMatrix
CIRM CENTRO INTERNAZIONALE PER LA RICERCA MATEMATICA
REPRESENTATION THEORY IN ARBITRARY CHARACTERISTIC by R. Gow, B. Huppert, R. Knorr, 0 . Manz, W. Wi llems
CASA FDlTRlCE DOTT. ANTONIO MILAN1
1993
CONTENTS
Foreword Introduction Semisimple modules and algebras Decomposition of algebras Groupalgebras Projective KGmodules The number of simple and indecomposable KGmodules Cliffordtheory Brauer's permutation lemma Symmetric algebras Modules of solvable (psolvable and pconstrained) groups padic numbers Reduction mod p and decomposition numbers Blocks of defect zero Examples The theorem of Fong and Swan
More on Brauercharacters and the Cartanmatrix
FOREWORD
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 Representationtheory of a finite group G over a field K of characteristic 0 (often assumed to be algebraically closed) is usually called "classical representationtheory". 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 representationtheory". It is distinguished from the classical theory by the fact, that the groupalgebra 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 KGmodules is in most cases out of question  there are too many of them  one concentrates on classes of KGmodules which contain only a finite number of isomorphismtypes, 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 JordanHolder about composition series of modules of finite length. (2) The KrullSchmidttheorem about unique direct decomposition of modules in
indecomposable ones (NagaoTsushima, p. 27). ( 3 ) Schur's lemma: If V is a simple Rmodule, then HomR(V,V) is a divisionalgebra. 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 rightmodule 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 groupalgebras KG. I n particular, we prove some results about the Jacobsonradical of
groupalgebras. 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 groupalgebras. In
5 we determine the number of simple modules
and show that there are infinitely many indecomposable KGmodules if char K = p and the Sylowpsubgroups 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 KGmodules from KNmodules. § 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 psolvable groups, which do not hold for all groups.
The sections 10 to 1 5 are devoted to the fundamental connection between representationtheory 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 padic numbers and not with algebraic numbers. This guarantees also the vital theorem of KrullSchmidt (see 11.7). The connection between the Cartanmatrix and the decompositionmatrix (see 11.12) is a powerful tool to study the idealstructure of the group ring. In
3
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
psolvable groups and prove the important liftingtheorem of FongSwan. 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
1.1
General Assumvtions.
All rings R considered here are algebras of finite dimension over a field K. All Rmodules are finitely generated Rmodules, hence also are of finite Kdimension.
1.2
Definition. a) An Rmodule M
#
0 is called simple (irreducible), if 0 and M are the only
Rsubmodules of M. b) An Rmodule 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 rightRmodule, is semisimple. 1.3
Definition. We define the Jacobsonradical J(R) of R by
J(R) = {r 1 r E R, Mr = 0 for every simple Rmodule M)
.
Obviously J(R) is an ideal (twosided) in R. 1.4
Theorem. a) J(R) is nilpotent.
b) J(R) contains all nilpotent rightideals of R. c) J(R) is the intersection of all maximal rightideals of R. Proof: a) As dimK R < w , there exists a composition series R = M >MI> 0
... > M k = O
of R as a rightRmodule. As Mi/Mi+l is simple, s o (Mi/Mi+l) J(R) = 0 . This shows MiJ(R)
C
Mi+l
. Hence
b) Let r be a nilpotent rightideal and M a simple Rmodule. Then M r is a submodule of M, hence is 0 or M. But M r = M implies the contradiction M=Mr
n
=O.
Hence M r = 0 and therefore r C J(R). c) If r is a maximal rightideal of R, then (R/r)J(R) = 0 ,hence J(R) C r . Let A be the intersection of all maximal rightideals of R and M a simple Rmodule. Then M
= R/r for some
maximal rightideal 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 rightleftsymmetry concerning J(R). 1.6
Lemma. (Nakayama) Let M be an Rmodule 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 Rmodule, then every submodule of M is semisimple.
Proof: Let N '
5
N
5
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).
1.8
Theorem. Equivalent are:
a) M is a semisimple Rmodule.
k b) M = o Mi with simple Rmodules M.. 1 i=l
k c) M =
C
Mi with simple Rmodules M.. 1
i=1 Proof a)
b): As dim
M = M1 o M '
K
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=nn'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 Rmodule is semisimple. Proof. a) a b): By 1.4 c), the intersection of all maximal rightideals of R is 0. As dimK R
1
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 Rmodule. 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
i=l
n J(R) = o MiJ(R) = 0. i=l 1.11
Corollary. Equivalent are:
a) M is a semisimple Rmodule.
Proof. a) b)
b): is trivial.
a a): As MJ(R)
= 0, M becames an R/J(R)module by
m(r
+ 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 Rsubmodules 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 2sided ideals different from 0 and R. n b) R = o e R and e R is a simple Rmodule. 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)
E
I a..1J E K, a..1J = 0 if j 2 i) .
K o ... o K (as algebra and as module). How many isomorphismtypes of
simple Rmodules does R have? c) Determine the largest semisimple Rsubmodule of R. How many types of compositionfactors 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 rightRmodule.
k
4) Let Mi(i=l,
... ,k) be submodules of M with semisimple M / M i Then M/ fl Mi is i=l
semisimple.
52 2.1
Decompositions of Algebras
Theorem. a) Let R = PI
0
... 0 Pn
be a decomposition of R into rightideals 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
I
1
.
b) Suppose conversely, that
1= e
1
+ ... + 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
where
...+ e nR = P l o ... o P n ,
e.R C P. implies P. = eiR. 1
1
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 KrullSchmidt 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 2sided ideals B. which cannot be written as a direct sum of proper 2sided ideals. J
a) If
1= f
1
+ ... + 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
1
1
.
b) If R = I 1 o I 2 is a decomposition into 2sided 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 Rmodule, 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 rightideal 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 +...+
rfk
and f.r , r fi E Bi implies f.r = r fi , hence fi E Z(R). 1
1
Finally Bi = fiR = R fi follows as in 2.1 a). b) We have
c
with twosided 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 rightRmodule. From
k M = o M f .1 i=l
follows the conclusion. d) By c), there exists exactly one fi such that
2.4
Lemma. Let e and e ' be idempotents in R.
a) HomR(eR,e ' R) s e ' Re (as Kvectorspace)
.
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 divisionalgebra. (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
R
. The mapping
a , e n is obviously Klinear. 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 Rhomomorphism 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 divisionalgebra. 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
e.;.J
= 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 pnl
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 = 3pn1 (eo)  2pn1(eo)  eo = 3e0  2e:  eo = 0 (mod J(R)) and
2 J(RIzn = 0 , then p (e ) = p (e ) n o n o
If
If e a = ae 0
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,
..., eml
+ J(R) = 6..1 1 Let em = a + J(R) and put
and e.
m
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.
J
1J'
Then be. = e.b = 0 for j = 1, J J It follows from a + J(R) = b
..., m1 .
em
and
+ J(R)
=a
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
... en1 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 n1 = 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 Rmodule, hence e.J(R) is maximal in e.R. 1
1
2.6 Remark. In the study of integral representNions over rings of padic 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 padic integers.)
2.7 Definition. An Rmodule P is called projective, if the following property holds:
If M is any Rmodule 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

P.
Theorem. Let P be an Rmodule. Equivalent are:
a) P is projective. b) P is a direct summand of a free Rmodule F. (If P is finitely generated, then F can be chosen to be also finitely generated.) c) Suppose M and N are Rmodules. 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 =

P.
b): Let F be a free Rmodule and a an epimorphism of F onto P.
Then F = ker a o P ' , where P ' b) 3 c): Now suppose F = P
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
(i E I) .
Define y' E Horn (F,M) by fiyl = mi and let y be the restriction of y' to P.
R
For p E P we obtain
c ) : We apply c) for
Hence ya =
L
. 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 Rmodules 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
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 KruIlSchmidttheorem, 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
2
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 compositionfactor isomorphic to 6.R. 2.10
I
Proof: If V < U 5 e.R and U/V J Hence 0 z Uei C e.Rei .
c
e i R , then 0 z (U/V)ei = (Uei
+ V)/V.
J
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 compositionfactor 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
2.11
if there exists a chain '  e.R J
e.R

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
S
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 Rrightmodule, is a direct sum of n isomorphic simple modules, RVo
... o V s a y . n
b)R is antiisomorphic to the matrix ring (D) over the divisionring D = HomR(V,V), where n V is the only simple Rmodule.
Proof: a) As R is simple, this follows from 2.12. b) By 2.4 b), HomR(R,R) is antiisomorphic 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 algebraisomorphism 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 Rmodules 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 Rmodule R. By 2.13, we obtain that Bi is antiisomorphic to (D.) I
n.1 for some
skewfield Di. As K is algebraically closed, Di = K and Bi is isomorphic to (K),,, for (K),. is 1
I
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
b)
e..R = o K e.. 11 j=l 1J ' e..J(R) =
0
j=l
K e.. 'J
and K eil is the only simple submodule of eiiR. c)
R has only one block.
d)
Determine the compositionfactors of e..R.
e)
R has (up to isomorphism) n simple factormodules, but only one simple submodule.
I1
At first we have to decide which groupalgebras are semisimple. 3.1
m Lemma. Let U be a subgroup of G with cosetdecomposition G = U Uti and let V be a i =1
KGmodule. 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 KUmodule.) m
Prooff: We have G =
U tilu
. Suppose g E G
1 1 and gti = t i , ui with ui E U
.
i=l
Then
(1 .) is a permutation. Hence
3.2
Theorem. Let V be a KGmodule and W a KGsubmodule of V. Suppose U r G such
that ( 1 ) C h a r K ) c IG: UI
.
(2) VU = WU o W '
with a KUmodule W '
.
Then V = W o W" with some KGmodule 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 KGprojection and
n' = W 0 ker n1
with a KGmodule 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 KGsubmodule of KG has a complement. Hence KG is semisimple.
I
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 2sided ideal with i2 = K G i2 = O . By 1.4 b), therefore
J(KG) 2 i z 0
.
Next we consider an extreme case:
I
3.4 Theorem. Suppose G ( = pn and Char K = p.
a) Up to isomorphism there exists only one simple KGmodule, namely K with trivial action of G.
and dim J(KG) = IGI1.
c) The only simple KGsubmodule of KG is K
g. In particular, KG is an indecomposable gEG
d) Every projective KGmodule is free. Proof: a) Let V be a simple KGmodule and 0 z v E V. Let Ko = GF(p) be the primefield in K. Then G operates on vKoG. As vKoG is an epimorphic image of KoG (as a
I
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
;t
0 for all g E G and Kw is a KGmodule. 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 KrullSchmidttheorem every projective KGmodule is free. 3.5
Theorem (L.E. Dichon). Suppose Char K = p and I GI =
where p
4m. If V is a
projective KGmodule, then pa divides dimKV. Proof: Let P be a Sylowpsubgroup of G. If G = U tip, then i =1
is a free KPmodule. Hence every free KGmodule is a free Wmodule. Therefore by 2.8 every projective KGmodule is a projective KPmodule, hence is a free KPmodule by 3.4 d). This implies I PI
1 dimKV.
If in particular the trivial KGmodule K is a direct summand of KG, then by 3.5 pa divides dimKK = 1, hence p
4 I GI.
3.6
Remark. Let K be an algebraically closed field of characteristic p and let P be an
I
indecomposable projective KGmodule. 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 pcomplement, 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 KGmodule V we define ker V = {g
I
g E G, vg = v for all v E V ) .
Obviously ker V a G. a) If Char K = o, then
n
kerV=O.
V simple b) If Char K = p, then fl
V simple
ker V = 0 (G) P
is the largest normal psubgroup 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 KGmodule. 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 KGsubmodule 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
Hence
n ker V s 0 (G) V simple P
.
We now describe a way to construct KGmodules from modules for a subgroup H of G.
m 3.8
Definition. Suppose H
5
G and G = U H ti. Let W he a KHmodule. We form the i= 1
induced module
WG becomes a KGmodule by (w
ti)g = w o t.g = wh. o t . ,
Q
I
I
I
if t.g = h.I t i ' E H t i , . Hence application of g permutes the Ksubspaces W I
Q
t i in the same
way as the cosets H t. are permuted. I
3.9
Theorem (Nakayama's reciprocity theorem). Suppose H
5
G and let W be a KHmodule
and V a KGmodule. We have the following isomorphisms as Kspaces: a)
G HomKG(V,W )
HomKH(VH,W)
9
where VH is the restriction of V to H. b)
G HOmKG(W ,V)
5
H(3mKH(W,VH).
m
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
c
i=l
v a . o t.tyl = v a . o 1 + ... 1
J
I J
= (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 Klinear 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 =
1
i=l
G Hence a E HomKG(V,W ) and therefore a
1
(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
(W
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
1
3.10
I
dules of M isomorhic to V is by 1.9 a semisimple submodule
1
of M. If a E HomKG(V,M), then V a
1
/3 /31 is bijective. I
Remark. Let V be a simple KGmodule and M any KGmodule. The sum of all submo
S. Hence
k HomKG(V,M) = HomKG(V,S) =
@
HomKG(V,Vi).
i=l
i
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
L
,
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 KGmodule, 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
5
G, V is a simple KGmodule and W a simple KHmodule. 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 KGmodule 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 KGmodule, 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 KGmodule, 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)
S(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
3.13
Theorem. Suppose K is algebraically closed and KG semisimple. Then KG=
o
(dimKV)V.
V simple
In particular
IG I
H
( d i m K ~ ) L. V simple Proof: Now KG = S(KG). By 3.12, the multiplicity of any simple KGmodule

