The Representation Theory of Finite Groups (North-Holland Mathematical Library 25)

Furthermore cfi O(1) as cfi O. Hence cfi Vcc U OP). Let H be a p'-subgroup of G and let V be the R[G] module induced by the trivial rank one R -free R[II] module. By (111.2.6) I. LI V. Thus (11.2.9) implies (iii) and (iv). E THEOREM 4.16. Let B be a block of defect d. Let D (' be the decomposition matrix corresponding to B and let C`) be the Cartan matrix corresponding to B. Let A = (a 1 /p") where xs, x, range over all the irreducible characters in B. Then there exists a unimodular matrix X with rational integral entries such that

XAX' =

(Cc)-1

o

o

Furthermore (i) Every elementary divisor of C" divides p d and exactly one elementary divisor of C" is equal to p d . (ii) If C" has n elementary divisors p d-1 then B contains at least n +1 irreducible characters of height O.

PROOF. By (3.7)(iii) as,

= P (X,

x, Y = pd dsid,aq.

Thus A = p dD °(C") -I (D7. By (3.11) all the elementary divisors of D are equal to 1. Hence there exists a unimodular matrix D, with rational integral coefficients such that A

pdD,() °)D, 0 \ O

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Let X = D (i) Since p d (CI' has integral coefficients every elementary divisor of C° divides p d. By (4.7)(i) A has rank one. Thus p d (C) -1 has rank one and exactly one elementary divisor of C° equals p d . (ii) Choose x, E B of height 0. For x„ E B define a,,

e,, =

x,(1) a„ xr(i) p

d

for s r,

has height By (4.7) each e„ is a rational integer and en = 0 (mod p 2) if hs > 0. Hence if B contains m + 1 irreducible characters of height 0 then (es,) has at most m rows with entries not all —= 0 (mod p 2). This implies that A, and hence p d (0 -I , has at most m + 1 elementary divisors not divisible by p 2. Thus C° has at most m + 1 elementary divisors divisible by p d-1. Hence by (i) n m. 0

More precise results than (4.16) can be proved in special cases, see e.g. Fujii [1980].

COROLLARY 4.17. Suppose that for each integer d, G contains exactly m d conjugate classes of p'-elements which have defect d. Then G has at most m d blocks of defect d. Furthermore G has exactly md blocks of defect a. PROOF. Let d, be the defect of C,. By (3.11) {p d, } is the set of elementary divisors of C. The result follows from (4.16)(i). D

The next result will be improved later. See (VII.10.14) below. For an interesting application see Ito [1962a]. THEOREM 4.18 (Brauer and Feit [1959]). Let B be a block of defect d. For each integer h let kh denote the number of irreducible characters in B of height h. Let k = E k h denote the number of irreducible characters in B. Then k E khp 2„

,

Ap

2a +1

h

and if B contains an irreducible character of positive height then k p 2d-2 . Furthermore the height of every character in B is 0 for d = 0 or 1 and is at most d — 2 for d 2.

PROOF. By (4.1) (iii)

Es a',= pu a„.Thus if x, has height 0 in B then by (4.7)

2

( pa :_rd

d

(k0

1) + E

11=1

k,,p2d

E ( po - d

=P

di

a„

4,a-d )•

159

CHARACTERS IN BLOCKS

Hence k kr,p 2h pu — u 2 +1 for some u. Since p` u — u 2 1,p' for p' + 1. Thus in particular E'L., k h all u it follows that h..o khp 2° Ip

2d-2

4

Suppose that x, has positive height. Then by (4.7) )2 (

d

a„ k aP2

p % ) = P d 13'

Thus kop 2 pd p

?)

( pa .%) '

0<

a„

0 (mod p'). Hence the second inequality implies By (4.7) a„ I pc that < p° and so h + 1 < d. Hence h d — 2. The first inequality implies that {c op' p ° 14 — u 2 p 2d for some u. Thus d

k = k0 +

E

k,

P 2d - 2

A. P 2d

2d - 2 =

E

I

In case G is a p-solvable group Haggarty [1977] has found some better bounds for the possible heights of irreducible characters and has shown that his bounds are of the correct order of magnitude. LEMMA 4.19. Let B be a block of defect d. Let k be the number of irreducible characters in B and let I be the number of irreducible Brauer characters in B. Then 1 k. Furthermore the following are all equivalent. (i) d = O. (ii) k = I = 1. (iii) k = 1. (iv) det = 1, where C° is the Cartan matrix of B. (y) = (1). (vi) All R [G] modules in B are projective. PROOF. Since C° = (D ())'D ° where D' is the decomposition matrix of B the nonsingularity of C° implies that I k. (i) (ii). By (4.18) k <2 and so 1 = k =1. (ii) (iii). Clear. (iii) (iv). By (3.11) det = (det = 1. (iv) (v). By (4.16)(i) = (1). (v) (vi). U = L, where L, is the unique irreducible k [G] module in B. (vi) (i). v(cp,(1))= a for all cp, E B. Thus d = 0 by (4.5). 0

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LEMMA 4.20. If v(x,(1))= a then xs is in a block of defect 0 and xs(y)= 0 for all p-singular elements y E G.

PROOF. By (4.18) X. has height 0 and so is in a block of defect O. The last statement follows from (2.5) and (4.19). 111 4.21. Let B be a block of defect d. Suppose that B contains a unique irreducible Brauer character. Let k be the number of irreducible characters in B. Then k p d . If furthermore k = p d then every decomposition number for B is 1, every irreducible character has height 0 and the unique Cartan invariant is p d. LEMMA

PROOF. Let cp = cp, be the unique irreducible Brauer character in B. Then (cu ) is the Cartan matrix of B. By (4.16)(0 c• p d. Hence E = p d. If xu E B then d„,/ 0 and so 1. Hence k p d and if k = p d then d„, =1 for all xu E B. This implies the result. CI If k < p d in (4.21) then the situation is a good deal more complicated. This is clear by considering the case that G is a p-group.

LEMMA 4.22. Let p = E a,co, with a„ E K such that p(c,)E R for all] and 0 is a central character of ft[G]. Let e be the central idempotent of R[G] so that 'e is centrally primitive in R[G] and p corresponds to elf B is a block, let p B = E,g asco„. Then p (0,) E R for all j. Furthermore p B = 0 unless B is the block B (e) corresponding to e, and p'`') = 15 is a central character of R[G]. PROOF. Let f be a centrally primitive idempotent in R[G] and let B(f) be the block corresponding to f. By definition p") (0; )= p(f) for all ]. Thus p 8() (0,) E R. Since p(0,)= (e0,) it follows that p (Je). Thus (3 ") = 0 if e/ f and p B( ') = (5. Lii

/3 B(f ) (

0,) = p(f Of ) =

The next three results in this section which are of a more special nature are due to Brauer and Tuan [1945].

LEMMA 4.23. Let P be a Sp -group of G. Assume that the principal block of G contains a faithful irreducible character x = x, with x(1)< 2p. If G = G' then C o (P) is a p-group. PROOF. By (4.2)

C, lx(x,)1x(1)=- 1C1 1 (mod 77-) for all j. Suppose that

CHARACTERS IN BLOCKS

161

x E C, is a p'-element in CG (P). Then I C, 1 0 (mod p) and so x(z)=. x(1) (mod 7r) for all z (x). Since (x) is a p'-group this implies that (x<x), t()= 0 for every nonprincipal linear character fl, of (x). Since x(1) < 2p this implies that x(z )= n + pp, (z ) for all z E (x ) and some linear character pt, of (x) where n is the smallest nonnegative residue of x(1) (mod p). Let f be a K[(x)] representation which affords x(„) • Then det f(x) = (x). Since G = G' this yields that ii(x) P = 1. Thus p,(x)= 1 since (x) is a p'-group and so x is in the kernel of x. Thus x = 1 since x is faithful. Hence CG (P) is a p-group. 0 LEMMA 4.24. Suppose that Z(G)= (1). Let P be a Se -group of G and let 1Z(P)i = pb• If G has a faithful irreducible character x = x, with x( 1 ) = P" for some integer n then x is in a block of defect d a — b. In particular if P is abelian then x is in a block of defect O. PROOF. Let Z =Z(P). For z E Z let C, be the conjugate class of G containing z. Then la IXO (mod p) for z E Z and so x(z)lx(1) is an algebraic integer. Since x(z ) is a sum of x(1) roots of unity a well known argument of Burnside implies that either x(z) = 0 or I x (z )1= x(1). Since x is faithful and Z(G) = (1) the latter possibility can occur only for z = 1. Thus x (z ) 0 for z E Z, z 1. Let B be the block of G containing x and let d be the defect of B. Choose x, of height 0 in B. Then by (4.2)

I CI Ix, (z) x,(1) for

z E Z, z

jx(z) x(1)

0 (mod 7r)

1. Thus x(z) = 0 (mod 'up') for z E Z, zi 1. Hence

E xr(z)----)6(1) (mod 7rli a-d

)

zEZ

and so v(EzEzx,(z))= a — d. Since that a — d b as required. 0

EzEzx,(z)=- 0 (mod p t') this implies

LEMMA 4.25. Let 1G 1 = p'ego where p and q are distinct primes and (pq, go)= 1. Assume that G contains no elements of order pq. Then the following hold. (i) Let B(p) be a fixed p-block of G and let B(q) be a fixed q-block of G. Then Ex,(1)x s (y)=- 0 (mod (l b ) for every p-singular element y in G where x, ranges over all the irreducible characters which lie in B(p) and also in B (q). (ii) If pq11G1 then there exists a nonprincipal irreducible character which is in the principal p -block and also in the principal q -block.

162

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be a p-singular element in G. Let 0(x) = where x, ranges over all irreducible characters in B(p). By (3.14)(ii) 0 vanishes for all p'-elements in G. Hence by assumption 0 vanishes for all q-singular elements in G. Thus 9 A `" ) vanishes for all q -singular elements in G by (3.14) with q in place of p. Thus by (3.13) 0"' ) (1)—= 0 (mod e) proving (i). (ii) Let B(p) be the principal p-block and let B(q) be the principal q-block. The result follows directly from (i). PROOF. (i) Let y

The next result was first stated explicitly by Lusztig [1976].

LEMMA 4.26. Let D be a p-group with D i G. Let G" = G ID. Let ço be an irreducible Brauer characterof G. Let 0,0" be the principal indecomposable character of G, Go respectively which corresponds to cp. If y is a p'-element in G then

(Y)I 00(y0 0( vl ,—I ICG ) = cGo ol

CG

(y) n D

In particular 0(1)=113010"(1).

Every irreducible Brauer character of G has D in its kernel by (111.2.13). Thus if {cp,} is the set of all irreducible Brauer characters of G, it is also the set of all irreducible Brauer characters of G". Hence if a is a linear combination of the ço, it follows that (0, a)= (49 (), Choose a p'-element y in G. Define a by PROOF.

if x is conjugate to y

{1 a(x)= 0

otherwise

then 1 1G 0

G

= (0, a)G = (0 (), a) G0 1

= Got

0

(Y )IG

o

CG°(01-

Consequently the first equation is proved. The second equation follows from the fact that CG0 CG (Y)/CG (y) n D as D is a p-group and y is a p'-element.

COROLLARY 4.27. Let D be a p-group with G = DCG (D). Let {ço,} be the set of all irreducible Brauer characters of Go and let c,, e, be the Cartan invariant of G, Go respectively corresponding to cp,, cp,. Then c,,

163

CHARACTERS IN BLOCKS

Let 0, 0;) be the principal indecomposable character of G, G ° respectively which corresponds to (p,. If y is a p'-element of G then (4.26) implies that 0,(y)= ID10;) (y °). Therefore PROOF.

=FEE 0i(Y -1 ) 0)(Y)=--

»

I c,.y

If S is an integral domain let H„,d (S) be the space of homogeneous polynomials of degree d in n indeterminates x i , x,, with coefficients in S. The group GL„ (S), and hence any subgroup, acts on .1-4,,d (S) in a natural way. The next result is due to E. Cline. LEMMA 4.28. Suppose that F is a field and char F = p > 0. Let I be the ideal of F[x i , x n ] generated by all x°, where x E F[x,, x„] and the constant term of x is 0. Then I fl H „,(p_ o„(F) has codimension 1 in H „,(,_, ) „(F) and is mapped into itself by GL„ (F).

n 0 „(F), are mapped into themselves by GL„ (F). Since char F = p, I is the k space spanned by all monomials 4' • • x;'," with max {a,} p. The result follows since I--/„.(p _, )„(F) contains only one such monomial which does not satisfy the inequality, namely xr • xP„ -I . 0 PROOF. Clearly I, and hence I

THEOREM 4.29 (Thompson [1981]). Let G be a finite group and let V be an n-dimensional C[G] module. Let d(G) be the smallest integer d such that 11„,d (C) contains a one dimensional invariant subspace. Then d(G).(p —1)n, where p is any prime which does not divide I G I.

There exists an algebraic number field F0 and an Fo [G] module Vo such that Vo OF0 C = V. It may be assumed that F C K. Let V, = 1/0 OF° K. Since p I G I it follows from (111.3.3) that up to isomorphism there exists a unique R -free REG] module W with W 0R K = VI . Furthermore H= H ,(p_)(R) is completely reducible. Thus by (4.28) has a one dimensional subspace which is preserved by G. Thus by (111.3.3), H„,( _, ) „(R ) has a pure R [G] submodule of rank 1. This implies the result. El PROOF.

The following consequence of (4.29) which is due to Thompson can be proved by making use of the main result of Feit and Thompson [1961].

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COROLLARY 4.30. Let G,V,n,d(G) be defined as in (4.29). Then d(G)---. 4n 2 .

PROOF. Induction on n. It may be assumed that G is irreducible. Suppose that V = W G for some C[H] module W and some subgroup H of G with IG:HI= k >1. Let n = km. By induction there exists y E Hm, (C) with do 4m 2 , v0 such that the one dimensional space spanned by y is fixed by H. If x,} is a cross section of H in G and w = fl (vx, ) G H rrzk,dok (C) then G preserves the one dimensional space spanned by w. The result follows by induction. Thus it may be assumed that V is not induced by any C[H] module for any subgroup H of G with H G. This implies that any normal abelian subgroup of G is central. By Bertrand's postulate there exists a prime p with 2n + 1

By using better estimates on primes the number "4" can be decreased in (4.30). This is perhaps not too interesting. However it is plausible that there exists a constant c such that d(G)--.5- cn for all finite groups G. One cannot do better than this in general as the following example shows. Let G be a nonabelian group of order p 3 for some prime p then G has a faithful irreducible character of degree p. It is easily seen that d(G)= p. The next two results will be needed in Chapter X. LEMMA 4.31. Let Ô be a subgroup of G. Suppose that for a fixed i, (xu ) G is irreducible for all u with 0. Then (cp,),-; is an irreducible Brauer character.

PROOF. Let ças be the Brauer character afforded by the irreducible k [6] module f Let 0, be the projective R[6] module corresponding to L. and let 1, be the character afforded by 0,. Suppose that fs is isomorphic to a submodule of (L,)G. Then is is isomorphic to a submodule of ( U,) G . Hence 0, (u . Thus .

)G

(i), (1) =

E

E du,xu (1)= 0, (1),

(4.32)

where Is denotes the appropriate decomposition number of O. Let e,, be the multiplicity of L, as a composition factor of (L,) G . Then

SOME OPEN PROBLEMS

165

du, da,s for all u. Hence (4.32) implies that e, = 1 and Î. (1) = I (1). Hence a = (Ui )o. This implies in particular that the socle of (L,)G is irreducible and isomorphic to Ls. Apply the same argument to ç' i'. Thus (I) = C - and /1- ! is isomorphic to the socle of (L')0. Hence there exists an exact sequence (LO G --> Ls O. As e, = 1 this implies that ts (LOG and so (14 )6 ---- Ls. 0 COROLLARY 4.33. Let G be a subgroup of G. Let B be a block of G. Suppose that (x.) 0 is irreducible for every x. in B. Then (cp,)G is an irreducible Brauer character for every (ID, in B. PROOF. Clear by (4.31). 0 As a consequence of (4.33) we will prove the following results of Isaacs and Smith. For various refinements see Isaacs and Smith [1976], Pahlings [1977]. COROLLARY 4.34. Let P be a Sr -subgroup of G and let N = NG (P). Let B be the principal block of G. The following are equivalent. (i) G has p-length 1. (ii) If x. is in B then (x )N is irreducible. (iii) If ço, is in B then (ç', )N is irreducible. PROOF. (i) (ii). Since G = 0,(G) the Frattini argument implies that G = O(G)N. If x„ E B then 0.(G) is in the kernel of x„ by (4.12) and so (x.)N is irreducible. (ii) (iii). This follows from (4.33). (iii) (i). P = 0„ (N) ig in the kernel of irreducible Brauer characters of N and so P is in the kernel of (p, for every cp, in B. Thus P C 0„ , ,„ (G) by (4.12).

5. Some open problems

In this section we list some open problems in the theory. A discussion of these and related questions can be found in Brauer [1963]. Some nontrivial examples of modular character tables and Cartan invariants can be found in James [1973], [1978]. These can be used to illustrate the problems below.

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CHAPTER IV

(I) If IC,Icp,(x,)1 cp, (1) E R is it true that IG I ci,,(x , )1q)1( 1 )= (), where is the central character of R[G] corresponding to the block B?

In answer to an earlier question, Willems [1981] has pointed out that 1 (x)/ 1 (1) need not be in R. As an example let p = 2 and let G = the smallest Janko group. If x, is an element of order 3 and p chosen suitably then ç, (l) = 56 and cp,(x,) = — 1. Thus cp,(x,)1(p, (1) = I

I C.

-

— 209/2 0 R. See Fong [1974]. (II) Is IG IN, (1) E R for all j?

Willems [1981] has shown that the answer to (II) is yes if and only if I C, (x, )/ I; (1) E R for all i, j. In case G is p-solvable both (I) and (II) have an affirmative answer. See Chapter X, section 6. A result related to these questions can be found in Torres [1971]. Added in proof. J.G. Thackray has shown that McLaughlin's simple group G has an irreducible Brauer character of degree 2° - 7. Since a .52-group of G has order 2 this shows that (II) has a negative answer in general. (III) Does the character table of G uniquely determine (cp, (x 1 )), the table of irreducible Brauer characters? (The uniqueness can of course only be expected after a fixed monomorphism from IC, to the complex numbers has been chosen.) (IV) Does the character table of G uniquely determine the set of integers

Virtually nothing is known about these questions in general. An affirmative answer even to the weaker question (IV) would be very useful in the study of the structure of finite simple groups. Chapter X, section 6 has a related result in case G is p-solvable. (V) Can the number of blocks be characterized in terms of the structure of G?

By using the first main theorem on blocks, (111.9.7), it is possible to describe the number of blocks of positive defect. Thus the question is really about blocks of defect O. A weaker question is the following. (VI) What are some necesary and sufficient conditions for the existence of blocks of defect 0?

167

SOME OPEN PROBLEMS

In connection with problems (V) and (VI) Wada [1977] has proved the following and related results. In case the class C in (5.1) consists of involutions, the result had originally been proved by Brauer and Fowler [1955].

LEMMA 5.1. Let P be a p-group contained in G. Suppose that C is a conjugate class of G such that xy -1 P — {1} for x, y E C. Let 1P1 = p° and let 1C 0 (X ) = p" for x E C. Then there exists an irreducible characterx of G with p°' I x('). If furthermore Pis a S,-group of G then Xis of height 0 in a block B with defect group D, where D is the defect group of C. PROOF.

Let x E C. Define =

x(x)x(x -I

)

X( 1 )

A'

where x ranges over all the irreducible characters of G. By assumption 0(z) = 0 for z E P — {1}. Since OW = CG (X )1 it follows that !cG (x)i - (op, p

1p)p

x(x)x(x-i) 1 ) P. C"P

X

As p°' is the exact power of p dividing the denominator of the left hand side and x(x ), (xp, 1 p )p are algebraic integers, there exists x with x(1).

Suppose that P is a Sr -subgroup of G. Then

E

G : CG (X)1X(X)X(X -I ) (Xp,1p)p X( 1 )

As IG:P XO (mod p ) and I G :CG (x )1 x(x)/x(1) is an algebraic integer there exists x with G:CG (x)lx(x)

x(1)

(x

0

(mod rr)

for

Tr a prime in a suitably large p-adic field. Let B be the block containing x. By (4.4) D is a defect group of B and x is of height 0 in B. 0

In case G is solvable Ito [1951a], [1951b] has also obtained results in this connection. See Chapter X, section 6 for some of these. Related results are proved below. See (VI.5.6), (X.6.3) and (X.6.5). Other conditions for the existence of characters of defect 0 can be found in Ito [1965c], Ito and Wada [1977]. For related results see Gow [1978], Willems [1974

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[5

(VII) Is there a function of p and d which bounds c,, for cp„ cp, in a p-block of defect d? If G has a p -complement, in particular if G is p-solvable, then it follows p°, where p° = Fong [1961] has shown that for from (4.15) that G p -solvable c„ p d See (X.4.6). At one time it had been conjectured that p d for all G. This conjecture was shown to be false by Landrock [1973] for the group Suz(8) and p = 2, where p° = 64 and c,, = 160. The following results proved by Chastkof sky and Feit [1978], [1980a], [1980b] show that Landrock's result is not an isolated example. Let p = 2, let P„, be a S2-group of G„, and let c;;) be the Cartan invariant of G„, corresponding to the principal Brauer character. If G„, = Suz(2"' ) or ) then .

limc;TVI

= 1.

If G„, = SL3(2- ) or SU3(2'" ) then 1P„, =- 8' and I1M

C (IT ) 19 "' %. 1.

For information about the decomposition numbers of Suz(2- ) and 5U3(2") see Burkhardt [1979a], [1979b]. The Cartan matrix of a block with a cyclic defect group can be computed in terms of the Brauer tree. This result will be found in Chapter VII. In general it seems to be difficult to compute Cartan invariants of simple groups even when the character table is known. For instance the Ree groups 2 G2(32 "±') all have a S2-group which is elementary abelian of order 8. Yet the Cartan matrix of the principal 2-block has only recently been computed and shown to be independent of n. This computation involves some very delicate and complicated arguments. See Fong [1974], Landrock and Michler [1980a], [1980b]. For a similar result concerning the smallest Janko group see Landrock and Michler [1978]. A general approach to questions about irreducible and principal indecomposable k[G] modules for Chevalley groups in characteristic p can be found in J.E. Humphreys [1973b], [1976]. However the groups SL2(p") constitute the only class of Chevalley groups for which much is known about Cartan invariants. For instance every Cartan invariant is a sum of at most two powers of 2 and c 1 , = 2" for SL2 (p" ) in characteristic p. See J.E. Humphreys [1973a], Jeyakumar [1974], Srinivasan [1964c], Upadhyaya [1978].

169

SOME OPEN PROBLEMS

The most complete results concerning Cart an invariants for an infinite class of groups of Lie type are due to Alperin [1976c], [1979] for the groups SL2 (2") for p = 2. Before stating these results some notation is needed. Let cp be the Brauer character afforded by the natural 2-dimensional representation of SL2 (2") over the field F of 2" elements. Let o- be the Frobenius automorphism. Let S be the group of integers modulo n. For i E S define ço, = çor''' and for I C S let cp, = R E! ço,. Then every irreducible Brauer character of SL2 (2") is of the form cp, for some subset I of S, and conversely these are all the irreducible Brauer characters of SL2 (2").

THEOREM. CI,/ = 0 unless for each i E S with i EInJ and i +1 I r-1.1 we have i +1 0 I and i +1 0J, in which case c, For decomposition numbers of the groups SL2 (p") see Srinivasan [1964c]; Burkhardt [1976a], [1976b]. Question (VII) is also related to an old conjecture of Frobenius as

follows. LEMMA 5.2. Let IGI=p°go with (p, go) = 1. Suppose that G contains exactly g o p'-elements. If c 1 p° then G has a normal p-complement. PROOF. By the Cauchy—Schwartz inequality and (3.3) 192 = ( 0 1,

711)2

(i3O1)07i,

= cirp'

p 2 ".

Thus î and 0, are proportional. Hence 0,(z)= 0,(1) if and only if z is a p'-element in G. Thus the set of all p'-elements in G is the kernel of o, and this kernel is a normal p-complement. COROLLARY 5.3 (Brauer and Nesbitt [1941]). Let I G I = puer` where p, q, r are distinct primes. If G contains exactly q br` p'-elements then G has a normal p-complement. PROOF. By (4.15)(iv) (VIII) Is

p°. The result follows from (5.2). D

p`l whenever x„, cp, lie in a block of defect d?

By (X.4.6) (VIII) has an affirmative answer for p-solvable groups. (IX) Is it true that every irreducible character in a block B has height 0 if and only if B has an abelian defect group?

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CHAPTER IV

By (4.18) blocks of defect d 2 have no characters of positive height and so (IX) has an affirmative answer for d 2. In case G is p -solvable Fong [1960] has shown that if the defect group of B is abelian then every character in B has height 0 and the converse holds at least for the principal block. See (X.4.3) and (X.4.5). Knôrr [1979] has recently proved a result which implies that if V is either an R -free R[G] module such that VK is irreducible or V is an irreducible ft [G] module and V lies in a block with an abelian defect group D, then D is a vertex of V. Some consequences of an affirmative answer to (IX) are mentioned in Brauer [1962b]. (X) Let k(B) be the number of irreducible characters in the block B. Let k o(d) be the minimum of k(B) as B ranges over all blocks of defect d in all groups G and let k(d) be the maximum as B ranges over all blocks of defect d in all groups G. (i) Is lim e,— k o(d)= 00? (ii) Is k(d) - -Ç. p a ?

Nothing seems to be known about (X)(i). (4.18) yields an upper bound for k (d) which will be improved below. It can be shown that if the defect group of B is abelian and generated by two elements then k (d) generalizing the known fact that this inequality holds for d 2. See (VII.10.17). P. Fong has observed the following result in which the function k(d) arises naturally.

LEMMA 5.4. Let H be a d-dimensional linear group over the field of p elements. Assume that H is a p'-group and FI X (1). Then H has at most k(d)-1 conjugate classes.

PROOF. Let P be an elementary abelian p -group of order p' 1 . Then P is isomorphic to a d -dimensional vector space over the field of p elements. Let G = HP with P
6]

HIGHER DECOMPOSITION NUMBERS

171

5.5. Suppose that H satisfies the same conditions as in (5.4). Then H has at most ip" conjugate classes.

COROLLARY

PROOF. Clear by (4.18) and (5.4). E This result can be improved slightly by making use of (VII.10.14) below. Nagao [1962] has shown that if the number of conjugate classes in the semi-direct product of H with the underlying vector space in (5.4) is always at most pd then (X)(ii) has an affirmative answer for p -solvable groups. (XI) Let B be a block with defect group D and let 1 be the block of N o (D) with G = B. Does the number of irreducible characters of height 0 in B equal the corresponding number for É? (XII) Let m, (G) denote the number of irreducible characters of G whose degree is not divisible by p. Ism, (G)= m, (N o (P)) for a S„-group P of G?

It is clear that an affirmative answer to (XI) implies that (XII) has an affirmative answer. McKay [1972] first noticed that (XII) has an affirmative answer for many simple groups and raised the question in general. Alperin [1976a] conjectured that (XI) always has an affirmative answer and showed this to be the case for G = GL,, (q) with q a power of p. The assertions that (XI) and (XII) have affirmative answers are known as the Alperin—McKay conjectures. Macdonald [1971] showed that (XII) has an affirmative answer for all primes if G is a symmetric group. Related results can be found in Macdonald [1973], Pahlings [1975b]. Olsson [1976] has verified that (XII) has an affirmative answer for all primes if G = GL (q). Green, Lehrer and Lusztig [1976] proved a result which implies that both questions have an affirmative answer if G is a reductive group of Lie type with a connected center over a field of characteristic p. In case G is p -solvable it is known that (XII) has an affirmative answer. See Isaacs [1973], Wolf [1978], Dade [1980], Okuyama and Wajima [1979], [1980]. The results of Chapter VII show that (XI) has an affirmative answer if B has a cyclic defect group.

6. Higher decomposition numbers Let y be a p-element in G and let {q} be the set of all irreducible Brauer characters of Co (y). If is an irreducible character of C o (y) then

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(yx)= e(x) for all x E CG (y) where e is a path root of unity since y is in

the center of CG (y). Thus x, (yx ) = E,d, q,;(x) for all p'-elements x in (y) where each d is an algebraic integer in the field of p° th roots of unity over the rationals. The linear independence of {go ;'} implies that d is well defined. The numbers d's', are the higher decomposition numbers with respect to y. Clearly the higher decomposition numbers d;, with respect to y = 1 are the ordinary decomposition numbers. Suppose that z = y" for some u E G. Then CG (z)= CG (y)". Furthermore after a suitable rearrangement q);(x")= q(x) for all p'-elements x E CG (y). Since CG

X, (zx ) = Xs ((Yx )" ) = (Yx ) for x E C G (y) it follows that (zx u) = / d go: (x u ). Hence d= d Thus the set of higher decomposition numbers with respect to y depends only on the conjugate class containing y. If ci is an automorphism of the field of p"th roots of unity over the rationals then it follows directly from the definition that a- permutes the set {d } where i, y are fixed. The proof given here of the next result is due to Nagao [1963]. Aside from Brauer's original proof other proofs may be found in Iizuka [1961] and Dade [1965]. THEOREM 6.1 (Second Main Theorem on Blocks) (Brauer). Let y be a p-element in G. Suppose that cfl',/ 0 for some t, i. Let fl be the block of C G (y) containing gp;. Then fiG is defined and x, E BG. Furthermore y is conjugate to an element of the defect group of fiG. PROOF. Let P = (y) and let H = CG (P)= CG (y). Let J be the centrally primitive idempotent in R[H] with fit = B(e). By (111.9.4) fit' is defined. Let x, E B = B(e) where e is a centrally primitive idempotent in R[G] and let V be an R -free R[G] module which affords x,. Let s be the Brauer mapping with respect to (G, P, H) and let 0 be the character afforded by VH sa (e). By (III.4.11) and Nagao's theorem (111.7.5), x, (xy) = 0 (xy) for all p'-elements x in H. Since ca/ 0 this implies that cp;' is afforded by an irreducible constituent of VH so (e). Hence s o (e)è = J. Therefore fiG = B by (111.9.4). The last statement follows from (2.4). Ill For any p -element y the p-section containing y or simply the section containing y is the set of all elements in G whose p-part is conjugate to y. Clearly the p-section containing y = 1 is the set of all p'-elements in G. Let

6]

173

HIGHER DECOMPOSITION NUMBERS

{x;} be a complete set of representatives of the p'-classes in CG (y). It is easily seen that {yx} is a complete set of representatives of the conjugate classes in G which lie in the section containing y. Let CY = ( C) denote the Cartan matrix of CG (y).

LEMMA 6.2. Let B be a block of G. Let y, z be p-elements in G. (i) If y is not conjugate to z in G then for ail i, j (d)d=o. EB

(ii) /, (dL)*cPs; = c for all i, j.

(iii) Exse (d)*d ; = 0 unless ‘ip' and (pi lie in the same block t Of and = B. In that case Exs ED(d)*d-= c.

CG

(y)

PROOF. By definition (x;))=

(YxN * ' (Xs (z-x;))

= (E Xs(Yxr) * X(zx;))-

If y is not conjugate to z then E„ys(yx )*xs (zx I) = 0 for all i, j. By (3.6) (c/4'(x;')) is nonsingular. This implies one of the equations in (i). Since

E xs cyxo*xscyxJ)— Ic. cyx015,i --- cc,-,,,,(x it follows from (3.8) that (Yx

(Yxj)) =

Hence the nonsingularity of (c,o(x)) implies (ii). Fix i, j. Let 13- 1 , 132 respectively be the block of CG (y )7 CG (z ) respectively 0 for some s then by (6.1) I = with (1) E P, and cp; E 1-32. If (d)* and A's E B. Thus

E

(d)*.c4=

E (d)*d.;

if

x,EB

otherwise.

=0

This completes the proof of (i), and by (ii)

E )(s

e/3

(d)d=c if

=0

otherwise.

B

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CHAPTER IV

Since c = 0 if Ê

.

B 2 , (iii) follows.

[6

E

The next result is a refinement of (3.14). See Osima [1960b].

LEMMA 6.3. Let B be a block. If 0 = bsx, with b, E K define 0 8 = E , b,x,. (i) If 0 vanishes on a p- section of G then 0 8 vanishes on that p- section. (ii) If z i and z 2 are in different p- sections of G then , E

E

0

x EB

.

(i) Suppose that 0 vanishes on the p-section of G which contains the p -element y. Then E,,, (x)= 0(xy)= 0 for all p'-elements x in CG (y). Thus E, b„ciL = 0 for all i. Hence by (6.1) Ix, en b„d',',= 0 for all i and so 0 ° (xy)= b4 P r (x)= 0 for every p'-element x in CG (y) as

PROOF.

required. (ii) Let 0 =E,x,(z x,. Then 0(z) = 0 for every element z in the p-section containing z 2 . The result follows from (i). (6.3) can be used to generalize (4.23)(ii) as follows.

LEMMA 6.4. Let p and q be distinct primes. Suppose that y is a p-element in G and x is a q-element in G such that no conjugate of x commutes with any conjugate of y. (i) Let B(p) be a fixed p-block of G and let B(q) be a fixed q-block of G. Then E x, (x (y = 0 where x, ranges over all the irreducible characters which lie in B(p) and also in B(q). (ii) There exists a nonprincipal irreducible character which is in the principal p-block and also in the principal q-block. )

PROOF. (i) Let 0 =1,EB( , ) x,(x)x s. By assumption no element of G is in the p-section containing y and also in the q-section containing Thus by (6.3)(ii), 0(z)= 0 for every element z in the p-section containing y. Hence by (6.3)(i) 0 B( P )(y)= O. (ii) Immediate by (i). LI The following notation will be used in the rest of this section. B,, B 2 ,... are all the blocks of R {y,} is a complete set of representatives of the conjugate classes in G consisting of p-elements. Thus ly,xN is a complete set of representatives of the conjugate classes in G.

6]

175

HIGHER DECOMPOSITION NUMBERS

k is the number of conjugate classes in G. k(B,„) is the number of irreducible characters in B. 1(y,) is the number of conjugate classes in C G (y,) consisting of p'-

elements. 1(y,, B,„) is the number of irreducible Brauer characters of CG (y, ) which lie in blocks E of CG (y, ) with fi G = B„,. x = (xs (y,x)) is a character table of G. If y is a p -element in G then cif/ = (cp(xN is a table of irreducible Brauer characters of CG (y). tit, is the direct sum of the matrices tpY ,. D = (4) where s is the row index and (i, j) is the column index. D,„ is the submatrix of D where s ranges over all values with x, E B„, and for each i, j ranges over values for which (i);' is in a block ij of CG (y, ) with EG = B„,. Thus D„, is a k(B)x (1, 1(y„ B„, )) matrix. If T is a ring of algebraic numbers in K and S is a subset of G define Ch T (S)da a =I a,x, as E T,a(z)=0 ChT (S, B„,) ={a a =

EEB„,

for z ESI.

a,E T, a (z) =0

for zES}.

In case T is the ring of rational integers write Ch(S) = Ch T (S) and Ch(S, B„,)= ChT (S, B„,).

LEMMA 6.5. (i) 1(y ; )=E,„1(y„ B) for all i. (ii) k =

k (B„,) =I, I (yi ).

(iii) x, cp,D are all k X k matrices and x =Dtp if the rows and columns are

suitably arranged. PROOF. (111.2.8) and (6.1)

imply (i). (ii) and (iii) are clear by definition.

0

LEMMA 6.6. (i) D is the direct sum of the matrices D. after a suitable arrangement of rows and columns. (ii) Let T be a field of algebraic numbers in K. Then for all m

dim T (ChT (G, B„, )) = k(B„,)=

E 1(y„ B,„)

(iii) D„, is a nonsingular k(B)x k(B„,) matrix and -± (det D„, )2 is a power of p for each m. (iv) Let y be a p-element. There exists a block E of C G (y) with ir = B„, if and only if y is conjugate to an element of the defect group of B„,.

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CHAPTER IV

(i) Immediate by (6.1). By definition dim, (Ch, (G, B„,))= k (B„,). By (6.5) D is a square matrix. As D is nonsingular each D„, is a square matrix. Thus in view of (i) both (ii) and (iii) will follow once it is shown that -±- (det D„, ) 2 is a power of p. The mapping which sends every element of G into its inverse permutes the rows and columns of D and sends D„, to D. Thus (det D„,) 2 = -±- det(D'D, ). By (6.2) D*,' D„, is the direct sum of Cartan matrices. Hence ▪ (det )2 is a power of p by (3.9). (iv) Suppose that y is not conjugate to any element of the defect group P of B„,. By (2.4) d;',= 0 for all xs E B„,. Since D„, is nonsingular, this implies that B„,/ lj G for any block 1 of CG (y)• Suppose that y is conjugate to an element of P. It may be assumed that y E P. By (III.6.10) and (4.2) there exists a p'-element x E CG (P) such that for Xs E B„„ (.0.,(C)/ 0, where x is in the conjugate class C. Choose Xs of height 0 in B„,. Thus xs (x) O. Since x ECG (P) it follows that x E CG (y) and x,(xy) 0, and so x s (xy ) O. Hence cR,/ 0 for some i. The result follows from the second main theorem on blocks (6.1). El PROOF.

6.7. Let T be a subring of K which consists of algebraic numbers. Let S be a union of p-sections in G. Then (i) Ch T (S)= IEB,„ ChT (S, B„,). B„,) for all m. (ii) If T is a field then dim,- (Ch,

LEMMA

PROOF. (i) Since G — S is a union of p-sections this is clear by (6.3)(i). (ii) Suppose that E., asx, E ChT (S, B„,). Then E„ ad ço (x)= 0 for all y = y, S and all p'-elements x E CG (y). Hence E ad; = 0 for all y = y, S and all j such that cp;' is in a block É of CG (y) with BG = B,„. Thus the vector (a,) satisfies E,s 1(y,, B„,) homogeneous linear equations. By (6.6) (iii) these equations are linearly independent. Therefore by (6.6)(ii) dim T (Ch T (S, B„,

k (B,)—

E hES

LEMMA

1(y„ B„,)=

E

/(yi, B,„).

El

Y,ZS

6.8 (Suzuki [1959]). Let S be a union of n conjugate classes in G. E S for all integers h such that (h,1 = 1.

Assume that if z G S then z Then rank(Ch(S)) = n.

PROOF. Clearly rank(Ch(S)) n. Thus it suffices to exhibit n linearly independent elements in Ch(S). Let { ai,} be a normal basis of the field of I G Ith roots of unity over the

177

HIGHER DECOMPOSITION NUMBERS

rationals, where h ranges over all the integers with 1 h J GI and (h,I G I) = 1 and a„ is the image of a, under the automorphism which sends th root of unity onto its hth power. Let z,, z„ be a complete set of representatives of the conjugate classes in S. For i = I, n define 0, = as,x, where c4, = /,, a,,x, (z , ). Thus each a„ is a rational integer. If z E G and z is not conjugate to z:" for any m with (m,I G)= 1 then (z)= Et, a„ Es (z (z ) = 0 and so 0, E Ch(S). Suppose that E, bp, =0 for rational integers b,. Then for each i

1

h

where in the last sum (j, h) ranges over all pairs with z, conjugate to z,h . Since {a„} is a basis of the field of G I th roots of unity over the rationals this implies that b, = O. Thus {0,} is a linearly independent set. D The following refinement of (6.6)(ii) is due to W. Wong [1966]. LEMMA 6.9. Let T be a subfield of K consisting of algebraic numbers. Let S be a union of p-sections in G. Assume that at least one of the following assumptions holds. (i) T contains a primitive (I G Oth root of unity. (ii) If z ES then z h E S for all integers h with (h,IGI)= L Then dimT (Ch T (S, B„, )) = Ey, Es (y„ B„,) for all m.

In view of (6.8) dim,- (Ch T (S)) = Iy,Es (y,) under either assumption. Thus if the result if false then for some PROOF.

dim, (ChT (S, B„))< E 1(y„ B„, ) y, ES

by (6.7)(ii). Hence by (6.7)

dirnT (chT (s)) = E dirnT (chT (S, B„,)) < E E l(y i, B,„) y, ES

= E l(y,)= dim,- (Ch,- (S)). y, ES

This contradiction establishes the result. x, is a p-conjugate of x, if x, = x for some automorphism ci of the field of GIth roots of unity over the rationals which leaves the p'-th roots of unity fixed. Two p-conjugate characters have the same irreducible Brauer

178

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characters as constituents and so lie in the same block. x, is p-rational if it has no p-conjugate other than itself. LEMMA 6.10. Let M be a group of automorphisms of the field of IG th roots of unity over the rationals such that the p'th roots of unity are in the fixed field of M. Distribute the irreducible characters in B„, into families of conjugates under the action of M Let n, be the number of irreducible characters in the ith family. Then the following hold. (i) The number of orbits of the columns of D,„ under the action of M is equal to the number of families in B„,. (ii) If p X 2 or if the 4th roots of unity are in the fixed field of M then after a suitable relabeling the ith orbit of columns of D,„ consists of n, columns. PROOF. If p X 2 or if the 4th roots of unity are in the fixed field of M then M is cyclic. Since D„, is nonsingular by (6.6)(iii) the result follows from a combinatorial lemma of Brauer. See Brauer [1941c], Lemma 2.1 or Feit [1967c], (12.1). El COROLLARY 6.11. Suppose that p X 2. The number of p-rational irreducible characters in B. is at least as great as the number 1(1, B„,) of irreducible Brauer characters in B„,. PROOF. Since the ordinary decomposition numbers are rational D,„ has at least 1(1, B„,) rational columns. The result follows from (6.10). 0 Some results concerning the fields generated by the entries of x, tin and D can be found in Reynolds [1974].

7. Central idempotents and characters Some of the formulas in this section have been used by Osima as the starting point for an alternative development of much of the theory of blocks. For this approach see Osima [1952b], [1955], [1960a], [1964], Iizuka [1956], [1960a], [1960b], and Curtis and Reiner [1962]. The following result is well known and easily verified. LEMMA 7.1. Let e, be the centrally primitive idempotent of K[G] corresponding to cos . Then

CENTRAL IDEMPOTENTS AND CHARACTERS

e, — X,G( 1 )

179

xs (z ')z.