=
in KG
is dimKV. 3.14
Theorem (Clifford). Let V be a simple KGmodule and N a G. Then VN is a
semisimple KNmodule.
Proof: Let W be a simple KNsubmodule of VN. Then also Wg for any g E G is a simple KNmodule, for if x E N, then 1 W g x = W gxg g = W g .
Hence
1
Wg is a KGsubmodule z 0 of V. So by simplicity of V we obtain
gEG
v= 1
Wg. gEG Hence by 1.9 a) VN is a semisimple KNmodule. 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.
3.15
Lemma. Suppose N 4 G and Char K
4 I GIN 1. If W is a simple KNmodule, then W G
is a semisimple KGmodule. ti Proof: By 3.2 it suffies to show that (W )N is semisimple. But if G = U Nt, then 
t€T
W
G
= o
Wot.
t€T If w E W and y E N, then
G Hence W o t is a KNmodule, 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 Sylowpsubgroup P and G = U Pt, t€T
where T is a transversal of P in G. Then J(KG) = J(KP)KG = < (yl)t
I y E P, t E T >
and
dimKJ(KG) = (IPI~)IG/PI= IGIIG/PI
.
Proof: a) If V is a simple KGmodule, then by 3.14 we have 
with simple KNmodules W.. Hence VJ(KN) = o W.J(KN) = 0 . J j=1 J This shows J(KN) C J(KG). b) Suppose G = U
tET
1
Suppose a =
Nt, hence KG = o KN tET
. t.
at t E J(KG), where at E KN. Let W be a simple KNmodule. By 3.15, W
G is
t€T a semisimple KGmodule. Thus W
G
a = 0, hence for all w E W
This forces w at = 0, so W at = 0 for every simple KNmodule W. This shows at E J(KN), so
c) By b) and 3.4 b) now J(KG) = J(KP)KG = < (yl)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
5
m G and G = U H ti. Consider K as the trivial KHmodule 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 permutationmodule for G:H. Obviously I i' I
Kw with w =
1 i= 1
v.1
and
are KGsubmodules 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 KGmodule. 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
@
A.
G IS . the "natural" permutationmodule for G of degree 4.
Suppose at first Char K + 2. Then by a) K
G
= Kw o A.
If A is not simple, it has a Idimensional 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) =
I
K
ifK
= 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
]
s
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 KGmodule of dimension 1. Hence K has only the compositionfactor K, with multiplicity 4. But if g E G and o(g) = 3, then K
G
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 KHmodule. 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 compositionseries 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.

and
Already the groupalgebra 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 = b1
a) Suppose that Char z 3 and K contains a primitive third root
E
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
I
1
(i=0,1,2)
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 1bla
= cle a.) As
1
E L
El, we see easily that elR is a simple
Rmodule. The same is true of e2R. Case 1: Assume Char K
#
2,3. Then R is semisimple. We already know three simple
Rmodules, namely K
l
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 signumrepresentation is the trivial representation. AF now J(R)
;t
0 and we know already simple modules of dimensions 1
and 2, we obtain R/J(R)
P
(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
Therefore
b) Now we suppose Char K = 3. By 3.15 c), dim J(R) = 4. We put e l = 1 (l+a) , e2 = 1 (la) . 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 = (1a)(l+b+bL) =
1
sgn g.g .
gEG As b1, bL1 E J(R) (see 3.6 b), so
Obviously, w+g = w + and wg = sgn g wfor all g E G. We put
we have
Then
Therefore elR/elJ(R) is a module for the trivial representation. Furthermore
Hence w
+ = sl(lb) E elJ(R) 2
and the module
belongs to the signumrepresentation. 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 KS4module V of dimension 3. If
V
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 KGmodules of dimension 1 if Char K = 3 resp. Char K z 3. Hence if Char K + 2,3, we obtain R = KG
E
(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 KGmodules). 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
1
i=1,2,3. As eoR has at the top the 1dimensional module K, and as there is now (Char K = 3)
;
only one 1dimensional and no 2dimensional simple module, eoR has only the composition
I
factors K,K,K.
1
b) Now let Char K = 2 and let
E
be a primitive third root of unity in K. Put
1 2j2 e. = 3 (1+EJ 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 KGmodule by 3.5, the e.KG are
J
indecomposable. We analyze eOKG:
As eob = eo7 we obtain e KG = (eo,e a 0
eOa2 eOala2) (as Kspace).
By 3.16 c) follows
.
1
eoJ(KG) = eo (h (1al) , bl(la2) , bl(lala2)) = (eo(lal) , eo(la2) ea(lala2))
.
As Char K = 2, also
e (1al)(la 1a 2) = eo(la2)(1a 1a 2). e 0(1al)(la2) = eO(l+a1+a 2+a 1a 2)  0 This shows 2 eOJ(KG) = KeO(l+al+a2+ala2) = S(eOKG) . e0(1al)b = eobeOba2 = eO(la2) , e (1al) eo(la2)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(1a2) + eo(l+a 1+a 2+a 1a 2) .
(7 I)
, which has eigenvalues
i,
r
 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 KGmodules. As G/V
ir
2,3. We want to find the
S3 (V the Kleinsubgroup), 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
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) @
w2
@
(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,
then
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 psolvable, 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).
P
Exercises 6 ) Let G = (gl
a) KG s K[yl
,..., gd) be an elementary abelian pgroup 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 pgroup. The minimal number d of generators of G is given by
I G/HG)I
d = p (Burnside). Let G = (gl d
,..., gd). Suppose again Char K = p.
A
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)
i+l
in terms of the structure of the pgroup G; see BlackburnHuppert 11, p. 259). 8) Suppose K is algebraically closed, A
that dimKM
5
5
G and A abelian. If M is a simple KGmodule, show
I G:A I .
9) Let G = (g) be a cyclic group of order I GI =
contains a primitive mth root of unity
a . a) eiSP = r'e.I
E.
with p
4m. Suppose Char K = p and K
Form
.
b ) l = e o + ...+ e
m1
ande.e.=d..e.. ' J 'J'
c) dimKeiKG = pa and eiKG is an idecomposable projective KGmodule. d) KG has m blocks. e) Show J(KG$ = ( g r n  1 $ ~ ~ and hence a J(KG)P
a 3 J(KG)P =
o.
10) Let G = GL(n,q) and V = V(n,q) the natural Gmodule. Show: If n > 1 and (n,q)
* (2,2),
then V is not a projective KGmodule. 11) Let P be a Sylowpsubgroup of G and K with Char K = p the trivial module for KP. If V G G is any simple KGmodule, 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 1dimensional subspaces of the Zdimensional vectorspace over GF(3). The kernel of this permutationrepresentation is (E) and GI(E)
e
b) Simple KGmodules with E in the kernel have dimensions 1,1,1,3.
A
4'
1
C)
If E is not in the kernel of a simple KGmodule V, then dim V is even.
d) KG = (K) I
@
(K)
@
6)(KI3 @
@
(Q2
@
(KI2 @ (KI2
.
13) Let Char K = p. Equivalent are: a)
(GI
= pn.
b) KG is an indecomposable KGmodule.
c) KG/J(KG) is a simple KGmodule. 14) Suppose that V is a KGmodule, H s G and Char K
4 1 G : HI. I f VH is a projective
KHmodule, then V is a projective KGmodule. 15) Let G = GL(n,p) with n > 1 and let V = V(n,p) be the natural KGmodule
(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 KGmodules 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 isomorphismtypes of simple KGmodules. 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 KGmodule 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 KGmodule and y an epimorphism of P onto M. Then P
c
P(M) o P'
Proof. a) Let n be the natural epimorphism of P(M) onto k
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).
i=l
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
5
P(M)J(KG)
.
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)
.
4.3
Definition. Let V and W be KGmodules. Then V
W becomes a KGmodule 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!)
4.4
Theorem. a) Let F be a free and V an arbitrary KGmodule. Then F o
KGmodule in rank F

V is a free
dimKV free generators. V is projective.
b) If P is a projective and V an arbitrary KGmodule, then P o Proof. a) It obviously suffices to show that KG o 
V is free, for the tensorproduct is distri
butive. Let V be the Kvectorspace V with trivial action of G (vg = v for all v E Vo , g E G). 0 Then KG o V0 is a free KGmodule in dimKV free generators. We define a Klinear mapping
t
of KG o
V onto KG o
V0 by
(gov)t=govg Obviously
t
1
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
V
r
KG o
1
= ( ( g o v)t)h
Vo as KGmodules.
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
G.
I
a) If W is a projective KHmodule, then b) Suppose Char K = p and p
wG is a projective KGmodule.
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)
5
2
P(K G) o P ' . In particular
1G:HI
.
c) Suppose that H is a pcomplement of G, hence I G:H 1 = pa and p 1 I HI. Then P(KG)
E
(KHIG is the permutationmodule for G:H. (By Phillip Hallf s theorem all
psolvable groups have pcomplements.)
Proof. a) Let F be a free KHmodule such that F 
w
~
~
w
~
~

~
W e W ' . Then
F
~
=
F
~
,
~
K
G
is KGfree, for KHoKHKG=KG. Hence WG is a projective KGmodule.
.&I H 1,
b) As p
so every KHmodule is projective. Hence by a) ( K ~ is)KGprojective. ~ If a is
the epimorphism of KG onto KG, given by
then
L Q
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 KGmodule, then P(M) is isomorphic to a direct summand of
P(KG) o
M.
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
IGI =
with p ) r m .
a) dimKJ(KG)
2
pa 1.
b) dimKJ(KG) a I GI 
&A. G
If in particular G has a pcomplement, then dim J(KG)
s
I GI

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
k
dimKP(Mi)
.s dimKP(KG)
'
dimKMi
.
This implies k
k
i=l
i=l
( GI = dim KG = o dimKMi dimKP(Mi) 5 dim P(KG)
(dimKMi)
2
But from k KG/J(KG)
e
o dimKMi
i=1

Mi
(see 3.13)
follows
Therefore
4.8
Remark. Equality in 4.7 is extremely rare:
a) dimKJ(KG) = palholds if and only if either G is a pgroup (ct 3.4 b)) or if G is a
I
Frobeniusgroup with Frobeniuskernel 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 Sylowpsubgroup. 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)
5
I GI