LEMMA 7.2 (Osima). Let B be a block of R[G] and let e be the centrally primitive idempotent in R[G] corresponding to B. Then e

i)lz

1 I =G E, E B IG

E

I Li-

„En

(1)(p1 (z -1 ) z.

PROOF. X, E B if and only if co, (e) X O. Thus by (7.1)

e= E

es

x,EB

1 I

E E xs(1)xs(z-)1

PEG x,EB

Thus the first equation follows from

= In particular this implies by (2.5) that the coefficient of any p-singular element in e is zero. If z is a p'-element then

E

As

(1)x ( z — ').=

=

EE

cp,EI3 x,EB

E 13

q),(z - ') ds.xs(i)

q, (z -1 )0,

This proves the second equation.

( 1 ).

0

LEMMA 7.3. Let e be a centrally primitive idempotent in R[G]. Let B be the block corresponding to e and let À be the central character of k[G] corresponding to B. For each j let C, denote the conjugate class of G with a, E R. If ii,A(6,) 0 then B and C, have a x, E C. Then e = a1 • common defect group.

a,e, with a, E R. Let D be a defect group of B and PROOF. By (7.2) e let P, be a defect group of C,. If 4# 0 then P, CG D by (111.6.5) since e EZD (G;G). If À()P4 O then D C G P, by (III.6.10). CI The next result is due to Osima [1966] in case a is the principal Brauer

character.

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CHAPTER IV

180

THEOREM 7.4. Let e be a centrally primitive idempotent in R[G] and let B be the block corresponding to e. Let x, be a character of height 0 in B. Let a be a rational integral linear combination of {(I), I E B } with v(a(1))= (x, (1)). Let 77 „ be defined as in (1.2) and let n c, = Es bsx,. Define

1 x, (1) 0 71

`' (z-I)z

Then f, E R[G] and e — E J(Z(G :G)). PROOF.

By (4.7) f„, E R [G]. By (4.1) and (7.1)

bs x, (1) J1) I G ,' 1G

,(1)

f = x 6, .

_

x,(1) b, e b, xs (1) K*3

where es is the centrally primitive idempotent in K[G] corresponding to (os. Thus if x, B then cos (e f„) = (o, (e) (os (f„)= O. If Xs E B then —

—

(1) (os (e — f„)= cos (e)— (o s (f,)= 1 X' b, x s (1) .

Hence by (4.7) (o,(e f,) = 0 (mod 7r). Thus cos (e f,)=- 0 (mod 7r) for all s and so .1(e — f) = 0 for every central character of R[G]. D —

—

—

The next result is due to Iizuka and Watanabe [1972], who also prove further results of a similar nature concerning blocks of positive defect. This result generalized earlier results of Brauer [1946a], [1956].

THEOREM 7.5. Let Z be the center of k[G] and let M be the subspace of Z spanned by all C.', such that C, is a conjugate class of defect O. Then M is an ideal of Z and the number of blocks of defect 0 is equal to dirn, M2 . PROOF. By (111.6.2) M is an ideal of Z. Let be all the centrally primitive idempotents of R f G] corresponding to blocks of defect O. By (7.3) e, E M for i = 1, . . n. Let eu = e,. Then M = meo m(i e0). We next show that M2 = (Me0)2 . Let C, C, be classes which have defect O. Let e be a centrally primitive idempotent of R[G] which corresponds to a block B of positive defect. Then by (7.1) -

lc; ix (x)x (ox (x,) os , 5

XED

GI X(1)

181

SOME NATURAL MAPPINGS

where x, E C„ X 1 E C1 , X. E C. If I G =p" then1C,I=- 10 = 0 (mod p') but I G x (1) 0 (mod p' ). Hence C,Cie =0. Therefore {M(1 — JOY = O. Since Ma4(1 — e)= 0 this implies that /14 2 = (MO'. Since e; corresponds to a block .131 of defect 0, B, contains a unique irreducible 1:[G] module and Zei = Re,. Hence Me, = ze; = Re, and so MJ, = Re,. Therefore —

1142 = e 17W =

E

8. Some natural mappings To a large extent this section consists of results which are reformulations of some results proved earlier in this chapter. We will follow Broué [1976b], [1978b] quite closely. See also Serre [1977]. The notation introduced in section 6 after (6.4) will be used with the following modifications. T will denote an arbitrary subring of K. Instead of Ch T (S, B) we will write Ch T (G :S, B) to emphasize the dependence on G. We will also write Ch T (G: G) = ChT (G). Greg denotes the set of all p'-elements in G. The Grothendieck algebra AC,(K[G]) is isomorphic to Ch(G) in the obvious way. By (3.12) Ch(G : Greg ) is isomorphic to the Grothendieck algebra A 9,(R[G]). Let Ch T,,,„, (G) denote the ring of all T-linear combinations of {0,}. Let Ch T,r„, (G : B)= ChT.,,(G) n chT (G : B). As usual the subscript T will be omitted in case T = Z, the ring of rational integers. Define the linear map b = to' , c = c, d = cr as follows: where (0, ) =1„ b :ChT,,„,„(G)—> ChT (G), c ChT (G : Greg ), where c(0, )= where d ()= d : ChT (G)—> Ch r (G : Greg ), By (3.1) the following diagram is commutative. C hT,,,,„(G)

VN Ch T (G)

Ch T (G : Greg )

It is clear from the definitions that b, c, d respect blocks. In other words (d(0)) 8 = d(06 ) for 0 E Ch T (G). Similarly (1)(0)) B = b(013 ) and (c(0))' = c(0 8 ) for 0 E ChT,,,„, (0). Furthermore if x E Ch T (G) and 0 E then by (3.4)

[8

CHAPTER IV

182

(8.1)

(d(), 0 ) = (X, NO).

This situation will now be generalized. Let y be a p-element in G. Assume that T contains a primitive (y )Ith root of 1. If 4' is an irreducible character of CG (y) then there exists a ,(y)th root of 1, wc (y) such that yx) = wc (y)(x) for x E CG (y). Define the linear map wy of ChT (C G (y)) by wy (‘)= coc (y)4" for any irreducible character 4" of C G (y). Then for any 4 E Ch T (C G (y)) and any x E C G (y) it follows that (wy ())(x) = 4. (Yx)For any subgroup H of G let Rea Ch T (G)—>Ch T (H) be defined by Res(0) = OH. Let Inca Ch T (H)—> Ch T (G) be defined by Ind(0) = O G. Let c Y.' denote the map c = ccG( Y ) for the group CG (y). Define bY .G = Inaa (y) ° (COy ) -1 ° b CG ( Y ) ,

d

= d wy -ResC?G(y) .

LEMMA 8.2. (i) The following diagram is commutative. Ch T,„„, (CG (y)) c7.G

Ch T (C G (y):CG (Y)reg)

ChT

(ii) If T1 E Ch, pr„, (C o (y)) then bY .G (77) vanishes outside the p-section which contains y and it is a class function such that (b" (77))(yx)= Ti(x)

for x E CG (y)reg .

(iii) If 0 e Ch r (G) then d" (0) is an R-valued function on CG (y) defined by (dY' G (0))(x)= 0(yx)

for x E Co (Y) , ,g-

PROOF. Statements (ii) and (iii) are direct consequences of the definitions. Then (i) follows from (ii) and (iii). LII LEMMA 8.3. If 0 E Ch T (G) and

T1

E ChT„,„, (CG (y)), then

(d" (0), n)cG(y) = (0, b»6 (y)) G .

PROOF. By (8.1) and the Frobenius reciprocity theorem (111.2.5) (d' (0),17)c„(y) = (Res20(y) (0), (WY = ( 0, V G (r1))G.

E

° b CG(Y) (y ))c' ay)

SOME NATURAL MAPPINGS

183

Let B be a block of G and let A A2,... be all the blocks of C, (y) with ii ° = B. If 4" is a class function defined on CG (y) let = E, The next result is equivalent to the second main theorem on blocks (6.1). ,

THEOREM 8.4. Let 0 E Ch, (G) and let n E ChT,„„,(C G (y )). Then the following hold. (i) (GB) = (bY' (0)) B . (ii) d (0 8 ) = (d" (0)) 8 .

PROOF. By (8.3)(i) and (ii) are equivalent. Thus it suffices to prove (ii). Since x“(yx ) = E, dL',(,p,(x ) for x E CG (y )„g it follows from (8.2) (iii) that dY' (x“) = E, cacp,. The result is now seen to be a reformulation of (6.1). 0

Let Z, (G) denote the center of T[G]. If S is any subset of G let Z, (G : S) denote the set of all a =E„EG a„x E (G) with a„ =0 for X E S. In the remainder of this section we will only be concerned with the case T = K. Observe that Ch K,„,.,(G)= ChK (G :Greg ).

By definition Ch K (G) is the dual space of ZK (G). Let B be a block and let e be the centrally primitive idempotent in R[G] corresponding to B. If 0 E Ch K (G) then O B = 00e, where (0 e )(a) = 0(ea) for a E ZK (G). Let S be a p-section of G. For 0 E Ch K (G) define ds (0) by

(ds

0(x )

if x E S,

0

if x E S.

(0))(x) = {

For a --- E„EG ax E ZK (G) let 8,9 (a)= E„ ES a.a. Then ds is the transpose of Ss. In other words 0 (Ss (a)) = (ds (0))(a) for a EZ K (G), 0 E ChK (G). LEMMA 8.5. Let S be a p-section in G. Let B be a block of G and let e be the centrally primitive idempotent of R[G] corresponding to B. Then the following hold. (i) If y is a p-element in S then bY'od = d, (ii) If 0 E ChK (G) then (ds (0)) B = ds (B B ). (iii) If a E ZK (G) then Ss (ea) = eSs (a).

CHAPTER IV

184

PROOF. Since T = K, (i) follows from (8.2)(ii) and (iii). Then (8.4) implies (ii). Since 8.9 is the transpose of ds, (ii) implies (iii). D

Observe that (8.5)(ii) is essentially equivalent to (6.3)(i), and (8.5)(iii) is equivalent to the fact that if e is an idempotent in ZK (G), S is a p-section of G and a = IxEs a,rx E Z K (G) then oc = Ex Es bx x. In this form the equivalence of (8.5) (ii) and (8.5) (iii) was first proved and used by Iizuka [1961].

Let y be a p-element in G. Let bY = bYG, dY = d'. If E Ch K (C G (y): CG (y)„,) then by (8.2) (ii) bY( ) vanishes outside the p-section which contains y, and it is a class function such that (bY(71))(yx) = 71(x) for x E CG (y)„ g. If 0 E ChK (G) then by (8.2)(iii) (dY (0))(x)= 0(yx) for x E CG (y)„,. Define pY, SY analogously as follows. For S a p-section of G which contains y, (y):

pY

CG (y)„,)--->ZK (G :S),

where xEc G (y),

and z runs over a cross section of 8Y

ZK

(dY (0))(a)=

ZK (CG

(y ):

(y) in G.

where

(G)--> Zic (CG (y)),

LEMMA 8.6. (i) If a E

(ii) If 7-1

CG

CG

SY

E

ax)

xeG

E

x eCc-,(y)m g

ax.

(y )reg ) and 0 E ChK (G) then

(Y)1 0 (P Y (a)).

e chK (CG (y): CG (y)re g)

(bY (n))(a) =IG:C G (y

TI

and a

e ZK

(G) then

Y (CO.

PROOF. Direct consequence of the definitions. LI LEMMA 8.7. Let y be a p-element. Let e be a centrally primitive idempotent in R[G] and let s be the Brauer map with respect to (G, (y), CG (y)). Let so(e) be the central idempotent in R[C G (y)] such that so(e)= §(e). Then the following hold. (i) 13Y (so(e)a)= eBY (a) for a E ZK (CG (y): CG (y)„,). (ii) (ea)= so (e)5Y (a) for a e ZK (G). PROOF. In view of (8.6) this is the dual of (8.4) if y is replaced by y it follows from (8.4). 0

and so

9]

SCHUR INDICES OVER

Qp

185

9. Schur indices over Q,,

Let cp be an irreducible Brauer character of G. Let F be a finite field such that cp(x)EF for all x E G. By (1.19.3) there exists an irreducible F[G] module which affords ço. The situation is more complicated for fields of characteristic O. Let F be a field of characteristic O. Let x be an irreducible character of G and let F() = F (x (x), x E G). Then there exists a positive integer m F (x) such that mF (x)x is afforded by an F(x)[G] module and if nx is afforded by an F(x)[G] module then m, (x) I n. The integer m, (x) is the Schur index of x with respect to F. The proof of the next result as well as the existence of the Schur index can for instance be found in Feit [1967c], section 11. LEMMA 9.1. Let F be a field of characteristic O. Let x be an irreducible character of G. Then the following hold.

(i) mF (X)= mF(,)(X)(ii) If 0 is afforded by an F[G] module then mF(X)1( 0, X)(iii) If F Ç L then mF(X)ImL (X)[L(X):F(X)i(iv) There exists an extension L of F(x) with [L:F(x) ] = mF (x) and

mL (X) = 1 -

Suppose that F(x)— F. Let V be an F[G] module which affords the character mF (x)x and let E( V) be the endomorphism ring of V. Then E(V) is a central division algebra over F with [E(V):F]= m,(x)2. Furthermore if L is an extension field of F then mL (x) = 1 if and only if L is a splitting field for E(V). See for instance Curtis and Reiner [1962], Chapter X. The structure theory of division algebras whose center is a finite extension of the field of p-adic numbers, Q, is well known. Schur indices over Q, are studied for instance by Yamada [1970]. We state the following without proof. THEOREM 9.2. Suppose that F is a finite extension of Q,,. Let x be an irreducible character of G. If L is any finite extension of F(x) with m,(x)1[L :F(x)] then m L (x)=1. In particular (9.2) implies that L can be chosen to be unramified or totally ramified. On occasion one or the other of these choices is convenient. The next result is due to Brauer [1945]. See also Brauer [1947], Gow [1975].

186

CHAPTER IV

THEOREM

[10

9.3. Suppose that d.,/ O. Then InQ,, (X. )1 d1 1Q, (x.,

49,

):

Q,

(x.)1.

Let Z,, be the ring of p-adic integers in Q. By (1.19.3) there exists an irreducible 4, (cT),)[G] module which affords cp. Let V, be the corresponding projective indecomposable 4,, (cp, ) [G] module. By (1.13.7) there exists a projective Z p (cp,)[G] module U, which affords 0, with a = Thus 0, E Q, (cp,). Hence rrtQp () (x„) d by (9.1)(ii). As mQ (X.) = mQ(x.)(X.), by (9.1)(i) the result follows from (9.1)(iii). E

PROOF.

COROLLARY 9.4. If x. is irreducible as a Brauer character then mQ, (x.)= 1. PROOF. Immediate from (9.3). 0 COROLLARY 9.5. If p ri'lx?„ (X. ) = 1 -

,e !GI

and x is an irreducible character of G then

PROOF. Clear by (9.4). E For related results concerning Schur indices see Broué [1977].

10. The ring

,2

(1[G])

21 ° ( G)— A(1 7Z[G]) is the ring of all integral linear combinations of generalized Brauer characters of G. The ideal ,z0, ) (G) of A ° (G) can be identified with the ring of integral linear combinations of characters afforded by projective R[G] modules. The main result of this section gives a criterion for an element to be an ideal generator of A ), ) (G), and hence in particular a criterion for A > (G) to be a principal ideal. The argument is from Feit [1976]. See also Alperin [1976e].

THEOREM 10.1. Let x,, x, be a complete set of representatives of all the conjugate classes of G consisting of p'-elements. Let i)(1 CG (x,)1) = a; for 1 i m. Let n E ANG). The following are equivalent. (i) For 1 i m, -q(x,) =pu, for some unit u in some algebraic number field. (ii) H 71 (x) ± 11:, p. (iii) For all a E A )1) (G) there exists y, E AC(G) with n ya, = a. Furthermore if 71 satisfies (i), (ii), (iii) then y, is uniquely determined by a.

101

THE RING

A(fi1G1)

187

Before proving (10.1) we deduce some consequences. COROLLARY 10.2. Suppose that P is a p-group with P < G and there exists E A (GIP) satisfying (i), (ii), (iii) of (10.1). Then there exists n E A, > (G) satisfying (i), (ii), (iii) of (10.1). PROOF. Clear by (4.26). 0 COROLLARY 10.3. Let P be a Sr -group of G. Suppose that P < G. Let cp, be the Brauer character of G with yo,(x)= 1 for all p'-elements x in G. Then 0, satisfies (i), (ii), (iii) of (10.1) with u, =1 for all i. PROOF. Clearly 1 G/ p E ,2 (°, ) (GIP) and satisfies (i), (ii), (iii) of (10.1). The result follows from (4.26). 0 A much more important class of examples arises from the work of Steinberg [1968]. He showed that if G is a simple group of Lie type then the Steinberg character of G satisfies condition (i) of (10.1) with u, = ±- 1 for all i. This motivated Ballard [1976] and Lusztig [1976], who independently showed that the Steinberg character satisfies condition (iii) of (10.1). Their work led to (10.1). It is not always true that A > (G) is a principal ideal of A °(G). It can be shown by explicit computation that for p = 5 and G = PSI-, (9) --- A6 there is no n in A > (G) which satisfies the conditions in (10.1). Also if G = the first Janko group, then it is not difficult to verify that no n in ,z01) (G) satisfies the conditions of (10.1) for any prime p111,1. The proof of (10.1) is based on the following elementary result.

LEMMA 10.4. Let 0 be a complex valued class function on G. Let y,, y,„ be a set of pairwise nonconjugate elements of G such that 0(y,) 0 for i m. Let X be the complex vector space consisting of all complex valued class functions f on G such that f(y)= 0 if y is not conjugate to any y,. Let L(0) be the linear transformation on X defined by L(0)f = Of. Then dim X = m, the characteristic roots of L(0) are 0(y,),..., 0(y„,) and det L (0) = Z,11 e(y,). PROOF. Define f, E X by f(y ; )= 8• . Then {f} a basis of X and so dim X = ni. Furthermore L(0)f, = 0(y,)f, for all i. Thus L(0) has the required characteristic values and so the required determinant. 0

188

CHAPTER IV

[11

PROOF OF (10.1). Let Pi be a S„-group of C G ()I f ). Since 71 E i0, ) (G) it follows that ii (y , ) „p, E A )((Yi)x Pi). Thus -r1(3),)= pu ;for some algebraic integer m. Therefore RI, (yi )= m p. Clearly RI, 71(3) ; ) is rational valued and so RI, u; is a rational integer. The equivalence of (i) and (ii) is now clear. Condition (iii) holds if and only if 712V(G) = A ?1 ,(G). Since 712V(G) C A ?1,(G), this holds if and only if I A' ) (G): TiinG)1= iC(G): ,V1) (G)1. By (10.4)

I /0(G):7-1A Û(G)1= idet L(y)I= ±fl n(yi)= By definition I A"(G): 2V o (G) = det C, where C is the Cartan matrix of G. By (3.11) det C = p. Thus condition (iii) holds if and only if 11:1, ui = ± 1, which is equivalent to condition (i). If 77 satisfies (ii) then (3),) 0 for all i. Thus yr,(3),)= a (y i )Ti (yi)' and so yr, is uniquely determined by a. 0 11. Self dual modules in characteristic 2 Throughout this section F is a field of characteristic 2 and V is an absolutely irreducible F[G] module which affords a real valued Brauer character. Equivalently V = V. V is of quadratic type if there is a nondegenerate G-invariant quadratic form on V. V is of symplectic type if there is a nondegenerate G-invariant alternating form on V. We will show that V is essentially always of symplectic type and then give some criteria for V to be of quadratic type. The results in this section are primarily due to Willems [1977]. However the first two results are older. See e.g. Fong [1974] Lemma 1 or James [1976] Theorem 1.1. We will first introduce some notation. A is the representation of G with underlying module V. E is the space of all matrices on V where X —> A (x)' XA (x ) for x E G where the prime denotes transpose. Thus E is an F[G] module with E =VOV=VSV* by (11.2.8). S={XEEIX=X1,

T = {X + X 1 IXEE}.

Clearly T C S and S,T are both F[G] modules. Vo(G) is the F[G] module with dim, V o(G)= 1 and Inv G (Vo (G))= V(G).

SELF DUAL MODULES IN CHARACFERISTIC 2

11]

THEOREM 11.1.

189

If V V o (G) then V is of symplectic type.

PROOF. Since V V* there exists M E E with M-1 A (x)M = A (x - ')' for all x E G. Thus M'A (x )'( MY' = A (x -1 ) and so A (x -1 )' = (Mi )' A (x)M' for all x E G. Hence M'M-I A (x) = A (x)M1t4 -1 for all x C G. Thus by Schur's lemma M' = cM for some c E F. Thus M = cM' = c 2M. Hence c' =1 and so c =1. Therefore M' = M. Furthermore A (x )'MA (x) = M for all x E G. Thus M defines a G-invariant symmetric bilinear form on V. Let W be the subspace consisting of all isotropic vectors in V. Thus W is a G-invariant subspace and so W = (0) or W = V. If W = (0) then dim, V = 1 and V = Vo (G). If W = V then M defines a nondegenerate symplectic form on V. I=1 11.2. Let tp be a real valued irreducible Brauer characterof G in characteristic 2. Then either cp =1 G or ço (1) is even.

COROLLARY

PROOF. Let U be an F[G] module afforded by cp. If U is of symplectic type then ço (1) = dim, U is even. The result follows from (11.1) with U = V. 0

For i <j let X„ denote the matrix in E with 1 in the (i, j) and (j, i) entries and 0 elsewhere. Let X„ denote the matrix with 1 in the (i, i) entry and 0 elsewhere. Then {X,, I i j} is an F-basis of S and {X,, I i < j} is an F-basis of T. By (1.19.3) there exists a finite subfield Fo of F and an Fo[G1 module Vo such that V = (V0)F. Let VP denote the F0[G] module derived from V by applying the automorphism of F0 which sends a to a'. Let V(2) = (V )F. LEMMA 11.3. (i) There exists a submodule E0 of E with EIE 0 -= Vo(G). (ii) EIS T. (iii) SIT -= PROOF. (i) Let E, denote the space of all matrices on X —> A (x - ')XA (x) for x E G. Then as FIG] modules

V where

E,=-VOV*-=VOV-=E.

Let E, be the set of all scalar matrices in E. Then E2 .-= Vo(G) is a submodule of E. The result follows as El = (ii) Define f: E —> T by f (X) = X + X'. Then f is an of E onto T with kernel S. homrpis (iii)Without loss of generality F may be replaced by its algebraic closure.

190

[11

CHAPTER IV

Thus the map sending a to a' is an automorphism of F. Let x e G and let A (x)= (a,,). Then A (x )' X„A (x ) = (a, for some X E T.

si

=Ea

+X

Cl

Let M = (m11 ) be an upper triangular matrix in E, i.e. mo = 0 for i > j. Define the quadratic form QM on V by QM

Define fM

(v) =

E mixix; where y = (x 1 , ..., x„). by

E Hom, (S, F)

f m ((x; ; ))-

E m,,x,,.

LEMMA 11.4. Let M be an upper triangular matrix in E. Then QM is G-invariant if and only if fm E FlomF[G] (S, F) where F V o(G).

PROOF. Let y = (x,, Qm(vA)=

x„ ) E V and let A = (a,,) E E. Then

E

Thus QM (vA) = Go m (v) if and only if the following two equations are satisfied for all s < t.

E infaisais = mss, E ,.;

+ afs a„) =

(11.5) (11.6)

For all s
E m„a,sa,s, i, j

fi,,f (A' X„A ) =

E m,,(a,sa„ + aua,$ ).

Thus fm (A ' XA ) = f (X) for all X E S if and only if (11.5) and (11.6) are satisfied. Consequently fi,4 (A' XA)= f i,4 (X) for all X E S if and only if QM is A -invariant.

11]

SELF DUAL MODULES IN CHARACTERISTIC

2

191

11.7. V is of quadratic type if and only if there exists a submodule S o of S with S/S 0 - -- V o(G).

THEOREM

PROOF. Every quadratic form on V is of the form C),,, for some upper triangular matrix M. Every map in Hom, (S, F) is of the form fr,,, for some upper triangular matrix M. Hence the result follows from (11.4). LI COROLLARY 11.8. If V is not of quadratic type then there exists a nonsplit exact sequence 0—> Vo (G)—> W—> V—>0.

PROOF. By (11.7) Vo(G) is not a homomorphic image of S. Thus by (11.3)(i) and (ii) there exists an exact sequence 0—> To —> T— V0 (G)—>0.

Thus by (11.3)(iii) there exists an exact sequence 0—> Vo (G)—> S IT o —> V (2) —> 0.

As Vo (G) is not a homomorphic image of S it follows that the sequence is not split. Hence there also exists a nonsplit sequence of the required form. 0

COROLLARY 11.9. If Vis not in the principal 2-block then Vis of quadratic type.

PROOF. Clear by (11.8).

LI

COROLLARY 11.10. If G is solvable then V is of quadratic type. PROOF. Induction on GI. If I G I = 1 the result is clear. Suppose that I G I > 1. If V is not in the principal 2-block the result follows from (11.9). If V is in the principal 2-block then 02,,2 (G) is in the kernel of V by (4.12). As 02,,2(G) (1) the result follows by induction. El

CHAPTER

V

The notation and assumptions introduced at the beginning of Chapter IV will be used throughout this chapter. Also the following notation will be used. If B is a block of G then AB is the central character of REG] corresponding to B. v is the exponential valuation defined on K with v(p) = 1. xi, x2, .. are all the irreducible characters of G. q)2, ... are all the irreducible Brauer characters of G. (0, is the central character of K[G] corresponding to X s. If H is a subgroup of G and h:Z(K;H:H)-->K is linear, define h' : Z(K;G G)—>K by h' (0) = h(C n H) for any conjugate class C of G. Some of the results in this chapter can be proved without the assumption that K and k are splitting fields for every subgroup of G. See for instance Broué [1972], [1973]; Hubbart [1972]; Reynolds [1971]. 1. Some elementary results For the results in this section and the next see Brauer [1967], Fong [1961] and Reynolds [1963].

LEMMA 1.1. Let H be a subgroup of G and let be an irreducible characterof H. Let e =Es asxs and let co be the central character of K[H] corresponding to Then

E .2„x„ (ow, =

r (ow

G.

SOME ELEMENTARY RESULTS

1]

193

PROOF. Let { L,} be the set of all conjugate classes of H and let z, E L. Let ‘(x) = 0 if x E G — H. If C is a conjugate class of G with z E C the definition of induced characters implies that

Thus

asx, (1)co, (C) =

( E`ICG a IGI (z)I X"z ' s

= IG :1114- (1)

E

L, çC

w(L,)= r(l)co G (C).

LEMMA 1.2. Let H be a subgroup of G and let Bo be a block of H. Assume that B o contains an irreducible character such that r is irreducible. Then Bô is defined and r E Bô .

PROOF. Let 4-G = X s and let co be the central character of K[H] corresponding to By (1.1) cos = co'. Thus 65 0 = Co, is a central character of R[G]. 0 LEMMA 1.3. Let B o be a block of the subgroup H of G for which Bô is defined. Let 4' be an irreducible character in B o and let r =Escio(s. If B is a block of G then

E

avx , ( 1)) > vQ-G (1))

Eel3 avx,(0)= ,(4-- (1))

if B/ if B

Bô,

= K.

xs

PROOF. Let to be the central character of K[H] corresponding to 4- . Let e be the central idempotent in R[G] corresponding to B. Then to, (e)= 1 if xs E B, and cos (e)= 0 if xs 0 B. Also (.5 G (e) = 0 if B B ô and (.5 G (e)= 1 if B = B. The result now follows from (1.1). LI COROLLARY 1.4. Let B o be a block of the subgroup H of G for which Bô is defined. Assume that B o and Bô have the same defect. If 4- is an irreducible character of height 0 in B o then some character x s of height 0 in B ° as a constituent of r with multiplicity a 5 0 (mod p). PROOF. Clear by (1.3).

El

194

CHAPTER

V

[1

THEOREM 1.5. Let B be a block of G with defect group D. Suppose that AB (07 ) 0 for a conjugate class C of G. Then there exists z E C with z C CG (D) such that the p -factor y of z is in Z(D). PROOF. By (111.9.7) there exists a block Bo of NG (D) with defect group D such that Bô = B. Let Ao = À. Since A ô(0)/ 0 there exists a conjugate class Co of NG (D) with CO C C and A 0 (00) # O. Choose z E Co . By z C CG(D). If is an irreducible character in Bo then ‘(z)# O. Hence by (IV.2.4) y is conjugate in NG (D) to an element of D and so y E D. Since D C CG (Z) C CG (y) it follows that y E Z(D). 0 THEOREM 1.6. Let B o be a block of the subgroup H of G for which Bô is defined. Let Do be a defect group of B o and let D be a defect group of Bô . Then (i) (ii) (iii) (iv)

Every element in Z(D) is conjugate to an element in Z(D0). The exponent of Z(D) is at most equal to the exponent of Z(D0). If B q has defect 0 then Bô has defect O. If p IIGI then H# (1).

PROOF. (i) Let x, be a character of height 0 in Bô . By (III.6.10) and (IV.4.8) there exists a p'-element x such that D is a Sr -group of CG (x) and x,(x)(o, (x 0 (mod IT). If y E Z(D) then D is a Sp -group of CG (xy) and x, (xy ) x, (x ) 0 (mod IT). Thus v (a), (xy))= v (0), (x )) = O. Hence U), (xy ) 0 (mod ii- ) for all y E Z(D). Let be an irreducible character in Bo and let to be the central character of K[H] corresponding to Then = Co G. Hence if y E Z(D) there exists a conjugate z of xy in H with (Ti(z)# O. By (1.5) the p -factor of z is conjugate to an element of Z(D0). (ii) and (iii) are immediate by (i). (iv) If H = (I) then Bô has defect 0 by (iii) and so contains exactly one irreducible character xr . Hence if 4' is the unique character in Bo then r = x (1)xs and so by (1.3) .

v(IG1) = v(e (1)) = v(x,(1) 2)= 2v(IG1).

Thus v (IGO= 0 contrary to assumption.

0

LEMMA 1.7. Let B be a block of G and let u be an automorphism of G such that x = x, for all x, E B. If y is an element in a defect group of B then y is conjugate to y in G.

PROOF. By (IV.4.8) there exists a p'-element x and a character xr of height

2]

195

INERTIA GROUPS

0 in B such that x, (x)6), (x -I ) 0 (mod ir) and a S„-group D of CG (X ) is a defect group of B. Replacing y by a conjugate it may be assumed that y E D. Hence xr (yx) = x, (x ) 0 (mod 7r). If y and y ° are not conjugate in G then by (IV.6.3)

EE B1,6(yx)1 2

-=

E xxyxnx.(yx)-o

x EB

Since Ix, (yx )1 0 for all s this implies that x, (yx ) = 0 for all x, E B. Hence in particular x,(yx) = 0 contrary to the previous paragraph. E 2. Inertia groups

Throughout this section Ô is a fixed normal subgroup of G. In general a tilde sign will be attached to the quantities associated with Ô. For instance • are all the irreducible characters of Ô. 05s is the central character of K[0] corresponding to is . Let A be a subset of Ô which is a union of conjugate classes of G. Let 6 be a complex valued function defined on A which is constant on the conjugate classes of Ô which lie in A. If z E G define 6 by 6' (x) = 6(x' - ') for all x EA. Then 6' is defined on A and is constant on the conjugate classes of Ô which lie in A. The inertia group T(ô) of 6 in G is defined by T(0) = {z I z E G, 6' = 6 } . Clearly Ô c T (0 ). If iV is an irreducible 17 [ 6] module which affords 0, then it is easily seen that T( W) = T( 1 ). Similarly if is an R -free R[Ô] module which affords L. then it is easily seen that T( 'K) = T However T (V') need not be equal to T( -/-K). Let A be a block of O. The inertia group TO ) of . in G is defined by T(A) = {z I z E =13-}• LEMMA 2.1. Let 1-4 be a block of Ô. If i„ c-,6; T(is )C T(i4 ).

En

then T(0 1 ) C T(ñ) and

PROOF. Clear by (IV.4.9). E THEOREM 2.2. For any group H with Ô CHC G let ÇA, (H) be the set of all irreducible characters ofH whose restriction to Ô have as a constituent. Let 9J(H) be the set of all irreducible Brauer characters ofH whose restriction to have 45 , as a constituent. (i) If T(CP,)C H then the map sending B to 0 G defines a one to one correspondence between 2t(H) and 91 (G).

196

CHAPTER

V

[2

(ii) If Vis an irreducible P[G] module which affords a Brauer character in W(G) then a vertex of V is contained in T(Cp,). (iii) If T( S )CH then the map sending 0 to O G defines a one to one correspondence between 2L (H) and ?Is (G). (iv) If x i E 9.i s (G) then there exists an R -free R[G] module which affords xi and whose vertex is contained in T ( s ).

PROOF. (i) It suffices to prove the result in case T(Cp,)= H. Let 0 E 91?(H). By (11.2.9) (e ) ' , = fi + B where Cp, is not an irreducible constituent of Ob. Let i be an irreducible constituent of q1G with yin = i + 0 1 . By (111.2.12) = e(p, for some integer e > 0 and i (1) =G : tpG (1). Hence = i E 9f?(G). Furthermore if 0,, tif2 E W(H) and 0; = 0? then ( 143)ii = (OH and so 01 = 02. Therefore the map sending 0 to O G is one to one. It remains to show that it is onto. Let cp, E 2l?(G). Let V be an R[G] module which affords 4), and let IV, be an 1 [6] module which affords (p,. By (111.2.12) %id is completely reducible and so /(V6, l)/ O. Thus by (111.2.5) /(VH, 0 and so W)#0 for some irreducible constituent W of Ii7 ;1 . Since (1' / 1, 1)6 = : GIW, it follows that Vi7, is a constituent of IV,. Thus if (Is is the Brauer character afforded by W then 0 E 21?(H). Since /( V, W G ) = /(VH, W) #0 the irreducibility of V and WG implies that V = W G and hence cp, = oG as required. (ii) Clear by (i). (iii) If p is replaced by a prime not dividing IG1 this follows from (i). (iv) Clear by (iii). Lii

co=

A block B of G covers a block 14 of O if there exists xs E B and is a constituent of (x)o.

E /E-3

such that

LEMMA 2.3. The blocks of Ô covered by the block B of G form a family of blocks conjugate in G. If I4 is covered by B and xs E B then some constituent of (Xs)6 belongs to P. If t E f3 there exists x s E B such that 1 is a constituent of (xs)o. PROOF. This is a reformulation of (IV.4.10). 111 LEMMA 2.4. xs and xt belong to blocks which cover the same block of Ô if and only if there exists a chain x,, = X s, x12 , =xi such that for any m either x and Xj, are in the same block of G or (x),„)0 and (x,,)G have a common irreducible constituent.

2]

INERTIA GROUPS

197

PROOF. If such a chain exists then by (2.3) consecutive characters belong to blocks which cover a fixed block of Ô. Suppose conversely that x, and xi are in blocks which cover a fixed block of O. Let be an irreducible constituent of (Xs )ô. By (2.3) there exists X, in the same block as x, such that i„, is a constituent of (x,)0. The chain xs,x„x, satisfies the required conditions. 0 THEOREM 2.5 (Fong [1961], Reynolds [1963 1 ). Let fl be a block of Ô. The map sending b. to bG defines a one to one correspondence between the set of all blocks of T (li ) which cover ii and the set of all blocks of G which cover Furthermore if b. is a block of T(B) which covers B the following hold. (i) The map sending 0 to O G defines a one to one correspondence between the sets of all irreducible characters, irreducible Brauer characters respectively, in b and bG. (ii) With respect to the correspondence defined in (i) b. and !I' have the same decomposition matrix and the same Cartan matrix. (iii) b. and b. ° have a defect group in common. (iv) ii is the unique block of 6 covered by Ê.

PROOF. Let b be a block of T(ñ) which covers b and let be an irreducible character in By (1.2), (2.1) and (2.2) (iii) if3 G is defined, E b. ° and ÊG covers ii. Suppose that B is a block of G which covers Let x, E B. By (2.3) (xi )6 has an irreducible constituent is E Û. By (2.1) and (2.2) (iii) xi = r for some irreducible character of T(b) such that is an irreducible constituent of 4"6. Let b be the block of T(A) with 4- E B. is Then Ê covers El and i4 G = B by (1.2). Let 0,,02 be irreducible Brauer characters of T(b) in blocks which cover ii. If Cp E ii and 43 is an irreducible constituent of (e)G then Cp is a constituent of OA )d for i = 1,2. Thus if i= tg then (4/ 1 )0 and (02 )6 have a common irreducible constituent which is in and so 01 = 02 by (2.2). Therefore the map sending 0 to 0G defines a one to one correspondence between the sets of all irreducible characters, irreducible Brauer characters respectively, of T (li ) and G which lie in blocks that cover ii. This map clearly preserves decomposition numbers and hence also Cartan invariants. Since the decomposition matrix of a block is indecomposable it follows that a character or a Brauer character O of T(b) is in Ê if and only if 0 G E Ê G Thus the map sending Ê to Ê ' is one to one as required. Furthermore (i) and (ii) are proved. (iii) Let 15 be a defect group of Ê. By (111.9.6) 15 C D for some defect group D of Ê. By (i) and !I' have the same defect. Thus D = D.

r

.

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(iv) Since E = E for all z E T(f3) this is clear by (2.3).

[3

D

COROLLARY 2.6. Let B be a block of G and let E be a block of Ô covered by B. Then T(b) contains a defect group of B.

PROOF. By (2.5) B = if3 G for some block E of T(b) which covers E. The result follows from (2.5) (iii). 0

3. Blocks and normal subgroups The notation of the previous section will be used in this section. For the results in this section see Brauer [1967a], [1968]; Fong [1961]; Passman [1969]. LairmA 3.1. Let a E Z (R;G : G)n R[6]. If L is a central character of 1 [6] then (a) = (a).

PROOF. Clear by definition.

D

LEMMA 3.2. Let {B- 1„,} be the set of all blocks of Ô where for some z E G if and only if i = j. Let Jim be the centrally primitive idempotent of R[6] with = B(é,„). Let e, Then fe 1 1 is the set of all primitive idempotents in Z(R; G : G)(-1 R[6]. PROOF. Any idempotent in Z (R ; G: G) n R[6] is a sum of centrally primitive idempotents in R[Ô] and is invariant under conjugation by elements of G. The result follows. D LEMMA 3.3 (Fong [1961]). Let E,„, and

be defined as in (3.2). For each i let be all the blocks of G which cover some and let e,, be the centrally primitive idempotent in R[G] with B„, = B(e,„). Then e, =

I rrt

Jim

= /m

eim

•

By (3.2) {e,} is a set of pairwise orthogonal central idempotents in R [G]. Thus it suffices to show that e,„,e, X 0 for all i, m. Let V be a nonzero R[G] module in B 1,,,. By (2.3) (V 6)e, O. Hence Ve, = Ve,„ e, 0 and so e, X O. D PROOF.

LEMMA 3.4 (Passman [1969]). Let B be a block of G and letE be a block of

199

BLOCKS AND NORMAL SUBGROUPS

Ô. Then B covers 1j aEZ(R;G:G)nR[6].

if and

only

if

Ab (a)= AB (a) for

all

PROOF. By (3.2) and (3.3) B covers 1 if and only if Ab (ë) = AB (e) for every idempotent in Z(R;G : G) n 1t[6]. By (I.16.1) ft is a splitting field of Z(R;G : G) n R[6]. The result follows since two central characters of a commutative algebra are equal if they agree on all idempotents. 0 LEMMA 3.5. Suppose that GIG- is a p-group. If E is a block of 6 there is a unique block B of G which covers E. PROOF. Suppose that B, and B 2 are blocks of G which cover E. Let A, = AB, for i = 1,2. If C is a conjugate class of G consisting of p'-elements then E Z(R; G :G)n R[6]. Thus by (3.4) A i (0) = (0= A2(0). Hence by (IV.4.2), (IV.4.3) and (IV.4.8) (iii) B 1 = B 2 . 0

e,) =

A block B of G is regular with respect to 6 if AB ( 0 for all conjugate classes C of G which are not contained in Ô. B is weakly regular with respect to 6 if there exists a conjugate class C of G with C C 6 such that AB (0') / 0 and the defect of B is equal to the defect of C. In case 6 is determined by the context the phrase "with respect to 6" will be omitted. By (III.6.10) a regular block is weakly regular. Furthermore if AB (0') / 0 and the defect of B is equal to the defect of C then B and C have a common defect group.

LEMMA 3.6. Let B be a block of G. The following are equivalent. (i) B is regular. (ii) B = EG for every block E of 6 which is covered by B. (iii) B = PG for some block E of Ô. PROOF. (i) (ii). Let B- be a block of 6 which is covered by B. Let C be a conjugate class of G. If CZ 6 then (AA )'(0) = 0 = AB (0). If C C 6 then by (3.4) AiJ (0') = AB (). Hence by (3.1) (AEO G =AB. (iii). Clear. (i). If C is a conjugate class of G with CZG then AB (0) = 00-() = 0 and so B is regular. D (iii)

LEMMA 3.7. Let B be a block of G which is regular and let E be a block of Ô. Then B covers È if and only if B =

200

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V

PROOF. If B covers Pi then B = 13 G by (3.6). If B = 13- G then by (3.1) AA (a) = (AO' (a) for all a E Z(R; G G) n R[G]. Thus by (3.4) B covers E LEMMA 3.8. Let 13 be a block of Ô. Then there exists a block of G which is regular and covers 1 if and only if 13 G is defined. In that case 13- G is the unique block of G which is regular and covers

P.

PROOF. If ÉG is defined then 13- G is regular by (3.6) and É G covers 13- by (3.7). The converse follows from (3.7). The last sentence is clear. D 3.9. Let B be a block of G with defect group D. If CG (D) C B is regular.

LEMMA

6 then

PROOF. If AB (C) 0 for some conjugate class of G then by (III.6.10) D CCG(Z) for some z E C. Hence z E CG (D) C 6 and so C C 6. 0 LEMMA 3.10. Suppose that P is a p -group with P< G and CG (P) C Ô. Then every block B of G is regular with respect to Ô. For every block 13 of 6, the block 13 G is defined and is the unique block covering P. PROOF. Let B be a block of G and let D be the defect group of B. By (111.6.9) P C D. Thus C G (D)C CG (P) C 6 and B is regular by (3.9). Let be a block of G. By (3.8) ÉG is defined and is the unique block covering 1. Ll 3.11. Suppose that P is a p-group with CG (P) ci P < G. Then the principal block is the unique block of G.

COROLLARY

PROOF. By (3.10) with 6 = P, 13- G is the unique block of G where 13- is the unique block of O. Hence G has exactly one block which must necessarily be the principal block. 0 THEOREM 3.12 (Fong [1961]). Let 13 be a block of 6 and let B 1 ,13 2 ,... be all the blocks of G which cover B and let B = B, for some j. For each i let e, be the centrally primitive idempotent of R[G] with B, = B(e,) and let e =Ee,. The following are equivalent. (i) B is weakly regular. (ii) A defect group of any block B, is conjugate to a subgroup of a defect group of B. (iii) The defect of any B, is at most equal to the defect of B.

BLOCKS AND NORMAL SUBGROUPS

201

(iv) Let e =Eace where C ranges over all conjugate classes of G and ac E R. There exists a conjugate class C consisting of p'-elements with C C Ô, cic X 0, AB (0) X 0 such that the defect of B is equal to the defect of C. PROOF. (i) (ii). There exists a conjugate class C of G with C C Ô and AB (C_) # 0 such that B and C have a common defect group D. By (3.4) AB, AA AB ) X 0 and so by (III.6.10) a defect group of B, is a subgroup of some defect group of B. (ii) (iii). Trivial. (iii) (iv). Since AB (e) = 1 it follows that d cAB (6) / 0 for some C. By (IV.7.2) C consists of p'-elements. By (3.3) C C O. If e, =Ea,,c e" for each i then for some i a1# O. Hence by (IV.7.3) the defect of C is equal to the defect of B, and so is at most equal to the defect of B. Since A (0) / 0, (III.6.10) now implies that B and C have the same defect. (iv) (i). Trivial. 0

(e) =

(e) =

(e

COROLLARY 3.13. Let Li be a block of O. The following hold. (i) There exists a weakly regular block B of G which covers 1. (ii) Let I be a block of T(13- ) which covers P. Then Li is weakly regular if and only if AG is weakly regular. PROOF. (i) Choose a block B of maximum defect among those which cover A. By (3.12) B is weakly regular. (ii) By (2.5) A and AG have the same defect. Thus Ê has maximum defect among all blocks 1 of T(b) which cover if? if and only if ./f3 G has maximum defect among all blocks Ê. The result follows from (3.12). 0 THEOREM 3.14 (Fong [1961]). Let Li be a block of O and let B be a block of G which covers 1 and is weakly regular. There exists a defect group D of B with D C T (1 ). For any defect group D of 13 with D C T(1) IT(Li): DOI 0 (modp) and D n O is a defect group of t.

T(Li) contains a defect group D of B. Thus by (2.5) and (3.13) (ii) it may be assumed that G = T(A). Let é" be the central idempotent of R EG] with ij = B(é). Since G = T (1 ), =Eace with ac E R and C ranging over the conjugate classes of G. By (3.3) and (3.12) (iv) there exists a conjugate class C of G with C C O, dc X 0, AB (0) X 0 such that D is a defect group of C. Let z E C with D C CG (z). Let L be the conjugate class of O with z G L. Thus C = U Lx where x ranges over a cross section of NG (L) in G. By (3.4) PROOF. By (2.6)

202

CHAPTER

AR

( 0) = Ao ( 0) =

AB

V

(i')=

AL -1 ( 1 )-

Since T(13- ) = G this implies that

T(fi):No (L)1,ka (Z)= AB (0) O. Thus T(/4 ): NG (L 0 (mod p) and /lb (f..) O. Since NG (L)-- CG (z)Gand D is a Sr -group of CG(Z) it follows that I T(fi):DO I 0 (mod p). Since dcvka (L)= Oit follows from (IV.7.3) that fi and L have a common defect group. Clearly D n G is a Sr -group of Co (z) and so is a defect group of L. E LEMMA 3.15. Let B be a weakly regular block of G and let 1-4 be a block of covered by B. Let Xs be of height 0 in B and let it be a constituent of (X.)6 with it E E. Then the ramification index of xs is not divisible by p height 0 in Pi and IT(b):T(X 7,)1 0 (mod p).

,

has

PROOF. By (2.5) and (3.13) it may be assumed that G = Let D be a defect group of B. Let d = v (I D I) and let d = v (I D n Op). By (3.14) d is the defect of E. By (111.2.12) (x, )d = e 4; where {z} is a cross section of T(it) in G and e is the ramification index of x, Since 6D/0 --- DIG n D it follows from (3.14) that v(I GI)— d=- v(IOD:DO= v(IO: 6-

n D 1) = v(161)- d.

Thus v(xs ( 1 )) v(1 6 1) — d. Since v (it ( 1 )) v — d and x, (1) = eIG:T(X- )lit (1) it follows that e I G : T(x- )I 0 (mod p) and X', is of height 0 in b. O COROLLARY 3.16. Let kJ be a block of Ô. There exists

it E /3- of height 0 such

that T(it ) contains a Se -group of T(13- ).

PROOF. By (3.8) there exists a block B of G which covers E and is weakly regular. Let x, be of height 0 in B. By (2.3) some constituent X", of (xs )ô is in E. The result follows from (3.15). El

4. Blocks and quotient groups The notation of the two previous sections will be used in this section. Furthermore G ° = GIG- and in general a superscript ° will be attached to the quantities associated with G`). Every R [G1 module may be considered

BLOCKS AND QUOTIENT GROUPS

203

to be an REG] module in a natural way. Similarly every character or Brauer character of G' is a character or Brauer character of G. LEMMA 4.1. Let B ° be a block of G'. Then there exists a block of G which

contains B ° and covers the principal block of Ô. Conversely if B is a block of G which covers the principal block of Ô then B contains a block of G'.

Two R[ G °] modules in B ° are linked by a chain of R[G1 modules. Hence they are also linked as R[G] modules and so lie in the same block B of G. Thus B ° C B. If x, E B ° C B then the principal character of Ô is a constituent of (x)6. Thus by (2.3) B covers the principal block of Ô. Conversely suppose that B covers the principal block of Ô. By (2.3) there exists Xs G B such that (x, )d contains the principal character of Ô as a constituent. Thus by (111.2.12) Ô is in the kernel of Let B ° be the block of G' with x, e B ° . Then B ° C B by the previous paragraph. 0

PROOF.

LEMMA 4.2. Let B ° be a block of G' and let B be a block of G with B ° C B.

Let D be a defect group of B. Then D contains a Se -group of 6- and D' = 16contains a defect group of B". In particular ifd is the defect of B and d' is the defect of B ° then d' d — v(1G I). Furthermore if B ° contains an irreducible character or an irreducible Brauer character of height 0 in B then D' = DG - 16 is a defect group of 13'. '

PROOF. Let P be a Se -group of Ô and let P be a p -group with P c P such that P° = 16 is a defect group of B". Let V be an R -free R[G1 module which affords an irreducible character of height 0 in B". Thus P' is a vertex of V. Hence V considered as an R [G] module is R [P]-projective. Thus by (IV.2.2) P is a vertex of V. Hence P C G D. If v(ço, (1)) or v(x, (1)) equals v(I G 1) d then v(1 G 1) d v (I G : Ô 1)- cr and so d') v(1 I). Thus d" = d — v(1 Ô I) by the first part of the statement. E —

—

—

LEMMA 4.3. _Suppose that Ô is a p'-group. Let B, ,

, B„, be all the blocks of G which have Ô in their kernel and let D, be a defect group of B, for j = 1, , m. Let 13 7 , , be all the blocks of G'. Then m = n and after a suitable rearrangement B, = 13; for all j. Furthermore D = D,G - 16 is a defect group of 13; . PROOF. Since Ô is a p'-group the principal character is the only irreducible character in the principal block of Ô. Hence a block of G covers the principal block of Ô if and only if it has Ô in its kernel. Thus by (I V.4.11)

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any two modules in B, are linked by a chain of R [GI modules and so B, is after a suitable a block of G ° . Therefore by (4.1) m = n and B. rearrangement. Since Ô is a p'-group B, and B`i have the same defect and so by (4.2) D(,) is a defect group of B?. 0 LEMMA 4.4. Suppose that Ô is a p-group. Let B be a block of G and let D be a defect group of B. Then Ô C D and there exists a block B ° of G ° such that C B and D ° = D/G- is a defect group of B ° . PROOF Let d = v( D I). Choose cp, E B with v(q), (1)) = v (I G I) — d. By (111.2.1 3) G is in the kernel of cp,. Let B ° be the block of G° with cp, E B ° . By (4.1) B ° C B. Let d ° be the defect of B ° . The result follows from (4.2). CI 4.5. Suppose that Ô is a p-group which centralizes all p'-elements in The inclusion B ° C B defines a one to one correspondence from the set of all blocks of G ° onto the set of all blocks of G. If D is a defect group of the block B of G then D ° = DG- /6 is a defect group of the block B ° of G ° where 13 ° C B. LEMMA

G.

PROOF. In view of (4.1) and (4.4) it suffices to show that if B is a block of G and B?, B? are blocks of G° with 13° C B and B °2 C B then B7 = B ° Let xs e B° and xr e 13'2'. If x is a p'-element in G then Ô c CG (x). Thus CG ()CVO = CG 0 (xl where x ° is the image of x in G° . Since x„ x E B this implies that

I G ° 1 X.±x) _ ICGo(x °) I Xs ( 1 )

_ (x

(x)

_ G° I C G 0(01

(x) xi ( 1) (mod ,n-)

for all p'-elements x in G. Hence by (IV.4.3) and (IV.4.8) B; = .13?. El

4.6. Suppose that D is a p-group and G = DC G (D). Let B be a block of G with defect group D. Then B contains exactly one irreducible character x, which has D in its kernel and B contains exactly one irreducible Brauer character cp,. Furthermore x, (x)= ça, (x) for all p'-elements x in G, has height 0 and (I), (x) = I D I(P, (x) for all p'-elements x in G. COROLLARY

PROOF. By (4.5) there exists exactly one block B ° of G/D with B ° C B. Furthermore B ° is a block of defect O. Let xs, cp, respectively, be the unique irreducible character, irreducible Brauer character respectively, in B ° . Thus x, (x) = (x) for all p'-elements in G. Since every irreducible Brauer

205

BLOCKS AND QUOTIENT GROUPS

character in B has D in its kernel, cp, is the unique irreducible Brauer character in B. Then x, has height 0 in B by (4.5). Since (c„) is the Cartan matrix of B, (IV.4.1 6) (i) implies that c„ =ID I proving the last statement. El If G = DCG (D) for some p -group D and B is a block of G with defect group D then the irreducible character Xs constructed in (4.6) is the canonical character of B.

THEOREM 4.7. Suppose that G = DCG (D) for some p-group D. Let B be a block of G with defect group D and let x = x, be the canonical characterof B. Let 4-1 , , be all the irreducible characters of D. Define O. follows : ify0D,

01 (z)= 0 = (Y)X(x)

if Y E D

where y is the p-part of z and x is the p'-part of z. Then {0, j j = 1, .. .,n} is the set of all irreducible characters in B.

PROOF If z E G then 47 = for all j. Thus if xr E B then by (111.2.12) (xt)D = 41 for some j and some integer e > 0. Let z E G, let y be the p-part of z and let x be the p'-part of z. If y D then x, (z ) = 0 by (111.6.8) and (IV.2.4). Suppose that y E D and let H=D x (x) . Then (x( ),, = 6a for some character a of (x). Thus x, (xy ) = (y)a (x) and

"J ( Y )) X, (xY )= ci ( 1 X, (x)=

(Y1 ) d1 (x) =

dt;

w e; (xY

O. where cp; e B. Consequently x, = [cla/6 Let A be the set of all p'-elements in G and let A ° be the image of A in G° = GID. Since x is a character of G° which vanishes on all p -singular elements we see that ( 1 )]

G ,,k-i(y)12,,,,lx(x)1 2 =- I D EG„Ix(z)1 2 =1. - IG1 . 1 ,E Thus

1= 11x, = d:; ) 2 1 0,112 =d( '; ) 2. Hence d, = (1) and so x, = O.

f xr,lx(x)12-

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CHAPTER V

[5

Therefore there exists a subset S of 11 , , /21 such that {Of jj E SI is the set of all irreducible characters in B. If x, = Of then d,i = 6 (1). Thus I DI=c11= E This implies that S =

E

(1)2.

iEs

,

0

5. Properties of the Brauer correspondence

The results in this section are due to Brauer [1967]. An alternative approach has been given by Passman [1969]. See also Brauer [1971a], [1974a]. The presentation given here is a modification of these two methods. Recently Alperin and Broue [1979], Broil& [1980] have studied the Brauer correspondence in a more general setting. Their approach generalizes and gives an alternative treatment of the first main theorem on blocks and much of the material in this section and the next. P be a p-group, P C G. Let N = NG (P) and let H < N with (P) C H. If Bo is a block of H then Br,' is defined.

LEMMA 5.1. Let CG

By (3.10) Ir is defined. Thus by (111.9.2) and (111.9.4) Br); = (13(;')G is defined. 0 PROOF

p-group, D C G. Let N = NG (D) and let H < N with CG (D) C H. The map sending B o to B 0' defines a one to one correspondence between the blocks of G with defect group D and representatives B o of N-conjugate classes of blocks of H which satisfy (i) D n H is a defect group of B o . (ii) v (1 DII j) v(IT(B 0)1), where T(B 0) is the inertia group of Bo in N. LEMMA 5.2. Let D be a

PROOF. By (111.9.7) it may be assumed that N = G. By (2.3) and (3.10) the map sending B o to BI;I defines a one to one correspondence between blocks B of N and representatives B o of N-conjugate classes of blocks of H. If B = .K has defect group D then by (3.14) D n H is a defect group of Bo and v DH = T(13 0)1)Suppose that Bo satisfies (i) and (ii). Let D, be a defect group of .13 0" . By (3.14) and (i) D n H and D, n H are both defect groups of B0 . Thus By (3.14) and (ii) 7)(1 DH v( T(B0)1)=

5]

PROPERTIES OF THE BRAUER CORRESPONDENCE

207

I). Thus IDI=1 D, I. Since D1N D CD f . Therefore D = D i . ,

Let D be a p-group, D C G. (D,B) is a block pair in G if B is a block of DCG (D) with defect group D. If (D,B) is a block pair in G then by (111.9.4) B A is defined for every group A with DC G (D) ÇA C G. Let (D,B) and (D*,B*) be block pairs in G. (D*,B*) weakly extends (D,B) if D
D* .

LEMMA 5.3. Suppose that (D*,B*) is a block pair which extends (D, B). Let H = D *CG (D). Then (D*,B *) weakly extends (D,B) and D* is a defect group of B" =B*" . PROOF. Let Do be a defect group of B*" with D* CDo . By (3.14) Thus Do n DC G (D)= Do n DC G (D)= D. Do CG (D)= H and D*ID. Hence 1D*I= 'Doi HIDC G (D)--D* nDC G (D) and Do/D and so D* = Do . By (3.10) and (3.14) T(B)= H. Hence (D*,B*) weakly extends (D, B). 0 THEOREM 5.4 (Third Main Theorem on Blocks) (Brauer). (i) Let Go be a subgroup of G and let B o be a block of Go with defect group D. If

CG (D) C Go there exists a block pair (D, B) in G with B 00 = B o . (ii) Let (D, B) be a block pair in G. Let Do be a defect group of B ° . Then Furthermore 11)1
PROOF. (i) Since D = D n DC G (D) this follows from (5.2). (ii) By (111.9.6) I D I !Doi

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Suppose that (D * , B *) is a block pair which properly extends (D, B). Let H = D*CG (D). Then B G = (B" )G = B *1)G and so ID <1 D* (

by (111.9.6). Assume conversely that I D I v(I DC G (D) I). Thus D# D* and so (D* ,B*) properly extends (D,B). (iii) Let A be a group with CG (D *) C A i D *C G (D*). Then D*C G (D*)/D* — A/A n D*. Since D* is in the kernel of 4' * this implies that 4- A is irreducible. Hence in particular 4' t r,( co) is irreducible. Let 0 be an irreducible character of A which has D* n A in- its kernel such that D . Then A = CG (D*)(D* r) A). Thus 4',1; is CC G (D-) is a constituent of a constituent of 0 as D* n A is in the kernel of both 0 and CA. Therefore is the unique irreducible character of A which has D * n A in its kernel such that G (D.) is a constituent of its restriction to CG (D *). Since D is in the kernel of 4' it follows that D is in the kernel of every irreducible constituent of i" DC G (D•) • Thus if a is the multiplicity of (D *) as a constituent of i" CG (D) (111.2.5) implies that -

G

aG

o

a = (c 0 (D), a G (D.))

—

(4"Dc 0 (D.),

Ct, CG (D"))

= {ac G (D.rc G (D)).

Since D *C G (D *) n D CG (D) = DCG (D *), (11.2.9) implies that a = (‘,{4- PDcG (D.)1DcG (D) )= (4,{C H }DcG(D))

where H = D *CG (D). Let 4' * H = O + O o where 0 E B" and 00 is a sum of irreducible characters none of which is in B". Thus ((0o)p cG( p ), 4') = O. Since D is in the kernel of 0 it follows that 4' is the unique irreducible character of D CG (D) which is a constituent of 0_ DcG (D) • Hence a — and so

(C,{C *H }Dc G (D))—

0 DC G (D) —

(4", ODcG (D))

a. Therefore 0(1) =

4 (1) .

However

v( ( 1)) = v(IDC G (D): D =

= 1,(IH:D*C G (D*)14- *(1))= v4' *" (1)). Consequently v(0(1)) = v(a)+ v(C H (1)). Thus (1.3) implies that B" = B*" if and only if v(a)= O. E

6]

BLOCKS AND THEIR GERMS

209

6. Blocks and their germs Let B be a block of G. Suppose that B = Bô for some block Bo of a subgroup Go of G. In this section connections between B and Bo are investigated. Some of these results are of importance for application to questions concerning the structure of groups. The proofs in this section make use of the three main theorems on blocks. The first three results are due to Brauer [1967]. The results in the rest of the section are related to work of Reynolds 1 1974 Let B be a block with defect group D. By (111.9.7) there exists a unique block B, of NG (D) such that B = B. The block B, is called the germ of B with respect to D. If D is determined by the context B, is called the germ of B.

LEMMA 6.1. Let B o be a block of a subgroup Go of G and let Do be a defect group of B o . Suppose that

CG

(Do)C Go . Then B = Bô is defined and

Z(D)C Z(D 0 )cl Do C D for some defect group D of B. In particular if B has an abelian defect group then D o is a defect group of B.

PROOF. By (5.4) (i) B = fiô is defined. Let d be the defect of B and let d0= v(1 D01) be the defect of Bo . The proof is by induction on d — do . If d — do = 0 then by (111.9.6) Do is a defect group of B and the result follows with D = Do. By the third main theorem on blocks (5.4)(i) it suffices to take Go = Do CG (DO. Suppose that d — do> 0. By (5.4)(ii) there exists a block pair (D* ,B*) in G which properly extends (D o , B o ). Since B* G = Bô = B induction yields the existence of a defect group D of B such that Z(D)cl Z(D*) C Co• (Do) = Z(DO) C D0 c1 D* C D.

This proves the first statement. The second statement is an immediate consequence of the first. LI A more direct proof of the next result can be found in Brauer [1964a] Theorem 3. See also Brauer [1971a], [1971b]; Osima [1964]; Hamernik and Michler [1973]; Hamernik [1973].

THEOREM 6.2. Let B o be a block of a subgroup Go of G and let Do be a defect group of B o. Assume that CG (DO) C Go. Then B o is the principal block of Go if and only if Bô is the principal block of G.

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[6

PROOF. By the third main theorem on blocks (5.4)(i) B o = 13() for some block B, of Do CG (D0 ). Thus it suffices to prove the result in case G0= DoCG (Do). For any subgroup A of G let B,(A) be the principal block of A and let 6 (A) be the principal character of A. Let D be a defect group of B and let P be a Se -group of G with D C P. Let do = v(I Do ) and let d = v( D 1). The proof is by induction on d — d0 Suppose that d — do = O. By (111.9.6) it may be assumed that D = Do. By (3.10) Bo = B (Do CG (Do)) if and only if B' = B 1 (N) where N = N G (Do). Let V be the R -free R[G] module which affords 6(G). Let f be the Green correspondence with respect to (G, P,N G (P)). Thus f ( V) affords 6 (NG (P)). Thus by (111.7.7) and (111.9.7) if B is a block of NG (P) then B,(G)= B G if and only if B = B, (NG (P)). Suppose first that B,; = B, (G) then Do = = P and so B,;' = B 1 (N). Assume conversely that Bf; = B, (N). Thus Do is a Se -group of N and so Do is a Se -group of G. Hence Do = D = P and /3 ( = (B) G = B, (G). Suppose that d — do> 0. By (5.4)(ii) there exists a block pair (D*,B*) in G which properly extends (Do,B0). Assume that /3 ( = B, (G). Thus B" = B,(G) and so by induction B* = B,(D*C G (D *)). Thus by induction BY = B*" = Bi (H) where H = D *C G (D). By (3.7) B(' covers B o and so B o = Bi(DoCG (D0)). Assume conversely that B o = BI(DOC G (DO)). Hence 6 (Do CG(Do)) is the canonical character of Bo . Thus by (5.4)(iii) (D*C G (D*)) is the canonical character of D *C G (D*). Therefore B* = B ,(D* CG (D *)) and so by induction B = B *G = B (G). E .

6.3. Suppose that y is a p-element in G such that CG (y) has a normal p-complement. If x = x, is in the principal block of G then x(yx)= x(y) for every p'-element x in CG (y). COROLLARY

PROOF By (IV.6.1) and (6.2) x(yx) = E, ' (x) where cp ; ranges over the irreducible Brauer characters in the principal block of CG (y). By (IV.4.12)(ii) );(x) = ; (1) for all j and all p'-elements x in CG (y). LEMMA 6.4. Let H be a normal p'-subgroup of G. For any subset A of G let

A denote the image of A

in 6 = G/H. Let B o be a block of a subgroup Go of G for which 13 ( is defined. (i) His in the kernel of B (? if and only if H n Go is in the kernel of B o . (ii) Suppose that Go n H is in the kernel of B o . There exists a block i40 of Go H with H in its kernel such that B o corresponds to 1jo by the natural

6]

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BLOCKS AND THEIR GERMS

isomorphism G 0H1H G0/G0 n H. Furthermore A is defined and B,?= IV. Also 1H: Go n HI 1 (mod p).

PROOF. The isomorphism G0/G0 n H ----defines a one to one map —> from the set of all irreducible characters of Go with Go n H in their kernel to the set of all irreducible characters of G0 H with H in their kernel. Thus C--G„ = Let T be the union of all blocks of G which have H in their kernel. (i) Suppose that H is in the kernel of B. Let 4- be an irreducible character in B0 . By (1.3) there exists an irreducible character x in B with xC Thus 4' C xG„ by Frobenius reciprocity (111.2.5). Since Go n H is in the kernel of xG„, it is in the kernel of C. Since C was arbitrary in Bo this implies that G o n H is in the kernel of Bo . Suppose conversely that Go n H is in the kernel of Bo . By Frobenius reciprocity (111.2.5) 4' c Co". Since H < G a cross section of G0H in G is a complete set of representatives of the (Go, H) double cosets in G. Thus the Mackey decomposition (11.2.9) implies that

r.

G G )11 =

= Efc(1)60,-,0H,

=

where x ranges over a cross section of G0 H in G. Therefore (1,,,(C G ) H ), = I G:Go HIC(1)=IG:Go Hl4-- (1)= G (1) .

Thus 4-- is the sum of all the irreducible constituents of C G which have H in their kernel. Hence (C G )T = If B is a block with B CT then v((C G r (1)) v(C G (1)) by (1.3). Since (1) = G0 H: Gole - ( 1) it follows that v(C (1)) = v(4- (1)). Since (C G = 47- this implies the existence of a block B, C T with v((C G )81 (1)) = v(C G (1)). Hence B 1 = .10 by (1.3) and so H is in the kernel of B. (ii) The existence of B- 0 follows from (4.3). Let C be an irreducible character in Bo. Let w, Co be the central character corresponding to respectively. Thus corresponds to B- 0 . Let x i , x2, ... be all the irreducible characters of G and let bs = (xs ,‘ G ). By (i) (C G ) T = c Thus by (1.1) (i)w- = E b. (1)(.0„ G MCC) G = x

E,ET b,x,(1)w s .

Since B O °C T it follows from (IV.4.22) that (C G (1)/e G (1)* G = G . Evaluating at 1 we get that ( z- mu 1 ),= 1. Therefore co = G . Consequently È 1 is defined and fg= B. Furthermore -

IH:

n HI

(

‘G(1)14--

-

(

1) = 1

(mod 71-).

212

Hence

CHAPTER

: Go

n

1 (mod p).

V

[6

111

THEOREM 6.5. Let IT be a set of primes with p E 7r. Let B be a block of G with defect group D. Let N = NG (D) and let B, be the germ of B. Assume that 0,(N) is in the kernel o f B, . Suppose that Go is a subgroup of G, B o is a block of Go with defect group D o, CG C Go and .13 (? = B. Then 0,(G 0 is in the kernel of Bo. )

PROOF. By (111.9.6) it may be assumed that D o C D. By (5.4) Bo = .130 for some block B2 of DOCG (D0). Let d = v(ID I), do = v(jDo I). The proof is by induction on d — do. Suppose that d — do = O. By (111.9.2) .13? = B,?= B and .13? = Thus B t2'j = B, by (111.9.7). By (3.10) B, covers B2. Therefore 0,,(N)= 0,(DC G (D)) is in the kernel of B2. Thus 0.,(G0) n D CG (D ) is in the kernel of B2 and so 0„,(G0) is in the kernel of Bo by (6.4). Suppose that d — do > O. By (5.4)(ii) there exists a block pair (D*,B*) in G which properly extends (D o, B2). Let H = D *CG (D o ). By induction 0,(H) is in the kernel of B = B* 11. Since BV covers B2 it follows that 0„ , (H) = 0 , (Do C G (Do)) is in the kernel of B2. As 0„ , (G0) fl DoCG (DOC 0„ ,(D o C G (Do)), (6.4) implies that 0,(G0) is in the kernel of Bo = B. 0 6.6. Let IT be a set of primes with p E 77. Let B be a block of G with defect group D. Assume that OAN G (D)) is in the kernel of the germ of B. Suppose that y is a p-element in G such that CG (y) has a normal S,-subgroup. If x = Xi E B then x(yx)= x(y) for every 7r'-element x in

COROLLARY

CG(Y)•

PROOF. By (IV.6.1) and (6.5) x(yx) = d;cp;'(x) where cp ranges over blocks of CG (y) which have O,. (C G (y )) in their kernel. Thus (p),' (x) = cp; (1) for every 7r'-element x in CG (y). THEOREM 6.7. Let IT be a set of primes with p E

77. Let B be a block of G with defect group D. Let Do be a p-group contained in G and let B o be a block of D o C G (Do) such that .13, = B. (i) Suppose that whenever D o C D' for some z E G, O,' (CG )) is in the kernel of B o. Then 0,(N G (D)) is in the kernel of the germ of B. (ii) Suppose that 0,(CG (Do)) is a S,-group of CG (Do) and 0, (C G (Do)) is in the kernel of B o. Then 0,(N G (D)) is in the kernel of the germ of B.

6]

BLOCKS AND THEIR GERMS

213

PROOF. By (6.4) it may be assumed that D o is a defect group of Bo. (i) Let d = v(ID1) and let do = v ( Do I). The proof is by induction on d — do. If d — do = 0 then Do = I:» and by (3.10) the germ of B with respect to I:» covers B o. The result follows since 0, (C G (Dz ))= 0,(NG (Er )). Suppose that d — do > 0. By (5.4)(ii) there exists a block pair (D*, B*) in G which properly extends (D o, Bo). Let H = D *C G (Do). Suppose that D* C Dz. Let Ho be the kernel of Bo. By (IV.4.11) Ho is a p'-group. Furthermore H o is the kernel of B' = B* 11 and 0(C G (Dz ))C Ho . It follows from (6.4) that 0, , (CG (Dz ))C Ho n D * CG (D*) and so is in the kernel of B*. The result follows by induction. (ii) Since the kernel of Bo contains all 71-'-elements in CG (Do) the assumptions of (i) are satisfied. Thus the result follows from (i). 0 The following example shows that the assumptions in (6.7)(ii) that 0, , (C G (Do)) is a S„.,-group of CG (Do) cannot be dropped even in case

IT = {P }. Let (y ) be a cyclic group of order 4. Let S = S, be the symmetric group on 5 letters and let A = A, be the alternating group. Let G = (y)S where (y ) < G and y z =y for z E A, y z = y ' for z E S — A. Let p = 2. Let 4' be the irreduciblt character of degree 4 of S which is a constituent of the permutation representation of S on 5 letters. Then A is irreducible. Thus 4' A is in a block of defect O. By (3.5) the block of S containing 4' is the only block of S covering that of A. Thus it is weakly regular. Hence by (3.14) there exists an involution x ES—A such that (x) is a defect group of the block of S containing 4'. Therefore x is necessarily a transposition and Cs (x) = (x) W where W is the symmetric group on the three letters fixed by x. Consider 4' as a character of G which has (y ) in its kernel and let B be the block of G which contains By (4.5) D = (x, y) is a defect group of B. Thus in particular B is not the principal block of G. Let Do = (y). Then Do CG (Do) = (y ) A and by (3.10) B = B (? for some block Bo of Do CG (D0). 07(130CG (DO) = (1) is in the kernel of B0 . It is easily verified that NG (D)= (x, y) W. Thus W' = 0 2 (NG (D)). Hence by (4.3) the principal block of NG (D) is the only block which has W' in its kernel. Thus by (6.2) 02 , (N G (D)) is not in the kernel of the germ of B. H2 be subgoups of G with LH, C MI-I2 and Hi , H2 p'-groups. Assume that [M,H2] C H2, [L, H,] C HI , L CM and 112 n H, L C H1 . Let B be a block of L with defect group D such that Cm (D) C L. Assume that H, n L is in the kernel of B. Let i be the block of LH, which has H, in its kernel and corresponds to B by the natural

THEOREM 6.8. Let L, M, H,,

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[6

isomorphism from L/L n H, to . Assume furthermore that Ii""2 is defined. Then B " is defined and H2 n M is in the kernel of B". If É " is the block of MI-I 2 corresponding to B " by the natural isomorphism of M/M n H2 onto MI-I2! H2 , then É"", = P. The following diagram illustrates the statement.

= B"H=

MH2

LH,

É

B L PROOF. By (111.9.5) B" is defined. Since M n H2 < M and M n H2 n L C L n Hi it follows from (6.4)(i) that M n H2 is in the kernel of B". Since [L, H2] C H2 and L n H2 ç L n H, is in the kernel of B, there exists a block 1-32 of LH2 with H2 in its kernel such that 112 corresponds to B by the isomorphism L/L n H2 LH 2/H2. By (4.3) 11 2 has D as a defect group. The Frattini argument implies, that C.,/,/,(DH2/1-12)= CM (D)H2/H2 and so C., (D)C (D)H 2 C LH2 . Thus by (111.9.5) 1-3;1 is defined for LH 2 C H C MH2 . We will next show that 21qH

2=

(6.9)

Let 4" E B and let E P2 with 4; L = 4. Let {x s } be the set of all irreducible characters in /3- 12'2 and let as = ( x„ 4' 1' 12).2 . By (1.3) v(4'2(1)) = v (Iasx, (1)). As 4- is the unique irreducible character of LH, with C 4--1 and H, is its kernel and since H, is in the kernel of each x„ it follows from Frobenius reciprocity (111.2.5) that

e).,=

a, = ((xs)L„,, -

((x,)L, ‘), =(( X)M,

m )m-

Since H2 is in the kernel of each Xs by (6.4)(i), each (xs)m is irreducible. Thus there exists a block B' of M such that {(x,),,,,} is the set of all irreducible characters in B'. Hence f3 ' = EM2. However

E as (x,),, (1)) = v (.4

(1)) =,(1mH2: LI-1,1x (1))

) =1)(IM:L14- (1))= 1, (r ( ).

Hence B' = B" by (1.3), and (6.9) is proved. Since H2 n H, L C H, it follows that H, H2 n LH, = H, . Consider the natural isomorphisms

7]

iSOMETRIES

215

L/L n H, LH I /H, LH/LH fl H, H2 LH,/ LH, n H 112 . LH, H 2/ H, H2 Let LI, be the block of LH, H2 with H, H2 in its kernel which corresponds to B. Let A be the block of LH 2 with LH2 n H,H2 in its kernel which corresponds to P3 . Let be an irreducible character in B, let 4 be the unique irreducible character in P2 with 6_ - = and H2 in its kernel. Let 0 be the unique irreducible character of LH, H, with H, H, in its kernel such that OL = ‘. Then 0 E B3 by definition. Furthermore Ou E Since it follows that B = P2. Thus P 2 corresponds to P3. (COL = = Since .r3 is defined, (6.4) implies that f3r , "2 = 13 3 and so by (6.9) " = B- z . By (6.4) fin = /3- ""2 . Therefore 13 M" = B "

r":

In (6.8) it is essential to assume that ii MI!, is defined. This does not follow from the other conditions. For instance let M = L = H, = (1) . Let tp be the central character of (1), let {1} be a conjugacy class of H2 and let C' = {x I x E C}. Then tir", (e')(if" , ((") = 0 but q'(ÊÊ') = C O.

7. Isometries Let X be a set of characters of the group N and let A be a union of conjugate classes of N. The following notation will be used in this section. M(N) is the ring of all generalized characters of N. V(N, A : X) is the vector space of all complex linear combinations of characters in X which vanish on N — A. If X is the set of all irreducible characters in a union S of blocks of N let V(N, A : S) = V (N, A : X). If X is the set of all irreducible characters of N let V(N, A) = V(N, A : X). Let V(N,N : X) = V(N : X) and let V(/V,N) = V(N). M(N, A : X) = V(/V, A : X) n m(N). Let Let M(N, A : S) --V(N, A : S)n M(N) where. S is a union of blocks of N. Let M(N, A) = V(N, A ) n M(N) and let M(N : X) = V(N : X) n M(N). The inner product of characters defines a natural metric on M(N, A : X) and on V(N, A : X). If Y is a set of irreducible characters of G and a = Las x., with complex as let a =I,„. 1-a,xs . If Y is the set of all irreducible characters in a union S of blocks of G let a S = Y . It is evident that if a is a generalized character of G then so is a Y. IT is a set of primes and is the complementary set of primes.

216

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[7

If p E 7 then S„(7r,N) is the union of all p-blocks B of N such that if D is a defect group of B then 0 „. (NN (D)) is in the kernel of the germ of B. If z E G then z, denotes the 7r-part of z. The 7r-section of G containing z is the set of all elements in G whose ii- -part is conjugate to z r . (N , A : X) = {a a E V (N, A : X), a (yx ) = a (y) for any element y E A and any element x E 0, , (C H (y))}. M„ (N, A : X) = V, (N,A :X)n M(N).

Similarly the subscript Tr applied to any of the sets defined above means that it is intersected with V„ (N, A : X). In this section we will be concerned with the following two hypotheses. HYPOTHESIS 7.1. (i) N is a subgroup of G. A is a subset of N which is a union of 7r-sections of N such that any two 7r-elements in A which are conjugate in G are conjugate in N. (ii) If y is a 7r-element in A then CG (y) = C H (y )0, ,(C G (y)).

HYPOTHESIS 7.2. (i) (7.1) is statisfied. (ii) 0 „.(C G (y)) is a S„-group of CG (y) for all 7r-elements y in A.

We first prove some elementary results. LEMMA 7.3. Let H
Define T = {(y„)z t I z ranges over a cross section of CH (Y„) in H, t E CH (y„)z 1.

Then T C y„H andITI=1111. Thus T = y,H. Since H< G it follows that (yx )„ E y,H and so (yx),, = (y,)z t for some z E H and some 7r'-element t E CG (y f, ) . Thus t=1 and (yx),.= y As H < G, (yx )„. = y,. x0 for some xo E H. Hence (yx).'„ , ' = (y„ , xo)z = y„ u for some u E H. Since y„. u centralizes (yx)z.„' = y„ it follows that u E CG (y„)(-1 H. Therefore (yx)z = y, ((yx).T)=

u = yu.

LEMMA 7.4. Suppose that p E V,. (G). Let y be a 7r-element in G. Let x E CG (y) such that x,, y have relatively prime orders and let x 0 E 0„ , (CG (y)). Then p(yx)= o(pcx0).

7]

ISOMETRIES

PROOF.

Let C =

CG

Cc (x, )

217

(y). Since y, x, have relatively prime orders,

CG

(yx,) =

CG

((yx )„) C

C.

By (7.3) (xx0)' = xu for some z E C, U E 0, , (

c) n cc (x,r) = 0,, (c) n CG ((yx),)C 0,,(C G (yx)„).

Therefore f3(yxx o) -= p ((yxx oy)-= p(yxu) = 0(yx)

by the definition of V, (G). E 7.5. (i) a E V„(N,A : X) if a E V(N,A : X) and for y E A, a cN(y,) is a linear combination of irreducible characters of CN (y„) all of

LEMMA

which have O. (CN (y„)) in their kernel. (ii) Suppose that (7.2) is satisfied. Then a E V„(N,A : X) if and only if a E V(N,A: X) and a is constant on 7r-sections in N. (iii) Suppose that (7.1) is satisfied. If z E G and z„ EA then z„, = z 1 z 2 with z i E CN (z„) and z 2 E 0(CG (z„)). If furthermore zE A for some xE G and z = z z with z;EC N (z;`,) and z E 0, , CG (z.:,)) then a(z,z,)= a(zz0 for all a E (N,A). (

(i) and (ii) are immediate by definition. (iii) The first statement follows directly from (7.1). If z:r E A then by (7.1)(i) there exists y EN such that z z,. Then z7z?-= (z ) (z. Since xy C CG (Z,), z P' and (z are in 0„,(CG (z,T )). Let xy = uv with u E N and v E 0,, (CG (z,r)). Then PROOF.

(zz0Y

zw (zOY (z,z I )"

z,z7 (z r z ,)"

(mod O w, (CG (z„))).

Since u,y,z:r zÇ and z„z, are all in N it follows that = (z,z,)"z o for some zo E 0„,(C G (z, )) n NCO „, (CN (z„)).

(z

Therefore a(z„z i )=a((z,z i )z o)=a((z:r z))=a(z:r z).

D

Suppose that (7.1) is satisfied. If a E V, (N, A) define a E V„ (G) as follows. If z r is not conjugate to an element of A let a' (z)-= O. If z E A for some x E G then z- E CG (z) and so z:,,= zi z 2 where E CN (Z.:,) and z2 E 0„, (C G (z)). Let a (z) = a (z By (7.5)(iii) a' is well defined. Thus T is a linear mapping from V„(N,A) to V,(G).

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V

LEMMA 7.6. Suppose that (7.1) is satisfied and A is a union of p-sections for some p E r. If y is a p-element in A, x 1 a p'-element in CN (y), x2 E 0, , (C G (y )) and x = xi x 2 then aT (yx)— a (y.x1).

PROOF By (7.4) dr (yx) = a (y.x 1 ). Since yx, E A the definition of T implies that a (yx 1 ) -= a (yx 1 ). E LEMMA 7.7. Suppose that (7.1) is satisfied. (i) If a E V, (N, A) and 13 E V, (G) then (dr,P)G -= (a, PON. (ii) If al , a2 E V, (N, A) then (al, a2)N = (a T, «;)G. (i) If y is a n--element in G let {y}„G , {y }N, respectively, be the ii- -section of G,N respectively, containing y. Let y = a 73*. Then 'y(z) 0 if z E {y},G and y is a u--element of G which is not conjugate to any element of A. If y is a r-element in A then CG (y)=- CN (y )0, , CG (y)). Thus there exists a cross section { x,} of CN (y) C CG (y) with { x,} 0, , (C G (y)) such that every r'-element in CG (y) is of the form zx, for some i and some r'-element z E CG (y). Since y E V, (G) this implies that PROOF.

(

ycyx)=

le

E, y(z) z.(y),

‘.7

.ECG(r)

L N

I

(V) .ECN (Y)

y(yx)

1

(z)

INI

where / 1 means that x ranges only over r'-elements. Thus if y ranges over r-elements in A which form a complete set of representatives of all r-sections in G which meet A then 1

1 z

(a/S)G

(ii) Let

p = a;

,, y;; Y(z)= INI

EE Y

y(z)= (a, f3N)N.

z E{YY:

in (i). Then

(ceT,c0G = (cel,(«)N)N

E = 1 .EA IN

E al 1 , EN

=— 1N

1 INI E

ai(z)«;(z).

zEN

(z)a 2 (z)*

(Z )a2 (Z )* = a2)N •

0

The existence of T is a very useful tool for the investigation of certain questions concerning the structure of finite groups provided M, (N,A)T C

71

ISOMETRIES

219

M(G) or equivalently T maps generalized characters in V, (N,A) onto generalized characters of G. A discussion of some of these applications can for instance be found in Feit [1967c] or Smith [1974], [1976a]. In case IT is the set of all primes it is easily seen that aT —a G for all a E V, (N, A) and so M, (N, A )r C M (G). This case, which gave rise to the whole method, was first studied by Brauer and Suzuki who were the first to realize the importance of results of this type. It is not known whether M„ (N, A)' C M(G) whenever hypothesis (7.1) is satisfied. However Reynolds [1963] and Leonard and McKelvey [1967] have shown that M„ (N, C M(G) provided hypothesis (7.2) is satisfied. This result and its proof is a generalization of an earlier theorem of Dade [1964] who proved the same conclusion under the stronger hypothesis that (7.2) holds and CN (y) is a 77--group for every 77--element y in A. Dade's result in turn generalized a result where the conclusion was proved under still stronger hypotheses, see Feit and Thompson [1963], section 9. All these results are of course generalizations of the case that IT is the set of all

primes. In this section we will primarily be concerned with showing that under suitable additional hypotheses T is related to the Brauer correspondence. The basic approach in this section is due to Brauer. Some of the material can be found in Brauer [1964a], Gorenstein and Walter [1962], section 11, Niccolai [1974], Reynolds [1967], [1968], [1970], Walter [1966] and Wong [1966]. Glauberman [1968] Remark 3.2, first pointed out that Brauer's approach yields Dade's result in case IT = {p} for some prime p. In this case a proof that M„ (N, A)T C M(G) will be given below.