E l pa
I
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 psolvable b group and pa I GI. If M is a simple KGmodule, then dimKM divides [ G I . If dimKM = p m 4.9
T
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 KGmodules and G arbitrary if and only if G has a normal Sylowpsubgroup (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).
I
For general rings, besides the projective modules the injective modules play an important role. For groupalgebras over fields projective and injective modules coincide. We use duality theory of finitedimensional vectorspaces to prove this. 4.10
*
*
Definition. Let V be a KGmodule. We put V = HomK(V,K). Then V becames a
KGmodule 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 KGmodule, so is P . Proof. As HornK( ,K) is additive, it suffices to show 
I
We define the Let { a g E K ) be the Kbasis 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
9
.
Hence A is a KGisomorphism. Theorem. Let P be a projective and V any KGmodule. If a is a monomorphism of P
4.12
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 KGhomomorphism. (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 KGmodule W. Dualizing again (and using dualitytheory 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*)'
2
Pa
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'
.
4.13
Theorem (BryantKov5cs)
a) Suppose that for every 1 it g E G there exists a KGmodule 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 Gmodule. Then IGI1 0
j =I
v o ... o
v
j
has a direct summand isomorphic to P for any indecomposable projective module P. In particular every simple KGmodule is isomorphic to a submodule of some V o ... o V Proof. a) A KGmodule 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
I
g E G ) is linearly independent over K.
Now we prove: Let U,V be KGmodules 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
Then
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) /GI1 (V1 o V) = o n. (V o ... o V) (for some n.) 1# ~ E G j=1 j J
o
contains a submodule W isomorphic to KG. By 4.12, W is even a direct summand. By the KrullSchmidttheorem, 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. 70571 1). 4.14
Lemma. Let K be algebraically closed and eKG with eL = e an indecomposable
projective KGmodule with simple head M = eKG/eJ(KG). Let V be any KGmodule. Then dim HomKG(eKG,V) = dimKVe is the multiplicity of M as a compositionfactor of V. Proof. Obvoulsy HomKG(eKG,V) = Ve (as Kvectorspace) by
a + e a = e 2 a = (ea)e
for a E HomKG(eKG,V). Let W be a maximal KGsubmodule 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) =
otherwise
Theorem. Let K be algebraically closed and MI, M2 be simple KGmodules. 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
5
dim Mi, then MI o
P(M2) is a direct summand of
k
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.
I
lm
M2
1
Exercises.
16) Define the concept of an injective envelope of a KGmodule and prove the dual of 4.2. 17) Let V. (i=1,2,3) be KGmodules. 1
a) HomK(V1,V2) becames a KGmodule by (.g)(v,) for a E HomK(V1,V2)
= (4vlgl))&
. And then
* = V1 Q
HomK(V1,V2)
V2
*
as KGmodules. What corresponds to HomKG(V1,V2) in V1
Q
V2 ?
b) Show HomK(V1 o V2V3)
* = Hom K(V2' V 1 Q V3)
as KGmodules and
18) Let M1 and M2 be simple KGmodules and K algebraically closed. a) The multiplicity of the trivial KGmodule K in the head of M1 o
0
otherwise
M2 is
.
b) The multiplicity of K in the socle of M1 o
M2 is
*
1
ifM2=M1
0
otherwise .
*
c) Define h E HomK(Ml o M1,K) by
*
Then A is a KGepimorphism of M1 o M1 onto the trivial KGmodule 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
1
Show wg = w. e) The trivial KGmodule K is a direct summand of M o M* if and bnly if 1'K 1 Char K dim MI
4
19) Let M be a KGmodule. If M is simple (indecomposable), so is M
*
9 5 The number of simple and indecomposable KG modules
5.1
Definition. For a Kalgebra A, we define the center of A by Z(A) = {z E A
I
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 Ksubspace of A. Let A = A1 s ...s Am with 2sided 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 semisimple Kalgebras 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
2
0.
a) We denote by R1 ,..., R the conjugacy classes of G. Fix an element gi E Ai in each class h
ak(k
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 semisimple, 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 Kbasis of YKG).
+ T, ... ,gh + T) is a Kbasis of KGIT. Moreover: If g E A., then g + T = g. + T.
b) {gl
J
J
5
h) be
4 I GI, i.e.
Proof: a) Clearly,
 are linearly independent. If g zl, ... ,Ah
ffi E Z(KG). Conversely, if a =
1
E G, then glffig =
Ri and thus
ax x E Z(KG), then
xEG
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 Gconjugate, say y = gxgl. Then
n  y = x  gxgl = gl(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)xl yx) = 0 for all x,y E G and consequently qi(T) = 0 h
1
(i=l, ...,h). If
h ai(gi+T) = 0 (with n i E K), then
i= 1
1
oigi E T and
i=l
h a. = q?. ( J
J
2
i=l
aigi) = 0 for all j
.
5.5 Corollary: Let KG be semisimple (i.e. char K = 0 or char K
4 I GI) and K algebraically
closed. Then the number of isomorphism types of simple KGmodules equals the number h of conj ugacy classes of G. Proof: By Wedderburn's Theorem (2.14), KG = (K), 
0...0
1
where m is the number of
(K), m
simple KGmodules. 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 2P2 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)clc1(c2  cJ) P l 1 2
.C
r
212 (mod T).
ai(aibibiai) 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
.
I
In order to show tP E T, it thus suffices to assume that t = abba (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 n1 n1 n1 (a+blP = aP + bP + t for some t E T But then (1) yields
.
n n (a+blP = aP
n
n
n
+ 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
ET
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
1
(a+blP = aP
1
+ bP
E
.
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)
J
n j
. We fix a E So and
i a 0 such that aP E T. We
also decompose


a
a = a o...o 1 rn
with
a.J E. A.J ' . a.J E K G . I
1
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
1
. 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
a?J
1
is equal to 0. Since the trace is the sum of the eigenvalues and since char K = p, it I
1
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
k
ffl
,..., ffk of
G. Then {gl
+ S ,..., gk+ S) is a Kbasis of KGIS.
Proof (1). For g E G, let g = ux be its unique factorization into an element u of ppower order I
and a p'element x such that ux = xu. If up = 1, then I
I
I
(gx)P = gp  xp = 0 and 5.7 yields gx E S. By 5.4 b), we also have xg. E T
J
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
1
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
(,z
Let m be the pfpart of I G I. We choose integers a and b such that ap
1
+ bm = 1. Since m is a
I
1
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 pfclasses 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 KGmodules 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
mand KGmodules. 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 psubgroup 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/Qmodule. Now Alperin's weight conjecture may be stated as follows: The number of simple KGmodules is equal to
1
{ [Q,V)]
1
Q a psubgroup of G, V a simple H = NG(Q)module which is projective as a KH/Qmodule }
I.
This means that the number of simple modules can be computed plocally. In the meantime, the conjecture has been proved for several classes of groups, including psolvable groups. If the underlying field K is not assumed to be algebraically closed, the question about the number of simple KGmodules 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 heavyhanded  consequence of 5.9 is that a pgroup 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 socalled 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 Ksubspaces Vm consisting of the i mi ( i=O,...,m ) is a Kbasis 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 (xiymi)A = (ax + b ~ ) ~ ( + c xdy)mi . It is easy to check that this turns Vm into a KGmodule. We show that Vm is simple for m = O,l, ...,p 1. To do so, let W be a nonzero submodule of Vm and pick n
Orf=
j mj a.xy E W , wherea z O and n s m . n i=o J
l
For 0 r t E GF(p), we consider the matrices S(t) =
(A
:) E G. Then
. n sit) = 2 a.(x+tyyymJ= fj(x,y)J E w j=o J j =O n
f
for suitable polymomials f. E K[x,y]. Observe that fo = f and f = a y J n n Kvector space,
p1
1
t=l
. t'(f.s(t)) E W for all i E Z .
Using the fact that p1 . t=l
we obtain
1 i f p1 dividesi 0 otherwise
m
. Also, since W is a
Since 0 s n s m s p1, this sum equals  fn if n c p 1, and  fo fn if n = m = p1. m As fo = f E W and fn = anym (an L 0), we conclude y E W in any case.
4
We now play a similar game with the matrices T(t) = (:
E G. Namely
p1 i m. p1 . & t (y T(t)) = & t  ' ( t x + ~ )E~W for all i E Z t=l t=l and this is equal to p1
1
t'
t=l
=
i
. m
m ( T ) ~ J ~ ~ 1 J ( ~I =) j ~ mj y p& j=O j =0 t=l
1
(y)xiymi
+
XP'
di=
&
j=o p1 1 ji
(m) x iy mj J
for 0 r i srnc p1 and O c i < p1 = m
ypl for m = pI, i = O
.
i mi E W for all Recall that for m = p1 we have already seen that yPI E W. It follows that x y i=O,l, ...,m. Thus W = V
m
and V
m
is simple for m = 0,...,p1.
b) In fact the Vo ,..., V
are all the simple KGmodules. This follows from Brauer's P1 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
*+2
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 mi
xy
i mi x y
if m is even if m is odd
.
Thus Z(G) acts trivially exactly on the modules Vg, Vt, V4 ,..., Vp3
, Vpl. 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
1
a) Consider G = SL(2,7). Then GI = 2 3.7, and by 5.10, there exists a simple KGmodule
V with dimKV = 5. Thus
dimKV
4 [GI .
b) The phenomenon of a) does not occur for psolvable groups G. This is an immediate
consequence of the FongSwanTheorem (see 14.2). In 9.3 however we give an easier proof for the fact that if G is solvable, then dimKV
I I GI
for all simple KGmodules V.
Theorem 5.9 tells us that the number of simple KGmodules is finite. On the other hand, the number of indecomposable KGmodules 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 KGmodules of arbitrary large Kdimension. In particular, there exist infinitely many isomorphism types of indecomposable KGmodules.
I
b) If P is cyclic, there exist at most GI nonisomorphic indecomposable KGmodules.
To prove theorem 5.13, we need several preparations. 5.14 Lemma. Suppose U
5
G and let T be a transversal of U in G such that 1 E T.
a) If V is a KUmodule, then
V o l = { v o l l vEV) G is a direct summand of (V )" with V b) Suppose Char K
of
(w~)~.
s
V o 1 as KUmodule.
4 I G:U I. If W is a KGmodule, 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 KUmodule, 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 KGmodule. 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 skewfield.
Proof. Direct decompositions V = V o V 2 as KGmodule 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 skewfields) 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 identitymatrix 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 matrixcalculation we obtain B = 0, A = D and .AN = NA. If we choose now
then AN = NA implies
If V is a KGmodule for D, then HomKG(V,V) corresponds to all matrices
The mapping of this matrix onto a l is obviously an algebrahomomorphism 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 KGmodule.
5.17 Theorem. Let G be a pgroup and char K = p. a) If G is not cyclic, there are indecomposable KGmodules of every even dimension.
I
b) If G = ( g ) is cyclic and GI = pa, then there are (up to isomorphism) exactly I GI
indecomposable KGmodules, namely
with dim V. = i 1
Vi = KG/(~a (i = 1,...,p ).
KG
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) = (g1)KG. Then also KG/(~I)'KG has only one maximal submodule, therefore is indecomposable. Let conversely V be an indecomposable KGmodule. Then the canonical Jordanform 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 KGmodule Vi with dim Vi = i for 1 6 i
5
pa.
5.18 Proof of 5.13. a) Suppose P E Syl G and P not cyclic. By 5.17 a), there exists an indecomposable KPmodule P G If V of dimension 2k (k=1,2,...). By 5.14 a), Vk is isomorphic to a direct summand of (Vk)p.
k
with indecomposable KGmodules W., then by the KrullSchmidttheorem 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 KPmodules Vi, where dim V. = i (i=l,..., /PI). We show that every indecomposable KGmodule 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 KPmodules Ui, by the KrullSchmidttheorem W is isomorphic to a G direct summand of some U i , where dim U.
1
5
dim W. The number of nonisomorphic 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.
I
Hence there are at most G:P
I I P I = 1 GI
indecomposable KGmodules.
5 6 Clifford Theory 6.1 Definition. Let N a G and W a KNmodule.
a) For g E G, we consider the KNmodule WogLWoKNKG. 1
Observe that (w o g)n = wng
og
W o g is called the gconjugate module of W. b) The subgroup
I(W) = {g E G ( W
= W o g as KNmodules)
is called the inertia group of W. Since W o n = W
Q
1 = W for n E N, we see that N
5
I(W).
As promised in 3.13, we study in more detail how a simple KGmodule restricts to a normal a G. subgroup N 
6.2 Theorem (Clifford). Let N a G and V a simple KGmodule.
1 Wg and Wg a W o g is simple. gEG In particular, VN is semisimple and all composition factors of VN have the same dimension.
a) Let W be a simple KNsubmodule of VN. Then VN =
b) Let VN = V1 o ... o Vt be the decomposition of VN into homogeneous components Vi. Then rightmultiplication by the elements of G transitively permutes the Vi. In particular, all V. have the same composition length and we may write 1
V.
1
. Wi o ... o Wi = eW.1
for a simple KNmodule 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
e
,
V. o a(,i)
KG
(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
t
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.
c
c
Let w.l.0.g. {V
J k J Vm} be the Gorbit generated by V1 (m r t). Then V1 o ... o Vm is a
1
KGsubmodule 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
t
2
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
I
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
r
W. I
(i=l,...,t)
.
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 3dimensional Kvector space with Kbasis {vl,
" "Y vlg=vl, v h = v
v ). IVe define a KGmodule 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
I
a,b,c € K
)
and thus 0 and 1 are the only idempotents of EndKG(V). Therefore, V is an indecomposable KGmodule.
 G. Then v, = ( u ) o (v,w)
We consider N = ( h )
Q
and a similar argument as above shows that (v,w) is an indecomposable KNmodule. 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 KGmodule. Then
k
where e E PI, W is an indecomposable KNmodule and ti € G. Proof. Since V is projective indecomposable, there exists a KGmodule X such that V o X = KG. We decompose
into projective indecomposable KNmodules Wi. Since
it follows that V o X = W
G o ... o WG .
By the KrullSchmidt Theorem, we may assume that V is a direct summand of W:
and
set W = W1. Let T be a transversal of N in G. Then WG = o W o t tET
with KNmodules W o t
.
Since W is an indecomposable KNmodule, all conjugate modules W o t as well are indecomposable. We write
with indecomposable, painvise nonisomorphic KNmodules Vi and mi E N. Since VN is a
G direct summand of (W )N , it again follows from the KrullSchmidtTheorem 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 KGmodule, we have
with nonisomorphic indecomposable KNmodules Vig. Comparing the multiplicity of V. in
J
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 KSmodule. Set I = I(V), and
decompose
vI = v1 s . . . e V m into indecomposable KImodules V.. I a) Then V? is an indecomposable KGmodule. b) If V. is even simple, then VG. and V are simple as *ell. 1
c) V?
1
= 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
s
Q
t
with KSmodules V s t
.
V o t for all t E T, and by the KrullSchmidt 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 KNmodules).
Consequently,
a) Suppose now that VG = W o X with KGmodules W We first observe that
o l+rER
z
Vi o r is a KImodule. Since
0 and X.
and since V. o 1 G V. is an indecomposable KTmodule, we may assume by the KrullSchmidt 1
1
theorem that Vi is a direct summand of WI. Set WI = Vi o W ' for a KTmodule W '
.
If we restrict further down to N. we obtain
But W was chosen as a KGmodule. 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 KNmodules V o r (r E R) in W N is at least n..I By (*), G therefore (V . ) I
N
is isomorphic to a direct summand of WN.
In particular, dimKW a dimKvY = dim W
K
+ 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
G
1
.
By Nakayamareciprocity (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
G
niV. By Clifford's Theorem (6.2), WN is
semisimple and hence the indecomposable KNmodule V is simple. This proves the second assertion of b). Since W is a KGmodule, 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
=
0
(Vj o r)N =
I t r ER
1srER
and (Vi)N = ni V. By KrullSchmidt 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 wellunderstood, 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
KNmodule 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) KGmodule 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
n
= 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 KNmodule. 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) = plp(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
+i
.
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).
8
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))
e
K, hence
Homm(V,V) = K1 o J(Homm(V,V))
.
Therefore by (*) p(hn)cc" = a1
+y,
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) = Blp(x)B and
(**I
p(hn)pn = 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
P"
by (**). Hence
I
and
I
Now put as in a)
I
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 KNmodule. a) If there exists a KGmodule 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 KGmodule W. By 6.2 e), it follows that
t
V
= 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 KGmodule W:
Let G = Q8 or D8 and N = Z(G). Let K be any field of char K
#
2 and consider the ldimen
iional KNmodule V on which N acts nontrivially. Clearly, I(V) = G holds. If there exists W
such that WN
s
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 KNmodule 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 1dimensional KNmodule with I(V) = G, then V is extendible to G. Proof. Let p : N + AutK(V) be the representation afforded by the KNmodule 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
s
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 welldefined, is a representation of G and satisfies aN= p.
Let V be a simple Amodule for a Kalgebra A so that D := End (V) is a skewfield. A Jacobson's Density Lemma states that given Dlinear 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 KNmodule. Suppose that V
is extendible to the KGmodule W. a) If X is a simple K(G/N)module, then W o KX is a simple KGmodule (with module structure given as in 4.3). b) If X,Y are simple K(G/N)modules and W o KX
s
W o KY, then X
= Y.
Proof. a) Let U be a nonzero KGsubmodule of W e KX, and 0 * u E U. We may then write n
1
u=
w.ox. 1 i=1 1 for a Kbasis {wl, ... , wn) of W and suitable elements xi E X; without loss of generality, x 1 z 0.
By hypothesis, WN
I
e
V is a simple KNmodule, 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 =
1
ahh (ah E K), and using xh = x for all
hf3
x E X, h E N, we obtain (wi o xi)a =
1
ah(wi o xi)h =
hEN
Thus ua = w o xl and W o xl
1
hEN
a (w.h e x.h) =
1
hEN
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) = dimKWdimKY , K K hence dim X = dimKY. We fix Kbases {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
,x
For h E N, we have m
1
(wh)a.. o y. = (wh o x . ) a = (wh e x.h)a = (w e x.)ha 'J
j=1
J
= (w o x.)ah = (
1
j=1
I
m
wa.. o y.)h = 1 (wa..)h e y. 'J J j=1 'J J
Comparing coefficients of y., we obtain that J
1
1
m
1
.
Thus a . = c i j l W for some c . E K (i,j = 1,...,m) 11 1J
.
Let (aij(@) and (b..(g)) be matrixrepresentations of the action of g E G on X and Y with IJ
respect to the bases {xl ,..., xm) and {yl ,...,.)y,
Now
and this is equal to
Comparing the coefficients of yP this implies
1
j=1
~ . ~ ( w g ) a ~=~ (1 g ) c..(wg)b (g) for all w C W , J 1 J ~k
and therefore 
1