THEOREM 7.8. Let p E Ir. Assume that (7.1) holds and A is a union of p-sections of N. Let Bo be a p -block in S, N) such that V(N, A : B o)/ (0). Let D be a defect group of B o and let o be the germ of B o with respect to D. Then

(D)CG (D) =NN (D)0„.,(CG (D)). (NN (D)) is in the kernel of no and there exists a unique block no of NN (D KG (D) with 0,,(CG (D)) in its kernel such that no and n o (i)

NN

(ii)

as blocks of NN

(D)CG (D)

0, , (CG (D))

coinde NN

(D)

0, , (N N (D)) .

(iii) Let y EA n D and let B(y) be a block of CN (y) with B (y)" = B o . Then 0, , (CN (y)) is in the kernel of B(y). If B(y) is a block of CG (y) with 0, , (C G (y)) in its kernel such that B(y) and B(y) coincide as blocks of

220

CHAPTER

CN (y)

V

[7

CG (y)

0, , (CN (y)) — 0 „, (C G (y)) then B. (y ) G = ijô

PROOF. Let y EA nD and let B(y ) be a block of C N (y ) with B(y) N = B0 Let D, be a defect group of B (y). By (5.4) there exist block pairs (D„ B,) in N for i = 1, n such that (D B , ±1 ) properly extends (D„ B,) for i = 1, n — 1, BN ( Y ) = B (y) and D„ is a defect group of Bo . The following assertion will be proved by induction. For i = 1 , . . . , n CG (D,)= CN (D,)0, , (C G (D,)) and B, is in Sp(7r, D, CN (D,)). Assume that the statement has been proved for i < m. If m = 1 let P = (y). If m >1 let P = D — 1 . Hence by (7.1) or induction CG (P) = z E CG (D„,)C CG (P). Thus z = z 1 z 2 with CN (P)0,, (C G (P)). Let z, E C N (P) and z 2 E 0„, (CG (P)). Since D,„ C NN (P) it follows that D,„ {0, (C G (P))n N} is a group and [D„„ zd = [D„„ z 1 ] is contained in O,. (CG (P )) n N As D„, is a Sr-group of D„, {0 „, (CG (P)) n N} there exists z3 E 0(CG (P))n N with z, z I E Nis/ (D,„). Thus .

.Z3

z 2 E NG (Drn)

n O,

(CG

(P)) ç CG (D„,) n O,. (CG (P)) Ç O,. (CG

(a,))

.

Since z =(z z 3 1 ) (z 3 z 2) this implies that CG (D,„) = CN (D,„)0,, (CG (D„,)) as the opposite inclusion is trivial. By (6.5) B, is in Sp (7r, D, CN (D,)). The statement follows by induction. (i) Since V (N, A : B o ) (0), (IV.6.2) implies the existence of y E A n D and a block B (y) of CN (y ) with B (y ) N = B o. Since D„ is conjugate to D in N the result follows from the previous paragraph. (ii) Immediate by (i). (iii) Let Jj, be a block of D, CG (D,) such that 0(D, CG (D,)) is in the kernel of B- , and B with if?, as blocks of D, C N (D,)

D,

CG

(D,)

0,, (CN (D,)) 0, , (C G (D,))

for i = 1 ,

n.

Then D, is a defect group of  and so (D„ ,) is a block pair in G. Suppose that BN = B o for i 1 , n. It will first be proved by induction on n — i that E= i. Suppose that n — m = O. Replacing D„, by a conjugate in N it may be assumed that D = D„,. By the first main theorem on blocks (111.9.7)

7]

221

ISOMETRIES

B = Lio . Thus /-3 1:" (D)c G(D) = /40 by (6.8) with M = N, (D), L = DCN (D), H1 = H2 = 0(CG (D)), and so /42= ii c?. Suppose that n — m> and the result has been proved for m < i n. Thus 13- 2 +,= 1. Since 0 (D B extends (D,„ B,,) it follows from (6.8) that (D„,,,,

extends ( ( )„„ii„,). Therefore B = = K. This proves the statement in the previous paragraph. Since B"tY ) -= B(y) it follows from (6.8) applied to CN (y) and CG (y) that ifi. (Y ) = ij- (y ). Thus b(y)G =

THEOREM 7.9. Let p E Ir. Assume that (7.1) holds and A is a union of p-sections of N. Let {A},

be the set of all elements in G whose 7r-part is conjugate to an element of A. For any p-block B o of N such that V(N, A : B o) (0) and B o C S (7r-,N) let b. ° be defined as in (7.8) (ii). Let B be a p-block of G with B C S, (7r, G) and let /301, B02,... be all the p-blocks of N such that V(N,A (0), B o, CS,(7r,N) and ijg= B. Then the following hold. (i) If a E V, (N, A :U,B0,) then a' E V, (G,{A},: B). (ii) V„ (N, A : U,B 01 c (iii) If a E V„ (N, A :U,130,) then a' -= (a G . (iv) If B C S. (7r, G) then M, (N,A :U,13 0,)" C M„ (G,{A},?:B). )

PROOF. If y is a p-element in G and 0 is a function defined on CG (y) let By be the function on the set of all p'-elements x in CG (y ) defined by PY

(X

) = g (YX )-

+ O. It suffices to show that By = 0 for all p-elements y in G. Let y be a p-element in G. Let S, be the union of all p-blocks Ê of CG (y) with .6 G = B and let S2 be the union of the remaining p-blocks of CG (y). For t = 1,2 let V1(C G (y ): S, ) be the space of all complex linear combinations of irreducible Brauer characters in S. By (IV.6.2) (a")).13 E V' (CG (y): S 1 ) and Oy E V' (C G (y): S2). it suffices to show that a; E V' (CG (y): S1) because then (i) Let a = (a")B

6, = a ; — (a ')1y3 E V'(CG (y): S 1 )

n v'(c G (y): S 2) = (0).

If y is not conjugate to an element of A then a; = 0 by definition. Thus it may be assumed that y E A. Let a -Ea, where a, is a linear combination of irreducible characters in B o,. If y is not conjugate to an element of a defect group of Bo, then (a, ), by (IV.2.4). If x is a p'-element in CG (y) then x = x, x2 where x, C CN (y) and x2 E 0(c. (y)). By (7.6) a ;(x ) = ay (x,). Thus a; = ay is a linear combination of irreducible

222

CHAPTER

V

[7

Brauer characters of CG (y )/0„ , (CG (y)) CN (y )/0„ , (CN (y )). Hence (4.3) and (7.8)(iii) imply that a; E V'(C G (y): S i ) as was to be shown. (ii) Clear by (i). (iii) Let y be a p -element in G. If y is not conjugate to an element of A then both a and a G vanish on the p-section of G which contains y. Hence by (IV.6.3) (a ) 8 also vanishes on the p-section which contains y. Thus it suffices to show that if y is a p-element in A then a' and (a G ) 8 agree on the p-section of G which contains y. Let C = CG (y). By the Mackey decomposition (11.2.9) a G (yx) = /, {at,. n }C (yx) for every p' -element x in C, where NzC ranges over all the (N,C) double cosets of G. If yx E cA z then y E c A and so by (7.1)0 there exists z, E N with z, z E C. Thus NzC = NC. Hence {a = {(a Nnc )C l y . Therefore by (IV.6.1) {(a G )i3 } y = ({(a N ,,c) c } s ) y where S is the union of all blocks B, of C with B = B. By (7.4) (a Nnc) y is a linear combination of Brauer characters of N n C all of which have o,,(N n C) in their kernel. Since C/0(C) N n C10(N n C) there exists a linear combination g of Brauer characters of C all of which have 0,(C) in their kernel such that (a N ,-,-) y and p agree as functions on the set of p'-elements in C/0,(C). Since y E Z(C) it follows that {(a Nnc)l y = {(aNn c)y } C . Thus if 0 is the character afforded by an indecomposable projective module of C which has 0,(C) in its kernel then (fa NnO c ly7 1% ((a Nnc),

(1) NnavnC

(0 Nnc, Nnc)vnc — (0, (P)C

where the prime indicates that the summation in the inner product ranges over p'-elements. Thus by (IV.3.3) {(a } y — p is a linear combination of irreducible Brauer characters of C none of which have 0„ , (C) in their kernel. Hence by (6.5) ({(a Nncf} s ) y = ps • By the second main theorem on blocks (IV.6.1) (a N ,-, c)y is a linear combination of Brauer characters in blocks B(y) of N n C with B(y)" C U,1301 . Thus g is the corresponding linear combination of Brauer characters in blocks B(y) of C. Since 13- = B it follows from (7.8)(iii) that 13 s = 6. It remains to show that 0 = (a . Let x be a p'-element in C. Then x = x, x2 with x, a p'-element in N n C and x 2 E 0„, (C). Therefore since B is in S,, (7r, G) it follows from (7.6) that )y

,6(x) = ay (x 1 )= a(yx,)=- a' (yx) = (cr")y (x).

Thus = (a ")y as required. (iv) Clear by (iii).

7]

ISOMETRIES

223

LEMMA 7.10. Let p E 7T. Assume that (7.2) holds and A is a union of p-sections of N. Let Bo be a p-block of N such that V(N,A :Bo )/ (0) and B0 CS,(7r,N). Let ijo be defined as in (7.8)(ii). Then f3 (?CS, (7r, G). PROOF. There exists an irreducible character x in Bo such that x does not vanish on the p-section which contains some p-element y E A. Let D be a defect group of B0 . Thus D is a defect group of /Jo . By (IV.2.4) it may be assumed that y E D. Since 0, , (C G (y)) is a S„—group of CG (y) it follows that 0(D CG (D)) is a S„,-group of DC G (D). There exists a block pair (D,B 2) in G with B-^ (D)c ' (E)) = ijo Since b o covers B, it follows that 0, , (D CG (D)) is in the kernel of B2. By (6.7) (ii) /j (? = BC S, (7r, G). 0 •

THEOREM 7.11. Let p E 7T. Assume that (7.2) holds and A is a union of p-sections of N. Let {A}. be the set of all elements in G whose 7r-part is conjugate to an element of A. Let B i , B„ be all the p-blocks of G in Sp (7r, G). For each i let B 1 Bi2, . . . be all the p -blocks of N in Sp (7r, N) such that V(N,A :13 11 )/ (0) and = 131 where is defined as in (7.8)(ii). For each i let Si = U) B ii . Then the following hold. (i) V(N, A :B 11)= V, (N, A : B 11) for all i,j. (ii) V(G, {A } : /31 )= V, (G,{A} :13,) for all i. V, (N, A : (iii) V, (N, A) = V, (N, A : U7- 1 Si) = (iv) V, (G, {A })= V, (G, {A}, : U Bi) = (1)% I V3.3 (G,{A} : (y) V, (N, A : Si ) = V, (G, {A}: BO for i = 1, n. (vi) If {A}, n N = A then M, (N,A :Si ) = M,. (G, {A BO for i = 1 n. (vii) Suppose that for every p-element y in A and every integer h with (h,1 G 1) = 1, y E A. The for i = 1, ... n ,

,

•

I

rank M, (/V, A :S1 ) = rank M, (G,{A} :

= dim V, (G, {A }:B1 ) .

PROOF. (i) and (ii). Since 0, , (C G (z„))C 0, , (C G (4)) for all zE{A} these results follow from (6.5). (iii) The second equality follows from (IV.6.3) and (i). Clearly V,(N,A : Û S i ) C V,(N, A). Suppose that a E V, (N,A). For y a p-element in A, let B be a block of N with (a ')j O. If B i / B2 then by the second main theorem (IV.6.1) no irreducible Brauer character of C ry (y) is a component of both (a B, ), and (a °) ,. By (7.4) (a n ) y is a linear combination of irreducible Brauer

224

CHAPTER

V

[7

characters of CN (y ) with 0„,(CN (y)) in their kernels. Hence B C Sp (7r, N) by (6.7) and (7.2). Thus PG cs,(71-,G) by (7.10) and (iii) follows. (iv) and (v). The second equality in (iv) follows from (IV.6.3) and (ii). Let m = dim VT (N, A). Then m = dim VT, (G, {A}). Since V, (N, A : S ‘ y C V, (G,{A},?: B,) by (7.9)(i), both (iv) and (y ) follow from (iii). (vi) By (7.9)(iv) M,(N, A :S) C M„(G,{A},:B,).

Let a E M, (G, {A}, : B,). Let p = a,. Since {A } fl N = A it follows that /3 E V, (N, A) and a = p-. By (iii) 13 =f3 with 13, E V, (N, A : S,). Thus by (iv) and (v) p, 0 for i/ j. Hence p E V, (N, A : S ‘ ). Since p E M(N) the result follows. (vii) Since A is a union of p -sections it follows that if z E {A} and (h,I G I) = 1 then z h E {A }. Thus by (IV.6.9) rank M, (N, A : S, ) = dim V, (N, A : S,) and rank M, (G, {A},? : B,) = dim V, (G, {A},? : B,). The result follows from (v). COROLLARY 7.12. Suppose that (7.2) holds and then a is a generalized character of G. PROOF. Clear by (7.9)(iii) and (7.11)(iii).

IT

= {p}. If a E M„(N,A)

E

Suppose that (7.2) is satisfied. Let X be a set of characters of N/0, , (N) with M(N, A : X) C M, (N, A). The set X is coherent if T can be extended to a linear isometry from M(N: X) into M(G). If 4- is an irreducible character in X and T is coherent then 11C I12 = 1. Thus -±- Ç is an irreducible character of G. The next result yields some information concerning the values assumed by Ç on certain elements of G. This result was stated without proof in Feit [1967c], p. 175. Unfortunately the fact that A has to satisfy assumption (i) of (7.13) below was omitted in the statement. THEOREM 7.13. Assume that (7.2) is satisfied and 0, , (N) is a S„-group of N. Let X be a set of irreducible characters of N/0„ ,(N). Suppose that the following assumptions are satisfied. (i) If p E IT then the set of p-singular elements in A is a union of p-sections of N. (ii) X is coherent. (iii) V, (N, A :X) = V, (N, N — (N): X). Then the following conclusions hold. (1) If 4- E X then (C,a')G = (ÇN,a), for all a E V, (N,A :X).

7]

225

ISOMÉTRIES

(ii) Let X = { s } and let 6 =16, (1)4's . For each t there exists y, E M(N) such that (y„ 6,)= 0 for all s and an integer a, such that

a, ( 1)

()N

Furthermore there exists y E V(N) and a rational number a such that

C(z)=

(z)+

(1){4(z)+ y(z)}

for all z E A.

PROOF. (i) Let in be the set of all primes p E 7r such that if a E V „(N, A : X) then a vanishes on all p-singular elements in N. Let 7r2 = — in. For each p E 7r2 let S (p) be the union of Sp (1 r, G) and all p-blocks of G of defect O. We will first show that if /- E X and p E 7r2 then EC E S(p) for E = 1 or — 1.

Choose p E 7r2 . Let 4" EX and let E = 1 or —1 such that EC is a character of G. Let k be the set of all p-singular elements in A. By assumption (i) k is a union of p-sections and (7.2) is satisfied if A is replaced by A. Furthermore Vs, (N, Ap) C V„ (N, A) and T defined on (N, AO is the restriction to Vs, (N, A) of T defined on V, (N, A). Let X = { s } with = . For each s let as = (1)6 — . By assumption (iii) {a s } is a basis of V,(N,A : X). Since p E 7r2 there exists t such that a, does not vanish on A. Let a, = 13 + 02 where p, vanishes on N — Ap and 02 vanishes on A. Thus pi 0 and pl -= wy(P) by (7.9)(iii) and (7.11)(iii) applied to (3, E Vs,(N,Ap). Therefore by (IV.3.4) aT -= 0 + il where 0 E V(G:S (p)) and n is a complex linear combination of characters afforded by projective R[G] modules not in S (p). Furthermore 0/ 0 and a; = 4; (1)6T — (1)‘; . Thus either EC E S (p) or (1)C vanishes on all p-singular elements of G. In the latter case C vanishes on all p -singular elements of G and so EC is in a p-block of defect O. Hence in any case EC` E S(p). Let 4- E X. Define tir as follows: ç&(z) = =0

(z)

if z is p -singular for some p E 71-2 and the 7r-section of z in G meets A,

otherwise.

Since EC E U ,,,,S(p) it follows from (6.5) that C is constant on p-singular 7r-sections which meet A for all p E 77-2 . Thus t/f E V,. (G). Hence if a E V,(N,A :X) (7.7) implies that (C, a T)G -= (qr, a )G -= (ON, a )N = (CN , a)N

226

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[8

(ii) Let ()N = 6 + Ea,1 6 + y, where (6, y, ) = 0 for all j and t. Then (i) implies that for all i, j, t 6 (1)au —

(1)aq = ((6`)N, 6 (1)6 — 6(1)6) N — (6, 6 (1)6 — 6(1)6)N

= (41, WC — 6(1)C)a — {6 (1) 8. Therefore 6 (1)a, = =

at;

6(1) 8p} = O.

(1)au for all i, j and t. Hence in particular

6 (1) an for all 6 (1)

j and t.

The first equation in the statement of (ii) follows by setting a, = au. If z E A then 4", (1)6 (z)— 6 (1)C(s) = (1)6 (z)— 6 (1)6 (z) for all s and t. Thus the first equation in the statement of (ii) implies that (1) {

a a ` 6(z)+ y,(z)1 — (1) { 6 (1 ) 6 (1)6(z)+ (z)} s

for all z E A. Thus the second equation in the statement of (ii) follows by setting Y=

1

and

a—

al

6(1)

2.

8. Tr-heights

In this section we will use the notation of Chapter IV, section 8. Let v(1 G i) = a. Let e be a centrally primitive idempotent in R[G]. Let B = B(e) be the block corresponding to e. Let D = D (B) be the defect group of B and let d = d (B) denote the defect. Thus v (I G : D (B )1) = a — d (B). In this section we will primarily be concerned with ChR (G,B) and ZR (G)e. Observe that Chic (G,B) is the dual space of ZK (G)e. See Broué [1976b], [1978b], [1979], [1980] for the results in this section and further related results. LEMMA 8.1. If a E Z R (G)e and O E Ch R (G, B) v(0, (a))

then

P(I G D(B)i).

PROOF. By (111.6.6) e = Tr(c) for some c E R[G]. Since a E Z, (G) it follows that ae = (ac). Let {x,} be a cross section of D in G. Since O is a class function on G it follows that

8]

227

Tr-HEIGHTS

0 (a) = 0 (ae) = 0 (Ix: ' acx;) =IG:D 10 (ac).

This implies the result. 111 For a E ZR (G)e let s (a ) be the largest integer such that 77- ' (° ) I 0(a) in R for all 0 E ChR (G,B). Define the 7r-height of a, h„ (a) by (7r h- (" )) -=- (71-'1G : D

where the parentheses denote a fractional ideal of K. For 0 E Ch R (G,B) let t(0) be the largest integer such that R for all a E Z R (G)e. Define the 7r-height of 0, h,(0) by (7r h- ())= (77- '

0(a) in

G:D1 -1 ).

By (8.1) h„ (a) and h,. (0) are nonnegative integers for a E ZR (G)e and 0 E ChR (G,B). Suppose that x. E B. By (IV.7.1) 1 , (x. ( 1 )) = 1'(X. (e )). For a E ZR (G) X.(a)lx„(1)= co u (a)E R. Therefore (X. ( 1 )) = (X. (e)) (

»IG

(X. ( 1 )).

Thus if ( 71- ') = (P) it follows that h,. ()/m is the height of xu . In particular x„ has height 0 if and only if it has 7r-height O. This argument also shows that h„(e)— O. Since e is a primitive idempotent in ZR (G) it follows that ZR (G)e is a local ring. Thus ZR (G)e J (Z R (G)e) R as K was chosen large enough. Let A be the central character of ZR (G )e. By (IV.4.2) A = rou for any x„ E B. LEMMA 8.2. Let a E ZR (G)e. The following are equivalent. (i) For every 0 E ChR (G,B) with h„ (0)= 0, v(0(a))= v(IG :D I). (ii) There exists 6 E Ch R (G, B) with v(0(a))= v(IG (iii) h„(a)=O. (iv) a J (ZR (G)e).

PROOF. (i) (ii) and (ii) (iii) are immediate. (iii) (iv). Suppose that a E J(ZR (G)e). Then x„(a )Ix u (1) = ro„ (a) = 0 for all E B. Hence 0(a) = 0 (mod 77-p"' ° I)) for all e E Ch R (G, B) and so ii,(a) O. (iv) (i). Let 0 E ChR (G, B) with v (0 (a)) > p', where y = v(I G: D I). Let 0 = Ecuxu. Then

0(a) z cu x,_,(g)_

(x41. ») wr, (a).

228

CHAPTER

[8

V

By assumption (0(a)Ipu ) = O. Since a 0 J (ZR (G)e) it follows that cor (a) = (a) # 0 for all u. Therefore

E c (x“ (1) "

)

=0

(mod 7r)

and so

E cu x“ ( i)

0(1)

Thus h, (0)

O.

0

(mod 7rpu).

0

LEMMA 8.3. Let 0 E ChR (G,B). The following are equivalent. (i) For every a E ZR (G)e with h„ (a)= 0, v(0(a))= v( G: (ii) There exists a E ZR (G)e with v(0(a))= v(1G (iii) h, (0)= O. PROOF. (i) (ii) and (ii) (iii) are immediate. (iii) (i). This follows from the fact that (8.2) (iii) implies (8.2)(i). E LEMMA 8.4. Let 0 E ChR (G,B). Then h,(0)= 0 if and only if v(0(1))= v(IG:D1).

PROOF. If v(0(1)) = v(1 G : D 1) then clearly h, (0) = O. Suppose that h„, (0)= O. Since h,(e)= 0 and v(0(e))= v(0(1)), the result follows from (8.3). 0 LEMMA 8.5. Let y be a p-element in G and let S be the p-section which contains y. (i) If y is not conjugate to an element of D then Z R (G :S)e = (0). (ii) If y is not conjugate to an element of Z(D) then ZR(G:S)e C J(ZR (G)e). (iii) If y is conjugate to ZR (G : S)e J(ZR (G)e).

an

element of

Z(D)

then

PROOF (i) By (IV.2.4) 0(x)= 0 for all x E S and all 0 E Ch R (G,B). Since Chic (G, B) is the dual space of Z R (G)e this implies the result. (ii) This follows from (1.5). (iii) It may be assumed that y E Z(D). By (III.6.10) there exists a p'-element x such that D is a Sr -group of CG (x ) and À 0 if C is the conjugate class containing x and À is the central character of R[G]

(»

8]

229

7r-HEIGHTS

corresponding to B. Let x = x be a character of height 0 in B. Let Co be the conjugate class of G containing xy. Since D is a Sa-group of CG (xy) and x (xy ) x (x) X 0 (mod 1r) it follows that v (x (xy )) = v(x (x))= 0 and SO

Therefore A (0 ) e) = (Ç)= co„ (Co) 00 e E Z R (G : S)e. LI

O. Thus C'o e

J (ZR

(G)e) but

LEMMA 8.6. Define a on G by a(x)= 1 if x is a p-element and a (x)-- 0 otherwise. Then a 8 E Ch R (G,B) and h,(a B )= 0 . PROOF. Let {q,} be the set of all primes distinct from p which divide I GI. Thus q» E R for all i. Hence by (IV.1.3) a E Ch R (G) and so aB E Ch R (G,B). Let 0 E ChR (G,B) with h, (0)= O. Let a = Eo(x - ')x. Let S = G reg and let Ss be defined as in Chapter IV, section 8. Then S s (a)e E ZR (G)e and by (IV.8.5) a B (8,(a)e)-- a(3,(a)e)= a(3,(ae))= a(3,(a))=0(1).

Thus h„(a8 )-= 0 by (8.3).

D

8.7 (Broué [1978b]). Let B be a block of G with defect group D. Let e be a centrally primitive idempotent corresponding to B. Let y E Z(D) and let S be the p-section which contains y. Let a = ae =E„ Es a,,x. Then v(a) -- V (I CG (y):D 1). Furthermore v(ay )= v(1C G (y): D I) if and only if a J(ZR (G)e).

THEOREM

PROOF Let a be defined as in (8.6). By (8.1) v(a (a)) v (I G : D I). By (8.2) equality holds if and only if a = ae J (ZR (G)e). Since a B (a)-= a(ae)= a(a)-=a I G :CG equality holds if and only if v(ay ) = v(I CG

): D

I)*

D

COROLLARY 8.8 (Brauer [1976aj). Let B be a block with defect group D. Then

v (E (0 2) = v (E (,o, (1) 4), (1)) = v G :DI IG where x„, cp, ranges over all irreducible, irreducible Brauer characters respectively in B.

230

CHAPTER

V

[9

PROOF. Let e be the centrally primitive idempotent of R[G] corresponding to B. By (IV.7.1) and (IV.7.2) e E ZR (G : Greg) and e = ax with 1Gja, =Ex. (1 = E,p,(1)0,(1). Since eJ(ZR (G)e) the result follows from (8.7). E )

9. Subsections

The results in this section are mostly due to Brauer [1968], [1971a] though the presentation here is based to some extent on Broué [197813]. The notation is the same as in the previous section and in Chapter IV, section 8. Let y be a p-element in G and let B be a block of G. If x. E B and d 1 j 0 then by the second main theorem on blocks (IV.6.1), cp J belongs to a block E of CG (y) with EG = B. For fixed y and j, the column of higher decomposition numbers d, is said to belong to the p -section S which contains y and is said to be associated with the block B. Let E be a block of CG (y) with f3 G = B. The set of columns d ) , as cp; ranges over the irreducible Brauer characters in if? is a subsection as to B or simply a subsection. It clearly depends on y and B. Denote it by S(y,b). A subsection S(y, IJ) is a major subsection if i4 and B = riG have the same defect. LEMMA 9.1. If S(y,IJ) is a major subsection then y E Z(.6) where D is a defect group of E. Furthermore

D

is a defect group of B = fi G

PROOF. Clear by (6.1). 0 Let y be a p-element in G and let if? be a block of CG (y). Define the linear map m'Y " ) :Ch K (G)—> Ch K (G : S) by rn ( “"(0)= bY ({dY (0)}),

for 8 E ChK (G),

where b Y and d are defined in Chapter IV, section 8. Define (y.

B)

Zic (G)—> Zk (G :S)

by tt ( " B) (a)= 0" (é8" (a))

for a

E

ZK (

G ),

where g Y and 5 Y are defined in Chapter IV, section 8 and ë is the centrally primitive idempotent of R [C G (y)] corresponding to

9]

231

SUBSECTIONS

By (IV.8.6) and (IV.8.7) m (Y.B).

9 9 tL (" ) for 0 E

(G).

The definitions of 13 Y and 5 Y imply that if a E ZK (CG (y): CG SY 0 OY (a) = a. Thus re' 6) and i' idempotent. Define x tY,B) —

then

1

oi (uY.

m (Y, f ( xu ) ,

) , eg)

X u (1) ) 6'

•

Thus x („Y'' ) = xu opt, (Y ' B) and co („Y.B) = cou ott (Y.B) . This implies in particular that d

1 (x)

for x E CG (y)„,

wj EB

and if a EZR (G) then f.t4Y. B) (a) E R. Furthermore if x. and x,, are in the same block then co (Y ,3) (a)

co,Y' n) (a) (mod 7r)

for a

E ZR ( G ).

THEOREM 9.2 (Brauer [1968], (4A), (4C)). Let D be a defect group of the block B of G. Let y E Z(D). (i) There exists a major subsection S(y,b) associated to B. (ii) Let S(y,b) be a major subsection associated to If?' =B. If x. E B then v(X (24B) (Y))= vdCG (Y): DO+ h(X.), where h(x.) is the height of Xu.

PROOF. (i) Let e be the centrally primitive idempotent in R[G] corresponding to B. Let MI be the set of all blocks of CG (y) with /f3 = B and let é, be the centrally primitive idempotent in R[C G (y)] corresponding to /i,. Let S be the p -section containing y. By (8.5) there exists a E ZR (G :S)e with a E J(ZR (G :S)e). By (8.7) v(ay ) = v(1C G (y):DI), where a =ax. By (IV.8.7) 8Y (a)= SY (ea) = EASY (a). The coefficient of 1 in 5 Y (a) is ay . Let ii, denote the coefficient of 1 in J,8Y (a). Then ay = Ed,. Thus there exists j with v(ay )= v(IC G (y): D I).

Let b, be a defect group of Thus (8.7) applied to the group

By (6.1) it may be assumed that b, c D. CG (y) with y replaced by 1, implies that

v(ICG(y):DI).--s.v(ICG(y):15,1).--s.v(a,),s.v(ICG(y):DI).

23?

CHAPTER V

[9

Hence equality must hold and so D = A as required. (ii) Let a be the sum of all elements in G which are conjugate to y. Thus a E ZR (G) and V (a) = 1. Hence the coefficient of y in p (''') (a)= 0Y (J) is the coefficient c of 1 in J . Since J J(ZR (C G (y))J), it follows from (8.7) applied to CG (y) with y replaced by 1 that v(c) = V (I CG (y):D 1). Thus ii,") (a) J (ZR (G)e) by (8.7). Hence if x. E B then B)

(a))

o,

(a)

0 (mod 70

which is equivalent to the assertion to be proved.

D

Let y be a p-element in G, let P be a block of CG (y) and let x., x„ be irreducible characters in /3- G . Define (Y . 8 )

m uu

1

-

I I—

(Y. B)

•

(y,b) ,

\*

xEco ( y )„g

where * denotes complex conjugation. In case y = 1 and P is a Sa -group of m (2,; B) = a u,„ where a is defined in Chapter IV, section 4. Thus G then Pi these numbers are a generalization to arbitrary sections of the numbers a 0 . Observe that if a is a field automorphism then (m (.Y,; B) )° = rn (2;» in for suitable y' and B' which do not depend on u or v. LEMMA 9.3. Fix u and v. Then E0,,R ) m („Y,; 13) = „„. PROOF. Clear by definition.

0

THEOREM 9.4. Let y be a p-element in G. Let A be a block of CG (y) and let xu , xu, x,„ E PG. Let d, d be the defect of B, B = .1j G respectively. (i) p d m;') is an algebraic integer and (m (Y ' B) (X.), X.)G =rnI,Y, ; B) =

E

Y“d0 *

where (yriy' is the Cartan matrix of A and * denotes complex conjugation.

(ii)E.m(oy,; B )(m(y;4 B ))*

m

y..v B

(iii) Let 1(.1i) denote the number of irreducible Brauer characters in È.

Then

m (iv) Suppose that x,„ has height O. Let hu , k, denote the height of x„, respectively. If hu then

9]

233

SUBSECTIONS

p d Xu (1)X,, Pd In(u):,B) =

(1)

(y

(mod ir

).

h d and equality can hold only if (y, b) is a Furthermore y (m major subsection and ho = O. —

PROOF. (i) The first equality follows from the definition. Thus (in (") (X“),

= I CG1(Y )1

E ,

p501EB

E60„„ dY (pY (x)

WO* cp,(x - ').

cc

By (IV.6.2) this yields the second equality. Since n has defect cl, (IV.4.16) implies that p ° (y) has integral entries. Thus po d InV3) is an algebraic integer. (ii) By (i) = (in (Y.B) (X.), m (Y.13)(Xw

=E

(Y•B)(xu),x.)G on(Y.B)(x.),xu)t

(iii) By definition m (2',;' ) = (m)*. Thus if M = M" ) = (m (uY,) it follows from (ii) that M2 = M. Hence the trace of M is the rank of M. By definition Eu m(uY,;') is the trace of M. By (i) the rank of M equals the rank of the Cartan matrix of if? which is 1(b). (iv) By definition m (y

= X , (1) 2, IGI

(0“

(C)x. (Yx) *

where C ranges over the conjugate classes in the p -section which contains y and yx E C. As xu and x,, are in the same block this implies that

161 re, X. ( 1 ) —

=

1G1 m o, ij)

(mod 7,-)

(1) "

since 13 ° rn (L in is an algebraic integer by (i). As x, has height 0 this implies that (y. 13) pd m = p d Xu ( 1

) l

elB)

(mod Treu ).

Similarly d (y,

Pm

=P

d Xv ( 1 )

(1)

( B)

Zrn

(mod re-).

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CHAPTER V

[9

The required congruence now follows. Since v (p a m) 0 , (1)X , ( 1) ( y. 6))

v(pd

d

_ J.

Thus the congruence implies that 1)(m,Y,;' ))% h„— d and if equality holds then h„ + h, d — d — d. The result follows as d d. E 9.5 (Brauer [19681(5H)). Let B be a block of G with defect group D. Let y E Z(D). Let S(y,B- ) be a major subsection associated to B. Suppose that x., X G B and x has height O. Let x. have height hu. Then

THEOREM

v(m)

hu - v(I D I).

PROOF. For O E Chic (G) let

=

E O(x)x. GH EG

(0, 02). By (9.4) Thus Ô E Z, (G) and 04 0= (in'Y

' B)

(X.),X.)G = m" ) (X.)(i-)= X. (i-d Y' B

= Xr ) ( X..)

-))

= x ( 1 )(4 B) (X.., )

(WY•13) ( E x`v (x

=

(

1 )x))

_ (E IGI uj"

Xu ( 1 )

Since x,„ has height 0 it follows from (9.2) that v(XV3) (Y -I ))= v(X; B) (Y))= P(ICG(Y):D1)•

Thus by (8.7)

Ex ■■:. B) (X - 1 )X 0

(ZR

(G)). Consequently

=0. Therefore v( 111 ‘,13) )= v(X.( 1 )) — v(I GI) = h. -

1).

9.6. Let B be a block with defect group D. Let B. be a block of DC G (D) with É' = B and let T(f3) be the inertia group of n in N G (D). Let Y CZ(D) be a complete set of representations of conjugacy classes of G

THEOREM

9]

235

SUBSECTIONS

which meet

Z(D).

(T(f3),

n CG (y)) double coset

a complete set of representatives in NG (D). Then each subsection S. S(y,(f3") c. ( Y )) with y E Y and x E My is a major subsection. Conversely every major subsection associated to B is conjugate to exactly one subsection S y.x. NG

(D)

For y E Y let My denote

PROOF. If y E Z(D), then D CG(D)C CG (y). Thus if y E Y and x follows from (111.9.2) that

fo (y))G (fix

G

E

my it

B.

Since D
Y=Y

This implies in particular that z CG (y'). Thus )

co (Y

)

CG

= ix Y G(Y') (y

)

(9.7)

•

= CG (y')

or equivalently

CG

(y)z =

(fr)C,

The block of CG (y') on the left has defect group D" = D', while that on the right has defect group D'' = D. Thus D is conjugate to Dz in CG (y'). Thus in (9.7) z may be chosen to be in NG (D). Hence the first equation in (9.7) implies that y = y' since y, y' E Y. Hence z E NG (D) n CG (Y). Now (9.7) becomes (1}—)c o (Y ) = (É )c .. Thus (5.2) applied to CG (y) implies that É" is conjugate to É.' in NG (D) fl CG (y). Thus x' E T(b)x (NG (D) n CG (y)) which implies the result. D Let B be a block of G with defect group D. Let É be a block of DC G (D) with If?' = B and let T(f3) be the inertia group of É in NG (D). Then e = I T(b): DC G (D)I is the inertial index of B or the index of inertia of B. Clearly the inertial index e of a block B is an integer depending only on B. By (5.2) eO (mod p).

236

CHAPTER

V

[9

COROLLARY 9.8. Let B be a block with defect group D. Let 1 be a block of DC G (D) with B- G = B and let T(B- ) be the inertia group of 13- in NG (D). Let e be the inertial index of B. Let y E Z(D) and let C be the conjugate class of G which contains y. Let n denote the number of subsections associated to B of the form S(y', f3') with y' E Z(D) n C. Let n B denote the number of conjugate classes of major subsections associated to B. Then n YE, is the number of T(É) conjugate classes of Z(D) n C. Furthermore

IZ(D)1-.5. nB

PROOF. Let Y and My for y E Y be defined as in (9.6). By (9.6) Wig = My 1. This implies the first statement. Each (T ( ), NG (D) n yEVflC CG (y)) double coset contains at most e cosets of NG (D) n CG (y). Thus

1My I ING (D): NG (D) fl CG (y)1 By (9.6) nB = y EY IMY I and Z(D)I = E y eY Thus riB --5.1Z(D)1 ,-5. en B . CI

e

IMy 1NG (D):NG (D)

ncG(Y)I.

LEMMA 9.9. Let B be a block of defect d and let h be the maximum height of an irreducible character in B. Let n B be the number of conjugate classes of major subsections associated to B. Then n B p a- h and if h >0 then nB .

PROOF. Let x“ be an irreducible character in B of height h. Let S(y,1J) be a major subsection associated to B. By (9.2) m O. By (9.4)(i)(iv) p d-h rn (j B) is a totally positive algebraic integer. By (9.3) Ep d-h m (uy,; 13) pd-h, where the sum ranges over all major subsections associated to B. Thus the arithmetic-geometric inequality implies that )

1

{n p

a-h m

} 1 in.

E p d-h

nB

,

-

nB '

Thus nB --- .19 41-h . If equality holds then m= p for all major subsections and in particular m= p -" -h ) and so v(m; 13))= h — d. By (9.4)(iv) h = 0. 0 COROLLARY 9.10. Let B be a block with defect group D. Let ID

= p d and let n be the largest integer such that p" e, the inertial index of B. Let h be the height of an irreducible character in B. Then

h n + d — ,(1Z(D)j).

9]

SUBSECFIONS

PROOF. By (9.8) and (9.9) 1Z(D )1 /e fin = p a-h . Thus p h I Z(D) p"". D Hence p h

237

pde <

p a ±n÷1 .

COROLLARY 9.11. Suppose that the notation is as in (9.10). Let pr be the index of the Frattini subgroup of D in D. If d > 0 then h
By (5.2) T(D)/D C G (D) is isomorphic to a p'-subgroup of GL, (p).

Thus e 1=1

as r>O. The result follows from (9.10). D COROLLARY 9.12. Suppose that B has an abelian defect group of rank r > O. Then the height of an irreducible character in B is less than ir(r +1). PROOF. Clear by (9.11). 0

The estimate in (9.11) is extremely crude and obvious improvements can be made in (9.11) and (9.12). However these methods do not appear strong enough to answer Question (IX) in Chapter IV. THEOREM 9.13 Let B be a block of defect d with an abelian defect group D. Let k = k(B) denote the number of irreducible characters in B. Suppose that the inertial index of B is 1. Then B contains a unique irreducible Brauer character, k(B)= pet, every irreducible character in B has height 0, every decomposition number is 1 and the unique Cartan invariant is pet. PROOF. By (9.8) there are at least pet subsections associated to B. Thus by (IV.6.5) k (B) . Hence by (IV.4.21) it suffices to show that B contains exactly one irreducible Brauer character. This will be done by induction on 'GI. If D CZ(G) the result follows from (4.6). Suppose that DZ Z(G). Choose y E D, y 0 Z(G). Let S(y,ii) be a subsection associated to B. Since D is abelian it is a major subsection. Let {x„ } be the set of all irreducible characters in B. As D is a defect group of 13- , the inertial index of LI is 1. As y Z(G), induction may be applied to C G (y) / G. Thus Jj has a unique irreducible Brauer character q. Hence the Cartan matrix of 13- is (13'). By (9.5) d;, 0 for all u. The arithmetic-geometric inequality implies that

238

CHAPTER

frIldY„,

12 1

÷

1/k

V

[9

fl};1 2 .

Therefore

'car

k

p°

Since there are p d subsections associated to B, (IV.6.5) implies that 1 has a unique irreducible Brauer character for each such subsection S (y, ). This is the case in particular for S(1, B). E If D is not assumed to be abelian in (9.13) the result is easily seen to be false. Consider for instance the case G = D. A suitable generalization of (9.13) has been proved by Brou6 and Puig [1980]. It is a good deal more complicated than (9.13). We next prove a technical preliminary result. LEMMA 9.14. Let (go ) be a positive definite symmetric matrix with integral entries. Let Q be the corresponding hermitian form. Let q be the minimum of Q(e,e) as e ranges over all nonzero vectors with integral coordinates. Let E be a primitive p"th root of 1 for some n and let tr denote the trace from Q(E) to Q. If a is a nonzero vector whose entries are algebraic integers in Q(E) then

tr(Q (a, a))

[Q(E ): ()] q -= p" - ` (p — 1)q.

PROOF Let t = (p — 1). Let 14:10 e,E ', where each e, is a vector with integer coordinates. Then Q (a, a) =1,:,J10 Q (e„ ei ). Since tr(1) = t, tr(E ` ) = — p"' if i 0 (mod p" -1 ) but i 0 (mod p") and tr(E ) = 0 otherwise, it follows that (-1

tr(Q (a, a)) =

As

(9.15)

,

where =—

E,.; Q (ei, ei ) +

Q(e1 e,)

with i, j ranging over all values congruent to s mod p" - ` and 0 i ,j < t. This can be rewritten as As =

E Q (e, — ei, e, — ei ) + E Q (e,, e,) i<)

with i, j as above.