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 a1 ~nstead, we see that y is invertible.
Hence X
= Y as KGmodules.
6.10 Lemma. a) Suppose U
s
G. Let V be a KUmodule and W a KGmodule. Then
(Wu m KV)
G
= W m K ~ G(as KGmodules).
b) Let N 4 G and W a KGmodule. If we consider K(G/N) in the obvious way as a KGmodule, then
Proof. a) We obviously have ( w U o KV)G = ( w U o K ~ o)K
u
e
W ~ U e~K(V e KUKG) = WU o KV
G
as Kvectorspaces 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
B
v) e t)a g .
G. b) This is the special case of a) where V is the trivial KNmodule, hence V IS isomorphic to K(G/N). 6.11 Corollary. Let N Q G, K algebraically closed and V a simple KNmodule. Suppose that V is extendable to the KGmodule W. a) Let U be any simple KGmodule which has V as a constituent in UN. Then there exists a simple K(G/N)module X such that U"W0
K
X.
b) If W ' is another extension of V to G, then W19woKX for a 1dimensional K(G/N)module X. Proof. a) By Nakayamareciprocity (3.8), 0 + Homm(V,UN)
ti
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

z
that U as well is a composition factor of VG
Let
(W )G
N
z
W
0
KK(G/N).
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
0
KSi)/(W 0
" W 0 K(Si/Sil)
is a simple KGmodule (i=l,...,m). The JordanHolder theorem thus yields that U a W 6x1 KX where X = Si/Sil for some i. b) Since dimKW1 = dimKW, a) implies dimKX = 1.
We conclude this section with some examples of how to construct simple KGmodules over an algebraically closed field K of characteristic p. If p
4 I GI, then KG n
o (dimKV).V is V simple
semisimple (3.3, 3.12) and the number of simple KGmodules equals the number of conjugacy classes of G (5.5). Moreover, constructing a simple KGmodule 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 socalled "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 pgroups. 6.12 Example. Let K be an algebraically closed field of characteristic p.
a) Suppose that G has a normal Sylowpsubgroup P = 0 (G). By 3.6 b), we know that P P s; ker V for each simple KGmodule V and V can be considered as a simple K(G/P)module.
I
Since p l, G/P 1, the above comments apply in this case. b) Suppose now that G is pnilpotent with normal pcomplement N = 0 ,(G). We decompose P m KN = o Wi into simple KNmodules Wi and again assume that these are known. Then i= 1
with projective KGmodules WG (2.8).
(1) Let W be one of the Wi. The W Let X
t
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) KNmodule U such that XN is a direct sum of some Gconjugates of U. Since by definition ( w ~ =) t ~~ W T
(T = transversal of N in G) ,
13t
it follows that U is Gconjugate 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 Gconjugacy classes of the KNmodules
W1
,.., Wm (r 5 m). Then Vi := H(W G) = W G/(W G)J(KG)
(i=l,...,r)
are representatives of the isomorphism types of simple KGmodules. Moreover,
I
I
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 KGmodule 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
JW
G
c
wG j
Wi, W. are Gconjugate. J
nonisomorphic.
Also, by Nakayamareciprocity (3.8), dimKHomKN(Wj,(Vi)N) = dimKHom
G
1 if Wi , W . are Gconjugate J
Thus (Vi)N is the direct sum of the Gconjugates of Wi, each occuring with multiplicity 1. In particular,
I
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 Gclasses contained in N.
c) Suppose that G is pnilpotent with abelian normal pcomplement N. Then dimKW = 1 for all simple KNmodules W and b) immediately yields: 1) The number of isomorphism types of simple KGmodules equals the number of orbits of
P E Sylp(G) on N. G 2) Each simple KGmodule V occurs as H(W ) for some simple (1dimensional) KNmodule W and dimKV = ( G:I(W)
1.
6.13 Example. Let K be algebraically closed of characteristic p. We consider a Frobeniusgroup G = FH with abelian Frobeniuskernel 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
Sylowpsubgroup, 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 nontrivial simple (Idimensional) KFmodule, 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 nontrivial simple KFmodule. Then WG s mG W , m are Gconjugate . I
Since I = F, the Gconjugacy class of W consists of H I KFmodules. We have thus constructed
1 s
simple KGmodules of the form w G , satisfying F s ker WG.
2) Let k be the prclass number of H. By Brauer's theorem 5.9, G/F
G
H has exactly k
different simple K(G/F)modules Vi. Considered as KGmodules, they satisfy F s ker Vi. Consequently, v. I

W
G
for all i, and all simple KNmodules W + IF. 3) In 1) and 2), we have found
+k
different simple KGmodules. The Frobenius
partition
yields that each pfclass 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 KGmodules.
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 Frobeniusgroup with nonabelian Frobeniuskernel will be treated in 7.8, using Brauer's permutationlemma. Exercises. 17) Construct the simple KGmodules 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 KGmodule and
where W is a simple KNmodule and { g ,...,gm) a transversal of I(W) in g. Then 1
where the U. are simple KGmodules of dimension 1 with N in the kernel. The V o J ~~j simple and pairwise nonisomorphic. In particular,
2 e rm = ( G I N ( . (It follows that the number em of simple components of V
N
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 Eulerfunction.
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,lentry of B'FB is
(Here we use the wellknown 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 ) .
1
7.2 Lemma (L. Kovacs). Let K be any field, let P and Q be permutationmatrices 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
i
I
J
Proof: We put w(P) = 2 o.(P). J j We let P and Q operate on a Kvectorspace 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 1dimensional fixedspace. 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 cycledecomposition. 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.
j
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 KGmodules (up to isomorphism). We define classfunctions 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 pcomponent 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,
..,an
are linearly independent on {gl, ...,gk). In particular, Vi is determined (up to isomorphism) by it's tracefunction
p..1
c) The (k,k)matrix caicgj)) is nonsingular. 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
1
Di(gp) = [la..:
*
I 1 bl O
and
n
D.(g ,) = l
P
*
..
.
n
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
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
1
ciPi = 0 on KG.
i= 1 But
k
and D. is the projection of KG onto the summand (K),,. We pick an element J J
a.
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
Pi

h
Pi
for h E H,
2) on the p'classes of G by +(gj) We assume that
Gh
(j=l,...,k).
hl $(gj) = @(gj ) for all iJ=l,...,k, and all h E H. a) If P(h) denotes the permutationmatrix of h on {PI, ..., $1 and Q(h) the permutationmatrix of h on {gl, ..., gk), then for all h E H,
P(h)T = T Q(h)
where T = (/3.(g.)) and det T z 0. J b) For every h E H, the number of cycles of length j of h on
{p
Pk) and {g
same. c) The number of orbits of H on {PI, ..., Pk) and {gl ,..., gk) is the same. Proof: a) We define permutationmatrices P(h) = (p..) and Q(h) = (q..) by 1J 1J
1 if/$ Pij =
=B,
0 otherwise
and
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(&hI
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
I
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 permutationmatr~xS such that for
P(h)S = S Q(h)
dll
h f H.
This is not true in general. I t 1s equl\alent u ~ t h1
k
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 pfconjugacy
{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 plclasscs of N by y  )g 1
D ~ ( Y=) Di(Yg We put
/3:
@y)
= trace
1.
DY. Then
= trace ~
1 1 ~ ) =( ~ ~ ~ ).g(
~
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 Frobeniusgroups.
7.8 Example. Let as in 6.13 G = FH be a Frobeniusgroup. We assume char K = p and
PY IFIThen H acts fixedpointfreely on F. By a very elementary argument, H acts also fixedpointfreely on the conjugacy classes of F (see Huppert I, p. 500, theorem 8.9 c).) Hence by 7.7, H acts also fixedpointfreely on the tracefunctions pi of F, hence by 7.4 b) on the simple KFmodules 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 fixedpointfreely 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 KGmodules with
m
rn
F not in the kernel. If k is the number of p'classes of H, there are k simple KGmodules 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 KGmodules from the simple KFand KHmodules. Suppose in particular, that G = FH is a Frobeniusgroup 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 KGmodules 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 KGmodule, {VI,..., vn) a basis of V and . .
vig =
I
(i=l,...,n).
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
If
p is the tracefunction
on V and
on V.then 6dg)= B ( ~  ' ) .
Proof: By definition 4.10 we have 
I
This shows n aig =
I
The trace
I
if
I
1
a..(g )a..
1x1 J'
J
f i ' of g on V ' is hence given by
p is the tracefunction on V
7.10 Theorem (R. Brauer). The number of selfdual simple KGmodules is equal to the
number of plclasses gG such that g G = (g1)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
I
we can apply 7.5. 7.11 Theorem (P. Fong). Suppose K is algebraically closed of characteristic 2 and V is a selfdual simple KGmodule. If V is not the trivial KGmodule K, then V carries a symplectic, nonsingular Ginvariant form. In particular, dim V is even. Proof: Suppose at first that V = Kv is 1dimensional, 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 KGmodule.
Suppose dim V r 2. At first we construct a bilinear form (.;)
on V by
(v1'v2) = (hv1)(v2)' where h is a KGisomorphism of V onto V 0 = (v 1,v2)  (hvl)(v2)
*
. If
for all v2 E V,
then hvl = 0, hence vl = 0. Therefore (.,  ) is nonsingular. For g E G we have
Hence (.;)
is Ginvariant. We put q(v) = (v,v). Then
q(vl + v2) = (vl + v2' v1 + v2) = (vl,vl) + (v1*v2) + (v2'v1) = 9(v1)
+
+
(v2'v2)
q(v2) + [v1'v21'
where the obviously Ginvariant bilinear form
[;I
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
[.;I
+ 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 Ginvariant subspace of \'. .%s \' is simple, so R = 0. Hence
[.;I
is nonsingular
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)
(*)
q(vl
;t
0 for some vl E V.
Then by (*),
U = {v
I
v E V, q(v) = 0)
is a Ginvariant 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 V1 > 0, contradicting the simplicity of V. 7.12 Theorem. Suppose Char K = 2 and 2 l, dim C' for every simple KGmodule. Then G has a normal Sylow2subgroup. Proof: (Okuyama). By 3.6 b), the maximal normal 2subgroup 02(G) of G is in the kernel of every simple KGmodule. 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 KGmodule K is the only simple KGmodule which is selfdual. Therefore by 7.10, the only selfinverse 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 dihedralgroup
and 1
(ij) = ji = (ij)
1
.
Hence the order of ij is a power of 2. In particular, for any g E G we obtain that < i,ig > is a 2group. By an elementary lemma of R. Baer this implies i E OZ(G), a contradiction.(See BlackburnHuppert 11, p. 500; a more direct proof of Baer's lemma is in M. Suzuki, Group Theory 11, p.195196.)
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 Chevalleygroups (G. Michler, Journal of Algebra 104 (1986), 220230.) 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 Sylowpsubgroup P, then by 3.6 b), P is in the kernel of any simple KGmodule V. Hence V is a simple module for the prgroup 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
1
dim PI.
Proof: Let P. (i = 1,...,k) be the projective cover of the simple KGmodule 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
o
Mi+ M i
1
If Mi = M*i
c
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
1
is divisible by 2"".
* dim Mi (Pi o P i ) .
e
Mi= M .
The same is obviously true for 
dim
* dimMi(PiaPi).
o Mi+ M
As 2'
T dim KG
this implies
za T dim PI.
.
*
7.15 Lemma. If Mi is a simple KGmodule. then Pf51
)
m
P(\li) .
Proof. If
then by dualitytheory
*
1;
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)
*
*
a
P(M i).
tj 8 Symmetric Algebras
8.1 Definition. Let K be a field. A finite dimensional Kalgebra 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 nonzero 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(ahl) = 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 antiisomorphic 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 nonzero 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 Kalgebra. Then A is a symmetric algebra if and only if there exists a nunsingular symmetric Kbilinear 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 Kbilinear, 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
*
nonsingular. 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 nonsingular. Thus ker i. contains no nonzero 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 .
b)
dim r
c)
R(L(r)) = r and L(R(I)) = I .
d)
The map r + L(r) is a duality of the lattice of right ideals of A onto the lattice of left
K
ideals of A, that is
and
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 nonsingular, 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 nondegenerate, we obtain by a wellknown 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
I
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 2sided 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 groupalgebras 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( 1e) 2 4 r ) . By a), we have
r
c S,(A)
= S1(A) = RIJcAI)
and by 8.5,
Hence
Let  denote the canonical epimorphism from A onto X = NJ(A). Now observe that AI[A(I~)+ J(A))] as left Amodules and EndA@) Since
s
 
& End&=)
ais simple, & is a division algebra. Hcnic
(antiisomorphism by 2.4 b). is also simple, since A is semisimple
as leftAmodule. (This depends on 1.5 b)!) Thus ueobtain
L(r) = A(1e)
+ J(A) .
By part a) of the Theorem and 8.5 follows
r = R(L(r)) = R(A(1e)
+ J(A)) = RiAt Iel) ? 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 Amodule. 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 Kspaces) 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 Kalgebra is called a Frobenius algebra if there exists a
*
h E A = HomK(A,K) which does not contain any nonzero right ideal in its kernel. Thus symmetric algebras are in particular Frobenius algebras. a) A finite dimensional Kalgebra A is a Frobenius algebra if and only if there exists a nonsingular associative Kbilinear 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 Amodule, 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 Kalgebra and let P1 ,..., Pk be a complete set of representatives of the isomorphism classes of projective indecomposable Amodules. 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 Amodules. Let c.. denote the multiplicity of V. as a 'J J composition factor of Pi. (This is well defined by the JordanHblder 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
1
+ ... + 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 Kalgebra. 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 Kvector spaces. Let (,) denote the bilinear form associated with the symmetric algebra A. For j
it
r and all a, a ' E A, we have
(e.ae., e a t e ) = (e.ae.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.ae., e Ja t es ) ~= ( era l es , e.ae.) = 0. I l j = ( e ra t ese l. , ae.) J
Thus
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'
5
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 KGmodule. Thus the simple KGmodules are obtained by inflation from the simple KS3modules. So, by 3.17, the trivial module K
G
and a Zdimensional module, say V, are the only simple KGmodules.
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 2complement 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
2
1. Otherwise c l
+ c~~ dim V = c 1 2
2
+
2c22
.
1.
= 8 and G induces on P1 a group of unitriangular
matrices, hence a 2group. Therefore A4 is in the kernel of PI. But this is not true as P1 is a faithful permutationmodule for G. Thus c12
2
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
force
I
ell = 4 , c 1 2 =
2 , cZ2 = 3 .
Therefore
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 KGmodule. 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
1
the representation on V i It is easy to see that o D+gj = I . Furthermore