(9.16)

9]

239

SUBSECTIONS

For a fixed s let N o — Nô be the number of i with 0 i < t, i s (mod p" -1 ) and e, = O. Let N = NI be the number of i with 0 i < t, i s (mod p" -1 ) and e, nonzero. Then No + N 1 — p —1. If No = p — 1 then A. = O. If No = 0 then the second term in (9.16) is at least (p — 1)q and so A. (p — 1)q. If 0 < No < p —I then the first term in (9.16) is at least No N i q N o q and the second term is at least N i q and so A. (p — 1)q. Since a X 0, some e, is nonzero. Thus A. 0 for all s and A. (p — 1)q for at least one value of s. The result follows from (9.15). 0 9.17. Let B be a block of defect d. Let S(y, If?) be a major subsection associated to B. Let k(B) denote the number of irreducible characters in B and let 1(B- ) denote the number of irreducible Brauer characters in B- . (i) Let be the Cartan matrix of B- and let 6, be the quadratic form corresponding to p d -1 . Let q(ii) be the minimum value assumed by (:) on the set of all nonzero vectors with integral coordinates. Then

THEOREM

q (B)k (B) p d 1( 1 ). (ii) If B contains no irreducible characters of positive height then ic (B)2_, p 2di(ft) . PROOF. Let {x.} be the set of all irreducible characters in B. Choose n so that if E is a primitive p"th root of 1 then m E Q(E ) for all u, v. (i) By (9.5) ntX 0 if x,„ has height O. Thus by (9.4)(ii)

m (uY,,' B) =

E

m

T>0

for all u.

By (9.4)(i) p dm Ê) t) (a, a), where a = (d ). Hence p" -1 (p — 1)q(13tr(p d ni) by (9.14). If this is summed over all u then (9.14)(iii) implies that p" -1 (p —1)q(13- )k(B)--tr(pdl(I - ))= p" -1 (p —1)p d l(b).

The result follows. (ii) By (9.5) p d mW 3) is a nonzero algebraic integer for all u, v. By (9.4)(ii) m(u):;')1 2 = m (Z; B) If k = k (B) and t = p"-1 (p — 1) the arithmetic-geometric inequality implies that •

fri p 2d

I in (Lifi)121. 1/kr

I

kt =1 kt

E

n 2d 1 m (Lti)o- 12 „ „2,1B)a

240

CHAPTER

V

where o- ranges over the Galois group of over y, (9.4)(iii) implies that k

EE

”2d m

(21”,B)a

[10 Q(E )

over Q. If this is summed

2d/(f1)__Iip 2d1(fi) .

10. Lower defect groups This section contains results of Brauer [1969a] and reformulations of some of these results by Iizuka [1972]. Related results can be found in Iizuka and Ito [1972]. Brauer [1970] has used these ideas to give an alternative proof of the first main theorem on blocks (111.9.7). Another approach to these results can be found in Broué [1979]. See also Olsson [1980]. The notation introduced in Chapter IV, section 8 will be used in this section. Furthermore e l , e 2 , ... are all the centrally primitive idempotents in R[G]. For each t, B, is the block corresponding to e, and k (BO denotes the number of irreducible characters in .131 . For a subset S of G, = E.Es x E R[G] and g is the image of g in R[G]. Observe that ZR (G) = and ZR (G)= (131, ZR (G). ZR (G)e, Furthermore ZA(G)J, = ZR (G)e, is an ideal of ZA(G) which is a local ring and Z R (G)e, is an ideal of Z R (G) which is a local ring.

LEMMA 10.1. With each block B, it is possible to associate k(B,) conjugate classes {C ;11 ---s. j k(B,)} such that the following conditions are satisfied. (i) Every conjugate class of G is associated with exactly one block. (G )e . (ii) {Ce} is an R-basis of

PROOF. Let WI be all the conjugate classes of G. Let {a ;} be an k -basis of Then

ZA (G)Jt .

= dim

(G)J, = rank R ZR (G)e, = k (B,)

Let 0, = a; and let A denote the matrix (aN )), where (t, j) is the row index and i is the column index. Since A is nonsingular it is possible to arrange the classes WI so that

l A= (A

*

A,

101

LOWER DEFECT GROUPS

241

where A, is a nonsingular k(B,)x k(B,) matrix and the columns of A, are indexed by (t, j) for each t. The result follows since cr;e, = 8„a;. D Let 43(G) be a complete system of representatives of the conjugate classes of p-groups in G. For P e q3(G) let V, (G) be the linear subspace of Z Ft (G) which is spanned by all e', where P is a defect group of C,. For P E q3(G) let 13(P G) = {Q (;°G P, Q E 13 (G)}.

Define U(P : G) =IGEW5(P• G) VG (G) and V (P : G) = (G)GU(P: G). Let V (P : G) = V(P : G) n zf, (G)J,. By (111.6.3) V(P : G) is an ideal of ZA (G). Therefore V, (P : G)= V, (P: G)e, and if U,(P : G)= U(P : G)r-1 V,(P :G)

then V(P : G)=

(P : G),

V(P : G)I U(P : G)

ED V, (P : G)IU,(P : G).

(10.2)

LEMMA 10.3. Let P E q3(G), let B = B, and let {C;} be defined as in (10.1). Let ep,„,.( p ) be the subset of {C;} consisting of those classes which have P as a defect group. Then the image of {,,e,11 Ls. j Ls. m B (P)} in V, (P : G)IU, (P: G) is an 1-basis of this space. The number mB (P)= mB (P: G) depends only on G, B and the conjugate class of P.

PROOF. It is clear that {era; J,11 j m8 (0), Q C G P} is a basis of V, (P: G) for all t and P by (10.2). Thus {p, j ë 1 j Ls. m B (P)} is a basis of V, (P: G)IU,(P: G). The last statement is an immediate consequence. 111 If m B (P)/ 0 then P is a lower defect group of B. It is said to occur with multiplicity inn (P). LEMMA 10.4. Let

P E q3(G) and let B = B1 . Then m B (P: G) = B- mB (P :NG (P)), where Ê ranges over all the blocks of NG (P) with = B.

PROOF. Let N = NG (P) and let § be the Brauer homomorphism with

242

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[10

V

respect to (G, P, N). Let ë be the primitive idempotent of Z, (G) corresponding to B and let .(e)= EÈ', where ranges over all the primitive idempotents in Z (N) such that jj G = B for B corresponding to ë By (111.9.8) is an 1-isomorphism from Vp (G) onto Vp (N). Thus by (10.3) .§ induces an 1-isomorphism from V, (P: G)IU, (P: G) onto V(P :N)§(01U(P:N).§(J). By (10.2) V (P :N).§(J)1U(P : N).§(J)

V (P : N) - U (P :

where ranges over all the primitive idempotents in Zp (N) that occur in the sum .(e)= Therefore mn (P : G) =

=

(V, (P : G)! U, (P : G)) dima (P : N ) e. 1 U(P

m(P:N).

COROLLARY 10.5. Let D be a defect group of B. Then D is a lower defect group of B. If P is a lower defect group of B then PCG D.

PROOF. By (III.6.10) D is a lower defect group of B as e EZA(G)E The second statement follows from (111.9.6) and (10.4). 0 ,,(G) be a complete set of representatives of the conjugate classes of p-elements in G. If y E 130 (G) let S(y) denote the p-section which contains y. Let W(y : G)= W(y) be the subspace of ZA(G) which is spanned by all e, with C, C S(y). Then ZR (G)= y W (y), where y runs over 130 (G). By (IV.8.5)(iii) W(y )ë C W(y) for all y and t. Thus if W, (y)= W(y) n Z, (G) then W (y)= W(y)e and W(y)= avv,(y). Furthermore Zn (G)E = eyE$ „(G) W (y). It is easily seen that if { C;} is defined as in (10.1) then an R-basis of W, (y) is formed by the set of all those C;ë, with C;C S (y). Thus the number of such classes depends only on G, B and the conjugate class of y. Let

Define W ,, y (P : G)= 'V, (P : G)n W(y) and T ,, y (P : G)= U, (P : G) n W(y).

Then V,(P:G)= fçp W, y (P:G), yE,,(G) (10.6)

W(P :G)IT,, y (P : G). y Esi3(G

)

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LOWER DEFECT GROUPS

243

10.7. Let B = B,. Let P c ( G ) and let y E $0 (G). Let {C;} be defined as in (10.1) and let Cy,p,,,.... C;,,,„,6( p ) be the subset of {C;} consisting of those classes which have P as a defect group and are contained in S (y). Then the image of { e- ",,p,,é, 1 j m (B)} in W ,. y (P : G)1 T(P: G) is an k-basis of this space. The number m YE,(P)= m (P: G) depends only on G, B, the conjugate class of y and the conjugate class of P.

LEMMA

PROOF. The first statement follows from (10.3) and (10.6). The last statement is an immediate consequence of the first. D

If m (P)# 0 then P is a lower defect group of B associated to the section S(y) which has multiplicity mL(P). LEMMA 10.8. (i) EyE430( ,m)n(P)= mn (P) for all P c (ii) Ep mG) M B (P) = k (B).

(G).

PROOF. Immediate from the definitions. D 10.9. Let y be a p -element in Z(G). Then mL(P)= mb(P) for all blocks B and P E $(G).

LEMMA

PROOF. The map which sends a to ay defines an 1 -isomorphism on Zk (G). It clearly preserves the ideal V (P: G) for all t, P and sends W(1) onto W(y ). Thus it sends W, (P : G)1 T ,,,(P : G) onto W, (P : G)1 (P: G) and the result follows from (10.7). E THEOREM 10.10. Let y E$ 0 (G), P E 43(G). Then (P : G)=

Eo Ep m]l(Q :CG (y)),

where Q ranges over the elements in 13(C G (y)) which are conjugate to Pin G and 1 ranges over all the blocks of CG (y) with IJG = B.

PROOF. Let C be a conjugate class of G with defect group P contained in S(y). Let So be the p -section of CG (y) containing y. Then C0 = C n so is easily seen to be a conjugate class of CG (y) which has a defect group Q that is conjugate to P in G. Conversely every such conjugate class Co of CG (y ) is of the form cnso for suitable C. Let be the Brauer homomorphism with respect to (G,(y), CG (y)). If C is a conjugate class of

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G in S(y) with defect group P then the definition of Ts implies that

g(t'') = 00 + C', where C' is a union of conjugate classes of CG (y), none of which is in So . Let B = B, and let e be the primitive idempotent of ZR (G) corresponding to B. Let §(e) =E,„4„, where each é,„ is a primitive idempotent of ZR (CG (G)). By the previous paragraph g induces an 1-isomorphism from W(y : G) n V, (G) onto EG W(y : CG (y)) n 170 (CG (y )), where Q ranges over all elements of 13(G) which are conjugate to P in G. Thus by (10.6) : :G) onto and (10.7) § induces an 1-isomorphism from W Er VV, , y (Q : CG (y))IT, y (Q :Cc (Y)). The result now follows from (10.9). El

ED,

If y 1 then (10.10) implies that mL(P) can be computed in terms of information about local subgroups of G. We will now prove some results which yield some information about m L(P). THEOREM 10.11. Let S = S(1) be the section of G consisting of p'-elements. For a E R[G] let Tr(a ) = E EG ax. Let Cy,R, be defined as in (10.7) and for each t,P, j choose x,p,,E C,„ Then for B = B„

■

{0 .p.,1 P E 13 (G), 1 j

rilL(P)1

is an R-basis of Z R (G : S)e, and

{1CG

■

(x .P.)1 Êp, 1 e1

P E $(G), 1

j m L(P)}

is an R -basis of Tr(R [S])et . Furthermore

ChR (G: Greg, B,)1ChR.,

: Greg, B,)

ZR (G: S)e, /Tr(R[S])e,

as R modules.

PROOF. For a E ChR (G : Greg , B,) define a* = Ex ES a (x - ')x E ZR (G : S). Then the map sending a to a* is R -linear. By (IV.3.11) it is onto ZR (G :S). By (IV.2.3) ChR.pro, (G :G„g)* C Tr(R [ S ]) . For a conjugate class C let D(C) be a defect group of C. If x E C then Tr(x) = 1 CG (X)I Thus {D(C)C'IC C S} is a basis of the R module Tr(R [S]). Therefore ZR (G : S)ITr(R[S]) is a torsion R module whose invariants are {I D(C)11C C S). By (III.3.11) this is the set of elementary divisors of the Cartan matrix of G and so Ch R,„,„, (G : Greg)* = Tr(R [s]). By (IV.7.2) (a B, )* = a *et Hence Ch R (G : Greg, B)* = ZR (G : S)e, and Ch R,,,, (G: G reg, B,)*= Tr(R [S] )e. This establishes the required isomorphism.

11]

GROUPS WITH A GIVEN DEFICIENCY CLASS

245

It follows from (10.7) and the fact that R is a local ring that ZR (G :S)e, has the required basis. Let T, be the R module with basis {1C 0 (x

Eq3(G),1 j

m Ri (P)} ,

where B = B,. Then T, C Tr(R[S])er . Clearly

■

{1./3 (C .01IP E 13(G), 1 j

mh(P)}

is the set of invariants of the R module ZR (G :S)e r l'T,. Hence

EDi zR (G :S)e,IT; ZR (G : S)ITr (R[S ] ) ED, ZR (G :S)e,ITr(R[S])e, and so T, = Tr(R [S])e, for each t.

0

COROLLARY 10.12. Let B = B, and let m 0 be an integer. Then the multiplicity of p "' as an elementary divisor of the Cartan matrix of B is given by mL(P), where P ranges over all groups in 13(G) of order p'". PROOF. By (10.11) Ernh(P) is the multiplicity of p'n as an invariant of ZR (G: S)e,ITr(R[S])e t . The result follows from the isomorphism in (10.11). 0 COROLLARY 10.13. For y E 30 (G) let 1(y,B) denote the number of irreducible Brauer characters in blocks kJ of Co (y) with 1J G = B. Then

E

mYB(p).= 1(y,B).

Pe(G)

PROOF. By (10.10) it suffices to prove the result in case y = 1. In this case it follows from (10.12) as 1(1, B) is the rank of the Cartan matrix of B. 0 COROLLARY 10.14. Let D be a defect group of B. Let n be a block of DC G (D) with G = B and let T(É) be the inertia group of n in 1•10 (D). Then m(D) is equal to the number of T(13- ) conjugate classes in Z(D)n S (y ). PROOF. By (IV.4.16) and (10.12) m RI (D) = 1. The result now follows from (9.8) and (10.10). 0 11. Groups with a given deficiency class

The results in this section are due to Brauer [19644

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[11

Let r be an integer. The group G has deficiency class r for the prime p if G has no nonprincipal p -block of defect d r. The reference to p will be omitted in case p is determined by the context. If G has deficiency class r then clearly G has deficiency class r' for any integer r' r. The next two statements are immediate consequences of the definition.

LEMMA 11.1. If 19'

IGI then G has deficiency class r for the prime p.

11.2. G has deficiency class 0 if and only if the principal block is the only block of G. In particular a p-group has deficiency class 0 for the prime p.

LEMMA

11.3. Suppose that P < G and I P I = p". If G is of deficiency class r n then G is of deficiency class O.

LEMMA

PROOF. By (4.4) every block of G has defect at least n. Thus the principal block is the only block of G. 0 11.4. Let P be a p-group with I PI = p" and let G = PCG (P). Then G has deficiency class r if and only if G/P is of deficiency class r — n.

LEMMA

PROOF.

Clear by (4.5). 0

THEOREM 11.5. If G has deficiency class r then for every p-element y of G, CG (y) is of deficiency class r. Conversely if r > 0 and CG (y) has deficiency class r for every element y of order p in G then G has deficiency class r. Let y be a p -element, let Bo be a nonprincipal block of CG (y) and let d be the defect of B0 . By the third main theorem on blocks (5.4), • = B is defined and the defect of B is at least d. By (6.2) B is not the principal block of G. Thus if G has deficiency class r then d < r. Hence PROOF.

CG (y) has deficiency class r. Conversely suppose that G is not of deficiency class r with r > O. Let B be a nonprincipal block of defect d r and let D be the defect group of B. Since r >0 there exists an element y of order p in Z(D). By (6.2) and (9.2)(i) there exists a nonprincipal block of CG (y) of defect d and so CG (y) is not of deficiency class r. D An old result of Landau [1903] implies that for a given integer k, there are only finitely many finite groups which have at most k classes. Thus (IV.4.18) implies the next result.

11]

GROUPS WITH A GIVEN DEFICIENCY CLASS

247

THEOREM 11.6. There exist only finitely many groups G of deficiency class 0 whose S r -group has a given order.

of In case p = 2, Brauer [1974c] has proved the following generalization

(11.6). for the prime 2. THEOREM 11.7. Let G be a group of deficiency class r 0 abelian subgroup of Suppose that a S2- group P of G contains an elementary (a) depending only on a f bound order 2' . Let I P I = 2°. Then there exists a f (a). such that I G

The proof of (11.7) is not very difficult but depends on special properties of involutions. It is not known whether an analogous result is true for odd primes. If y is a p -element in G it is possible to apply (11.4) to CG (y). Thus (11.5) can be used to give an inductive characterization of groups of positive deficiency class. Such a characterization does not seem to be possible for groups of deficiency class 0 since the second statement of (11.5) is false in this case. For instance let q be a prime with q = 1 (mod p) and let G be a Frobenius group of order pq. Then G does not have deficiency class 0 for p but I CG (y)I = p for every element y of order p in G and so CG (y) has deficiency class 0 for every element y of order p in G. This construction shows the existence of infinitely many groups of deficiency class 1 with a Se -group of order p. It follows from (11.6) that an analogous situation

cannot occur for deficiency class O. It has been shown that a simple group of Lie type in characteristic p has deficiency class 1, Dagger [1971], J. E. Humphreys [1971]. This also shows the existence of infinitely many groups with deficiency class 1. Among these there are only finitely many with a given Se -group. A related result of a slightly different sort can be found in Tsushima [1977].

CHAPTER VI

1. Blocks and extensions of R The notation introduced at the beginning of Chapter III will be used throughout this chapter. The following assumptions and notation will also be used. K is a finite extension of Q,,, the field of p-adic numbers. R is the ring of integers in K. By (III.9.10) and (1.19.4) there exists a finite unramified extension k of K such that if T-Z is the ring of integers in k then the following conditions are satisfied. (I) is a splitting field of 14- [H] for every subgroup H of G. k is a splitting field for every p'-subgroup H of G. (II) Let B be a block of R [G] with defect group' D. Let (a) be the Galois group of k over K. Then there exists a block 13' of ./fZ[G] such that if V is an irreducible .g [G] module in B then r -1

v R

=0

V°1 with V E P.

If furthermore fj°) = b.' is the block of 1 [G] which contains %'" is a defect group of kJ ) by (111.9.1 0).

D

LEMMA 1.1. Let B be a block of R[G] with defect group D. Then every R[G] module in B is R[D]-projective and there exists an irreducible k[G] module in B with vertex D. PROOF.

Clear by (111.4.14). 0

LEMMA 1.2. Let B be a block of R[G]. Suppose that Ê contains a unique 248

2]

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RADICALS AND NORMAL SUBGROUPS

irreducible ïi[G] module up to isomorphism. Then B contains a unique irreducible R[G] module up to isomorphism.

PROOF. Clear by (1.19.4). 0 LEMMA 1.3. Let H < G and let U be a projective indecomposable R[G] module. Then U, where each U, is a projective indecomposable [H] module and for each i there exists x E G with U, = U. PROOF. Immediate by (1.13.6) and Clifford's theorem (111.2.12).

E

Suppose that H < G. The block B of R[G] covers the block b of R[H] if E covers 6(i) for some j. LEMMA 1.4. Suppose that H < G and the block B of R[G] covers the block b of R[H]. Let D be a defect group of B. Suppose that DCG (D) C H. Then the following hold. (i) B = b' is the unique block of R[G] which covers b. (ii) D is a defect group of b.

PROOF. (i) Choose the notation so that Ê covers 6, where 6 is defined as in (II). by (V.3.9) fi is regular and so by (V.3.7) 1i-3 = 6G is the unique block which covers 6. Then r3 (' ) is the unique block which covers 6('). This implies that B is the unique block which covers b. The definition of the Brauer homomorphism implies that B = b G. (ii) By (V.3.14) D is a defect group of 6 and so by (II) D is a defect group of b. 0

2. Radicals and normal subgroups Throughout this section H is a normal subgroup of G. The results in this section up to (2.7) are mostly due to Kniirr [1976]. Some related results have been proved by Ward [1968], Huppert and Willems [1975], Willems [1975]. LEMMA 2.1. Let V be an k [H]-projective R[G] module and let M= V I VJ(R[G]). Then M is R[H]-projective if and only if VJ(R[G])= VJ(k[H]).

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PROOF. Let J = J(1[G])and let Jo = J(k[H]). As JoR[G]= R[G]J, vfo is an R[G] module. Suppose that VJ = VJo. Let A = 17z[G], B = 1[I-1] and S = Jo. Then (1.4.11) implies that M is 1[1-1]-projective. Suppose conversely that M is 17z [Il]-projective. Then (V .// VJ o)H —> (V I VJ --->

--->

is a split exact sequence as (V/VJo)H is completely reducible. Thus --> VJ/ VJo --> V/ V.I0 --> M -->

is a split exact sequence as M is 1 [H]-projective. Hence V IV.T o = M e viivio and so VJ I VJo = (Vi VJ0)J = M J + (VJI VJ0)J = (VJI VJo)J.

Thus VJ/VJo = (0) by Nakayama's lemma (1.9.8) and so VJ = VJo .

E

2.2. Let U be a projective indecomposable k[G] module and let M= U IUJ(R[G]). Then H contains a vertex of M if and only if UJ(R[G])= UJ(R[H]). COROLLARY

PROOF. Clear by (2.1).

E

THEOREM 2.3. Let e be a centrally primitive idempotent in k[G] and let B be the block corresponding to e. The following are equivalent. (i) H contains a defect group of B.

(ii) VJ(k[G])" = VJ(R[H])" for all R[G] modules V in B and all integers n O. (iii) J(R[G])e = J(k[H]) R[G]e. (iv) UJ(R[G])= UJ(R[H]) for all projective indecomposable R[G] modules U in B.

PROOF. (i) (ii). By (1.1) every module in B is R [H]-projective. Thus (ii) follows from (2.1) by induction on n. (iii). This is clear as eR[G]= R[G]e is a module in B. (ii) (iii) (iv). Let fz[G]e = U„ where each Lf, is indecomposable. Each projective indecomposable fz[G] module in B is isomorphic to some By the hypothesis UJ(R[G])= k[G]eJ(k[G])= k[G]eJ(R[H])

=

u,J(k [H

])

.

2]

751

RADICALS AND NORMAL SUBGROUPS

As U,J(R[H])C U,J(R[G]) for all i the result follows. (iv) (i). By (1.1) there exists an irreducible R[G] module M in B whose vertex is a defect group D of B. As M U IUJ(R[G]) for some projective indecomposable module U, the result follows from (2.2). 0

COROLLARY 2.4. Let V be an irreducible R[H] module. Assume that every component of V 0 lies in a block with a defect group which is contained in H. Then V 0 is completely reducible. By assumption there exist centrally primitive idempotents {e,} such that a defect group of each e, lies in H and such that V Ge = 1/0, where e =Ie,. By (2.3)

PROOF.

V GJ(k[G])= VGJ(k[G])e = (V 0 fi[G]) J(k[G])e R[H]

= V 0 J(k[G])e RIF11 = V 0 R[H] =

J(R[H])R[G]e = VJ(R[H]) Ø R[G]e RII-11

(0).

Hence V0 is completely reducible.

0

COROLLARY 2.5. Suppose that p IG : HI. Let V be an irreducible R[H] module. Then V 0 is completely reducible. PROOF.

Since H contains a defect group of every block the result follows

from (2.4). The next result is quite old. See e.g. Green and Stonehewer [1969], Villamayor [1959].

COROLLARY 2.6. J (R[G])= J(R[H])R[G] if and only if p IG :HI. [1-1]If J(k[G])= J(k[H])R[G] then iTZ [G]1.1(R[G]) is projective by (2.1). Thus every irreducible fz [ G ] module is fz [H ]- -projective and so p G : HI by (1.1). If p I G : HI then every R[G] module is R[H] projective. Thus (2.1) implies the result. 0 PROOF.

-

If W is an indecomposable k[G] module let w ( W) = w,, ( W) denote

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the integer so that WH is a direct sum of w ( W) nonzero indecomposable k [H] modules. The integer w ( W) is called the H-width of W or simply the width of W. Observe that if U is a projective k [G] module then

w ( U) = w (U1 LJJ(R [G ])) = w (UI U. (R [H ]) )

by Clifford's theorem (111.2.12) and (1.13.7). THEOREM 2.7. Let B be a block of R[G] which covers the block b of R[H]. Assume that H contains a defect group of B. Then the following statements are equivalent. (i) Every indecomposable projective rt[G] module in B is serial. (ii) Every indecomposable projective I7Z[H] module in b is serial. If furthermore (i) or (ii) is satisfied then all irreducible modules in B have the same width.

PROOF. Suppose that (i) is satisfied. Let U be a projective indecomposable k [G] module in B of maximum width. Let UH = P„ where each P, is a projective indecomposable iTt[H] module. Then w = w(U)= w (U/ UJ(k [II])). Let n be a positive integer such that P,J(k [H])" (0). By (1.3) P,J(k [1/])" (0) for i = 1, . , w. Let V = UJ(k[H])" IUJ(172[H])"1 Then

v,, =

P,J(R[H])" I P,J(R[H])"

and so w ( V) w. By (2.3) V is irreducible since U is serial. Let Uo be the projective indecomposable 1:-t[G module which corresponds to V. Then w (U0) = w( V) w (U). Hence the choice of U implies that w (U0 = w. The indecomposability of the Cartan matrix (1.16.7) now implies that every projective indecomposable k[G] module in B has width w and so every irreducible 1:-t[G module in B has width w. Thus (i) implies the last statement. Furthermore P,J(172[H])" IP,J(1-Z[H])" ±1 is irreducible as w( V) = w. Hence each P, is serial. Since w is the width of every projective indecomposable [G] module in B, the previous paragraph implies that if U is any projective indecomposable [G] module in B then UH is a direct sum of serial modules. Hence (ii) holds. It remains to show that (i) follows from (ii). Let U be of minimum width among the projective indecomposable R[G] modules in B. Let n be a positive integer such that V = ]

)

]

3]

SERIAL MODULES AND NORMAL SUBGROUPS

253

UJ(.172[G])" IUJ(k[G])"±I X (0). Let UH = 1 p, where each p is a projective indecomposable R[H] module. By (V.2.3) and (1.3) each P, is serial. By (2.3) VH = (UJ(R[H])" IUJ(R[H]) ±1), = e P,J(R[H])" IP,J(R[H])"'I and so w ( V) w = w( U). The minimality of w now implies that V is irreducible and w ( V) = w. The result follows as the Cart an matrix is indecomposable.

It should be mentioned that in general not all irreducible fz[G] modules in V will have the same width. For instance suppose that p = 2, H = SL 2(8), G is the semi direct product of H with a cyclic group of order 3 and rz contains the field of 8 elements. Then the principal block of G contains an irreducible .17Z[G] module of degree 6 and width 3, and of course it also contains the one dimensional module of width 1. COROLLARY 2.8. Let B be a block of R[G] which covers the block b of R[H]. Assume that H contains a defect group of B. Then the following statements are equivalent. (i) Every indecomposable projective k[G] module in B is serial. (ii) Every indecomposable k[G] module in B is serial. (iii) Every indecomposable projective fz[H] module in b is serial. (iv) Every indecomposable k[I-1] module in B is serial.

PROOF. Clear by (1.16.14) and (2.7). 0

3. Serial modules and normal subgroups

Throughout this section the following hypothesis and notation will be assumed to hold. HYPOTHESIS 3.1. (i) H
254

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[3

3.2. (i) B = b' is the unique block of R[G] which covers b. (ii) b contains a unique irreducible ft[H] module V up to isomorphism.

LEMMA

PROOF. (i) Clear by (3.1)(ii) and (1.4). (ii) This follows from (3.1)(iii) and (1.19.4). 0 By (V.3.14) T(6)/H is a p'-group. Let a be an irreducible R[GIH] module with a7-( 6),,, faithful. By (3.1)(i) dinu a = 1. Let V be defined as in (3.2)(ii). If M is an R[G] module we will write Ma = MO a. In particular a' is defined by induction as a' = 0 a. THEOREM 3.3. (i) There exist integers s, e > 0 with e Ii T(6): HI and an irreducible 1 [G] module W such that W, Wa, , Wei' are pairwise nonisomorphic modules in B, W Wa and every irreducible k [G] module in B is isomorphic to some Wa'. Furthermore V s(e;=, Wa'). (ii) If 1 = ft then s = 1 and e = (6): H 1.

Suppose first that k = 1. Clearly T(b) = T(V). Choose x E T(b) so that T (b)= (x, H). Thus the image of x generates the cyclic group T(b)1H. Let s = land let e = IT (b): H . Let A be a representation of R[H] with underlying module V. Since T(b) = T( V), there exists a linear transformation M on V such that A (x -I zx)= M-I A(z)M for all z E H. Thus PROOF.

M—A (z)Me = A (x

= A (x -')A(z)A (x' )

for all z E H. Thus by Schur's Lemma Me = cA(xe ) for some c E k. Let F be a finite extension field of R with co E F such that c. Define A (x) = c 0.1 M. Thus A extends to a representation of T (b) over F. This representation is irreducible as its restriction to H is irreducible. Since R = 1 is a splitting field of T(b) there exists an R[T(b)] module X such that X, ----- V. Consequently Xa' is irreducible for all j and (Xa'),----- V. By (3.2)(i) Xa' is in b T('') for all j. Let [3 denote the trivial k [H] module. By (11.2.3) e

vT( 6,

(x, 0

—I

p T( 6 ) i =0

so

If Y is an irreducible R[T(b)] module in b T(6) then VI Y, by (1.4) and (VT , Y) 0 by (111.2.5). Thus Y Xa' for some j. If Xa' Xa' for some 0---i<j<e then by (111.2.5)

3]

SERIAL MODULES AND NORMAL SUBGROUPS

1 < (V T(b) ,

255

W = X G . Then Wa' =

Xa')= I(V, V) = 1. Let

(Xa' )G by

(V.2.5). The result is proved in case R = P. We will now consider the case of general R. The previous part of the proof implies the existence of an irreducible I [G] module lit in P. so that = (131_, Wa', where ê = T(b):H. Furthermore Wa = Wa' if and only if i j (mod i) and every irreducible module in P. is isomorphic to Wa' for some j. Let (a) be the Galois group of 1 over P. By (1.19.4) VØ1 ---(131:101 17 , where m is the smallest integer with Ver- ------ 1-7 . Let W be an irreducible R[G] module with W I W Ott-1. Let e, t be the smallest positive integer such that Wae W, (V G )'' respectively. Then ell T (6): . Also t f m as (VN ) ()N. Let U be the direct sum of all distinct P[G] modules of the form Wee'a). Then 034

Wal)0/4

Therefore

G

VG R

( 1

m

(v (on wai) t

R)

(V R

1=1

j1

By (2.4) and (3.2) VG is completely reducible. Thus (1.19.4) implies the result. 0

If M is an R [G] module then 1(M) will denote the number of composition factors of M. Let Soc(M) be the socle of M and let Soc" (M) be the inverse image in M of Soc(M/Soc —I (M)) for n 2. 7

COROLLARY 7

3.4. Suppose that V (2<), R is the direct sum of m irreducible

R[G] modules. Then the following hold. (i) If M is any irreducible P[G] module in B then M Ø TZ' is the direct sum of m algebraically conjugate irreducible 1 [G] modules. 7 (ii) If U is a projective indecomposable R[G] module in B then U OA R is the direct sum of m algebraically conjugate projective indecomposable P[G] modules. (iii) If M is an indecomposable 172[G] module in B then l(M k 14) = Mi(M). PROOF. Since a has values in R, (i) follows from (3.3) and (1.19.4). Now (ii) is a consequence of (i) and (1.13.6), and (iii) is a direct consequence of (i). E

256

[4

CHAPTER VI

THEOREM 3.5. (i) If U is a nonzero indecomposable projective R[G] module in B then U is serial and I(U)= ID I (ii) Let M be an indecomposable .ft[G] module in B. Then M is serial and Soc" (MH) = (Soc" (M)),, for all n. .

PROOF. Let P be a projective indecomposable 1 [H] _module which corresponds to V. Let I be a projective indecomposable fi module which corresponds to 17. By (IV.4.16) the unique Cartan invariant in 6 is I D l. Thus l(P)= l(P) = I D I by (3.4). By (2.8) U is serial. Thus I(U) is the smallest integer n with UJ(R [G])" = (0). Hence by (2.3) 1(U) is the smallest integer n with UJ(R [H])" = (0). Since P IL., it follows from (1.3) that 1(U) is the smallest integer n with PJ(R [H ]) " = (0). Thus I(U)= l(P)=ID as P is serial. This proves (i). Furthermore {I.JJ(k [G])"} f., = {I.JJ(k [G])0" and so Soc"(UH )= {Soc"(U)},., for all n. By (2.8) M is serial. Hence M = Sock (U) for some projective indecomposable ft[G] module in B and some integer k. Therefore

Soc" (MO = Socn +k (UH )= {Soc"± k (U)}14 = {Soc n (m)}„ for all n. 0

COROLLARY 3.6. Let M be an indecomposable k[G] module in B. Then M Øk 1 = itiv, where T ranges over a factor group of the Galois group of R over 1--t and each K4' is indecomposable with l(M)= l(M' ) for all T. Furthermore Soc(M) Ø R Soc(14)' By (3.5) M Soc"( U) for some integer n and some principal indecomposable R[G] module in B. Since U ØR1 or with 'ITJ a principal indecomposable fi[G] module, the result follows from (3.4)(ii). LI PROOF.

4. The radical of 17t[G]

This section contains various results concerning J(fi [G]). The material up to (4.8) is due to Tsushima [19714 See also Tsushima [19784 The following notation will be used in this section. 172, ... is a complete set of representatives of the isomorphism classes of irreducible R[G] modules. For each i U, is a projective indecomposable 1i[G] module corresponding to V. Let 47,, 0, be the Brauer character ,

4]

257

THE RADICAL OF R [G]

afforded by v., U, respectively. Choose the notation so that cp, = 1 G , the principal irreducible Brauer character. Let t, be the trace function afforded by v. The following hold. t,(x)= t

(x)

= t, (x)

.

for x a p'-element in G.

(4.1)

where x„ , is the p'-part of x in G.

(4.2)

Let E = 1(t (x), x E G) and let T, be the trace from F, to k. Choose the notation so that T TAO, . are all the distinct elements of the set {T, (t, )}. By (1.19.3) and (1.19.4) { T, (t,)} is the set of all trace functions afforded by_ irreducible fz[G] modules. Furthermore J(R [G])=

J (R[G] (g) k ).

4.3. Let {V, Ii E S} be a set of pairwise nonisomorphic irreducible R[G] modules. Let I = Is be the intersections of the annihilators of all y„ j E S. Let la, I j E SI be a set of nonzero elements in k. Then Ann(/) = [G]c, where c = Exec ales ajt, (X -1 )}X.

LEMMA

Let s = Eis ait,. Define t E Hom i (R[G],R) by (EEG b„x )1 = b1. By (1.14.9) t is symmetric and nonsingular. Clearly s is symmetric and nonsingular in Hornh (1:Z• [G11 R). Thus PROOF.

(

yeG

E

by) ct =

by

E a,t, (x - ')xy} t = E by E a,t, yeG

x.yeG

ait,( 1ES

E

yEG

byy )= (

(y)

les

by) s. YEG

The result follows from (1.16.17). 0

COROLLARY 4.4. (i) For all i,E„EG 'T,(t, (x -1 ))x E Ann(J(R [G])). (ii) Let a2, . be nonzero elements of fz. Then Ann(J(k[G]))= R[G]c, where c = ExEG {E, a,T, (t, (X -1 ))1X. It suffices to prove the result in case R = 1. Hence T, (4) = t, for all i. Then (i) follows from (4.3) with I = Ann( V ) V, and (ii) follows from (4.3) with / = J(.1 [G]). 0 PROOF.

4.5. Suppose that R = R. Let p" be the order of a Se -group of G. Let I be the annihilator of ED, Es v„ where S is the set of all j with p"+1 0, (1). Then J(R[G])C I and Ann(/) = R [G]c, where c is the sum of all p-elements in G. LEMMA

258

[4

CHAPTER VI

Clearly J(R[G])C I. Let 0, (1) = p"h, for each i. Thus h, is an integer for all i. by (IV.3.8) E, h*, (x) = 0 if x# 1 and E, h*, (1) = p — I G . Thus if a = p— I G I then a X O. Clearly ri,# 0 if and only if i E S. Thus (4.1) and (4.2) imply that for x E G PROOF.

E it (x) = E /t1 (x) = ies Therefore

E.EG fa

{0 aX 0

>,Es (t, (x -1 ))1x = c.

if x is not a p-element, if x is a p -element. The result follows from

(4.3). E

If x is a p'-element of G let c E R[G] be the sum of all the elements in G whose p'-part is conjugate to x. The ideal of k[G] generated by all c is the Reynolds ideal of k[G]. See O'Reilly [1974] for related concepts. THEOREM 4.6. Ann(J(R [G])) is the Reynolds ideal of R[G]. PROOF. Since cx E R[G] and J(14 [G])= J(R[G])OA Ê it suffices to prove the result in case k = 1. By (IV.3.10) there exists for each p'-element x of G an 1-linear combination Ea,t, of t h t2, . . . such that at (x) = 1 and Ea,t,(z)= 0 for all p'-elements z not conjugate to x. If y is a p-element in CG (x) then t, (xy )= t, (x). Thus (4.4)(i) implies that c x E Ann(J(k [GM. For each i, EyeG t, (y = Et, (x -1 )cx . Thus by (4.4)(ii) Ann(J(R [G])) is generated by all c.„ El

LEMMA 4.7. Let H be a subgroup of G and let I be a nilpotent left ideal of i[H]. LetP = Ix and let f = n xEG R[G]Ix. Then the following hold. (i) Ï is a nilpotent ideal of R[G]. (ii) rG(1)= ExEG rH (I)1 [G] and f = la {E.EGrH(IY ft[G ]} , where r, 1 denote right and left annihilators respectively. (iii) If I is an ideal of R[H] then = rLEG ITZ[G].

If y E G then fy C nxEGR[G] --- = Thus f is an ideal of i;[G]. Therefore f - ±1 c (k[G]l)= for any integer m 1. Hence by induction f"'' C PF" for any integer m O. Thus f is nilpotent. (1.16.15) (ii) By I = 1H (r H (I)). This implies that k [G]l = 1G (rH (I)I4G]). Therefore if x E G then PROOF. (i)

17Z [GI = 1G (rH. (I )R[G])= 1G ({rH (I)1"17Z[G]).

4]

THE RADICAL OF

.g [G]

259

Hence

Î=

n

H

R [G])

(ri-i(nrk[G])

Thus by (1.16.15)

(1)= r lQ

E x

EG

r,..,(I)"R[G])1=

E

xEG

rfi(I)R[G].

(ii) By (1.16.16) the left and right annihilators of an ideal in F[G] F[H] coincide. Thus by (ii) f

=

rig (T)R [G]) = rG ( G li,(1)'12[G])

/G

E

=

rG( xEG

n

Or

R[G]h,(.0-)

rG(1-4[Glio-n=

xEG

n

PR [G]. 111

xEG

4.8. For a Se -group P of G let Ep =Ex Ep x. Then Ann(J(k[G]))c EpEA[G], where P ranges over all the Se -groups of G.

THEOREM

Let P be a Se -group of G. Then J(R[P]) = {Eci„xlEa„ =0} and so Ann(J(k[P]))= EpR [P]. Apply (4.7) with I = J(F[P]). By (4.7)(i) PROOF.

J(R[P])C J(R[G]). The result now follows from (4.7)(iii). 0 COROLLARY 4.9 (Lombardo—Radice [1939], [1947]). (i) Let S be the set of all p-elements in G. If >aaxE J(k[G]) then for all x E G

E

yES

axy

E

ay„ =O.

yES

(ii) Suppose that for every Sp-subgroup P of G, all x E G. Then ZEG a„x E J (17Z [G]).

E y EP

=Iy Epay„ = 0 for

Let c be the sum of all the p-elements in G. Then EyEs azy -is the coefficient of z in (ExEG ax x)c = c(E„,Gaxx). Thus (i) follows from (4.6). Let Ep be defined as in (4.8). Apply the previous paragraph to P. Then (ii) follows from (4.8) and (1.16.15). PROOF.

Ey ESay

In general neither of the conditions in (4.9) is both necessary and sufficient for an element of k[G] to be in J(k[G]). This can be seen as follows.

260

CHAPTER VI

[4

Let G be a group with a Se -group of order p" such that p"±1 I (1., (1) for some i. For instance G = SL2(4) A, and p = 2. Then (I),(1)= 8 for some i. Let I be defined as in (4.5). Then J(/72[G]),i / but every element in / satisfies (4.9)(i). Let G = SL2(4) A, and let p = 2. Let P be a S2-group, let Ep = EEpx and let E = V=,(Ep), where Px , , , P", are all the S,-groups of G. Thus E is the sum in k[G] of all elements y in G with y 2 = 1. It is easily seen that Ep is in the annihilator of each V, and so E E J(R[G]). If z is an involution with z P then (zy)2 = 1 for y E P if and only if y = 1. Thus if E = >ax then Ey , aZ = az / O. Hence the condition in (4.9)(ii) is not necessary. However it will follow from (X.3.1) that (4.9)(i) is necessary and sufficient in case G is p -solvable. Section 6 below contains some conditions on G for (4.9)(ii) to be necessary and sufficient. The next result contains some information concerning dimAJ(R [G]).

THEOREM 4.10 (Wallace [1958], [1961]). Let p" be the order of a Sp-group of G. The following hold. (i) P" 1 dim, (J(k[G ]))

G (1— 1/i 1 (1)).

(ii) p" —1= dimA (J(R [G])) if and only if G is a p-group or G is a Frobenius group with a Sp -group as a Frobenius complement. (iii) Suppose that (I), is rational valued. Then dimA(J(R[G]))= IGI(1-1/0,(1)) if and only if G has a normal Se -group.

PROOF. It may be assumed that R = A. Since U,J(R[G])C J(R[G]) and dim, = 1 + dim, U,J(R[G]) it follows that p"

0,(1)-1-dimR(J(k [G])).