tr(D3 o D2)(g) = t r D3(gltr D,tg) = tr D3(g) = 1 , and therefore V3
I V
1
=# 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 pmjectixe 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 Idimensional module X. If X
z
V , then S1 opsrates on P1 as a group of unitriangular 1
matrices, hence has a nontrivial 3factor 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 =
1
(dimKVi)(dimKPi) = 3
+ 3 + 3 dimKP3 + 3  dimKP4.
i=l r3
+ 3 + 3.dimKV3 + 3.dimKV4=
24.
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),
i
~
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 Kalgebra. Suppose Sr(A) = r l o ... o r t with pairwise nonisomorphic simple modules r.. If t = 1 or I KI = 1
m,
then A is a Frobenius algebra.
25) Let V be a finite dimensional Kvector 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.
8
26) Let A be a symmetric Kalgebra w.r.t. to h E A
. Let M be a simple Amodule with
character f3. Furthermore, let {al ,..., an) and {bl ,..., bn) be dual Kbases 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
M=
a= eA/eJ(A). Then the right multiplication with z on eA is an Aepimorphism from eA
onto S(eA). 28) Use the duality * defined in 4.10 to prove that the scxle of an indecomposable projective
1
KGmodule is simple.
9 9 Modules of solvable, psolvable and pconstrained groups
I
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 KGmodule, then one of the following two cases occurs. VN is simple
(i) or
V=
(ii)
uGwith a simple KNmodule 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 KGmodule, 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
E
W o X with a 1dimensional K G/Nmodule 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 KGmodule, then the number of composition factors of VN divides I GIN
I.
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 =
I
v1 0 ... 0 vs
where s = 1 or s = G/M I and the V. are Gconjugate simple KMmodules. 1
Let t denote the number of composition factors of V
l I N. Since the V.J are all Gconjugate, 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
I
I.
IG/MI IM/NI = IGINI.
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 KGmodule, then dim V ( I GIAI. In particular, dim V
I I GI.
Proof. By Clifford's theorem, VIA = V1 e ... e Vs with s~mpleKAmodules 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 FongSwan 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
In
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 KNmodule, 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 KGmodule and let PG(V) denote its projective cover. Then
I
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 :
As
0 L Homm(U,VN)
G HomKG(U ,V), there is a KGepimorphism from UG onto V,
5
G hence from PG(U ) onto V. Thus PG(V) G by 4.5 a). Furthermore PN(U) has PG(U G)
I
I
G G PG(U ) by 4.2 b). But PN(U) is KGprojective
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.
I
I
Case 1: Suppose G/N = q
z
p.
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)
ti
5
PN(U)
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 KNfactor module of P G ( q N . But
I
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
r
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
qp.
I
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)
P
5
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 KGmodule with I GI
P
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 pcomplement H. If V is a simple KGmodule 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 pcomplement
H. Then for every simple KGmodule V there exists a simple KHmodule 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 psubgroup 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 psubgroup of G
with CG(N) C N. Then KG is an indecomposable algebra, i.e. KG consists only of the principal pblock 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 Gconjugacy class of x and x =
1
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
n
+ @1
n = fp E Z(KCG(N)).
Now observe that CG(N) is a pgroup by hypothesis. Hence 1 is the only nonzero 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 KGmodules).
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 2sided 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~morphluaCkarly (If)KG Ker a fl fKG = 0. Thus we obta~n tKG
9.13 Definition. A group G
I\
r
Imn = KG H
z;
Ker a and
proving b).
called pc~~rnstnincj ~f
CG(Opt,p(G).0 . ( G Ir P
T Op
.p(Gl
Note that psolvable 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 pconstrained, then the following are equivalent. a) KG is an indecomposable algebra. b) 0 ,(G) = 1. P Proof. a) =, b): Suppose H 
:= 0 , (G)
P
* 1. Put f =
Then KG = K G o (1f)KG yields a nontrivial 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 pconstrained 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 pconstrained, then
is the blockidempotent of the principal pblock. Proof. By 9.12, K G s KG10 ,(G). P
,(G) is pconstrained 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 pconstrained and let V be a simple KGmodule. Then the following are equivalent: a) V belongs to the principal pblock. b) Op,(G) acts trivially on B. (G) acts trivially on V. c, Opr,p 1
Proof: a) =, b): Since f = 
2
g is the block idempotent of the principal block,
we have V = Vf. Hence 0 ,(G) acts trivially on V.
P
b)
c): V may be regarded as a module for GI0
P
,(G). By 3.7, 0 (GI0 ,(G)) acts then P
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 pblock acts trivially on V, which means that V belongs to the principal pblock.
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
v
pnilpotent. 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
CB
... o KH. Thus the trivial
b) Put A = fl Ker V, the intersection over all composition factors of P. Then A a G and
v
H
5
A
5
G. Obviously A/H acts faithfully on P. S i m A acts trivially on each composition
factor of P,
A/H is a pgroup. 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
"n
Ker V
where the intersection runs through the set of simple KGmodules belonging to the principal pBlock Bo(G). Proof: By 9.12, 0 ,(G) Ker V for every simple KGmodule 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
I
~
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 KGmodule. If 6 is the set of 7
composition factors of PG(V)J(KG)/PG(V)J(KG), then
fl
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 pchief factor, if V is a pgroup. If V has a complement in G/H, we call V complemented. Let
5 (G) resp. @(G) denote the set of pchief factors resp.
P P complemented pchief factors of G in a given chief series of G.
Note that each pchief factor may be regarded as a simple F Gmodule. 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
P
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 chieffactor of G with E/H s HCG(P)/H. Then E centralizes all pchief factors of G above H. But pchief factors below H are covered by P, hence are centralized by HCG(P). Therefore E centralizes all pchief factors of G, which implies E
s
H, a contradiction.
9.23 Theorem (W. Willems) If K = IF , then the following statements are equivalent: P a) G is pconstrained. b) Each simple KGmodule belonging to Bo(G) is a composition factor of a suitable tensor product of pchief f.actors.
Proof: 
Put
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
P'~P
+
(G).
PI,P
Since h acts faithfully on 0 (G)/O ,(G) (note that G is pconstrained) and trivially on all P1,P P pchief factors, we obtain h E 0 (G), a contradiction. P'>P
(G) and M is a faithful K G/Hmodule. By the BryantKovacs theorem (see Thus H = 0 P',P 4.13), the regular K G/Hmodule regarded as a KGmodule is a direct summand of )G/H)1
o j=1
n. (M o .. . o M) for suitable integers n. (jtimes) 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
1
b)
+ a). By 9.19,
and by the assumption in b),
H=
n Ker V vGYP(G)
L
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 pconstrained. P The rest of this section is devoted to the proof of theorem 9.24. T o state it, let K = F and for
P
a simple KGmodule V, let m(V) denote the number of complemented pchief 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 psolvable. 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 KGepimorphism
where G acts by conjugation on I(H)/I(H)
2
.
Proof. Define /3 : I(H) + eI(H)KG/eJI(H) by (h1)P = e(h1)
+ eJI(H)
(h E H).
For h E H and g E G observe the following facts
(9
e(h 116 = e(h 1)g + e(16)(hg1)
(i i)
e(16)(hg1) E eJI(H)
(iii)
eg(1h) = e(lh) + e(gl)(lh)
The first two facts imply that /3 is a KGhomomorphism. Since I(H) KG = KG I(H), (iii) shows that (3 is an epimorphism. Thus we obtain a since I ( H ) C ~ Ker
P.
9.26 Lemma. Let V be an elementary abelian normal psubgroup of G. 2 2 Then V n I(V)/I(V) as KGmodules via the map v I (v1) + I(V)
Proof: Let V = (vl, ...,v n ) be elementary abelian of order p
n
. If vg, = v dil(g) ... v nd i n k )
n (g E G, 0 r dij(g) a p1), then g operates on the Kspace V = o Kvi (V written additively) by i=l
For v,w E V we have (vl)(w1) = (vw1)  (v1)  (w1) = 0 mod 1 ( v 2 . Hence vw  1 = (v1)
+ (w1)
mod l(V)
2
.
The action of G on the Kbasis {vi1 + 1 ( v 2
(v.l)g = vg  1 = I
Hence V
E
I
I(V)/l(V)
2
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 
s
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
G
It\')KT where Jo = J(KT). Since I(V)KT is a 2sided
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
(i)
J 2 = [ I ( V ) ~ K T+ I(V)Jo
+ Jolt
Ji
3
VI] i
2
where the last sum is direct since Jo C KT and
c
[ 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
nt
. ~ ~
= 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)
fl(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),
Thus
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
4
fJ
2
b) Consider the composition of KGmaps 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 KGsubmodule of eJ such that
Put T = {g
I g E G, e(g1) E L ) . \Vtt
c l a ~ mthat T is a complement of V.
For g, h E T, we have e(gh1) = e(gl)h
+ e(h1).
Suppose V C T. Then (eI(V)KG
+ eJL),eJ'
Hence T
5
G. 7
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(vl)g = evg  eg = egvg  eg = eg(vg1) = eigl)(vg1) E eI(G)I(V)
Hence It follows that
+ e(vg1).
+ 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
eI(v>
L.
Thus, for g E G, we have e(g1) = 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
4
("1) + I ( v ) ~we may write
+ u with u E I(V)2 . e(gv) = e(g1) + e(1v) = y + ew  ew + eu E L, w = (v1)
Hence
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 psolvable 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 pcomplement 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.)
33)
Let G be solvable and K algebraically ch=d '1 characteristic p > 0. The following statements are equivalent. a) The Sylow psubgroup of G is normal. b) p )i dim V for all simple KGmodules \ c) PG(V) = PG(KG) o V for all simple KGrr.kdulcs V. d) Let H be a pcomplement of G. Then \' H
34)
LS srmple
for all simple KGmodules V.
Let G be solvable and K algebraically c l ~ odi characteristic p. The following statements are equivalent a) G is of plength 1. b) p
I dim V for all simple KGmcdules \' f Bc{G).
$10 padic Numbers
Modular representation theory is  alas  a "plocal theory". Therefore it is only natural to work with padic fields instead of algebraic number fields. We give in this section a report on padic 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 padic 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:
(1)
v(ab) = v(a)
(2)
v(a+b)
2
+ v(b);
Min {v(a), v(b))
.
If we put
11 ab 11 = 2v(ab) then (3)
I(. (1 is a metric on Q, and from (2) follows the "strong triangle inequality" Ila+bll
5
Max {
II"Il7
Ilbll
>
llall
+
llbll
.
Now we can imitate the process of completion by Cauchysequences by using the metric The resulting set is again a field, the field every element a
10.2 Algebraic 
Q of rational padic numbers. We remark that P
+ 0 of Q can be uniquely written as P
where 0 5 a i < p and a.
z
0. (This series does converge in the sense of
extensions of
4
11  1) .)
4P.
of finite degree (K:Q ) = n. It is a nontrivial 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 integraldomain, the ring of integers of K. The only maximal ideal of I is p= {a
1
aEK,v(a)>O).
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 padic 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 padic 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 pmodular system for G, if the follou ~ n gconditions are fulfilled: (1)
K is a splitting field for every subgroup U
tti
G. uhich means
for some m. (depending on U). 1
(2)
K is a splittingfield for every U
for some n.. J
5
G. T h ~ hmeans thy definition) that
10.4 Remark. Without proof we state that a pmodular 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 splittingfield P G follows from a famous theorem by R. Brauer (see Huppert I, p. 591 or Isaacs,
Let K = Q for every U
5
p. 161). That also R is a splittingfield for every U s G is a less deep result of R. Brauer (BlackburnHuppert 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 charactervalues,
I
hence in the field of I G th roots of unity over GF(p). As there are no primitive pth 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
Q
P
supplies a pmodular 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 pmodular system for G. Let V be a KGmodule of finite Kdimension. a) We call an IGsubmodule M of V an IGlattice of V, if MK = V and M is a finitely generated Imodule. As I is a principal ideal domain and M is torsionfree, M is a free Imodule and rankIM = dimKV. If
then n V = o m. K. j=1
J
As M is an IGmodule, 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 IGlattice of V, we form the KGmodule 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 pmodular system for G. If V is a KGmodule, there exists an IGlattice of V. Proof. Let {v 
v n ) be a Kbasis of V. Put
Then M is an IGmodule, finitely generated and torsionfree as an Imodule. As I is a principal ideal ring, so M is a free Imodule. (Here it pays to use padic numbers and not algebraic numbers.) Let n M = o m. I. j=1 J Obviously MK = V and { m
m, ) is a Kbasis of V
11.3 Example. Let G = ( g ) be the group of order 2 and (K,I,K) a 2modular system for G. Let 
be the KGmodule with v1g=v2, v g = v 1' 2 We put W1
= v1
+ v2 '
W2
= v1  v2 .
As Char K = 0, so {w 1 , w 2) is also a Kbasis of V, and w l g = w1 ' w2g = 
W2 '
Now
M  v I o v 2 1 and M  w l o w I 1 1 2 I 2 are IGlattices of V. On M the group G operates trivially, but not so on M 2 1' Hence M l 8 M2 . But the compositionfactors of M are independent of the choice of the IGlattice M:
11.4 Theorem (R. Brauer). Let (K,I,R) be a pmodular system for G and V a KGmodule. Let MI and M2 be IGlattices of V. Then M I and
a2have the same composition factors as
KGmodules (including multiplicities). Proof.  Suppose n M  o m. I. 2j=1 J
As m. E V = MIK, there exists an s E D1 b ( 0 ) such J Hence M ns C MI. As the multiplication by 2
d
&at
r, an
S
lCisomorphism 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
I+
1 r.
5,)
.~IY.>
is Artinian, being a finitely generaled i  r n d u k . Thcn SfI M~IT'" is a finitely generated icmodule, hence is also Artinian. Therctllrr r k r h > r r r n of JordanHiilder holds for M ~ I M ~ ~ + ~ . We consider the diagram
We have the following IGisomorphisms:
+ M2)/MIn = M21(M1n fl M2).
(1)
(MI"
(2)
M1/M2 = Mln/M2n
(3)
(M7n+M2)/M2~Mlnl(MlnflM2).
(4)
(by multiplication with n).
The JordanHiilder 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 compositionfactors. Finally, (4) and (1) show that M I / M l n and
M2/M2n have the same compositionfactors,
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 padic field, I its ring of integers and
the maximal ideal of I. Let
n R = o a.1 j=1 J be an Ialgebra, which is a free Imodule. a) R = R/Rn
becomes a Kalgebra by
+ Rn) (a + In) = ra + Rx (rl + Rn) $ (r + R;r) = r I 2 (r

I, +
Rrr
for r,rl,r2 E R and a E I. b) The mapping
is a ringepimorphism of R onto I? uirh k z m l R.T. c) R is complete under the topolo:!
~nhcntci!ir:a
I
d) Suppose
T = e1 +...+ C n with
J
ER
' e.J = b'J. . 5.Then thsrc exrit dcrnpxents e.J E R with
and
I = e +...+ e 1
n'
s s I
]
= J . e.
.. .
and
("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)
1
J1
€
RJ'. r2  r J
J
5
RJ.
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
+Rn=r
m
+Rn=e
Further
Hence (rl,r 2...) is a Cauchysequence in R, which by c) converges to a limit e. Then
and 2 e 2  e = l i m ( r r.)=O. j+cc
J
J
Finally we construct the idempotents el, ...,en in R exactly as in 2.5 b). We need the KrullSchmidttheorem for IGmodules. We prepare the proof by
11.6 Lemma. Let I be a padic ring and R an Ialgebra, which as an Imodule is free and finitely generated. a) R x C_ J(R), where J(R) is defined as the intersection of all maximal rightideals 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 skewfield. c) Let 1 be a padic ring and M an IGmodule, which is finitely generated and free as an Imodule. If M is an indecomposable IGmodule, then HomIG(M,M) is a local ring. Proof. a) We show R n C r for every maximal rightideal 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/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 skewfield. c) If k
M=
t
]=I
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(\l.bl t a a finltcly generated free Imodule.
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
;i
padic ring and X! a n IGm~lifule.finitely generated and free as an
n Imodule. Then M = o M. with indecompsabls ICmodules M.. The isomorphism types of j=1 J J the M . are uniquely determined. J Proof.  M is a Noetherian Imodule, hence a Stxtherlan IGmodule. 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 KrullSchmidttheorem. (cf. AndersonFuller, 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 IGmodule 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 leftEmodule. We put
Y = {q
I
q E H , q HomIG(M,U) C J(E)
1.
This makes sense as H HomIG(M,U) L E
.
Obviously Y is an Esubmodule of H and J(E)H
Y. Hence H = H N is a vectorspace for the
skewfield E = E/J(E). If we show that dimg H = r, then r is uniquely determined. Let pi be an IGisomorphism of U onto Mi. We claim that Ebasis 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
I
E~
EE
.
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
q
5
1
E~
p i (mod Y)
i=l
.
which shows
Suppose
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 {
*i.
6. = 6 . p . p . E J(E), hence b = 0 . J J J J J ... , pr } is an Ebasis for HY.
11,
11.8 Remark The KrullSchmidttheorem is \ e n n r t l ) true for UGmodules. In exercise 33 we show that it never holds if G is not a pgroup.
If G is a pgroup, there exists an epimorphrsm of ZG onto the ring U[E] of integers in the cyclotomic field Q(E),where
E~
= 1
* E.
lip
2:
23.
i t LS known
that U[E] is not a principal
ideal ring. If a,b are nonprincipal ideals In B [ E ] . rhcn a o b = U [ ~ ]~
a
b
(see CurtisReiner, Representation Theory o i F ~ n l t cGroups and Associate Algebras, p. 150). This shows that the KrullSchmidttheorem is n~rttrue for Z [ E ] , hence also not for ilG. Hence the KrullSchmidttheorem can h d d at most for pgroups with p
5
19. (See also
CurtisReiner, 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 Jacobsonradical, defined as the intersection of all maximal rightideals of R. a) If r E J(R), then there exists s E R with (1r)s = 1. b) If M is a finitely generated Rmodule with M = MJ(R), then M = 0. c) (Nakayama) If M is a finitely generated Rmodule and N a submodule of M with
M =N
+ MJ(R),
then M = N.
d) If M is a simple Rmodule, then MJ(R) = 0. Proof. a) Suppose (1r)R C R. Then there exists a maximal right ideal 
r of R such that
(Ir)R 5 r. But then 1 = (1r)
+ r E r + J(R) = r C R,
a contradiction. Hence (1r)R = R and therefore (1r)s = 1 for some s E R. k b) Suppose M =
1
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 (1r,)s = 1. Then
a contradiction. Hence M = 0. c) We apply b) to the finitely generated Rmodule 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 IGmodules, finitely generated as IGmodules and free as Imodules. Suppose that P is a projective IGmodule. If M M is a projective IGmodule.
= P, then M = P. In particular, also
b) Let el, e2 be idempotents in IG. We put