(4.11)

If G is a p -group then clearly equality holds in (4.11). Suppose that G is a Frobenius group with Frobenius complement P and Frobenius kernel H, where P is a Se -group of G. Then any irreducible character of G which does not have H in its kernel is of defect 0 and cp, is the only irreducible Brauer character of G which has H in its kernel. Therefore dim, (.1( 1 [G ]) )=IGI — E ,P,(1)2 = p' —1.

Suppose conversely that dim, (J(k[G]))= p"— 1. Then (4.11) implies that dim, (J(k[G])) =00)— 1. Therefore U,J(k[G]) =(0) for i 1 and so cp, is in a block of defect 0 for i 1. Thus in particular cp, is the only Brauer character in the principal block. Hence G =0(G) by (IV.4.12). This proves (ii).

4]

THE RADICAL OF ./2

By (IV.4.15) 0, (1)

IGI= I

[G]

261

,(1)(p, (1) for all i. Thus (1)cp, (1)

,(1)

= cPOA G I — dimfi J (R[G])}.

(4.12)

This shows that the second inequality in (i) holds and completes the proof of (i). Suppose equality holds in (4.12). Then 15, (1) = ,(1)(p. (1) for all i. Therefore 0, = 0,cp, for all i. Since 0, is rational valued (IV.10.1) implies that 0 1 (y1 ) = p", for any p'-element y1 , where p°, is the order of a Se -group of CG (y,). For any p'-element y of G let c (y) be the number of elements whose p'-part is y. Then c(y)--- I 0,(y)1 for every p'-element y in G. Hence

'GI=

E

c(y) --- E1 0 1(Y)1E °I(Y)=1G1( 01,q) , )=IGI.

Therefore in particular c (1) = 0,(1)= p" is the order of a Se -group P of G. Hence P contains all the p -elements in G and so P
Hence dim A (J(R[ G])) = I G 1(1 lip"). As G has a p -complement H it follows that 01= (1H) G . Thus 0,(1) = p". —

The assumption in (4.10)(iii) that 01 is rational valued is probably unnecessary, but this remains an open question. In case G has a pcomplement H, cI = (1 kir is rational valued. Thus in particular 0 1 is rational valued if G is p -solvable. Other properties of the radical of a group algebra can be found in Clarke [1969]; Hamernik [1975b]; Koshitani [1977], [1978], [1979]; Morita [1951]; Motose [1974c], [1977], [1980]; Motose and Ninomiya [1975a], [1980]; Spiegel [1974]; Tsushima [1967], [1968], [1978b], [1979]; Wallace [1962a], [1962b], [1965], [1968]. Properties of the radical of the group ring of a p-group have been studied by Hill [1970]; Holvoet [1968]; Jennings [1941]. If G is p-solvable of p-length 1 see Schwartz [1979].

262

CHAPTER VI

[5

5. The radical of R[G] The notation of Chapter IV section 8 will be used in this section. Thus it is assumed that R = 1. For 0 E Chic (G) define

=

;G 0 (x

Thus ij E 4, (G). By (IV.7.1) {X„} is the set of centrally primitive idempotents in K[G]. Furthermore

I G,

EZ R (G).

Since 77-R[G]C J(R[G]) it follows that J(R[G]) = J(k[G]). In fact J(R[G]) is the inverse image in R[G] of J(k[G]). Let A denote the inverse image of Ann(J(R [G])) in RIG]. If e is a central idempotent in R[G] then J (R[G]e)= J(R[G])e and Ae is the inverse image of Ann(J(R [G]e)) in R[G]. Most of the results in this section are due to Broué [19764 [197813]. LEMMA 5.1. Let B be a block and let e be the corresponding centrally primitive idempotent in R[G]. Then Ae = irR[G]e +

E

G

R[G] (" "„)

x„c13

PROOF.

(1)

•

It suffices to show that A = rrR[G]+E R[G](I

G I v) Xu( 1 ) - u

or equivalenty that A=

G}

If x„, is the p' -part an element x in G then x. (IV.3.12), (4.1) and (4.2) imply that

E R[G] (xi .G01 ) "„)

R[G]

x. (x) (mod

7).

Thus

,

where {t, } is the set of all trace functions of irreducible k [G] modules. The result now follows from (4.4). E

5)

263

THE RADICAL OF R[G)

be a centrally primitive idempotent in R[G] and let B be the corresponding block.

LEMMA 5.2. Let e

(i) Z R (G)e CExuER RX. u . (ii) J (ZR (G)e) = (Exu.B 14.) PROOF.

n ZR(G)e.

For x E B, w u (X u )= 8. Thus

for

aE

Z,‘ (G)e,

a =

E„„EBcou(a)i„.

(i) If a E ZR (G)e then co„ (a) E R for all u. This implies the result. (ii) It is easily seen that J(Ex“EB Ri.)= 77-(E„ B Ri.). Furthermore -1(Ex..B 14.) consists of all a E E,EBRi. with wu (a) =- 0 (mod ,n-) for all x u E B. The result follows as J(ZR (G)e) consists of all a E ZR (G)e with w u (a ) 0 (mod 7r) for all x„ E B. Ell

I = Eu R ((I G I I x u (1))X u ). Let e be a centrally primitive idempotent in R[G] and let B be the corresponding block.

LEMMA 5.3. Let

(i) le = B R ((I G X. M)'O. (ii) le is an ideal in E x. ER and le is an ideal in (iii) For every positive integer n, Ie[J(ZR (G))e}" C

ZR

(G)e.

PROOF. (i) Clear by definition.

(ii) By (5.1) le C ZR (G)e. Thus it suffices to prove the first statement. This follows from the fact that = 8.4. and so

(xj.G0)1

1G() suoi“.

_x

(iii) Since Je is an ideal in E,E13 R, J (Z R (G)e)" C ir"

suffices to show that for n 0

E

x„EB

This follows from (5.2)(ii). 0 THEOREM 5.4. Let e, B,

I be defined as in (5.3). Let d be the defect of B.

(i) J(ZR (G)e)= (n-p-dIe) n ZR(G)e. (ii) For every integer n 1, J(ZR (G)e)" Ç (n- "p -die)n ZR(G)e. is an ideal of ZR (G)e. By the definition of I, i„ E p-dle for x„ E B. Thus E,,,EBRiu c p - qe. Hence by (5.2)(ii) PROOF. (i) Clearly (7rp -ale)( Z R (G)e

J (ZR (G)e)=

x„EB

Riu )n ZR (G)e C Trp -d Ie

n ZR (G)e.

264

CHAPTER VI

Let xo be a character of height 0 in B. Then ak

IGI s

tIGI \ X. ( 1 ) X V

X.( 1 )

Therefore w„(7rp-die)C 77-R and so 7rp —le n ZR(G)e is annihilated by the central character (70„ of 14 [ G]ë. Hence n-p -dIe n ZR (G)e C J (Z R (G)e). (ii) Induction on n. If n = 1 this follows from (i). Suppose the result is true for n — 1. By (5.2)(ii) J (ZR (G)e)" = J (ZR (G)e)J (ZR (G)e)' Ç

(

7r

E

,ER

n ZR(G)e.

1;4 u)

The result follows as le is an ideal in E,, EB R COROLLARY

by (5.3)(ii).

E

5.5. Suppose that R71- - = Rp. Let I, B, e, d be defined as in

(5.4). (i) J(ZR (G)e)"'" C 77-"Ie for all integers n 1. (ii) J(ZR (G)e)"' +' C 77-ZR (G)e. (i) This follows from (5.4)(ii). (ii) This follows from (i) as le C ZR (G).

PROOF.

LI

Broué has pointed out that the next result, due to Tsushima [1971b], can be proved by these methods. See also Reynolds [1972]. THEOREM 5.6 (Tsushima [1971b]). Let c E1 [G] be the sum of all the p-elements in G. Then c 2 is the sum of all the centrally primitive idempotents in 14 [G] which correspond to blocks of defect O.

Let S be the set of all p-elements in G. Let c o = / yE sy E R[G]. Thus -cT, = c. Since c is in the Reynolds ideal it follows from (4.6) that c E Ann(J(P[G])). Thus co E A. As wu (X' o ) = 6,,„ it follows that

PROOF.

co

= E ,..(c0x.(1)

= E (Du

GJ

u

(IGI

-\ x„(1) X“ I •

Since X„X'o = Suoi„ this implies that _

Nqcou (co)x„ (1)12 GI ( I G '4,71 I

1G

X. ( 1 ) \x. (1) X V •

(5.7)

6]

265

p-RADICAL GROUPS

By (5.1) and (5.7) a). (co)x. (1) 1 IGIER for all u. Let Bo be the union of all blocks of defect O. Since fi G11 x.(1)} / 0 if and only if x is in B o . This implies that c 2 = t(2) =

E

x.e8o

(0.(coi.•

If xu is in a block of defect 0 then xu (y) = 0 for y E S, y w. (c o) = 1. Hence c 2 = Ex.ER0 iu • 0

1. Thus

6. p Radical groups -

The material in this section is based on the work of Motose and Ninomiya [1975b] which includes earlier results of Deskins [1958]; Khatri [1973]; Khatri and Sinha [1969], and Motose [1974a], [1974b]. Much of this has been strengthened by Knôrr [1977]. Related results can be found in Broué [1978a], Isaacs and Scott [1972], Khatri [1974].

Define the following sets of subgroups H of G. G(G)= {II

I V G is completely reducible

for every

irreducible F-Z[H] module V}. G:11}. 91(G) = {H J(R [G]) C J(R[H]) R[G]l.

Since J(F[G]) = J (R[G]) F for an extension field F of fz it follows that 91(G) depends only on G and p. LEMMA

6.1. 91(G) = G(G).

PROOF.

Let H E 91(G) and let V be an irreducible /7? H] module. Then [

V G J(R[G]) = (1/ R(:[3,,) ] R[G]) J(R[G])= V Atli] J(R[G]) CV

REH]

J(R[H])R[G]

= VJ(R[H]) (3) R[G] = (0). R

[H]

Thus V G is completely reducible and so H E Cr(G). Therefore 91(G) C G(G)• Suppose that H E G(G). Let lx,} be a cross section of H in G. Then an arbitrary element a E J(R [G]) is of the form a = E, a,x, with a, E 1 [H].

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[6

If V is an irreducible R[H] module then V G a = 0 as V G is completely reducible. Therefore 0 = (V 1)a = E,Va, Ø x. Hence Va, = 0 for all i. Hence each a, annihilates every irreducible k[H] module and so a, E J(R[11]) for all i. Thus a E J(ft [H])ft [ G ]. Hence H E 91(G) and so (G)C91(G). 0 LEMMA

6.2. 91(G)CZ(G).

Let HE 91(G). Let V = ( V) be a one dimensional 1 [1-- ] module. By (6.1) H E(G) and so V G is completely reducible. Thus W V G, where W = Inv G (W) and dim A W = 1. If Q is a Se -group of H then Q is a vertex of V. Let P be a Se -group of G with Q C P. Similarly W has P as a vertex. Since W V G, we must have P = Q by (111.4.6). 0 PROOF.

A group G is a p radical group if 9i(G)= Z(G) or equivalently -

The next result shows in particular that G is p -radical if and only if the condition in (4.9)(ii) is necessary and sufficient for an element of I[G] to be in the radical.

THEOREM 6.3. The following conditions are equivalent. (i) G is p-radical. (ii) 91(G) contains a Se -group of G. (iii) J(17Z[G])= nxEGgfz[P'1)/7Z[G], where Pis a Se -group of G. (iv) Ann J(R[G])=Erepfz[G] where P ranges over all Se -groups of G and

Er

=Ex EpX.

ExEG ax E J(F[G]) if and only if Ey Epa„), =EyE payx = 0 for all x E G and every S e -group of G. (V)

(0 (ii). Clear by definition. (ii) (iii). Since P EN(G), J(ft[G])C J(R[P'])17Z[G] for all x E G. Thus J(R[GDC n„EG J(1[P1)1[G]. The opposite inclusion follows from (4.7)(i). (iii) (i). By (6.2) it suffices to show that Z( G)C at(G). Let H E Z(G) and let P be a Se -group of G with P C H. By (4.7)(i) nxE,J(R[P' )17Z[H] C J(R[H]). Therefore PROOF.

]

n

xcG

gR[P-DR[G]ç

n

J(R[P'DR[Irlk[G]

x EG

C r) J(R[Hx])R[G]. xEG

6]

p- RADICAL GROUPS

267

By (4.7)(i) CV G J(ft [H'I)R[G] C J(k [G]). By assumption J(R[G1) = n„EG J(R[P'DR[G]. Hence all these sets are equal and so J(R[G])=

n

J(R[H])R[G]C J(R[HDR[G].

xEG

Therefore H EN(G) and so Z(G)C9i(G). (iii) <> (iv). Since J(R[P]) = epft[P] each of the statements follows from the other by taking annihilators. (iv) <=> (v). Statement (v) is equivalent to the fact that a in 1 [Gi is in J(ft [GI) if and only if ae, = era =0 for all Se -groups P of G. In other words J(R[G]) = Ann(Er ejt [G]). Thus (iv) and (v) can be derived from each other by taking annihilators. E COROLLARY 6.4. Let P be a Se -group of G and let V =Invp V be an R[G] module with dim, V = 1. Then G is p-radical if and only if 1/ 0 is completely reducible.

By (6.1) and (6.3) G is p -radical if and only if P E (G). The result follows as up to isomorphism V is the unique irreducible R[P] module. 0 PROOF.

THEOREM 6.5 (Khatri [19731). Suppose that H < G. Then the following hold. (i) (ii) (iii) (iv)

H E Z(G) if and only if H E 91(G). If G is p -radical then G I H is p-radical. If H is a p-group then G is p-radical if and only if G 1H is p-radical. If H E Z( G) then G is p -radical if and only if H is p-radical.

Let P be a Se -group of G. (i) If H E Z( G) then H E G) by (2.5). Thus H E 91(G) by (6.1). The converse follows from (6.2). (ii) Let V be an irreducible k [PH/I-1] module. V G is completely reducible as an k [G] module if and only if V G is completely reducible as an 1[G /H] module. The result follows from (6.4). (iii) If G is p-radical then so is GIH by (ii). Suppose that GIH is p-radical. Let V be an irreducible k [ P] module. Since H C P, P is an irreducible i [P1111 module and so V G is a completely reducible R[G IH] module. Thus VG is completely reducible and G is p -radical by (6.4). (iv) Let V be an irreducible R[P] module. If V G is completely PROOF.

reducible then by Clifford's theorem (111.2.12) (V G ),., is completely reduc-

268

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[6

ible. Thus by the Mackey decomposition (11.2.9), V" is completely reducible. If V" is completely reducible then by Clifford's theorem V G is completely reducible. The result follows from (6.4). 0 COROLLARY 6.6. Let G = Gl x G,. Then G is p-radical if and only if G, and G2 are p-radical.

If G is p-radical then G, and G, are p -radical by (6.5)(ii). Suppose that G, and G, are p-radical. Let P, be a Sr -group of G i =1,2. By (6.5) J ( TZ [G,]) C J(R[P,])R[G,]. Since PROOF.

.

J(R[G])= J(R[G,])R[G2] + J(R[G,])R[G1]

as R[G]= [G1 ] Ø,1[G2] modules it follows that J(R[G])CJ(R[P,]) R[G]+ J(R[P,])R[G].

Let P = P x P,. Then P is a Se - group of G and J (k[P,])C J(R[P]) for i = 1, 2. Thus J(k[G]) C J(k[P])k [G]. Thus PE 91(G) and so G is p-radical by (6.3). D

CHAPTER VII

1. Blocks with a cyclic defect group

The purpose of this chapter is to study modules in a block with a cyclic defect group. In particular all indecomposable modules over a field of characteristic p in such a block will be described. In addition to this, some detailed information will be obtained concerning decomposition numbers, higher decomposition numbers and other properties of modules and characters in such a block. Since the situation for a block of defect 0 is very simple and has been described it will be assumed that the block has positive defect. A cyclic p -group has only a finite number of indecomposable modules in any field of characteristic p. (This follows from the Jordan form for linear transformations.) Thus a module in a block with a cyclic defect group has only a finite number of possible sources. Hence there exists a finite field such that any module over a field of characteristic p in a block with cyclic defect group is equivalent to a module over the given finite field. This implies that the following assumptions are not unduly restrictive. The notation introduced at the beginning of Chapter III will be used throughout this chapter. The following assumptions and notation will also be used. K is a finite extention of Q„, the field of p-adic numbers. R is the ring of integers in K. D is a cyclic subgroup of G.

ID! = p° > 1, For 0

i

D = (y).

a, D, is the unique subgroup of D with

C = CG (D,),

N, = NG (D,).

Therefore C < N, and 269

ID

: D, = p'.

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[1

Co CC,C—CC_ A CC=G, No CN1C•CN,CN a =G. B is a block of REG] with defect group D. As in Chapter VI, section 1, there exists a finite unramified extension k of K such that the following conditions are satisfied where R is the ring of integers_in K. (i) 1 is a splitting field of 1 [H] for every subgroup H of G. k is a splitting field of k[H] for every p'-subgroup H of G. (ii) There exist blocks Ê = fr),...,k—i) of R [G] such that D is a defect group of each /4' ) . Furthermore 7if V is any irreducible 14[G] V' (1) in 1 3' 0 ). module then V is in B if and only if V Ok R =E;_01 %' (iii) There exists an element o- in the Galois group of k over K such that after a possible rearrangement = flo) for j = 0, , r —1 and Jî' = B. Choose the notation so that fj- = f3 0 ). For any integer j' define P' co= fjo , where 0 j r — 1 and j' j (mod r). Since N = No c N for 0 j a, the First Main Theorem on blocks (111.9.7) applied to G and to N implies that for 0 i a there exists a unique block Ê of 1[N,] with B = Ê = A,• Thus also f3;sr, = Bk for O i k a. By (111.9.3) and condition (iii) above, this yields that = Ê for all j and 0 i k a. By the First Main theorem on blocks (111.9.7) D is a defect group of each block fa ). Let bk be a block of lz• [Ck ] which is covered by /I for 0 k a. For j = 0, , r — 1 define 6,(1, ) = 61". Then Ê,0,) covers 6v. Observe however that fri,'" need not be equal to 6k. By (V.2.3) the blocks covered by k for some z E Nk Bk are conjugate under the action of Nk. Thus 6T,' = I;r By (V.3.9) Bk is regular with respect to Ck . Thus by (V.3.6), 6 rk4, = Ê. By (V.3.14) D is a defect group of 6k The group Nk /Ck is a group of automorphisms of the cyclic p-group Thus either Nk /Ck is cyclic or p = 2 and Nk /Ck is the direct product of a group of order 2 and a cyclic group. In any case I Nk :Ck II (p — 1)p° -1 . Therefore by (V.3.14) I T(6k ): Ck I I (p — 1) where T(f)k ) is the inertia group of 6k in Nk. Furthermore T(6k )/Ck is cyclic. For 0 k a let bk be the block of R [Ck ] such that bk corresponds to the set of all R [Ck ] blocks 0;0 as in (111.9.10). Let Bk be the block of fi[Nk] which corresponds to the set of all 1 [Nk ] blocks {b`kl as in (III.9.10). Thus Bk covers V, for all x E Nk• If H is a group and V is an 14 [H] module define: I(V)= number of factors in a composition series of V. S( V) = S 1 ( V) = Soc(V) is the socle of V.

11

BLOCKS WITH A CYCLIC DF.FFF.CT GROUP

271

S"( V) is the inverse image in V of S( V/S —I ( V)) for n 2. T(V)= V/Rad V. Rad'( V) = Rad( V) and Rad" ( V) = Rad(Rad -1 ( V)) for n 2. Suppose that tt is a one dimensional representation of .14[H]. We will write V ti for the tensor product of V with the underlying module which affords tt. Thus in effect we are identifying ilk with its underlying module. If V1 , V, are fi[H] modules for any group H then IR (V,, V,) denotes the intertwining number of V, and V2. If V is an R[H] module then Vic = V OR K.

LEMMA 1.1. If 0 k a —I then (i) Co = ck n T(60), (ii) T(6k )= T(60)C, (iii) T(6k)ICk T(60)1C0. PROOF. (iii) is an immediate consequence of (i) and (ii). (i) Clearly C0 C Ck fl T(60). Suppose that x E Ck r) T(60). Then x E No and x d E Co for some integer d with d l(p —1). Since x E Ck C Ca -1 this implies that x E Co. (ii) Let Ck C H C Nk such that H/Ck is a Hall p'-subgroup of Nk/Ck. Thus H/Ck is cyclic and T(6k ), T(60)Ck are both contained in H. Hence it suffices to show that j T(6k ): Ck I = I 0)Ck C,. • Let t be the number of blocks of l'[Ck ] which are covered by 1 k4 . Thus t= : T (f)k )i . By (V.3.9) and (V.3.10) 6rNo covers a block '6 of [Co] if and only if I- I IN ° 6F-InNo. 0 By the First Main Theorem on blocks (111.9.7) this is the case if and only if 6H = (6c, )H = 6H, 0 or equivalently 6`, is covered by 1.4. Furthermore t is the number of blocks of 1 [Co] which are covered by f1rN. Thus t=iFinNo:T(6)1. Since I H: Ck I is the number of elements of D„_, which are conjugate to yPa ' in G is follows that H: C j =1H n N0: Col. Therefore (-

Hence

ri TOO:

Ck

=IH

(60)ck :Ck 1 = 1 7' (1 0):

I

H n No: col= t T(60): col. T(60)n

= T(6k). Ck

=

1 7' (1 0): C01

I.

For x E N 1 define a (x) by x -I r -l x = Thus a is a one dimensional representation of 1 [N„_1 ] whose kernel is G_,. By abuse of notation we will also use a to denote the restriction of a to any subgroup of N„_,. We will also denote the underlying one dimensional /7? [N,_ 1 1 module by a.

272

CI-IAPTER

[1

VII

In view of (1.1) {a' 10 j T(6 0 Co II is a set of distinct representafor 0 k a —1. Furthermore a 1'0'01 =- a ° is the trivial tions of R [T( one dimensional representation of every subgroup of T(6k ). ):

LEMMA 1.2. Suppose that for some k with 0 k a —1, bk' contains a unique irreducible I;[Ck ] module up to isomorphism. Then b k contains a unique irreducible R[Ca ] module up to isomorphism. Furthermore up to isomorphism b o contains a unique irreducible 17Z[Col module. PROOF. By (V.4.6) b.° contains a unique irreducible 1 [C01 module up to isomorphism. The result now follows from (VI.1.2). D THEOREM 1.3. Fix k with 0 k a —1. Assume that up to isomorphism bk contains a unique irreducible R[Ck] module V and bk contains a unique irreducible II [Ck J module V. (i) There exist integers s = sk , e = ek, with e T(6k): Ck , and hence module W such that e I (p — 1), and an irreducible k[Nk] W,Wa,..., Wa' - ' are pairwise nonisomorphic modules in Bk. Wa W and every irreducible fz[Nd module in Bk is isomorphic to some Wa'. Furthermore 1/Nk ---- s( 1 (ii) If R =R then s =1 and e =1T(b0 Co l. ):

PROOF. This is a direct consequence of (I.2), (1.16.13) and (VI.3.3). E The index of inertia of B or the inertial index of B is the maximum number of pairwise nonisomorphic irreducible R[N 0 modules in B o. By (1.2) the assumptions of (1.3) are satisfied for k = O. Thus the index of inertia of B is the integer e o defined in (1.3). In case R = 1 it is equal to T(b 0 Co l and so coincides with the index of inertia defined in Chapter V, section 9. Since N, C N 1 for 0 i a — 1, the Green correspondence defines a one to one map from the isomorphism classes of nonprojective R -free R[D]-projective R[G] modules to the isomorphism classes of nonprojective R -free R [D]-projective R[N,] modules. Such a correspondence is also defined for R[G] modules and R[N 1 1 modules. The following notation will be used ]

):

- Na

-

= Ba-l•

If V is a nonprojective R -free R [D]-projective R[G] module then V is the R[6] module which corresponds to V by the Green correspondence.

2]

STATEMENTS OF RESULTS

273

If V is a nonprojective [Di-projective 1[GI module, 17 is defined similarly. If V is projective define 17 = (0). By (111.7.7) V is a nonprojective R-free R[G] or k[G] module in B if and only if is in E. LEMMA 1.4. Let (D„ G), D(D,, G) be defined as in Chapter III, section 5 a —1. Then the following hold. (i) (D„ 6) . = {(1)} for 0 i a —1. (ii) If A E D(D„ G) then A n D = (1) for 0 i a —1. (iii) If V, W are R[D]-projective R -free R[G] modules or projective R[G] modules then (a) HA G, (1), V) ------ fr, (1), 17) , (b) HAG, (1), Hom R (V, HAG, (1),HomR W)). for O i

o-

PROOF. (i)(ii) Suppose that A E D(D,, 6). Thus A Can D; for some x E G — Ô. Since D„_, n = (1) it follows that A n D„_, C n D = (1). Thus A n D = (1) as D is cyclic. This proves (ii). Since 1(D„ 6)C D(D„ 6) and every element of (D„ 6) is a subset of D, (i) is an immediate consequence. (iii) This now follows directly from (III.5.10). 111 LEMMA 1.5. (i) Let V be a nonprojective R -free R[G] or a nonprojective k[G] module in B. Then V6 @ A , El) A2 where A, is projective, A 2 is a sum of indecomposable modules in blocks other than È and (A 2)D is projective. (ii) Let A-7 be a nonprojective R -free R[6] or a nonprojective R[G] module in 15-3. Then VG - v ED A for some projective module A. (iii) Let V, W be nonprojective R -free R[G] or nonprojective R[G] modules in B. Then HAG, (1), HAG- , (1), N7) and HAG, (1), Hom, ( V, W)) Hoo, (1), Hom R ( W)).

PROOF. (i) This follows from (11.5.3) and (1.4)(i)(ii). (ii) This follows from (111.5.4) and (1.4)(i). (iii) This is a special case of (1.4)(iii). fl

2. Statements of results

This section contains only statements of results. The proofs of these statements are given in the rest of this chapter.

274

CHAPTER VII

[2

The material in this chapter is an outgrowth of the work of Brauer [1942a]. In that paper he proved (2.11)—(2.19), (2.22) and (2.23) in case K = k and D I = p. He also defined the Brauer tree which is defined in section 6 and showed that the irreducible Brauer characters in B have height O. The methods he used do not generalize to handle the case that I D > p. In particular one of his results, (11.2) below, is in general not true for ID1> p as is shown by the examples at the end of section 11. A quarter of a century later Thompson [1967b] proved results analogous to those of Brauer for the case that D is a Se -group of G and CG (x) = D for all x e D, xL 1. In doing this he used the Green correspondence and proved (1.17.12) as well as a version of (5.6) below which is of critical importance for the whole development of this material. Almost immediately after this, Dade [1966] was able to combine Thompson's methods with the theory of blocks to prove (2.11)—(2.19) in general for the case that K = I. Then Janusz [1969a], [1969b] using Dade's work as a starting point gave a complete description of all the indecomposable k [G] modules in B in case K = k. Section 12 contains his results generalized to the case of general R. In particular for K = k he proved (2.2), (2.3), (2.20)—(2.22), (2.26). At about the same time Kupisch [1969] independently described the indecomposable R[G] modules in B for K = k. Earlier results in this direction had been obtained by Srinivasan [1960], Janusz [1966]. The k[G] modules which lift to R[G] modules were determined by Michler [1975]. In case ID I = p, (2.4) and (2.8) were proved in Feit [1969] under additional restrictions and generalized in Blau [1971a]. Lindsey [1974] proved (2.4) and (2.8) for k = 0 and R = 12 in case D is a Se -group of G. Feit [1969] suggested a method for simplifying some of the arguments of Dade and Janusz by making more systematic use of the Green correspondence and in particular by using (II.5.10). Also (2.5), (2.7) for ID! =p and (2.10) were announced there in case K = k, and shortly thereafter (2.21) was obtained in case K = k by these methods. It follows as a corollary of (2.10) that every irreducible Brauer character in Ê has height O. This result was first proved by Rothschild [1967] by applying a graph theoretic argument to the Brauer tree to prove (2.7) for K = k. Green [1974a] further simplified some of the above mentioned arguments by making use of (11.3.13). In particular he considerably simplified the proof of a result of Passman announced in Feit [1969]. He also obtained some results on projective resolutions for the case K = k which generalized earlier work of Alperin and Janusz [1973]. His results are contained in section 10 generalized to the case of arbitrary K. Peacock [1975a], [1975b],

2]

STATEMENTS OF RESULTS

275

[1977] pushed these methods further to get an alternative proof of (2.20) and some refinements for R = fi by an argument which was independent of the results for k[G] modules as opposed to i;-Z. [G] modules. In case IDI—p (2.20) had already been announced in Feit [1969]. Recently Michler [1974], [1976b] gave a proof of (2.1) and (2.2) for general K which is completely free of any "characteristic 0" results. In so doing he introduced the idea of using (1.16.13) which made it possible to

simplify considerably a very complicated portion of Dade's paper. The methods used in this chapter are an amalgam of those used in the various papers mentioned above. For instance in section 5, Dade's argument is followed quite closely though for most of the rest of the chapter his arguments have been replaced by simpler arguments as mentioned above. Frequently results are first proved in case K = k and then in general by Galois descent. Throughout this chapter e denotes the index of inertia of B, ê denotes the index of inertia of A. 2.1. B contains exactly e irreducible R[G] modules up to isomorphism.

THEOREM

2.2. B contains exactly ID le nonzero indecomposable R[G] modules up to isomorphism.

THEOREM

2.3. Let V be an indecomposable R[G] module in B. Then S (V) and T(V) are each the direct sum of pairwise nonisomorphic irreducible modules.

THEOREM

2.4. Every indecomposable 1 [6] module in A is serial. There exists an irreducible 1 [6] module W such that W, Woc, , W a' is a complete set of representatives of the isomorphism classes of irreducible 1[6]

THEOREM

modules in 2.5. Let L, M be irreducible R[G]. modules in B. The following are equivalent. (i) L M. (ii) S(L) S (11-1). T(M). (iii)

THEOREM

2.6. Let V be an indecomposable R[G] module in B. Then p° and Vis R [Dd-projective if and only if 1(1.1-)= 0 (mod p k ).

THEOREM

[2

CHAPTER VII

276 THEOREM

0<

2.7. Let L be an irreducible 17i [G] module in B then either e or p" —e l(L)
The next result is due to Blau and is a strengthening of an earlier result. THEOREM

2.8. Let 0 • k - a —1 and let V be an indecomposable fz[Nk]

module in Bk then the composition factors of V in ascending order are S(V), S( V)a S(V)a -2, ....

THEOREM 2.9. Suppose that for some k with 0 ,- k a some nonzero indecomposable R[Nk ] module in Bk is absolutely indecomposable. Then any irreducible k[Ar in B, is absolutely irreducible for 0 i a and any indecomposable k[N] module in B, is absolutely indecomposable. ]

Let L 1 ,. , I be a complete set of representatives of the isomorphism classes of irreducible 1-2- [G] modules in B. Let cp, be the Brauer character afforded by L, for 1 i e. Let W be the module defined in (2.4) and let tfr be the Brauer character afforded by W. —

THEOREM 2.10. Suppose that R = R. Then every cp is of height O. (1/1G : D 1)0(1) E R and (11 I G : D 1)0(1) # O. Furthermore there exist inte1 i e with 0
(mod p°).

For i = 1, . . . , e let U , be the principal indecomposable R[G] module which corresponds to L,. Let 0, be the character afforded by U. Let A be the set of all characters of No which are afforded by irreducible K[N0 modules in Bo and which do not have D in their kernel. ]

THEOREM 2.11. If K = k then l A I = (p° — only if l DI= p and IT(60): Col= p —1.

In particular I A I = 1 if and

Throughout the rest of this chapter it will be assumed that the following condition is satisfied.

IA I = (p° —

(*)

Condition (*) is a hypothesis that K must satisfy. By (2.11) it holds in case K = k. By (2.11) no two distinct elements of A are conjugate by a

2]

STATEMENTS OF RESULTS

277

field automorphism which fixes the elements of K. Observe that (*) can be satisfied with R an arbitrary finite field (of characteristic p) since there exists a purely ramified extension K of Q„ for which (*) is satisfied. It should be noted that (*) need not be satisfied for the extension of Q„ generated by all I D I th roots of unity. Consider the following example. G = No is a dihedral group of order 30 and p = 3. Let a„, denote a primitive m th root of 1 over Q, and let Ko = Qi(ai ). Let x be an element of order 15 in G. Then there exists a unique irreducible character 4' of G with 4- (x) = a ', + a 5. 1 . Thus 4- (x ) K o and A = + CI where a 7-5 = a Hence (*) is not satisfied for the block B of Ko[G]which contains 4' . In this case (*) is satisfied for any nonprincipal block of K[G] where K = Q„(I:=o a).

THEOREM 2.12. B contains exactly e AI characters which are afforded by irreducible K[G] modules. If IA I = 1 these will be denoted by x o, ,x„ If IA I /1 these characters are divided into two families x i , ..., x, and {x A À E AL In the latter case let xo= EA CA XA. If x is any p'-element in G then x,(x)-- x, A (x) for A, EA. The characters e A defined in (2.12) are called the exceptional characters in B. By (2.12) x, and x, agree as Brauer characters for A, E A. In particular they have the same decomposition numbers. Let do , denote the XA, cp, decomposition number for A E A.

THEOREM 2.13. d., = 0 or 1 for 1 i e and 1 u e. If K = k then d o , = 0 or 1 for 1 i e. If K/ k then (2.13) need not be true. This is related to questions about Schur indices. See (2.18). Let u = 0, , e and let A E A. By (1.17.12) there exists an R -free R[G] module X such that XK affords x. or x, and g is indecomposable.

THEOREM 2.14. Suppose that K = I. Let 0 u e and let A E A. Let X., X, denote an R -free R[G] module such that fC.,??, is indecomposable, (Xt. )K affords x„ and (XA )K affords x A . Then 1(X.)= 1 or p° —1 and 1(k)= e or pc — e. The value of 1(k) or 1(X,) is independent of the choice of the module X. or X A and depends only on x. or x A . Furthermore I()= 1(A) for À,a EA. If K = k define 8. = ± 1 for u = 0, . . . , e by 8. —= (k) (mod p") where X. is defined as in (2.14). If p' = 2 choose 6o# 5, at random.

278

[2

CHAPTER VII

2.15. Suppose that K = k. (i) Let 1 ,5 i e. If d„, 0, d,,/ 0 and u v then 5u + 8, = O. (ii) Let 1i e. There exist u, v with 8u + = 0 such that 011 , = x. (iii) Let x =- xo if I A I= 1 and let x = x, for some A E A if I A I 1. Then

THEOREM

80x (x)+

8.x. (x) = 0

for every p'-element x in G. 1 then 5, = • • = 8, =1 and 80 = (iv) Suppose that G = N. If IA I can be chosen so that 8, • • •= 8, = land 1. If I AI= 1 then the notation 80= —1. —

THEOREM 2.16. Suppose that K = k. Then every irreducible character in B is of height O. Furthermore x(i)

6 '1'(i) (mod p° )

for u

X A(1)=- — 800(1) (nod P °)

0, . . . , e.

for A

E

A.

The congruences in (2.16) are very strong if D is a Sp -group of G and get weaker as the power of p in G : DI gets larger. If for instance D1 2 11 G I they contain no information. Suppose that K= k. For 0 k --5a-1 let t/J, be the unique irreducible Brauer character in bk . Let {z } be a cross section of T(bk ) in Nk. Then {b ;,} is the collection of blocks of Ck covered by B k . Thus {0;‘ } is the collection of irreducible Brauer characters in blocks covered by Bk. If x E Dk Dk-hi then CG (X) = Ck. Let d (u, x, 414,) or d (A, x, Cc) denote the corresponding higher decomposition number for x. or x, respectively where u = 1, . , e and A E A or possibly u = 0 in case IA 1= 1. THEOREM

2.17. Suppose that K = k. For 0 k

such that for x E Dk Dk-hi the following hold. or for u =0 if d(u,x, ,pz) -= EA, for 1-- u

d (A,

-=

— ek80

V z.,

1,7

a —1 there exist e k =

I A I=

1.

(x) forA EA,

K.ET(bk)

where is an irreducible constituent of the restriction to D of an irreducible constituent of Ac in bk . Furthermore if 0 j a —1 and , Ej are the signs for the group C, then there exists y = 1 such that E:= ye i for 0 i j. G = C, for some j with 0 j a — 1 then e, • • • = Ea =1 unless

2]

279

STATEMENTS OF RESULTS

p = 2 and j = 0, in which case the notation may be chosen so that O i -5 a -1.

E, =

1 for

It follows from (2.17) that if IA I 1 and A E A then x, has a higher decomposition number not equal to ± 1 unless p = 2, e = 1 and À 2 = 1. In the latter case if x, is the nonexceptional character in B and a is the irreducible character of D with kernel equal to D i , we can define x", = XA for A EA U {1} and note that {x} satisfies (2.11)-(2.17) with , E.-1 unchanged but So replaced by - So. Then x„ = x is the 80, 81, E 1 , nonexceptional character. If A runs over A in (2.17) then it is easily seen that runs over a complete set of representatives of the T(b 0)-conjugate classes of nonprincipal irreducible characters of D. THEOREM 2.18. Let r be the number of blocks of k[G] which are algebraically conjugate to B. Then the Schur index mo of )6, over K is d0, for some i and is independent of A E A, and IIXA 11 2 = môr is independent of A E A. By (2.9) m = I (L,, L,) is independent of i. Let m o , r be defined as in (2.18). THEOREM 2.19. (i) If 1 u, i e then Qp() = Q, (C ) , mQ,(i„) = 1 and

1 xuI2= m. (ii) Given i with 1 i e then either there exists u,v with 1 ,-5 u, y e such that 0, = x + xi, or there exists u with 15. u e and 0, =

x u + e lemoho. If A = 1 and ê e then replacing K by a purely ramified' extension it may be assumed that mo = 1 and so xo can be recognized from the decomposition of the In this case it is called the exceptional character in B. Thus B contains an exceptional character unless I D 1 = e. Let a., = d u , for u 0 and let a0, = 1(em0) if d 0 . Thus if 1 i e, 0, = ax , + ai„xi, for suitable u, v. It follows from (2.19) that mo I (ele). Since ele= [Q„ (io, (A): Q,, (f(0)] this also follows from (IV.9.3). In section 13 it will be shown that equality holds if K is chosen suitably. In particular mo = ele if K is a field of minimum degree over Q„ which satisfies condition (*), or if K = Q„ E A). Suppose that d. 0 for some u, i. Let X. be the pure submodule of U, such that (X „,),, affords au,x 0 and S ()-Cu,) L„ and so fC„, is indecomposable. See (1.17.12).

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Observe that (2.19) implies that each irreducible constituent of occurs with multiplicity (1 D I - 1)/e for do, / O.

[2

,(0,

2.20. If d.1 / 0 then .gu, is serial. If furthermore d.,/ 0 for some + go and S (0)= g.; n X. u then Rad C./, =

THEOREM

,

2.21. Every indecomposable R[G] module in B is serial if and only if one of the following conditions is satisfied. is irreducible for all (i) D I - 1= e and after a possible rearrangement u with 1 ,5. u e and all i with (ii) I D I- 1 e. k, is irreducible for u with 1- u e and all i with COROLLARY

/ O.

(2.21) may be reformulated in terms of the Brauer tree defined in section 6 as follows.

COROLLARY 2.22. Every indecomposable R[G] module in B is serial if and only if the Brauer tree is a star and the exceptional vertex, if its exists, is at the center.

2.23. Let 0 u e. There is an ordering cp (I) , , (p ( s ) of all the irreducible Brauer characters which are constituents of x. such that if du,/ 0 and cp ( t ) = cp, then the composition factors of X„, in ascending order afford (p'` ),•• where the superscripts are read modulo s. Each constituent occurs equally often in X..

THEOREM

2.24. If xu is real valued then there exist at most two real valued irreducible Brauer characters which are constituents of x„.

THEOREM

THEOREM

2.25. Let 0 u e and let X be an R- free R[G] module such

that XK affords au,x., where du,/ 0 and )? is indecomposable. Then there exists j with cl„,/ 0 such that X X.,. Furthermore l (J) = 1 or p° -1 and l(k,)= 1(k,) if du„ du,/ O.

In view of (2.19) and (2.25), ô. = -1- 1 can now be defined for K in general as follows. If d„,/ 0 then 5. /(ku,) (modp°). If p° = 2 choose So 51

randomly. The next result is due to Janusz [1969b] as are (2.21) and (2.22). He proved these as corollaries of the classification of indecomposable R[G] modules in B. A proof of (2.26) is given at the end of section 12.

3]

281

SOME PRELIMINARY RESULTS

THEOREM

2.26. Let V be an indecomposable R[G] module in B. Then 11(S (V))— 1(T(V))1 1.

3. Some preliminary results Throughout this section it will be assumed that For 0 k a —1, isomorphism.

7

6k contains a unique irreducible R[Ck ] module up to

It will later be shown that this assumption always holds. Thus the results proved in this section will hold in general. Observe that any statement proved for Ô holds for Nk for 0 k a —1 since if Dk
LEMMA

PROOF. Clear by (1.2). 0 Let ë = e 1 be defined as in (1.3). Let W be the 1 [6] module defined in (1.3).