e. = e. 1
If
1
+ IGn.
el KG s 2 KG, then elIG e e21G.
c) Let M be a finitely generated IGmodule, free as an Imodule. If M is a projective KGmodule, then M is a projective IGmodule. d) Suppose p ); I G I . If M is an IGmodule, free as an Imodule, then M is a projective IGmodule. Proof. a) We consider the diagram 
Here
x is an isomorphism and v, is the epimorphism defined for m E M by m v2 = m
+ Mrrf
I\ \Ix=
s!.
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
exists
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 uhers u ' 5 Prr. Then t
O = u a = ( u ' a)x. As M is a torsionfree Imodule, this implies u'm = 0. But then
O = u'nv 2
U'Y
1
a
with
u'v
1
* 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 KGmodule is projective. Hence M is projective by c). We now fix our notations. 11.11 Definition. a) Let (K,I,K) be a pmodular system for G. Let h = h(G) be the classnumber of G and let V1, ...,Vh be the simple KGmodules (up to isomorphism). Let k = k(G) be the number of p'classes of G and let W1, ...,Wk be the simple KGmodules (see 5.5 and 5.8). By 11.2, there exists an IGlattice 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=
n
?KG i=l
with indecomposable KGmodules
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
KGmodule
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 compositionfactor (and direst summand) of eiKG is d... J J' b) If C is the Cartanmatrix for KG, then C = D ' D . whkh 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
J
c
i
= 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 IGlattice of Vr. Then h
is an IGlattice of e. KG. The multiplicity of Wi J Hence the multiplicity of Wi in N. is J
P
7 KG !5 J(KG) in Mr is by definition dri.
But also e.IG is an IGlattice 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 Cartanmatrix. 11 Now we can prove that for p'groups the classical and the pmodular representationtheory "are the same". 11.13 Theorem. Suppose p ./,
I GI
and let (K,I,K) be a pmodular system for G. Let V1, ...,Vh
be the simple KGmodules (up to isomorphism) and let Mi be an IGlattice of Vi. Then MI, ..., Mh are the simple KGmodules. 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 KGmodules 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.
1
E
W. (i=l,...,h). 1
..., h).
Renumbering the Wi, we
1 1
11.14 Remark. Every simple KGmodule W. appears as a wrnpositionfactor 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 Gset on which G operates b>permutation. We form the UGmodule 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 UGmodule. (Look at the trace of a projection rn Horn r
I
36) (A. Dress) Let GI =
'1
n I=
1
pl
1
ZGt LlQ?. \  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
and
a) If a is defined on V(Qi) by (Pig) cx = 1 then a E HomuG(M,?Z).
.)
(i=l .....r ~ .
b) Let
r
with bi E I. If we define
then
p E HomnG(U,M).
c) Show that M=KeraoimfisKeraon.
As U
V(Ri), hence the KrullSchmidttheorem ist not true for Wmodules.
9 1 2 Blocks of defect zero.
Throughout this section let
(K,I,R), be a pmodular 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 RGmodules. 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
G
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 RGmodule. uhich b p j e c t i v e and finitely generated over R. Then the following are equivalent: a) M is a projective RGmodule.
G
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 RGmcdule F. Since M is projective we have F = M
Q
M ' w lth a free RGmodule F.
Let n denote the projection of F onto M with kernel 31'. As n is RGlinear, it follows
G G T I ( ~ I M n ) =T 1 ( ~ I M ) x =id M ' b)
+ a): Let N be an RGmodule 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 Rprojective 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
Now
with T G l ( a n ) E HomRG(M,N)
12.3 Lemma. 
(y is RGlinear)
.
Let M be an IGlattice. If
{
q =ml + Mn, ... ,ms = ms + M n )
is a Rbasis of M, then ml,
... ,ms is an Ibasis 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
S
would be a contradiction. Hence M = o m.1 . i=l 1 12.4 Lemma. Let V be a simple KGmodule with IGlattice M. Let a E EndIG(M) and let tr denote the usual trace of linear maps. Then a) v(rg M)
5
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 Ibasis 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 KGmodule with IGlattice M. Suppose I GI I P
dimKV
.
Then a) M is a projective IGmodule
.
b) M is a projective simple KGmodule. Proof: Choose 0 + iii = m + Mn. By DefineaEEndR(M)by
12.3 let m = m
mln=ml
1'
and m a = O I
... ,ms be an Ibasis of M. forir2.
Now by 12.4 v(rgM) Since
lGlp
I
5
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.
too.
Furthermore,
hence
M=mIG.
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 KGmodules 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
KGmodules 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 KGmodule 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
I
g E G) generates M.
By 12.3, we may choose suitable vg which are an Ibasis 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)
5
G v(tr TA(y)) = V( G:A ) + v(tr y) = v( 1 G:A )
I
I
I
I I
Since this holds for all primes p we obtain ~ ( 1 ) G:AI
.
12.8 Theorem. Let V he a simple KGmodule with character X. Then the following conditions are equivalent a) I G l p
I
xCl)
*
b ) ~ ( g ) = O forall g E G with plo(g). c) ~ ( g =) 0 for all nontrivial pelements 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 1Gmodule. 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 KrullSchmidt 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 psubgroup 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 tsided indecomposable direct summand of KG (see 2.3). B is called of defect zero if B = (K)n as a Ralgebra. If we regard B as a KGright module. then BsnM with a simple projective KGmodule .\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 Cartanmatrix 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 KGmodule. By 2.11, Bi This proves Bi =
s
(dim Mi)Mi as a KGright module.
M. as a Kalgebra. 1
12.12 Theorem. Let V1,
.. ,Vh denote the simple KGmodules. Suppose pa = I GI p.
a) Let Wi be a simple KGmodule 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 IGlattice in VP then Mr = Wi. Also
$1
dim Vr
.
b) Suppose pa 1 dim Vr. If Mr is a lattice of Vr, then Mr is a simple KGmodule 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
ic
j, then o = c..=Cd . d sl sj  drj 'J s
This implies Mr
c:
.
W i by the definition of the decomposition numbers.
Since Wi is projective, Dickson's theorem (see 3.5) implies IGI
P
I
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
1=c
rr
=
C
2 dtr
forces dtr = 0 for t z r.
t= 1
12.13 Proposition. Let S be a Sylow psubgroup of G. If KG has a pblock of defect zero, then S fl
sg= 1 for a suitable g € G. In particular, 0P(G) = 1.

Proof: Let M be a simple KGmodule belonging to a pblock of defect zero. Thus MI
e R S for some e € l. As
om^^((^^)^, M) = ~
IS)
o m ~ ( 9Q
M is a factor module of the permutation module
m
e H O ~ ~ ~ ( K ~I , 0,R S )
(slG.
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
N.
Applying Mackey ' s formula, we obtain
= M1 asiS. ~ e . . .
(K,)GI
S
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!:
 S I , the KrullSchmidt theorem
sgn S = 1.
12.14 Remark. To prove the existence (rtsp. r*.wcrstencu)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 pblocks of &iaI u u long as p (One easily finds
x E Irrg(G)
with
1G
x : :sb?
2
5.
~nspectionof the Atlas.)
b) (Michler, Willems). If G is a group of Lier?pc. rhrn G has pblocks of defect 0 for all primes p. Here the proof depends on DeIigneLusaig 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 pblocks of defect 0 for p
2
5. Using the theory of modular f~~
asymptotically, which means: Given a prime p pblocks of defect 0 for all n all n
2
5.)
2
z
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
f
12.16 Example. We determine the simple KGmodules for G = SL(2,2 ) and Char R = 2. a) If E
L
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 Gmodule 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
[Y
:I.
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 Galoisconjugates 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
f
+ 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) =
10
if if
KGmodule 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 KGmodule M are the same, namely
A? the tracefunctions of simple RGrnidules 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,
I
. Then
... , js are distinct. But
LSi
8
kri contains the KGsubmodule of
1
1
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 KGmodules.
5
... < iS s; f a r e all the
12.17 Example. We determine the projective indecomposable KGmodules 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,
For
+
N = (1,

KGmodules. a)
VI
*
E
for all I :
VI
Hence tr A = a I I
+ aZ2 = tr Al. This proves Vi
simple modules. For a general I = { i
*
v I ( V .l1
1
*
0...@Vi) S
"
G
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
1
Since the trace functions of simple KGmodules are linearly independent over fi we see that Vi+l appears in V. o Vi with odd multiplicity, every other simple KGmodule 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
0
...
N o VN = VN o P(K) o ...
where P(VI) denotes the projective cover of V
I
.
'
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)
(V1
O
... O Vf) @I (V1
O
E
... O Vf) P (V 1 8 \.,)
has a section isomorphic to V1 o ... o Vf =
P(K)
@
VN =
 d V f @ Vf)
\'s.
P(V,) = \'$ a \'\
d)
8
ft3r
I *2
\'?; s \'\ .
Using c) we obtain
I GI = 1(dim VI) (dim PI \.I
)
s
Z I
t
I
Hence equality holds ever\\\ hcrc .inJ J
I,
4 I ZI
1
P
(f111))  2f
Then P(K) = (KI Z)
, the t r ~ v ~ Zrnodulr al KZ I \ prta.cir,\e Hence K G Z G Horn ( ( K J Z ) , k) a Hornu KG
Since
f+
LC.:
e) Let Z be the c y c l ~ csubgroup ~ ) Gt o r tv&r 2 Since 2
.
I
P(K) is a direct summand of (K 1 Z)
G
G:
15 projective,
by 4.5.
K;;IZ ) = K '
.
f f By d), dim P(K) = 2 (2 1) . Thus P t K ) =
IK
b
G
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 Cartanmatrix of SL(3.8) for p = 2 39) Let p
2
3 and q be primes with p ( q1. 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 Sylowpsubgroups S and S of G with S1 f' S2 = E. 1 2 p 1 , hence none of them is divisible by p2. b) The characterdegrees 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 ppower in characterdegrees of G is p k .
I
2) H has Sy16wpsubgroups 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 matrixring (I), 
over a padic ring I.
If
A = E (mod n I), then m is a power of p = Char b) Let
E
R.
be a primitive mth root of unity in the p a d i c field K, where Char R
Then E =
E
+ n I is a primitive mth root of unity
in
4m.
R
c) Let (K,I,K) be a pmodular system for G. Icr V be a KGmodule and M an IGlattice in V. If the trivial module is the only KGcompcaitionfactor of M, then G K e r V is a pgroup. Proof a) Suppose q
p is a prime dividing r n \Vc put B = A ~ ' ~Then . Bq = E and
i
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
13
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)
5
(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 KGmodule is W5 = Finally,
8 = ~ ~ $=11 )+ 7 d65 shows d65 = 1. Hence
Also now all simple KGmodules are red&ti.urs of IGlattices. (Observe det C = 3.) Char
R
= 7: As G
B
PSL(2,7), we obtaln fnrm 5.10 all the simple KGmodules
W1, W2' W3, W of dimensions 1,3,5.7. 4
s e ~mmcdiately
\\c
3 ~ 3 WZs (by 13.1,
anj
q
m
WJ
(by 12.5).
(Observe that we do not yet know the BrauercParactzr of W3 !) We have
As G has a 7complement (isomorphic ro Sir. uc ha\c by 1.5 C) 7 = dimP(W ) = 1
K
1
+ 6 dll
"
This implies d41 = 1, d61 = 0. Then 6
= ( 1~) = 1 + 3 d 4 2 + 5 d J j 4
forces
dJ,=O,
dq3=1.
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 KCmodule W3 is not the reduction of an IClattice. (Observe det C = 7.) 13.9 Example. Let G = SL(2,S). The charactertable is given by P
where
allZ =
The characters
9
and bli2 =
x 1 up to x5 come from G/(g2)
theory A, p. 235.)
= A5. (See L. Dornhoff, Group representation
Char
K = 2: As 02(G) = (g2) is by 3.6 in the kernel of every simpf KGmodule, we know
all simple KGmodules from 13.6, and we even know the first 5 rows of the decompositionmatrix D. We have simple KCmodules 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 Brauercharacters 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