LEMMA 3.2. W, Wa, , Wa -1 is a complete set of representatives of the isomorphism classes of irreducible k [6 ] modules in 1. PROOF. Clear by (1.3). D LEMMA 3.3. Let C„_, C H c 6- for some subgroup H. Let M be an ft[H] module in the block which covers b,. Then (M,) for all integers n, (i) (S"(M)) c C (ii) (Rad" (M)) c D Rada (M). . ,

PROOF. Let C, = C. The proof of both statements is by induction. If n = 1 both statements follow from Clifford's Theorem, (11.2.12). (i) It may be assumed that M = S"(M). As (M/S —1 (M)), is completely reducible, induction implies that Rad(Mc ) C (S -1 (M)) C S -1 (Mc ). Thus McÇ S "(Mc ) as required. (ii) By induction Rad -1 (Mc ) C (Rad" -I (M)) c . Thus Rad" (Mc C Rad(Rad-1 (M),) C (Rad"(M)), by induction. E )

CHAPTER VII

282

[3

LEMMA 3.4. Let C C H C O for some subgroup H. Let V be an irreducible R [H] module in the block which covers b„. Suppose that V Ok1 is the direct sum of m indecomposable modules. Then the following hold. (i) If M is any irreducible module of R[H] in the block which covers 7 then M Ok R is the direct sum of m algebraically conjugate irreducible R[H] modules. (ii) If P is a nonzero indecomposable projective k [H] module in the block which covers b„_, then P Ok .F'Z' is the direct sum of m algebraically conjugate indecomposable projective R [H] modules. (iii) If M is an indecomposable k [H] module in the block which covers then ,

1 (M (1p ,

= ml (M).

PROOF. This follows from (VI.3.4). THEOREM 3.5. (i) If P is a nonzero indecomposable projective 1 [6] module in El then Pis serial and 1 (P) = p°. (ii) Let M be an indecomposable 1 [6 ] module in b. Then M is serial and S"(Mcl, )= (S"(M))c._, for all n. PROOF. This follows from (VI.3.5). LII COROLLARY 3.6. Let M be an indecomposable R[G] module in b. Then M Ø1 =eiq, where 7 ranges over a quotient group of the Galois group ') for all T. of 1 over R and each 1(4' is indecomposable with l(M)= Furthermore

s(m)

s(My.

PROOF. This follows from (VI.3.6). 0 THEOREM 3.7. Let M be an indecomposable k[6] module in b. Then for 0 k a, Mis R[Dk ]-projective if and only if 1(M) 0 (mod p'). PROOF. If k = a this is a consequence of (3.5). Suppose that 0 k a — 1. Assume first that R = 1 By (1.3) e = (6,): C, for 0 j a and there are exactly e irreducible Brauer characters LP, in b. All of these have the same degree and hence are of height O. Thus if', GAÔ : DI is a unit in R. Therefore if M is an indecomposable R[Dd-projective module in El then /(M) 0 (mod p k ) by (IV.2.2).

3]

SOME PRELIMINARY RESULTS

283

By (3.5) k contains exactly ep°' pairwise nonisomorphic indecomposable modules M with /(M) 0 (mod p'). Thus k contains at most ep°' pairwise nonisomorphic indecomposable R[Dk ]-projective modules and it suffices to show that P contains at least ep° pairwise nonisomorphic indecomposable 1[D,,]-projective modules. We will show that for each j with k j s a, B contains at least e (p° — p° -") pairwise nonisomorphic indecomposable modules with vertex D. This clearly implies the desired result. Fix j with ksjs a. Let V be the unique irreducible module in 6, (up to isomorphism) and let Wce, 1 i -se be the modules in k, defined in (3.2). For 1- s [3' let X. be the indecomposable R[D,] module of 1dimension s. Then (X),,, ----- IN, : D, i X,. Thus X' and X,N' have no common indecomposable component if s t. Furthermore if s 0 (mod p) then every indecomposable component of ,C,N' has vertex D,. Since V C X;; it follows from (1.3) that Wa C C X sNi. By (3.5) the indecomposable components of X,N' in B, are all serial and must contain some Wa as a submodule. Thus B. at least e(p° -' pairwise nonisomorphic indecomposable modules with vertex D,. The Green correspondence between N, and now implies that contains at least e (p° — p° - ") pairwise nonisomorphic indecomposable modules with vertex D,. This completes the proof in case k Suppose now that fz is in the general case. Let M be an indecomposable nonzero 1[Ô] module in P. Let M be an indecomposable nonzero fz[G] module in /i such that M I M of, By (3.6) M -----kr where T ranges over a quotient group of the Galois group of 1 over R. FurtherMore l(M)= 1(14) by (3.4)(iii). Thus_it suffices to show that M is 1[D,,]-projective if and only if M is k [D k ]-projective. This is true by (111.4.14). LI

Gy,=,

LEMMA 3.8. e = J. 1:t contains exactly e irreducible 1 [6] modules up to isomorphism and n contains exactly elDi nonzero indecomposable 1 [6] modules up to isomorphism.

PROOF. By (3.2) k contains exactly ë irreducible 14[6] modules up to isomorphism. By (3.5) each indecomposable k [6] module V in P is uniquely determined by S( V) and I(V). Since I(V) may take any value between 1 and p° =IDI, it follows that P contains exactly é ID nonzero indecomposable [6] modules up to isomorphism. It remains to show that é = e. indecomposable 1[6] modules By (3.7) 1 contains exactly (p° —

284

CHAPTER VII

[3

with vertex D up to isomorphism. Furthermore B, contains exactly (p° — p" - ')e indecomposable [No] modules with vertex D up to isomorphism. By (111.7.8) 1j and B, contain the same number of indecomposable modules with vertex D up to isomorphism. Thus e = é. 0 THEOREM 3.9. Let 0 k

a —1 and let M be an indecomposable R[Nk] module in Bk. Then the composition factors of M in ascending order are S (M), S (M)a -1 , S (M)a -2, .

PROOF. Let V be an indecomposable k[Nk ] module in Bk with W S( V) and 1(V)= 2. By (3.2) T(V) ,---- Wa d for some integer d. Let P be the principal indecomposable k[Nk] module corresponding to W. Then V S 2 (P). By (3.2) and (3.8), {Pa' 10 i e — 1} is a complete set of representatives of the isomorphism classes of principal indecomposable modules in Bk. Thus if V, is an indecomposable R[Nd module in Bk with 1(V 0)= 2 then V, Va for some i and so T(V o)= S(Vo)a d . As every indecomposable R[Nk ] module in Bk is serial it follows that the composition factors of such a module M in ascending order are S (M), S(M)a d , . It suffices to show that a d = a -1 or equivalently that d =• —1 (mod e). By (3.6) it may be assumed that k = 1. Let z = yP° '• Then (z)= < Nk. Since W)R[(z)]R[(z))

it follows that dimk {Inv(z) (P)} = p°' dims W = dimk {SP° i (P)}. As (z) 1 Nk, Inv(z) (P) is a submodule of P. Thus Inv(z) (P)= SP — '( P) since P is serial. Let U = SP" ' +'( P). Then U(z — 1) = S(P) and T(U) -----S

1 (P)/S(P). Thus T(U) ---- S(P)a P" 'd .

Let x be a p'-element in Nk. Let {u, } be a set of characteristic vectors for x in U such that their images form an 1 -basis of T(U). Then u,z = u, + w„ where {w,} is an 1 -basis of S (P) as 1/(z — 1) = S (P). Let u,x = c,u, . Then for all i u; + a(x)w, =t4z"(x ) = upc -I zx = c 7' uizx

= c uix + c

1i/ix

= u + c wdc.

Thus wix = (x)w; for all i. Consequently {wi } is a set of characteristic vectors for x and 1,1/ 0 (x)= tif(x)a (x), where tir o is the Brauer character afforded by S(P). Therefore

3)

SOME PRELIMINARY RESULTS

S(P)a'1

285

S(P)a

and so -1p'dd

(mode).

M, and M2 be indecomposable 1 [0] modules in 1j such that l(M 1 )+l(M2)p . Then

LEMMA 3.10. Let

I-P(O, (1), Horru (M I , M2)) --- Hom (M M -R1G],-1, - 2)• PROOF. Let g ETr?, ) (HomA (M,, M2)). By (11.3.13) there exists an indecomposable projective k[G] module P and an exact sequence

M1±->' P j-> m2,0 with g = fh. Clearly P is in 1. Thus by (3.5) h(M,)C S'(P) and SP -1 '2) (P) is in the kernel of f. Since W 1 ) + p it follows from (3.5) that S'(P)C SP -1 '2)(P). Consequently g = fh = 0. This implies the result. LEMMA 3.11. Let V 1 ,

V2

be irreducible

k[G] modules in B. The following

are equivalent. (i) V,

(ii) S

V2.

S ( 172).

(iii) T(170 ,-, T( 17 2). PROOF. Clearly (i) implies (ii) and (iii). The modules Vi', Vf are in a block with defect group D. Thus once it is shown that (ii) implies (i), it will follow that (iii) implies (i) by duality. It remains to prove that (ii) implies (i). Suppose that S(171 )------ S(172) and V, V2. Then HomA EG JV„ V1 ) = (0) and so Fr(G,(1),HomA (y, y))= (0) for j1= {1, 2 } . Thus by (1.5)(iii) Fr(G,(1),HomA(V„17,))= (0) for {i, j} = {1, 2 } . Choose the notation so that /()---Ç. /( 72). Suppose first that /WO + /WO p'. Hence HomAtc1(1-/I, 172) = (0) by (3.10). By (3.9) 171 ---- S"(172) for some n contradicting the previous sentence. Suppose next that /(17,)+ /(172)> p". By (3.5) /(1-7,)--.Ç. p' for i = 1,2. Thus there exists a principal indecomposable module P and exact sequences

0—> 17,—)P—)X1 —>0

for i = 1,2. Therefore /(X2)---- /(X,), /(X,)+ /(X2)< p' and T(X1)='—'

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[3

T(X 2). By (111.5.12) and (3.10) HomR 1 0](XI, X2) = ( 0). However (3.9) implies that X, X i /S n (Xi ) for some n contrary to the previous sentence. El

LEMMA 3.12. B contains exactly e irreducible R[G] modules up to isomorphism and B contains exactly elDI nonzero indecomposable R[G] modules up to isomorphism.

The map sending V to 1-7 is one to one from the isomorphism classes of nonprojective indecomposable modules in B to the isomorphism classes of nonprojective indecomposable modules in A. Thus by (3.8) it suffices to show that B contains exactly e irreducible pairwise nonisomorphic modules. By (3.8) and (3.11) B contains at most e pairwise nonisomorphic irreducible modules. For i = 0, , e 1 let V, be an R[G] module such that V, Wa There exists an irreducible k[G] module L, with HomR [Gi ( y, / (0). By (111.5.13) applied to G and 6 and (1.5)(iii) HomAo i (Wa',L,) # (0). Thus Wa ' S(L,). Hence by (3.11) {L O i e 1} is a set of pairwise nonisomorphic irreducible modules in B. D PROOF.

—

—

7

THEOREM 3.13. (i) Let F be a finite extension field of R. Let 0 k a and let V be a nonzero indecomposable ftW k ] module in Bk. Suppose that V, is the direct sum of m algebraically conjugate F[Nk ] modules. Then for any j with 0 • j a and for every indecomposable 11[N1 1 module U in B„ UF is the direct sum of m algebraically conjugate absolutely indecomposable F[N] modules. (ii) Suppose that for some k with 0 k a, Bk contains an absolutely indecomposable nonzero k[Nk I module. Then for all i, every indecomposable R[N] module in B, is absolutely indecomposable.

Clearly (i) implies (ii) so that only (i) needs to be proved. By (3.6) is the direct sum of algebraically conjugate absolutely irreducible modules. Thus it suffices to prove the result for V = W and k = a 1. Suppose first that for some j with 0 j a 1, B, contains a nonzero indecomposable I[N1 ] module U such that U, is the direct sum of in algebraically conjugate indecomposable F[N1 1 modules. By (3.6) S(U) has the same property. By (3.2) every irreducible k[N1 1 module in B, has the required property and so by (3.6) every indecomposable ft[N,] module in B, has the required property. Thus the result holds for j = k = a 1. Suppose that 0 j a — 1. Let y bean indecomposable R[G] module PROOF. WF

—

—

—

3]

287

SOME PRELIMINARY RESULTS

in B, with vertex D, and let U be the 1 [6] module which corresponds to V under the Green correspondence. By (III.5.7)(iii) II, is the direct sum of m algebraically conjugate F[Nk ] modules. Thus the result holds for O j a —1 by the previous paragraph. By (111.5.7) (iii) the result holds for every nonprojective indecomposable R[G] module in B as it holds for every indecomposable 1[6] module in E. It also holds for the indecomposable projective k[G] modules in B as it holds for the irreducible R[G] modules in B. 0 3.14. Suppose that Ip(W,W)= m. Let 0 k a and let V,, V, be indecomposable R[Nk ] modules in Bk. Then b(V,, V 2)=-• 0 (mod m).

LEMMA

7 By (1.19.4) W R = (BC:, Wir; where 'T is an element in the Galois group of 1 over R, W o is absolutely irreducible, WII Vi/r,' if and only if i j (mod m). By (3.6) and (3.13) Vs ØR1 =1(V) for s = 1,2 and some absolutely indecomposable modules (V,)11 . Thus PROOF.

(y,, = i (y ,§) iL y, 1") =

?nip'

P, (v2

3.15. Let M M2 be indecomposable k[0] modules in m = I, (W, W). (i) If 1 (M 1 ).-e then Ifi(Mi,g).-- m for j} = {1, 2 } . (ii) If IA (MI, MI) = m then e.

LEMMA

11)

)„) .

E.

Let

Both statements are immediate consequences of (3.9) and (3.14). 0 PROOF.

3.16. Let V be an irreducible R[G] module in B. Then either l(17)---e or l(17)-?-- p" — e.

THEOREM

PROOF.

Let m = I,( W, W). By (111.5.13), (1.5)(iii) and (3.13) dim, Ir(G, (1), Hom, ( V, V)) = dim, Ir(G, (1), Hom, ( V, V)) = I, (V, V) = m.

Suppose that / ( Thus by (3.10) ./k (1-7 V) = m. Hence / ( e by (3.15). Suppose that l( V ) > There exists a principal indecomposable k[G] module P and an exact sequence ,

0—> -V—>P-1 X—>0.

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Thus /(X)= 1 (P)-1(177)
IR (X, X) = dimk Hoo, (1), Hom k(X, X))

= dimAHoo,(1),Hompi ( 17, If )) = m. By (3.15)

e and so /(77 )% p" — e.

D

LEMMA 3.17. Let V be an indecomposable 1[G1 module in B. (i) S (V) is the direct sum of pairwise nonisomorphic irreducible modules. (ii) T (V) is the direct sum of pairwise nonisomorphic irreducible modules. PROOF. The module V* is in a block with defect group D and T(V*)-----, Thus (ii) follows from (i) by duality. It remains to prove (i). S(V)*. If V is projective the result follows from (1.14.8) and (1.16.8). Suppose that Vis not projective. Let L be an irreducible 1 [G1 module in B. It will be shown that L occurs in S(V) with multiplicity at most 1. Suppose that 1(i.)+ p°. By (111.5.13), (1.5)(iii) and (3.10) Homk rG) (L, V) ----- Homk 1 60, 1-/). By (3.16) either /(L)--- -. e or e. In either case the result follows from (3.13) and (3.15). Suppose that 40+ /( '1-()> p". There exist principal indecomposable fz [6] modules PI , P2 and exact sequences

0--->X—>/) ,------> 0,

0—> Y

Thus 1(X)+ /(Y)< p". By (3.16) either 1(X) --.. e or 1(X)--- p" — e and so e. By (111.5.12), (3.14) and (3.15)

dimR I °(6, (1), Hom k(L, V)) = dimAH"(6, (1), Hom A(X, Y))

= (W, W). Thus by (111.5.13) and (1.5)(iii) IR (L, V) = I, (W, W). The result follows from (3.13) and (3.14). 0

The results proved so far in this section are sufficient for the proofs of (2.1)—(2.10). The next two results are needed for the proofs of (2.20)—(2.23). LEMMA 3.18. Suppose that M 1 and M2 are nonprojective indecomposable R[G] modules in B such that Rad(M, ) = S (M, ) V for i = 1, 2, T(M) V, for i = 1, 2 where V, V 1 , V, are irreducible. Assume that '1-( 1 17/2 and l("2). Then 2

)+

< p" <

(,)+

).

4)

PROOFS OF (2.1)-(2.10)

289

PROOF. By (1.5)(i) there exist exact sequences for i = 1,2

0—> 1,7 EBAIEBA 2 —>A7f, e,v,EDA->V,IEDA7(1)A->0, where A,, A are projective and A,, A A are sums of indecomposable modules in blocks other than ii. If this sequence is multiplied by the central idempotent corresponding to the block i then (1.15.6) implies the existence of projective 1 [O] modules P, such that

0—> 1.-/ —>ge P, — V1 —› 0 is exact for i = 1, 2. Suppose that /WO + /(1-/)

0 —> 17/

r. Then

o

is exact for i = 1, 2. Thus S(M,) S(M2) and so (3.9). Hence S S ( 72 ). Thus by (3.11) V, ---- V2 contrary to assumption. Suppose that 1(1-/2) + 1(1-i) p°. Since Al, is not projective 1(4)/ p'. Then there exist principal indecomposable modules Pi , P2 in 13 such that

S (Pi ) is exact for i = 1,2. Since Pi is not in the kernel of g, for i = 1, 2. Since f, (1.7) npi/(o) it follows that S (1-/) S (Pi ) for i = 1, 2. Thus S(') ----- T(17,) for i = 1,2 and so TWO— , T(1-/2 ). Hence by (3.11) V, ----- V2 contrary to assumption. CI LEMMA 3.19. Let M be an indecomposable [ G I module with S(M)= Rad(M) such that S(M) is irreducible but T(M) is reducible. Then T(M) is the direct sum of 2 nonisomorp hic irreducible modules. Furthermore l(S(M))/ 1 or pa - 1. PROOF. By (3.17) T(M) is the direct sum of pairwise nonisomorphic irreducible modules. By (3.18) 1(T(M))= 2. For any irreducible constituent V of T(M) 1 p° — 1. Thus if /(S(M))= 1 or p° — 1, (3.18) implies that 1(T(M))= 1 contrary to assumption.

4. Proofs of (2.1)—(2.10)

The proofs of (2.1) and (2.2) will be given simultaneously by induction on GI.

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VII

[5

Suppose that G = Co. Then e = 1. By (V.4.6) 60 contains a unique irreducible Brauer character. Thus the hypothesis of section 3 is satisfied and the result follows from (3.12). Suppose that G = Ck for some k with 0 k a — 1. By (1.3) e =- 1. If D = Dk then G = Co and the result follows from the previous paragraph. Suppose that D,,/ D. Let G = GID, By (V.4.5) there is a unique block f3 ° of G ° which is contained in and DID, = D' is the defect group of b`). By (1.3) the index of inertia of b() is 1. Thus by induction b" contains a unique irreducible Brauer character. Since every irreducible Brauer character of G has DK in its kernel it follows that Ê contains a unique irreducible Brauer character. Since the inertial index of /j, is 1 for all i it follows by induction that has a unique irreducible Brauer character for As C, = C , for k i a —1 we see that the hypothesis of section 3 is satisfied. The result follows from (3.12). Suppose finally that G/ C. Hence by induction 1;k contains a unique irreducible Brauer character for 0 k a —1. Hence the hypothesis of section 3 is satisfied and the result follows from (3.12). El —

As a consequence of (2.1) we see that the hypothesis of section 3 is always satisfied. Thus all the statements of section 3 are valid. (2.3) follows from (3.17). (2.4) follows from (3.2) and (3.5). (2.5) follows from (3.11). (2.6) follows from (3.5) and (3.7). (2.7) follows from (3.16). (2.8) follows from (3.9). (2.9) follows from (3.13). By (1.5)(ii)

1 IG:DI

. dim r, V -----

1

IG:DI

dimA V

(mod P")

for any R[G] module in B. Thus (2.10) is a direct consequence of (2.7).

El

5. Proofs of (2.11)—(2.17) in case K = Throughout this section it is assumed that K = k. LEMMA

5.1. In case G = Co, statements (2.11)—(2.17) are true with e =1,

8 1 =1, 80 = —1.

5]

PROOFS OF

(2.11)-(2.17)

IN CASE

K=

PROOF.

This is a direct consequence of (V.4.7). 0

LEMMA

5.2. Statement

291

(2.11) is true.

PROOF. By (5.1), T(b0)1C0 acts as a permutation group with no fixed points on the set of all irreducible K[Co] modules in Bo which do not have D in their kernel. Thus by (5.1) and e = [T(bo): Cd, bJ`b,,) has (p° — 1)/e irreducible characters which do not have D in their kernel. Thus A I = (p° — 1)/e by (V.2.5)(i). Since e (p — 1), I A = 1 if and only if e =p — 1

and

ID =P. LI

LEMMA

5.3. B contains exactly e+IAI irreducible characters.

PROOF. Let y 1. The number of conjugates of yS in D is IN, : CH where y E D, — D, 1. The number of blocks of C, covered by B, is(1/e)IN C, J. Therefore the number of irreducible Brauer characters of CG (y = C, in blocks which are mapped to B by the Brauer correspondence is 1/e times the number of conjugates of y in D. By (2.1) there are exactly e irreducible Brauer characters in B. Hence by (IV.6.6)(ii) B contains exactly e I irreducible characters. 0 LEMMA

5.4. In case G =

the decomposition numbers in B are all 1.

PROOF. By (5.3) there are p° irreducible characters in B = EL, By (2.1) there is a unique irreducible Brauer character in B. The result follows from (V.4.6). El LEMMA 5.5. Let X be an R-free R[C—] module in b such that g is

indecomposable. Then XK has no composition factor with multiplicity greater than 1.

PROOF. By (2.4) ,? is serial. Hence )? /Rad() is irreducible. Thus there exists an indecomposable projective R[C_,1 module P and a commutative diagram

X —> X/Rad(X)—> O.

Since 7rX C Rad(X) it follows that g /Rad(.) X/Rad(X) and Rad(X) is the unique maximal submodule of X. Therefore g is an epimorphism and the result follows from (5.4). D

?92

CHAPTER

VII

[5

The next result was first proved by Thompson [1967b] in a critical special case. LEMMA 5.6. Let 7/ be a character of G such that 0, = +n 1 for some i where = 0 or Tr is a character. Let characters of in Then 1(71c._,, Os) —

0,)1

be all the irreducible

1

for all s,t. IfO and n X 0, then (11c._,, 0s) X(7c, 0,) for some s, t.

PROOF. By (1.17.12) there exists an R -free R[G] module Y such that YK affords n and Y is indecomposable. If Y is projective then n = 0 and the result is trivial since is projective. Suppose that Y is not projective. By (1.5)(i) YG = I.-7E9A,E9A2 where A, is projective and A, is a sum of modules in blocks other than /j. By (111.5.8) and (1.4)(i) -ÇT is indecomposable. If 77, and 77 2 are the characters afforded by (A 1 ) K and (A 2 ) K respectively then ( (772 Os ) = 0 for all s and ((77 1 ) c„_,, 09 ) is independent of s. Hence it may be assumed that G = Ô and Y = 177. It follows from (3.5) that Y is serial and 17-c_, = 631 U', where U is an indecomposable k[C„_ 1 1 module in b„_, and x runs over a cross section of T(bc,_,) in N 1 . Thus Yc = Y0 El) A, where Vo U and A is a sum of modules in blocks other than bd _,. Hence it may be assumed that G = Now (5.5) implies that ( n„ ,, Os ) 1 for all s. This yields the result. 0 ),,

LEMMA 5.7. All decomposition numbers in B are 0 or 1. PROOF. Suppose this is not the case. Then there exists an irreducible character x with 0, = 2x + n' for some i, where n' = 0 or n' is a character. If 09 is any irreducible character of then (2,vc„ 09) is even contrary to (5.6). El

LEMMA 5.8. Let X be an R-free R[G] module in B such that XK is irreducible and )? is indecomposable. Then one of the following holds. (i) fCic is irreducible. (ii) There exists a principal indecomposable R[G- ] module U and an exact sequence

with YK irreducible.

5]

PROOFS OF

PROOF. /K (XK, XK) = 1.

(2.11)-(2.17)

IN CASE

K=

293

Since X is not projective. this implies that

dim R InG,(1),Hom R (X, X)) = 1.

Thus by (1.5)(iii) dimk H°(G, (1), Hom R (X, X)) = 1. =

Since X = X by (111.5.8) and (1.4)(i), it is serial. Thus there exists a principal indecomposable R[6] module U and exact sequences

As g is R -free, Y is a pure / (U) = p°. Hence /(g)-s_ p° /2 and let Yo = Y if / (V) p°/2. H° (G, (1), Hom R (Y0 , Y0)) is a

submodule of U and so V — V._By (2.6) or /(Y) - s. p° /2. Let_Y0 = g if /(g) p°/2 Thus in any case / (V0 ) p°/2. By (111.5.12) nonzero cyclic R module. By (3.10)

= Tr)(HoniR (Y0, Y0)) = 1. 1".(FlomR (Y0, Y0)).

Thus by (111.5.15) - (1),HomR ( 170, Y 0)) R. H°( G,

H o mREGJ (Y0,

Hence rank R Hom RtG) CY - 0, = 1 and so , - 0, Y Yo is irreducible. 0

(tY OK,

(Y0)1,)

=

1. Therefore

LEMMA 5.9. Let 0 k

a — 1. Suppose that G = Ck . Let G ° = Ck /D.-1. a —1 by 13,C H, and H, 1 Dc, _ 1 = CG o(D, 1 a_ 1 ). Then C, -(1H„ H,/C, is a p- group and TH,(b;)= C, for z e N. The following statements hold. (i) b <-* b defines a one to one correspondence from the set of all H,-orbits of blocks of C, which correspond to bk and the set of all blocks of H, which correspond to bk. (ii) Let z e N. Then O H, is a one to one correspondence between the irreducible ordinary or Brauer characters respectively in b; and those in Define

H for O

j

(b;)",. (iii) If x is an irreducible character in b„ x E D, — D,,, for i j and b; for some z E N, then

d("", x, tlf,) = d(xz ,

X,

(iv) Let x = be an irreducible character of G which has kernel and let x ° be in the block V, of G °. If x E D, — D,,, for 0 and z e N, then

d (x°,

(OD", = d(, X, tlf;).

j

in its a —1

294

CHAPTER VII

t5

PROOF. The first statement, (i) and (ii) follows from (V.2.5) and (V.3.14). The remaining statements are consequences of the Second Main Theorem on blocks (IV.6.1). 0 At this point (2.11) and (2.13) have been proved except for the fact that dA is independent of A. This will follow once (2.12) is proved. Observe that (2.15)(i) is an immediate consequence of (2.15)(ii). Also (2.15)(iii) is an immediate consequence of (2.12) and (2.15)(ii) since for 1 i e, cp, occurs with multiplicity 8„ + 8,, = 0 for suitable u, y in the generalized character =, 8.x.. 80x + The proofs of (2.12), (2.14), (2.15)(ii), (2.15)(iv), (2.16) and (2.17) will be given simultaneously by induction on J G I. If G = Cc, these results are true by (5.1). Suppose that G Co. Assume first that G = Ck for some k with 0 k a — 1. Without loss of generality k may be chosen so that G Ck_,. Since G C O 3ID1> p. Thus by induction all the results are true for Ck-i and also for G ° = Let ° be the subset of A consisting of all those characters in A which have 13o_ 1 in their kernel. By (V.4.5) there exists a unique block B ° of G ° with B ° C B. By induction there exist irreducible characters xi, x,„ A e A ° in B and these are precisely all the irreducible characters in B which have D, in their kernel. Let 0 1 , OA, A E A be the irreducible characters of Ck_, in bk-I. For any element x in G let xp, x denote the p-part, p'-part of x respectively. Let e,) , , ek be the signs defined in (2.17) for the group G ° . Let Eic—i be the signs defined in (2.17) for the group Ck-i. By induction and (2.15) the values of u = 0, , e are determined in all smaller cases. Let H be defined as in (5.9). By induction there exists y' = ± 1 such that the signs for Lq, multiplied by y' yield the signs for (b T' )° . Thus if a is the nonexceptional character of Hk_i/D-] in (b = (b°k then the generalized decomposition number of x, at for x E D — Dk is y'S times the generalized decomposition number of a, where 8 = 8, for b. If x E D then (5.9)(iv) implies that the sign E which equals the generalized decomposition number of a in (b kH 'f,')° at must equal the generalized decomposition number of a in b:'±V at x. By (5.9)(ii) a = t3 lik ', for f3 E bk _, and so by (2.17) applied to Ck _,, the decomposition number of f3 at x is equal to E. By the observation following (2.17) we may assume that = O. Hence there exists y = ±1 such that E; = ye, for 0 j k — 1. If AEA U {1} then induction applied to Ck-i yields that if is an

5)

PROOFSOF (2.11)-(2.17) IN CASE K =

295

irreducible constitutent of the restriction to D of an irreducible constituent of )t c in bk _, then 0

if x„

O xp)0,-,(x)

if xp E Dk-1,

G,,_, D,

if ; ED ; —

0 i k —2.

As C G (x) C Ck _, for x ED—D k this implies that

ax„)(//,?_1(x) 4- z (xp)tif‘z(xp')

if

( G D,

if x„

E

Dk,

if x„

E

D, — D,,, C

j

k — 1.

If x,, E D, — D,,, for some j k then by induction and (5.9) (iv) E,804- (xp)Ok (xp'),

A'A (x ) =

where 5 0 is defined by induction on G ° . If k < a — 1 then since G ° = Cd)(Dk /D‘,_,), induction implies that 3 0 = 1 and E, = 1 for k i
if xp O G D,

x,(x = )

if x,, E Dk ,

4- (xp )(Pk (x,,,)

-

I

6;80

' (x,,)( x

.)

if x,, E Di —

(5.10)

i

k — 1.

If p° = 4 let 50 = — 1, 6 = 1. In general 5, can be defined by induction and 61+ So = 0. Since e: = e,), for 0 i k — 1, comparison of the last two equations yields that for A e U {1} if x, EGDA., X.k(x) — Y81(n(x)= otherwise. For p. E A define

10

6(xp)[Xl(xp') — y66 (x)]

71,,.(x)=

where

6

if xf, CGDk, otherwise,

is an irreducible constituent of the restriction to D of an

296

CHAPTER VII

[5

irreducible constituent of Ac in bk_I. The previous equation implies that if A E A ° U {1} then XÀ= 1À + y6,0. For p. E A define x, = î + y8, 0,6 . It follows from the characterization of characters (IV.1.1) that 17,, is a generalized character. Thus also x„ is a generalized character. Direct computation shows that x, satisfies equation (5.10) with A replaced by p, and replaced by In particular x, (1) > 0. We next compute 11,36,112. Let yS e D, — D„, and let A be the set of all elements in G whose p-part is y '. If i k then EA I )(M. (X )1 2 = Z A I XI( X )1 2 by (5.10). Suppose that i
E A

(x)lz =

E

is

=

1

.P' -element in C,

I CEzENie z ()I s )0X ) i

z. Xisi

If z and z, are in different cosets of C1 then qi and t//,‘ , are in different blocks. Hence

E

EA

1,36. (x)I2

This implies that 11X, ,,112 =

E E, 16 cys)121101'2 =sEA E I xi(x)12 .

z

XI II 2 = 1. Hence x„. is an irreducible character for all /ICA U{1}. Equation (5.10) shows that each x„. is in B, the x, are all distinct and (x) (x ) for all p'-elements x and A,p. EA. Thus (2.12) holds. Furthermore (5.10) implies that (2.17) holds for G with signs, say E': for i k and E7 = e . 0 i k — 1. (Recall that for G = Ck, 6 1 = 1 and 8o = —1.) Since x, (1) = x,(1) is the degree of tik it follows that x„ is irreducible as a Brauer character. This implies that (2.14) and (2.16) hold. since AE,% „Jo , x, is the unique principal indecomposable character in B also (2.15) is true. Thus all the required statements have been proved in case G = Ck for some k with 0 k a —1. Assume next that G = N-1 . Let A 0 be the set of irreducible characters of Co in bo which do not have D in their kernel. Let 0 1 , 0,„ A E A o be the irreducible characters in b- 1 . It follows directly from (2.17) applied to C„_, that T(br,_,)1C, permutes the set {OA } in (p° — 1)/e orbits each of length e. Thus if p. = À N E A let x„ = 0. Then by (V.2.5) there are exactly I A I = (p° —1)/e irreducible characters in B whose restriction to has an exceptional character as a component. Since B has e +1/11 irreducible characters by (5.3), there must be e distinct constituents of In. Call these x,,...,x,. Since 0, is irreducible as a Brauer character, (1.3) implies that = x. and each x. is irreducible as a Brauer character. Furthermore

5]

PROOFS OF (2.11)-(2.17) IN CASE K =

297

xu (x) = cp„ (x) for all p'-elements x for u = 1, , e after a suitable rearrangement. If I A I 1 then the characters x,, A E A are the exceptional characters. It is clear that l(k) = 1 for 1 u e. Thus 8„ = 1 for 1 u e. If A E A then x,(x) = q , (x) for all p'-elements x, since 0 1 = 0, on the set of p'-elements and so O(x) = 9,?(x) for all p'-elements x. Thus — Xi + AAXAA for 1 i e. Furthermore /(go) = p° —1 and so 80 — — 1. This completes the proof of (2.12), (2.14), (2.15) and (2.16) in this case. Now (2.17) follows from the definition of induced character and the fact that (2.17) holds for C,. We may now assume that G N._,. Let x be a character in B such that x 0 for some principal indecomposable character (1). Let X, Z be R -free R[G] modules such that XK and ZK afford x and X and 2 are indecomposable. Then XK ZK and so gand 2 have the same irreducible constituents. Thus go and 26 have the same irreducible R[6] constituents. By (1.5)(i) there exist projective 1 [6] modules M1 , M2 in B- such that g Gin and 2 M2 have the same irreducible constituents. Hence 1(X)- I() (mod p °) by (3.5). Since 1 1(X), 1(Z) p° — 1, this implies that /(g)= 1(i) is independent of the choice of X. By (1.3)(i)

e

7

7

x(1) /(k)tp(i)

(mod p").

Suppose that x is irreducible. Let Y be defined as in (5.8). Then (5.8) implies th at one of the following occurs. (i) 1(X) = 1. XK affords a non exceptional irreducible character in 1. (ii) 1(*. )= p° —1. YK affords a nonexceptional irreducible character in B(iii) /(k)= e. ji"K affords an exceptional irreducible character in P. (iv) /(5-()= p° — e. YK affords an exceptional irreducible character in 1. Define an irreducible character in B to be exceptional if case (iii) or (iv) occurs with e/1D I— 1. Define it to be nonexceptional if case (i) or (ii)

occurs. Suppose that IA I X 1. Let 0,,„ À E A be the exceptional characters in ii. For A, m, E A, OA — O. vanishes on all elements whose p-part is not conjugate in 6 to an element in D — (1). Since 6 contains the centralizer of every element whose p-part is conjugate in 6 to an element of D and since two such elements are conjugate in G if and only if they are conjugate in 6, a simple computation shows that ((0, — 0, )°, I RUA

—

OF, ) G 112 = 11 14

=

— U 0,, — p,

112 = 2 for A X

CHAPTER VII

298

[5

and 40, — 0,) G 1(x)-= (O, — 04(x)

for x E

with x„ conjugate in Ô to an element of D. A standard argument (see e.g. Feit [196711] section 23) shows that there exists a sign 5 = -±- 1 and irreducible characters x,, À G A such that (OA —8(x, — x„). Since (x, x„ )(x ) = 3(6A — 0„)(x) for x E Ô with x„ conjugate in Ô to an element of D, it follows that the higher decomposition numbers for x, and tifk for some k with 0 k a —1 are not zero. Hence by the second main theorem on blocks (IV.6.1) x, is in B for all A E A. Furthermore (x,),.; = 80, + c E,E , 0„ + F where c is some integer and (F, 0,)= 0 for all p, E A. Possibly c and F depend on A. Let X, be an R -free R[G] module such that (X, )K affords x, and 5-C,, is indecomposable. Let Y, be defined as in (5.8). Then (-A )K affords OA if 5 = 1 and (Y,)K affords 0, if 8= — 1. (If IA = 2 then 8 is not uniquely defined. Thus 5 can now be defined so that 8 = 1 if and only if p°12.) Therefore x(1) 5et//(1)

(mod p°)

(5.11)

in any case. Since x, and x, have the same irreducible Brauer constituents, it follows from (1.5)(i) that /(X- ,)---- 1(fC„) (mod p°) and so

100 = 1 (k )

(5.12)

for A, p. E A Suppose that x is an exceptional character in B with x x, for all A E A. Let X be an R -free REG] module such that XK affords x and .g is indecomposable. Define Y as in (5.8). Define Y, similarly corresponding to X,. Then Hom R[G] (X, X,) = (0) for all A G A. Thus by (1.5)(iii) and (111.5.12)

H°( G-, ( 1), Hom R (g, g,))= WO, (1), Hom R (Y, Y,,)) = (0). Let Yo, = g, if /(k )= e and let Yo, = Y, if /(X---- ,)= p° — e. Then in any case l(f'0A) = e. Let Yo = X- if Yo, = g, and let Y o = Y if YoA = YA • Therefore / ( Yo) = e or p° — e. Thus / ( V0 ) + p°. Hence by (3.10) and (111.5.15) HomRroj(Yo, Yo, ) (0) and so ./K ((Y0)K, ( Yo, )K) = (0) for all E A. This is impossible as ( Yo)K must involve some constituent which affords an exceptional character. We have shown that B contains exactly I A I exceptional characters if IA I 1 and they all have the same degree. Define 50 = 5. By (5.3) B contains exactly e nonexceptional irreducible characters if I A I 1 and

61

THE B RA UER TREE

299

e + 1 nonexceptional irreducible characters if I A I = 1. If X0 is an R -free R[G] module which affords EA E XA, where I A 1 and 5-0 0 is indecomposable then (1.5)(i) implies that

I(fc.)-E 1(x-,)- so

(mod p°).

(2.12), (2.14), (2.16) and (2.17) are now immediate by (5.8), (5.11) and (5.12). It remains to prove (2.15). If e = 1 then (2.15) is clearly true. Suppose that e > 1. If (2.15) is false there exists i and u, v with 0 ---ç_u
6. The Brauer tree LEMMA Qp

6.1. For 1i 6, 1 u 6, Qp

A

QP ( C

) C Q, (cA). If I A

1 then also

).

PROOF. If is a p-rational character then Qp (i) is in the field generated by all ci)„ and so is in Qp (, ) by (2.9). The result follows from (2.17). 0

6.2. Let 1 ,5. u, y 6, u/ v. Suppose that x'u = r„ for some automorphism a of Qp (.). Then S u = Su and cannot have a common

LEMMA

constituent as Brauer characters.

PROOF. Let X u be defined as in (2.14) and let X„ = X. Then /(ku )= /(k) and so 8u =- Su (mod /3'). Since 6 2, p° >2 and so Su = 8u . The result follows from (2.15). 0 LEMMA 6.3. Suppose that I A I/ 1. Then Q p (i.), where 1 ,-5. u 6, À E A.

for any automorphism u of

300

[6

CHAPTER VII

PROOF. Suppose that Since o- permutes the irreducible Brauer characters of each CI, it follows from (2.17) that all the higher decomposition numbers of À are -± 1. Then (2.17) implies that p = 2, ê =1 and = Thus if x E D — D i then as 80 + 6, = 0, i(x) = e08,1No : Col (Po(1) = — and so

(x)'.

0

_(x).

Define the Brauer graph of B as follows. There is one vertex for each u = 0, e and one edge for each i = 1, . . . , e. The vertex corresponding to u is on the edge corresponding to i if and only if We will later define the Brauer graphs of a block of F[G] with cyclic defect group, where F is any finite extension of Q,„ and need not satisfy condition (*) of section 2. See section 9 below. LEMMA 6.4. If K = k the Brauer graph of B is a tree, i.e. it contains no closed paths. PROOF. Given i there are exactly two values of u with d, 0 by (2.15). Thus each edge has exactly two vertices. By (1.17.9) the graph is connected. As there are e edges and e +1 vertices there are no closed paths. D LEMMA 6.5. Let T be a connected tree and let o-, p be automorphisms of T which fix no edge. (i) ci fixes at most one vertex in T. (ii) If crp = po- and o- (P)= P, p(Q)= Qfor vertices P and Q then P = Q. PROOF. (i) Suppose that u fixes vertices P and Q of T with P X Q. Then u fixes the unique path from P to Q and so fixes some edge contrary to assumption. (ii) Since crp (P) = per (P) = p(P)it follows from (i) that P = p (P). Thus P=Q by (i) applied to p. THEOREM 6.6. If IA' 1/1 then Q, ("„) =Q. (Or ) for 1 --5 u ê, 1 i ê. If 1=1 then the notation can be chosen so that Q, = Q, (çô, ) for 1 u e, 1 i e. PROOF. By (2.9) M = Q,, (<6, ) is independent of i. Choose k with M C such that k is a Galois extension of Q,, and M is the maximal unramified

7]

PROOFS OF

(2.11)-(2.19)

301

subfield of k. Let 6- be the Frobenius automorphism of k and let a- denote the restriction of 6- to M. Thus (a- ) is the Galois group of M over Q,,. Let T be the Brauer graph of By (6.4) T is a tree. Let o- also denote the automorphism of T defined by o-. Hence o- fixes no edge of 'T if o- ' /1. By (6.5) there is a vertex P of T such that for all j with o- i# 1, the set of vertices fixed by aa is either empty or consists of P. If I Â / 1 then P corresponds to u = 0 by (6.3). If I A J = 1 choose the notation so that P corresponds to u = O. Thus for 1 u e. Hence Q() = M for 1 u e by (6.1). D

IA

COROLLARY 6.7. If the notation is chosen suitably in case = 1 then in any case for 1uê the Schur index m Q ,(X.)= 1 and for 1ue x. =„ where {X .,} is a set of pairwise distinct algebraically conjugate characters.

By (IV.9.3), (2.13) and (6.6) m Q, (î .) = 1. The second statement is an immediate consequence. El

PROOF.

COROLLARY 6.8. The Brauer graph corresponding to the block B is a tree. PROOF. The

Brauer graph is connected and has e +1 vertices. By (6.6) and

(6.7) it has e edges.

0

The Brauer graph will also be called the Brauer tree.

7. Proofs of (2.11)—(2.19)

By assumption (2.11) is true. Since (2.12) has been proved in case K = k it follows in the general case from (6.6). Since (2.13) has been proved for K = k it follows from (6.2) and (6.7) that du, = 0 or 1 for u# 0, 1 i e. Statements (2.14)—(2.17) apply only to the case that K = k and so have been proved in section 5. By (6.3) and condition (*) of section 2, is not conjugate to any other character of A for À E A. Let mA denote the Schur index of ,k'A over K. Let A = f3(1),...,ii(r) be all the /i [G] blocks conjugate to n. Then „ o) for E A, x, = m, Er,_,i,°) where x, is an irreducible character in nu). PROOF OF (2.18).