(i = 6, ... ,9) g, is rtpccscoted by E, hence cannot contain any
compositionfactor Wi ( i = 1,2,3,4) of
R
A;.
As
we obtain by 13.1 two further simple KGmcdulcs
% 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 KGmodules Wi are 1,3,5,2,4, where W1, W2' W3 are modules for
e
As. Obviously
%
is the simple module W4
Separating the modules with g2 in the kernel and using knowledge of As we obtain
D=
Here
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.
andC=
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 KGmodule is the reduction of some IGlattice. 13.10 Remarks. a) The determinant of the Cartanmatrix 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 compositionfactors
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 KGmodules are obtained by reduction of some lattices. We remarked already that this is true for all psol\dblc g r o u p
1 9 111. 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
it
dtws hold for all primes in the nonsolvable 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 KGmodule can be I~ticrlt o an IGmodule with prational character, then G is solvable. ( G. HiR, J. of Algebra 91 I 19R5). 1 1 1 (1987).)
(The values a
112
and b
112
in 13.9 are n~jt5nt~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 2group. (If g E Ker a and o(g) = 4, then we obtain an equation i = 1 + 2b, where i is a primitive 4th root of unity and b an algebraic integer.) 41) Using the charactertable of Aq, compute the decomposition and Cartanmatrices of A4 for Char K = 2,3. 42) Using the charactertable of SL(2,3) (Trento 1987, p.52), compute the decomposition and Cartanmatrices of SL(2,3) for Char K = 2,3. (There are simple KGmodules of dimensions 1,1,1 for Char K = 2; simple KGmodules of dimensions 1,2,3 for Char K = 3.) 43) Using the charactertable of GL(2,3) (Trento 1937, p.56), compute decomposition and Cartanmatrices of GL(2,3) for Char K = 2 and Char K = 3.) (Char
R = 2: Simple KGmodules of dimensions 1,2; det
5 C=2 ;
2 Char R = 3: Simple KGmodules 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
+b
over L = G F ( ~ ~Let ) . (K,I,K) be a pmodular system for G. Then G has the normal subgroups
T = { (x:b)
IbEL}
and X S = { (ax+b) I a, b E L, a
a) Show that T = 0 (G), 0 ,(G) = E and
P
P.
#
0)
.
hence I GIG'
I = p(p1).
b) The charactergroup of GIG' is of the form (a)x
( p ) , where o(a) = p
Obviously, KG has p1 simple modules of dimension 1 with trace functions
and o(p) = p1. 1
(i = 0,
... , p2).
c) G/T has t = pPll 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 pPl. All simple P KGmodules have been listed. e) Let M. be an IGlattice in W.. Then T contains every simple KGmodule with multiJ J J plicity 1.
Q The decomposition matrix of G is
where
F=
[: 1 :
is of type (p,l). Therefore
Calculation shows det C = p2p1 . 45) Let K be a field of characteristic p > 0 and let N be a normal psubgroup of G. Put J = J ( m ) and let G act by conjugation on the powers J'. Furthermore for a simple KCINmodule V let P resp. P denote its projective cover as a KG resp. KG/Nmodule. Prove a) P = P/PJ.
.
b) P/PJ a J ~ / J ~ +G'
I
c) dimKP = N I dimKP d) If N
I
Z(G), then CG = IN CGIN , where C
I
and GIN.
. G resp. CGIN are the Cartanmatrices w.r.t. C
$14 The Theorem of Fong and Swan.
The purpose of this section is the proof of the following important theorem: 14.1 Theorem (Fong, Swan). Let G be a psolvable group and (K,I,K) a pmodular system for G with algebraically closed K. Let V be a simple KGmodule. Then there exists a simple KGmodule V0 such that for every IGlattice M of Vo we have
fi = V.
An immediate consequence is
14.2 Corollary. Let G be psolvable and R a splittingfield for G of characteristic p. If V is a simple KGmodule, then dimR V divides I G I . (We remind the reader that the simple group PSL(2,7) has in characteristic 7 a simple module of dimension 5; see 5.11.) We show at first that a pmodular system with algebraically closed 14.3 Lemma. There exists for G a p m t d u l a r system (&I$) R is algebraically closed. Proof. We cannot take K itself 
to
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("
1
As K contains a primitive I GI th root of unity, it is a splitting field for all subgroups of G. 0 Adjoining an nth root of unity with p n defines a nonramified extension of Q P' ). Let K be the completion of KO. Then the Hence KO is a nonramified extension of Q ( E P IGI of valuation on K is still discrete. The residuefield R of K contains all primitive nth roots
4
unity for p &, n (see 13.1 b)), hence K is the algebraic closure of GF(p). (For details about padic fields see again J.P. Serre, Corps Locaux.)
14.4 Lemma. Suppose N 4 G. Let L be a splittingfield for G and N. Let V be a simple LGmodule such that
with a simple LNmodule W. Suppose that there exists an LGmodule U such that UN Then there exists a simple LGINmodule X such that V
e
W.
= U o X.
Proof. By Nakayamareciprocity (3.8) we have 0
z
G HornLN (V ,W)=HomLG(V,W ) .
By 6.10 we have
wG = (UN )G = U oLLG/N. If X is a simple LG/Nmodule, 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/Nmodule 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 counterexample 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 psolvable, 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 Tmodule, necessarily simple (as V is simple). As 1 dim  H < dim  V, there exists an I Tlattice M1 such that K 1 K
e
HI. Then
is an IGlattice in M oIK and obviously
This is impossible as (G,V) is a counterexample. 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
5
Z(G). But then ( G ) = N x 0 (G) = S = O , ( G ) .
P'>P
P
P
As G is psolvable, this implies G = 0 .(G).Then by 11.13 we do not have a P counterexample. 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 Slattice Wo with
0
= W and Wo 4 K is a
simple KNmodule. By 11.10 dl. \VO is a pojective I Nmodule. 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 = ua 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
Iautomorphism 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 plgroup and H is still psolvable. (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(y1l P(X) P(Y) = p(xY) , hence p(y) E ?Ly as p(y) 1
I
= 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)) = (xg1c We define a homomorphism
7c2
of H into AutI Wo by
(g,a)n2 = a:
.
Then Wo becomes an I Hmodule 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 KGmodule 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)) =
VX
.
Hence by (1) VNt By 14.4 we obtain V
e
e
Wo
... o W .
%o 17; Y with a KHmodule Y. As by (2)
dim  V = dim K K
O
dim
R Y > dimR Y ,
the pair (H,Y) is not a counterexample. As H is psolvable, there exists an Ifree I Hmodule 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 Hmodule Vo = Wo
Q
I Yo. Therefore only G = HIS operates on Vo and (G,V) is not a
counterexample. 14.6 Remark. Isaacs gave the following improvement of 14.1: The IGlattice M can even been chosen with prational character. If p > 2, then this character is uniquely determined. (Isaacs, Pac. J. Math. 53 (1974), 171188.) 14.7 Theorem. Suppose G is psolvable and W a simple KGmodule such that
I GI P I
dimRW. Then W is a projective KGmodule. (For solvable groups this is 9.7.)
Proof. By 14.1 we can lift W to an IGmodule Wo. As I GI I P
nos W are projective.
dimKWo, so by 12.5 Wo and
$ 15 More on BrauerCharacters and the CartanMatrix
As a motivation for the results to come, we begin with a very special case.
15.1 Example. Let G have a normal Sylowpsubgroup P. By the SchurZassenhaus theorem (Huppert I, p.126), then G = PH is a semidirect product. If "inflate"
(1)
t
to a 2 E Irr G with P
5
tE
Irr H, we can
ker 2 by ?(y h) = 4 h ) for y E P, h E H
or induce
(2)
T
to
G
T
.
By 3.6 b), P = 0 (G) is contained in the kernel of every simple KGmodule 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 KHmodules and the simple KGmodules; namely an I Hlattice M in the simple


KHmodule 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 KGmodule by 4.5. If T is a simple m  m o d u l e , then HornKG (W Therefore t.E
J
wG
G
,?) = Hornm
(W,T) =
aWT.
is the indecomposable projective KGmodule with head
?. If xi E Irr G
and
Irr H, then by 11.12 a) the decomposition number is
Hence we can calculate the decompositionmatrix
D and the Cartanmatrix
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
G
f' H = h
H
. Hence
C from. the
G ICG(h)l (h) =
t
T x€hGn
H
G This shows ( t )H = n t , where n is the permutation character of H on the Hset 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
I
I
is a power of p and IPI s d e t C s [ P I h(H) .
I
Furthermore, det C = I PI if and only if C (h.)] = 1 P '
I
for all hi
t
fixedpointfreely on P. While det C = P I h(H) if and only if G = P
1, hence if H operates x
H.
The question naturally arises which of these assertions remain true in general, i.e. if the Sylowpsubgroup is not necessarily normal.
We start by showing that C is an invertible matrix. For that purpose let si (i = 1,..., k)
I
be representatives for the plconjugacy classes of G and let ri = I CG(si) 'I2. As before, we denote the irreducible Brauer character of the simple KGmodule Wi by
r$i
Pi (i = 1,..., k).
Let
be the character of the projective I Glattice 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
'J
m,n=l
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 Kvectorspace of class functions on S. In particular, we form yij = (/3i,/3j)' and
T = (y..). 'J
15.4 Proposition.
r
= Z Z ' = CI
is the inverse of C. In particular, det C is a positive
integer. Proof. From 15.2 follows 
c = ( 2 ' )  1 z1 = ( z z ' )  l , hence the second equation. Also
= (p. P.)' = yij. 1' J Hence C" = r. As C is a regular matrix over
det
H, its determinant is a nonzero integer. Since
r = det Z det is positive, so is det C.
15.5
Corollary. a) Let
P
and
P'
be projective KGmodules which have the same
composition factors, including multiplicities. Then P = P ' . In particular, the Brauercharacters determine the isomorphismtype of projective RGmodules. b) (R. Swan). Let M and M ' be projective IGmodules. 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 KGmodules 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 Brauercharacter
M o I K
E
M'
I
E
P'.
of M is the character of M o I K, restricted to p'elements. Hence if
I K, then the projective KGmodules M and M ' have the same
Brauercharacter. By a) this implies M
e
M ' , hence M
M ' by 11.10 a).
15.6
Corollary. a) ( $ i , P . ) ' = 6 . . . J 1J b) {(Ii I i=l,..,k) and {Pi i=l, ...,k) are Kbases for the space of class functions from S to K.
I
Proof.  a) By 11.12 we have h
h
k
b) Suppose (on S). Then by a)
k
0=
(.Ia. (I., 6.)' = a. 1=1
1
J
J
(j=l,
...,k).
Hence the (Ii are linearly independent. The same argument works with the role of the (Ii and
Pi
interchanged. Since there are exactly k p'classes in G, the assertion follows.
(The linear independence of the
Pi
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. Hlinear 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 noncyclic Sylow subgroup; these groups are called elementary.) For two of the many proofs we refer to Huppert I. p. 586591 or Isaacs, p. 126131. 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 nelement and g n' p'part s are
g,
'
a n'element. All the elements of G with a fixed
I'
{U
CG(s)p}.
Here, by definition, G is the set of all pelements in G. (One should not confuse 1 GI and P P Gp 1 ; 1 G 1 is the order of every Sy lowpsubgroup of G, hence is a power of p, while in
I
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) =
0
= 1 P otherwise;
if g
and a3(g) =
I CG(g)pl "(g) 0
g
=
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
H
for every pfsubgroup H
of
G. Then
ai (i=1,2,3) is a
generalized character of G. Proof. 
a) For n l and rr
2
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 pgroup. Then
and
are obviously generalized characters of F.
b) We consider a3. If
x E Irr G, then 
("3'~)~

where
x1
II CG(s)p l a(s) ji(~)
&
E
3
u s )=
&
&
.,(us)
x,(us) = ( a l '
sES uEC (s) G P
FrsES uECG(s)P
stands for ( x ~ ) AS ~ . by a) al and
x1
sES
x1lG
7
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 decompositionmatrix, 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
Therefore
P'
I.
I GP , I .
divides
(It can be proved more generally that
15.1 1 Lemma. a) Let 
I GI
divides
I Gn(
/j be a Brauercharacter of
for any set n of primes.)
G. Then
pH
is a character for every
p ' subgroup H of G.
p is a Brauercharacter of G, there exists a generalized character q of G such that qS = 13. (If G is psolvable, by $14 7 can be choosen as an ordinary character of G .)
b) If
Proof. a) Let fi be the Brauercharacter of the KGmodule 
W. Then
&
is the
Brauercharacter of WH. As H is a p'subgroup, by 11.13 there is an I Hlattice M such that M
WH. By 13.4 c),
; I
pH
is the character of M.
b) By a),
pH
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 
xi
of G onto
.
d.. p. . ~ i l ~ = ~ j: ~ We extend it by Ulinearity to a map from the generalized characters to the generalized Brauercharacters. 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 psingular classes. Then
VI
is a Ulinear combination of the $i. And conversely, any $i vanishes on all
psingular classes. Proof. By 15.6 b) we have k I# =
1
j=1
a. $. (on S) J
J
for suitable a. E K. By 15.6 a) we obtain J a.J = (v7 Pj)' = (?, (Pj)1IG (since
I
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
I
I
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 Sylowpsubgroup of CG(P). P
0 (mod PI.
Proof. a) Clearly Z(P) = P
n cG(p) 5 c ~ ( P ) .
Let u be a pelement in CG(P). Then (P,u) is a pgroup, 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
P'
(G
P
,I
. 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 Brauercharacter 
pi
(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
I
b) If
is the trivial Brauercharacter, then
P
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.