302

CHAPTER VII

[8

There exists i with 1 i e and d o ,# O. By (2.13) do, = m,. Thus 1IXA ne,r = c1r. The result follows as do, is independent of A E A. D

12 =

PROOF OF (2.19). (i) This follows directly from (6.6) and (6.7). By (1.17.8) (ii) Let 0, =

7112

— x

d,

if 1

u e,

do; = m

IIXAII 2 d

By (2.13) and (6.7) L = 0 or 1 for 1 u e. By (2.18) do, = (m/mr)d o , and so d0 = 0 or m/m o r. Since oi is the sum of m algebraically conjugate principal indecomposable characters it follows that m = rê/e. Thus doi = 0 or élem o . The result follows from (2.15). 0

8. Proofs of (2.20)—(2.25)

LEMMA 8.1. Suppose that d„, 0, d„,# 0 for some i, u and S(U,)=

v. Then Rad U =

n

PROOF. i ( a)=1( )+1( x) as 0, = auix. + The existence of the Brauer tree implies that L, is the only common irreducible constituent of and Thus if (U1 ) .k,. fl f‘,,, then I, occurs with multiplicity at least 2 in both X, and and so u= y = 0 by (2.13). This contradicts the fact that u# v. Thus f‘,4 n = 5( U,). Hence /(.k, + X.,)= /(0)— 1 and D so Rad = + LEMMA 8.2. Let s t with du,/ 0, d.,/ 0 for some u. If d,„/ 0 then S 2(0; )/S (Q) Ls L, unless j = sort, d01 0 and do, = 0 for all i j. (The last condition is equivalent to the fact that x, (x)= p, (x) for all p'-elements x in G.)

PROOF. Assume that S 2 (0;)/S(0;) ----- L, ED L,. By (2.15) and (6.2) there exists v u with cl„,/ O. By (3.19) and (8.1) L, ED L, S 2 (X„,)/S(X„,)(1) 5 2 ()/5

Suppose that X., # S(X„) ). Then L, say, is a constituent of both X., and X., and so j = t by (2.19). Hence L, is a constituent of X., with multiplicity at least 2. Thus u = 0 by (2.13). Let A E .4. By (1.17.12) there exists a pure

8]

PROOFS OF

303

(2.20)-(2.25)

submodule of X (); such that (X,v ),, affords XA. Thus L ; C A C fCo; and so S 2(&) C VP-000. Since S 2()Z.1 )/S()Z. 1 ) (0), S 2(C01 ) has a composition series with composition factors L1 L 1 . Thus S 2(&) C S( 01 ) by (2.13). Hence fC,,i S(Z, J )— L ; and the second case of the lemma holds. Suppose that fC. = 4. By (2.19) either y # 0 or y =0 and e = ê =ID1— 1. In either case x. is the sum of m algebraically conjugate characters each of which is irreducible as a Brauer character. Let denote the irreducible R[G] module which affords X'. as a Brauer character, then [7 affords X' By (2.14) 1(/:;') = 1((iii r)= 1 or p" — 1. Since L, (DA by (2.9) it follows that 1(L1 )= 1 or p" —1 contrary to (3.19). 0 j

,

,EBL7

PROOF OF (2.20). In view of (8.1) it only remains to show that ,?„, is serial. Suppose it is not. Let n 1 be the smallest integer such that S" -"()/S"()Z. ) is reducible. Then fC,,,/S" -j () has a submodule M such that S(M) is irreducible and M/S(M) is completely reducible but not irreducible. Let S(M)L 1. By (2.19) M/S(M)— L s L, for some s/ t. Since d 1 0, dus # 0, du, X 0 (8.2) yields a contradiction as M is isomorphic to a submodule of Q. (2.21) AND (2.22). Clearly (2.22) is a reformulation of (2.21). Thus it suffices to prove (2.22). By (2.20) LT, is serial for 1 i ---5e if and only if there exists u with du, X 0 and X., irreducible. This is the case if and only if the Brauer tree of B is a star and the exceptional vertex, if it exists, is at the center. The result follows from (1.16.14). El PROOF OF

PROOF OF (2.23). If the result is false then by (2.20) there exist i, i', such that X. has a factor module M with S(M) -- L„ S 2(M)/S(M) -- L s and X., , has a factor module M' with S (M')— L„ S 2(MWS(M)— L, for s/ t. Then du, X 0 and du, X O. Furthermore S 2( Q)/S( L s L, by (3.18) contrary to (8.2). fl PROOF OF (2.24). Let d„, O. Since x. is real valued (X),, affords x. and fC*„, is indecomposable. Thus by (2.20) )-C., for some j with du, X O. Let cp ( ' ) ..... cp ( s) be the ordering of the irreducible Brauer constituents of defined in (2.23). Then by (2.23) 'p , cp" is a cyclic permutation of Thus if cp ( " ). = cp") then cp (" ) = cp ( n -k). Hence cp ( "" if and only if n — t t + 1 (mod s) or 2t n — 1 (mods). There are at most two solutions to this congruence which implies the result. E .

304

CHAPTER VII

[8

8.3. Let V, W be 17?[G] modules with S(V)-----. L„ If L, occurs as a composition factor of W with multipicity n then IA (W, nIA (4,4).

LEMMA

PROOF. It may be assumed that V C L7,. Thus by (1.16.4)

(W,

/g (W, U,)=-./g = nI (L i, Li ).

W*)= nIA(LI,L1)

0

PROOF OF (2.25). By (2.9) and (2.19) l()= l(*01 ) where is an P. -free [G] module which affords Thus l(5- ) -1()?,0 )is 1 or p — 1 by (2.14). There exists j such that L, is isomorphic to a submodule of We will first show that ,Z is determined up to isomorphism by 1()?) and j. If G = this is clear by (2.4). Thus it may be assumed that G/ = By (III.5.13)_ H°(G, (1), Homg (L„ .k))/ (0). Thus by ,?)) / (0). (1.5)(iii) 1-1 °(6, (1), Hom, Suppose that 1 ( ?)= 1. Then Homg (f„ ))/ (0). Hence .g T(i.:,). Thus by (2.5) )? is uniquely determined up to isomorphism. Hence also .g is uniquely determined up to isomorphism. Suppose that 450= p° — 1. Let P, Po be principal indecomposable 1[6] modules such that the following sequences are exact

0—> Y--)P—>k—*0 O --> Y0 -->

—> O.

Hence 1( Y) = 1. By (111.5.12), H 0 (6, (1), Homg ( Yo , Y)) / (0) and so Hom g ( Yo , Y) (0). Thus Y T( Y0). By (2.8) T(Y0) is determined by s(f,). Thus by (2.5), T( Y0) Y is determined by L,. By (2.8) SOO is determined by Y, and hence by L. Thus )? is determined up to isomorphism. Consequently )? is determined up to isomorphism. Thus in particular .k is serial. It remains to show that if )? ,Z„, then X By definition the multiplicity of SOO as a constituent of ,Z is 1 if u 0 and (1 D 1— 1)/e if u = 0. Hence (8.3) implies that

m

if u/ 0, }

IF (X , X) -,

m(IDI— 1) if u = 0,

= au,X “ 112 = rank, Hom REG) (X, X01 ).

This implies that I A (., fC)= rank R Hom R(GI (X, X u1 ). Thus there exists g E Hom REGI (X, X. 1 ) such that g induces an isomorphism g from X to k1. As X„; is serial, Rad X., is the unique maximal submodule of X.» Hence g

9]

SOME PROPERTIES OF THE BRAUER TREE

305

is an epimorphism. Therefore X., X/Z for some module Z. As both X., and X are R -free this implies that rank R Z = rank R X — rank R X„ = O. Consequently Z = (0) and so X X. 111

9.

Some properties of the Brauer tree

Let Ko be an arbitrary finite extension field of Qp. Let K be a field which satisfies condition (*) of section 2 such that Ko C K and K is a totally ramified extension of K 0, i.e. K, and K have the same residue class field. Let Bo be the block of K 0[G] which corresponds to B. Then the Brauer tree of Bo is defined to be the Brauer tree of B. Thus every block of K0[G] with a cyclic defect group has a Brauer tree for any field Ko with [Ko :Qp] finite. It is sometimes convenient to associate the irreducible K 0[G] module to a vertex of the Brauer tree rather than the character afforded by such a module. It is convenient on occasion to label the exceptional vertex if there is one. If x.,x 0 correspond to distinct vertices on the same edge then by (2.19) and (2.25) & + & = O. It follows from the results of section 2 that if G = then the Brauer tree is a star with the exceptional vertex, if any, at the center.

G =N

It will be shown in Chapter X that if G is p-solvable then the Brauer tree of a block B of G with cyclic defect group is a star. The converse of this statement is false. For instance the principal 13-block of Suz(8) has the following tree. The degrees are written by the vertices. 0

1

o

14

64

°14

o

35 ex

CHAPTER VII

306

[9

A more spectacular example is given by the principal 13-block of the automorphism group of Suz(8). 014 0

014

1

014 014 01

If p is any odd prime then the tree for the principal p -block of PSL,(p) looks as follows. There are l(p — 1) edges.

A-

p 1) 1

pl

- -

o

o

" •

4 (p +O l) ex

•

The sign is chosen so that the exceptional character has odd degree. It is not known whether every tree is the Brauer tree for a suitable block of some group. No tree has been shown not to occur though it seems likely that most trees will not occur. (Added in proof By using the classification of the finite simple graphs it can be shown that most trees do not occur as

Brauer trees.) Let e be an integer and let p be a prime with p 1 (mod e). Let G be the Frobenius group of order pe. Then the Brauer tree for the principal p-block of G is a star with e edges. Thus every star occurs as a Brauer tree. The only trees with 1 or 2 edges are stars. Thus they occur as Brauer trees. The principal 7-block of PSL 2(7) has the Brauer tree which is a line segment with 3 edges. Thus every tree with 3 edges is a Brauer tree. The next two examples show that every tree with 4 edges is a Brauer tree. The principal 5-block of S, has the following Brauer tree. 1

4

6

4

1

The principal 5-block of Suz(8) has the following tree. 014

9]

SOME PROPERTIES OF THE BRAUER TREE

307

The next two results were proved by Tuan [1944] in case a =1. See also Yang [1977].

THEOREM 9.1. Suppose that G has a cyclic S e -group. Then every 1i[G] module in the principal p -block is equivalent to an F[G] module where Fe is the field of p elements.

PROOF. As the principal p-block contains the principal character, the result is a direct consequence of (2.9). 0 THEOREM 9.2. The subgraph of the Brauer tree of B consisting of those vertices and edges which correspond to real valued characters and Brauer characters is either empty or is a straight line segment.

PROOF. Let an object be either a vertex or an edge in the tree. Suppose there is a real object in the tree. By (IV.4.9) the set of characters and Brauer characters in B is closed under complex conjugation. Thus complex conjugation defines an incidence preserving map of the tree which sends edges to edges and vertices to vertices. Two real objects in the tree are connected by a path. Thus they are also connected by the complex conjugate path. Since the tree has no closed path it follows that two real objects are connected by a path consisting of real objects. Therefore the set of real objects in the tree form a connected graph. By (2.24) this graph is a straight line segment. E The subgraph of the Brauer tree consisting of real edges and vertices is called the real stem of the tree. The next two results are due to Rothschild [1967] for the case that K = k. He used these methods to prove (2.10). The following notation is needed for these results. T is the Brauer tree corresponding to B. The edge corresponding to L, is denoted by E. The vertex corresponding to x„ is denoted by P. If 1 i e then T {E,} is the disjoint union of two trees. Let these be denoted by To(E,) and T(E) where 70(E, ) contains the vertex Po. n (E,) is the number of vertices in 7,(E,). Thus n (E1 ) is the number of vertices in T which are separated from Po by the removal of the edge E. d(P) is the number of edges on the unique path in T which joins P„

to Po. LEMMA 9.3. (i) For 0

u e, 8„ = (-1)`")80.

[10

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308

(ii) Let 1 j Ç.e and let P = Pu be the vertex on E, which is in T,(E,). Then one of the following holds. (I) 8o(-1)' ) = 8. =1; n(E1 )=1(I:1 ). (II) So( — 1)d ' ) = Su = —1; n(E,)= p`z — l(f ) ). (i) This is an immediate consequence of (2.19) and (2.25). (ii) This is proved by induction on n(E,). If n(E,)= 1 then .g,„ L, where P = Pu. Thus /(f. 1 ) - - S u (mod p') and the result follows from (i). Suppose that n(E,)-1. Let E 1 ,. . . , E. be all the edges in T(E1 ) which have P as a vertex. Hence E 1, , E„ E, are all the edges in T which have P as a vertex. Furthermore n (E,). By (1.5)(i) n (E,) = 1 + PROOF.

S. -=- /(f,)+

2 1(L

1 )

(mod p").

By induction and (i) =

Su

2 n(E,) -

1

Su {fl(E ; )— 1}

(mod p°).

=1

Therefore /(f,,)— Su n(E,)- 0

(mod p°).

The result now follows, as n(E,)--.Ç• e and by (2.7) /(f,) - -s. e or p' —e

e. Choose j with dS 0 and let 9.4. Suppose that ID L1,..., Ls be all the irreducible k[G] modules (up to isomorphism) which are constituents of X 0i. Then one of the following holds. (i) 50= —1. 1(L,)= n(E ; ) , e for 1 - i s and E;=, l(f,,)= e. (ii) 6 = 1. p° — 1(I. 1 )= n(E1 ) - e for 1 i - s and Es,_, fp' — /(f.;)} = e.

LEMMA

E1, . . , E. are precisely the edges of T which have Po as a vertex. Let P1 Po be the other vertex on E for 1 i Thus d(P,)= l and so ( — 1) a(1',) = — 1 for 1 i s. Since E:=, n (E,) = e, the result follows from (9.3)(ii). PROOF.

10. Some consequences 10.1. Suppose that G has a cyclic S„-group (x). Let I(x)I= p". Let V be an indecomposable R[G] module in B and let d be the degree of the minimum polynomial of x acting on V. Then p" - °1(V) - -s. d. If furthermore D .1 G then p (V) = d.

LEMMA

10]

309

SOME CONSEQUENCES

By (1.5)(i) it may be assumed that G = 6- and V = By (3.4) it may be assumed that K = k. By (3.5) it may be assumed that G = Then NG ((x )) = C0 ((x )). Hence Burnside's transfer theorem implies that G = (x)H with H = 0(G). Replacing x by a conjugate it may also be assumed that D C (x). By (111.3.8) every indecomposable R [G] module U in B is of the form U = U,? for some R[DH] module Uo in a block with defect group D. Hence V = V,?, where Vo is an R[DH] module in a block with defect group D. By the Mackey decomposition (II.2.10) (V) > —{(Vø)0 }. Thus the minimum polynomial of y on vo is (Y— 0'0, where d = do. Therefore it may be assumed that V = Vo and x = y. Since H is a p'-group, V,, = M„ where each M, is irreducible. Choose M = M, with MZ (Rad V), Then Ed,=o1 M(y —1) is an 1 [ G ] module which is not in Rad V. B contains a unique irreducible 1 [ G ] module W up to isomorphism by (2.1). V is serial by (2.4). Thus V= M(y — 1)'. Since MI WF., it follows that dim ', M dimA W. Therefore PROOF.

/ ( V)dirru W = dimk V

d dirri M

d dim W.

Hence / ( V) d. Suppose that D
THEOREM 10.2. Suppose that G is p-solvable with 0„ (G)= (1). Let V be a faithful R[G] module. Let x be an element in G of order p" and let d be the degree of the minimum polynomial of x acting on V. Let G,,= Then one of the following occurs. (i) d = p".

(ii) There exists a prime q and a positive integer a such that Go has a nonabelian S0 -group, p° —1 is a power of q and p" — 111 Go i. Furthermore (p" — 1) d < p".

310

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PROOF. Suppose the result is false. Let G be a counterexample of minimum order and let H = Op (G). Since CG (H)-1 G and 0 , (C o (H)) C H it follows that Co (H) C H. The minimality of G now implies that G = Go . Without loss of generality it may be assumed that K = k. There exists a prime q such that x"" - ' does not centralize any -group of H. By the Frattini argument (x) normalizes a S, -group Q of H. Thus the minimality of G implies that H = Q is a q-group. Let V = v, where each V, is indecomposable. Let A, be the kernel of ( V, ),. Since V is faithful ,n A, = (1). Hence Op (GIA 1 )=(1) for some j. Hence it may be assumed that V = V, is indecomposable. Let B be the block which contains V. Let a be the defect of B. Let W be an irreducible constituent of V and let A be the kernel of IV,. Then B covers the principal block of A. By (V.2.3) A is in the kernel of V as A is a p'-group. Thus A = (1) and so W is faithful as (G) = (1). Hence it may be assumed that V = W is irreducible. Suppose there exists an indecomposable 1 [(xP)111 module V, such that V V. By (111.3.8) V V Let d, be the degree of the minimum polynomial of x" in V. Then d = pd, and the result follows by induction. Hence it may be assumed that V 1/ for any R[(xP )H] module V. Thus in particular (x) is a vertex of V and so D = (x) is a defect group of B and n = a.

The inertial index of B is 1. Hence the Brauer tree for B has 2 vertices and one edge. Thus there exists an R -free R[G] module X with X -- V and X K absolutely irreducible. Consequently dimk V G j . Furthermore /( 't-/) = 1 or p° —1 by (2.14). Clearly d p". Thus d < p" as G is a counterexample. Hence Vo has no I[D]-projective summands. Thus Vo A./- by (1.5)(i). If 4%7) = 1 then 17 is irreducible and so x"" - ' is in the kernel of V. This contradicts the fact that V is faithful. Therefore l(f/)= p" —1. By (2.1) dimp V = (p" where W is the unique irreducible k [6] module in É. Therefore p" —1IIGI and so p" —1 is a power of q. By (10.1) d 1(1-/)= p" —1. This completes the proof. 0 The next result includes a theorem of Peacock [1979]. THEOREM 10.3. Suppose that G has a cyclic S e -group (x). Let 1(x)I= p". Let V be a faithful irreducible fz[G] module in B and let d be the degree of the minimum polynomial of x acting on V. Assume that 1.D1--p 2 . Then

dim i V d

(p° —

(p° — p +1).

10]

311

SOME CONSEQUENCES

PROOF. The first inequality is obvious. The last inequality follows from (1.3)(i). Suppose that the middle inequality is false. Then the minimum polynomial of y on V has degree strictly less than p° and so Vo = fi by (1.5)(i). By (10.1) 1 ( 7 ) < p° — e and so by (2.7) l( f/) e. By (2.4) every indecomposable module in É is serial. Thus there exists an indecomposable projective R [6] module P and an exact sequence P—* —> O. Let z = r-t . Then (z) = D„ _ 1 and /7
p >e

2 by assumption. This contradicts the previous paragraph.

E

The next result strengthens a Theorem of Lindsey [1974].

COROLLARY 10.4. Suppose that G has a cyclic S p -group (x) with 1(x )1= 19 " =p2 . Assume that (x)fl (x)z = (1) for z E G, z NG (KX». Let V be a faithful irreducible k [G] module. Then dirri i V p" — (p —1). PROOF. If V is projective the result is clear. If V is not projective then by (111.8.14), V is in a block with (x ) as a defect group. Thus the result follows from (10.3). E The next result is a generalization of (5.6) which generalized a result of Thompson [1967b]. In case D j = p this first appeared in Feit [1967b].

THEOREM 10.5. Assume that K = k. Let Y be an R -free R[G] module in B and let 17 be the character afforded by YK Suppose that f7 = MEl) • El) M,,, where each M is a nonzero indecomposable R[G] module. Let 9 ,, , Op . be all the irreducible characters of C- 1 in b-1. Then .

1( 71c.

0, )1

n

for all s, t.

PROOF. Without loss of generality it may be assumed that Y is indecomposable. If f is projective then eric 9,) is independent of s and the result is proved. Thus by (1.17.11) no M, is a projective 1 [G] module. By (1.5)(i)

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Y c = (j) A , (j) A 2 where A, is projective and A2 is a sum of modules in blocks other than b. If -rb is the character afforded by (A) R for i = 1,2 then ((-q 2)G„ 0s ) 0 for all s and ((7/ 1 ), 0,) is independent of s. By (111.5.8) and (1.4) 1'-7 — — 2t ED • 11-{„ and each /4, is indecomposable and nonzero. Hence it may be assumed that G = Ô and Y = 177. By (111.4.6) there exists an R[T(6„_,)] module Y. such that Y(? Y, (Y)T( b)= Yo e) A where A is a sum of modules in blocks which do not cover b„_, and Y. is a sum of n indecomposable modules. Thus it may be assumed that G = T(b„_,). By (1.3) every irreducible R[G] module remains irreducible when restricted to C„_,. Thus by (3.5)(ii) each (M,) G._, is indecomposable. Hence it may be assumed that G = It suffices to show that (.71, n for all s. Suppose this is false. Thus there exists 0 = 0, for some t with 0)= n, > n. Let M be a K[G] module which affords 0 and let Y, = y n n,M. Then Y, is a pure submodule of Y and so S(V,)C S(Y). Therefore /(S(Y,)) n since every indecomposable fz[C„_,] module in b„_, is serial. Thus Y, is a sum of at most n nonzero indecomposable modules. Thus it may be assumed that Y = Y. Hence YK affords the character n 1 0. Consequently ( n,

rank', Hom R[G ](Y, Y) =

IK (YK

) = n1.

Since HomR[G] (Y, Y) = Inv G Home ( Y, Y), it is a pure submodule of Hom R ( Y, Y). Hence = dirnk HomR IG] (Y, Y)

IK (YK,

) = n1.

Let W be the unique irreducible R[G] module in B. Then IA (M, Mj) = min{/(M), /(M)} as every R[G] module in B is serial. For 1 s p', O., is irreducible as a Brauer character. Hence /(Y)= n. Thus for 1 i n (M„ Y)=

J=1

(M„M; )%s-

/(M; )= /( Y) = n,. 1=1

Hence Y)=

nn,.

Thus n 1 n. This contradiction establishes the result.

D

The next result is due to Green [1974a] in case K = k. It generalizes an earlier result of Alperin and Janusz [1973]. THEOREM 10.6. Suppose that cl„,/ 0 for some u,i with 0 u e , 1 j There exists an infinite exact sequence

e.

10]

SOME CONSEQUENCES

p

,,

f2

p„_,, • •

313

(10.7)

such that each Ps is a principal indecomposable R[G] module in B and the following conditions are satisfied, where [V] denotes the isomorphism class of R[G] modules which contains V. (i) P — Ps , and fs (Ps )— fs+2e(Ps+2) for all s. dt, 01. (ii) {[fs (Ps )111 _ s -s.2e1 = (iii) {[P 2s_ Ili 1 s el = {[P2] I 1 s el = {[Ui l 11 j e). PROOF. There are exactly 2e modules X, up to isomorphism with du,/ 0 since by (2.19) there exist exactly 2 values of v for each j with du,/ O. Let 01 = a 1 X + a„,,x, and let X = U,I X,,,. Then T(g)----- L, is irreducible. Thus X is indecomposable. Since X i, affords a 1 x it follows from (2.25) that X — X,„, , for some j' with d,,/ O. Thus in particular if L, then 0, = au,x. + a1 X for some v. Therefore the following sequence is exact. U,

—

> Xu,

—

*0 .

(10.8)

Consequently the exact sequence (10.7) exists. Furthermore for all s, X,,, for some v, j with du, X O. As there are exactly 2e modules X, up to isomorphism with d,, X 0, it suffices to prove the result for any one of them. For convenience it may be assumed that /(k) = 1 by (2.25). Furthermore after a change of notation it may be assumed that fC,,, W as defined in (2.4). In view of (1.15.6) and (1.5)(ii) the exactness of (10.8) implies that is exact for some projective module 0 in B. . By (2.14) /(X„,)+ /()Z,,) < 2p". _Hence CI is a principal indecomposable module. Thus there exist principal indecomposable modules i's in E-1 and an exact sequence -->

-L> • • • —> .15

,

-

L.->

—> O.

Furthermore j (i) for all s. By (1.4) and (111.5.8) g,„ X all v, j with d 1 # O. Thus the following sequence is exact

where

314

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Choose the notation so that W, = Wa 1— for 0 i e 1. Let V(i, 1) denote the i [Ô] module in 13- such that S(V(i, l))= W, and V(i, /)) = 1. By (2.8) the following equations can be read off for all s. —

= f2s(P2..,)'---- V (s, p° — 1),

=

V(S, 1). (10.9) = , Thus {f s (Ps ) 1 s 2e) is a set of 2e pairwise nonisomorphic modules. Hence also {f„ (Ps ) I 1 s 2e) and (Ps )i 1 s 2e} are each sets of 2e pairwise nonisomorphic modules. Thus every module Xu, with du,/ 0 is of the form f, (Ps ) for some s with 1 s 2e. This implies (i) and (ii). Furthermore for each j there exist exactly two values of s with 1 s such that S(fs (Ps ).----- L,. The proof of (iii) will now be completed once it is shown that for 1 j e there exists a unique value of s with 1 s e such that f2s- I (P2,--1)

L.

By (2.5) and (j 0.9) there is a unique value of s with 1 s e such that Hom (L,, f2,-1(P2,-1)) # (0). Thus by (3.10) applied to O and (111.5.13) applied to G and by (1.5)(iii) there is a unique value of s with HomA(1,,,f2s-1(P2s_1))# (0). 0 Suppose that d, # 0 for some u, i. Green [1974a] has defined a function a on 11, , el as follows. Let P, be defined as in (10.6). If P2, Li, then P2s1-1 '''' U). By (10.6)a is a permutation on {1, ..., el. It also follows from (10.6) that if the notation is chosen suitably then there exists an exact sequence ---> U, -->

U,_,—> • • --> U1 ) —> U1—> U (, ) — X, —> 0.

It is clear that l(f,(P,))+

p°

for all s.

The permutation a = o-,,, depends on the choice of u and i. However the definition of a., implies that if du,/ 0 then au, = a„, in case 8. = 8„. If 60 then cr,,,au, = (1, 2 . e). COROLLARY 10.10. Suppose that d.,# 0 for some u, j. Let a = a,„. Assume that d,1 # 0 for some v, j with S. = Su . Then L, o) . PROOF. Clearly Su = 60 if and only if Xu, =f7 ,-(P 2,) for some s. By definition S(f2,,(P2,,1)) S(/52,) and 1 f2, (P2,±)) T(P2, S(P2,1). Thus if P2, U, then S(f2,41(P2,,_,))----- L, and 7tf2 1(P2s Lro). -

10

315

SOME CONSEQUENCES

]

In Green's terminology the ordered sequence /Jo. ( I),

U2,

U0(2) ,

,

U„ U,() ,

describes a circular "walk" around the Brauer tree accomplished in 2e "steps". Every vertex of the Brauer tree is reached at least once and each edge is transversed exactly twice. As Green also points out, a determines the Brauer tree as an abstract tree as follows. Consider the oriented circular graph as shown. The Brauer tree can be derived from this by identifying each pair of edges which carry the same lable in such a way that the orientations cancel out. a- (1) 1

/a(e) ./e . •

This process can be carried out for any permutation but will not necessarily yield a tree. For instance the identity always yields a star. If e =3 and a is a transposition then one gets a straight line segment. However if e = 3 and a = (1, 3,2) then one gets a triangle which is not a tree. LEMMA 10.11. Let B = Ê be an iZ[G] block with a cyclic defect group

D.

ID i= P d

Let C be the Cartan matrix of B and let Q be the quadratic form corresponding to p d C-I . Let q be the minimum value assumed by Q on the set of all nonzero vectors with integral coordinates. Let

.

Then q = e.

PROOF. By (IV.3.11) and (2.12) fx„ 1 u el is a basic set for 13. Let D ° , C° denote the decomposition matrix, Cartan matrix respectively with respect to this basic set. Let Ct be the form corresponding to p d (C1'. Then Q0 is integrally equivalent to Q and q is the minimum value assumed by Q ° on the set of all nonzero vectors with integral coordinates. By (2.12)

D°

[10

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316

where t = (p° —1)/e. Let J denote the matrix all of whose entries are 1. Then = (D')'D° = I + tJ.

Thus p d (Co)-1 = pdI — tJ. Hence if a = (a Q (a, a)=

E ,-1

then (10.12)

+ tE (a, — a) )2 .

Let No be the number of a, = 0 and let N,= e — N o. The last term in (10.12) is at least N oN, and the other term is at least N. Hence if a # 0 then N, 0 and so N,+ NN 1 1‘1,+ No = e.

Thus e q. If a, = 1 for all i then Q (a, a) = e and so q = e. D THEOREM 10.13. Let B be a block of fZ[G] with an abelian defect group D of order p". Let r be the rank of D and let k(B) denote the number of irreducible characters in B. If r = 2 then k(B)< p d. If r = 3 then

PROOF. Let y be an element of maximum order pc in D. By (V.9.2) there exists a major subsection S(y,É) associated to B. By (IV.4.5) there exists a unique block É° of CG (y)/(y) contained in B. Furthermore D/(y) is the defect group of É`) . By (111.2.13) y is in the kernel of every irreducible Brauer character in É. Hence É and fi ° both have /(É) irreducible Brauer characters. If q(ñ) and q(ñ0) are defined as in (10.11) then q (f3)= q (b ()) by (IV.4.27). If r =2 then A° has a cyclic defect group and so q(A)= I(A) by (10.11). Thus k(B)p' 1 by (V.9.17)(i). If r = 3 then 1(170 k (n o) < pd--` by the previous paragraph as A' has an abelian defect group of rank 2. Hence by (V.9.17)(i) D p 2d- c. This implies the result as c The next result was first announced in Brauer and Feit [1959]. THEOREM 10.14. Let B be a block of 1'[G] of defect d. Let k = k(B) denote the number of irreducible characters in B. d = 0, 1, 2 and k p' for d 3.

Then k p d for

PROOF. Induction on d. If d = 0 the result follows from (IV.4.19). If d =1 it follows from (2.1) and if d = 2 it follows from (10.13). Suppose

11]

317

SOME EXAMPLES

that d 3. If B contains an irreducible character of positive height the result follows from (IV.4.18). Thus it may be assumed that every irreducible character in B has height O. Let S(y, fi) be a major subsection associated to B. By (IV.4.5) there exists a unique block of CG (y)/(y) contained in B. Furthermore DI(y) is the defect group of É. By (111.2.13) y is in the kernel of every irreducible Brauer character in A. Hence n and b° both have 1(14.0) irreducible Brauer characters. Thus by (V.9.17)(ii) k 2 p 2d 1 ( n0 ) p 2a k (fo , where k(Jr) is the number of irreducible characters in ife. Since ifi ° has defect at most d — 1, induction implies that k ( bo) p 2(d-1)-2 as d =2d —2 for d = 2. Thus k2

and so k

p 2d+2(d- 0-2 = p 4d-4 p2d-2.

LI

11. Some examples The following result is due to Dade [1966] and shows that every possible sequence of signs Ek can occur in (2.17).

LEMMA 11.1. For O k a —1 let y k

= ±

1. There exists a group G and

a block B of G with defect group D such that if (2.17) then Ek = y k for 0 k a —1.

Ek

is defined as in

PROOF. For 0 k a-1 let qk be a prime with qk = —1 (mod p°) and 2< q,< • •
318

CHAPTER VII

[11

a-I

MO) x i (x).

X i(XZ

Thus by (2.17)

Ek =

p, .

The /3, can be chosen so that

a

=fl g,

for 0 k

a —1.

CI

DI=

11.2. Assume that I p. Let x be an irreducible character in B. Let Ko be an unramified extension of Q, (x) and let R o be the ring of integers in Ko. Let X be an Ro-free R0[G] module such that X.K. affords x. Then the following hold. (i) X is indecomposable. (ii) Every irreducible constitutent of X is absolutely irreducible. (iii) Let (s,1 (x)) be the matrix representing x in a representation with underlying module X. Let 77-0 be a prime in Ro. Then for given i/ j there exists x such that either s,1 (x) 0 (mod 77(J ) or sr (x) 0 (mod 71-0 ). LEMMA

PROOF. If k is decomposable the matrix representation can be chosen so that s,„ (x) = s,, (x) — = 0 (mod 7ro) for all x E G, where n = x(1). Thus (iii) implies (i). If some irreducible constituent V of )? is not absolutely irreducible then there exists an unramified extension K, of Ko such that if R, is the ring of integers in K, then 17 t, is a splitting field for V. Since V OA R, is completely reducible it follows that if K o is replaced by K, then it is possible to find a matrix representation with underlying module XR , which contradicts (iii). Hence (iii) implies (ii). It remains to prove (iii). Suppose that s,,(x)=s,,(x)= 0 (mod 7r0) for some i/ j and all x E G. The Schur relations imply that if o-,T are in the Galois group of Ko over Q, then

E 5.(x)ox - i)= 8„ I G I/x(1) xEC;

where 8,r = 1 if cr- and T coincide on 0,(x) and 8_ = 0 otherwise. Let Tr denote the trace from K o to the maximal unramified subfield K, of Ko. Thus Tr(s,, (x )) = Tr(s,, (x)) = 0 (mod p) for all x E G and xEC3

Tr(si; (x))Tr(s,, (x

Thus in particular

X(1)

11]

SOME EXAMPLES

IG1 I KO : K11 x(1)

319

0 (mod p 2).

Since D = p, I Ko : K 1 p — 1 by (2.17). Thus 1G 1/ x(1) O (modp 2). However as x is in a block of defect 1 this is impossible. 0 In Brauer's original treatment of blocks of defect 1, (11.2) and related results played an important role. See Brauer [1941c]. In case p = 2 and a > 0, B has index of inertia equal to 1. Thus every irreducible character in B is irreducible as a Brauer character. Hence (11.2)(i) and (11.2)(ii) also hold in this case. The next two results are needed for the proof of (11.5) which is concerned with the construction of examples that show that (11.2)(i) is false in all other cases. (11.5) below is also of interest because it shows that there appears to be no converse to (1.18.2). For the rest of this section K is a finite unramified extension of Qp and R is the ring of integers in K. — 1 } as an Let X be the R[D] module which has {(1 — y)' 11 i R -basis. Let Y be the unique submodule of X of index p. Then { z1 11 i —1} is an R-basis of Y where z,= p(1— y), z, = (1— y)' for 2 i p° —1. = V i e V2 where V, is the vector space over ft spanned by and V2 is the vector space over R spanned by {2,12 i p —1 } . LEMMA

11.3. If a >1 and p/ 2

PROOF.

By definition

then V I and V, are ft[D] modules.

f,(1— y)=z1(1 — y)=p(1— y)2 = 21 (1 — y)=z1(1 — y)=z1,,E

V, for

pa —2.

Thus it suffices to show that 2„._,(1 — y)E V2. Let Y, be the R module spanned by z, and let Y, be the R module spanned by lz, 12 i — 11. Thus = V, for ] = 1,2. pa , y)= E)(-1)s y s = E ( P ")(— osys s s=i p.—,

=

(Pa) (— 1)/(y —1)+1]s. s=,

[11

CHAPTER VII

320

Thus P°

ze_1(

1 y)= E s

r

=1

iy[s( y —1)+1]

( -

s

p--1 ( —1) (y —1)

E P )(-1)ss + E s s=, E

s=,

(

s

s

)(-1y

)(— lys + (1— iy° —(1-1)

P°

=- (y —1)

E()(-1)ss s

s=i

Let

(mod Y2).

,

r

f(x)=(1— x

)

= E P )(-1 x'. ( y

=c S

)'

Thus Pa

— p° (1— x)P — ' = f(x)= E P (—

Hence f'(1)=0 and zp —,(1— y)=- (y —1)[f(1) -± p ° ]

p ° (y —1) (mod Y2 ).

Since a >1 this implies that zp ._ 1 (1— y) E pY,+ Y, and so zp , i (1— y) E /12.

LEMMA 11.4. (i) V2 is an indecomposable R[D] module. (ii) Suppose that p/ 2 and a > 1. Let N = DE be a dihedral group with E = (x) of order 2. Then Y" = Z, Eel Z2 where = Wle W2, W1 is the trivial R[N] module, diMP W2 = p° — 2 and W2 has a one dimensional socle which is the R[N] module whose kernel is D.

PROOF. (i) By definition X/pX is a serial indecomposable 1 [D] module. There is a natural map of X/pY onto X/pX. This map necessarily sends Y/pY onto an indecomposable submodule A of X/pX of codimension 1. Hence dim A A = p° —2. Thus the minimum polynomial of y on A has degree p° —2 and so the minimum polynomial of y on Y/pY has degree at least p° — 2. This implies that V2 is indecomposable. (ii) Let f, =1(1+ x), f2 =1(1— x) in R[N]. Then f Y N C Y N for i = 1, 2 as Y is the unique submodule of X with dim A (X/ Y) =1 and X = J(R[D]). Let Z, f Y N. Then 4/0 for i =1,2. As (Y N )p Y it follows that (Z,) D = Y for i = 1,2. Hence (.7N)D

e

ni

SOME EXAMPLES

321

V, V, ED V2ED V2 where each summand is indecomposable by (i). Furthermore Z, e 22 — Y'" V"ei V.

As dimk Z, =p° —1, dimk 1= 2 and dimk 1/ 21"= 2(p° —2) it follows that 2, = Wile) Wo with dimk W 1 --- 1, dimk W2 = (p° —1) and W2 is indecomposable. Since 1/1"--- (X> exactly one of W,, say W,„ is the trivial k[N] module. Let W1 = W 1 , for j = 1,2. It remains to show that S(W2) is not the trivial krisil module. Let cp be the Brauer character afforded by W2. No irreducible constituent of (Y"), has D in its kernel hence no irreducible constituent of (Z,), has D in its kernel. Thus 1 + cp(x)= O. As q(x)=- —1 it follows from (2.8) that S(W2) is not the trivial ft[N] module. 0 THEOREM 11.5. Suppose that p/2 and a >1. Let q be a prime with q +1= - p° (modp" 1 ). Let G =PSL2(q). Then a Se -group of G is cyclic

of order p° • G has an irreducible character 0 of degree q (the Steinberg character), which is afforded by a Q[G] module. Furthermore there exists an R-free R[G] module x such that XK affords 0 and .)?= M, M2 where M, is the trivial k[G] module and M2 is an irreducible k[G] module with dimk M2 = q —1.

PROOF. The existence of 0 and the structure of the S„-group of G are well known facts. Let D be a Sp-group of G. Then CG (D) = D x H for some p'-group H and !NQ (D): CG (D)1= 2. In particular NG (D)/H N where N is defined in (11.4). Furthermore D n D (1) for z N G (D). Thus the Green correspondence between G and NG(D) is defined for modules with nontrivial vertex in D. Let k be the R[N G (D)] module with kernel H such that k as an R[N] module is isomorphic to Z. Let X be the R[G] module which corresponds to k. Thus k W1 ED vv2. Hence by (111.5.8) k = M ED M2 where A21, W, for i =- 1,2. Let B be the principal p-block of G. Then the index of inertia of B is 2. Since 0(z)= —1 for z E CG (D) — {1} it follows that 0 is in B, this implies that the Brauer tree for B is 1

q q-1

Let cp,,cp 2 be the irreducible Brauer characters in B where cp, is the principal Brauer character. Thus M, affords cp,. The principal block of

322

[12

CHAPTER VII

R[NG (D)] has a unique indecomposable module, namely W2 , which has no invariants and whose dimension over k is congruent to —2 (mod pa). It follows from (III.5.10) and (111.5.14) that a module which affords cp, corresponds to W2 . Thus M2 affords cp 2 . Hence in particular rankR X = q. Since fC has no invariants, neither does X. Thus the character afforded by XK is sum of exceptional characters and possibly B. As rank R X = 0(1) and the degree of every exceptional character in B is q —1 it follows that X, ‹ affords B. 0 The next result which is due to Benard and L. Scott, Benard [1976] will be used in section 13. In some sense it explains why it is necessary to assume that a >1 in (11.3).

LEMMA 11.6. Let R be the ring of integers in an unramified extension of Let X be an R-free R[D] module such that XK iS irreducible. Then

Qp.

.)-C is indecomposable. If furthermore Y is an R-free R[D] module with YK XK then Y----- X.

PROOF. If rank R X = 1 the result is obvious. Suppose that rank R X > 1. Consider the map f :X —> X defined by f(v)= v(1 — y). Then f is an R[D]-monomorphism and IX : f (X)I = det(1 — y). Since X K is irreducible, the characteristic values of y are all the primitive p"th roots of unity for some n. As rank R X> 1, n O. Hence det(1 — y) — = p, where ranges over all primitive p" th roots of 1. Therefore I X : f(X) I = I This implies in particular that .g is indecomposable. Furthermore f(X) is the unique maximal submodule of fC and so f(X) is the unique maximal submodule of X. Thus X is isomorphic to its unique maximal submodule. Suppose that YK X K. It may be assumed that YR = XK. If Y is replaced by a multiple it may be assumed that Y C X. Since PC Y is finite there exists a chain of submodules Y = X,,, C X„,_, C • • • C X 0 = X such that X, is maximal in X,_, for 1 i m. By the previous paragraph X, ----- X,_, for 1 i m and so Y X. LI .

:

12. The indecomposable k[G] modules in B

This section contains a description of all the nonzero indecomposable 1 [G] modules in B in terms of the Brauer tree of B and of its exceptional

12]

THE INDECOMPOSABLE R [G] MODULES IN B

323

vertex. The results in this section were all proved by Janusz [1969b] and Kupisch [1968] [1969] in case fz = fz. Given the results of section 2 for K in general, Janusz's arguments go through without any change. The structure of the principal indecomposable k [G] modules in B is described by (2.20). In this section I D e — 2e pairwise nonisomorphic nonzero indecomposable k [G] modules in B are described. These modules are neither irreducible nor projective. Thus (2.1) and (2.2) imply that up to isomorphism all the indecomposable R[G] in B are accounted for. The fact that the number of modules described is equal to I D e — 2e involves a counting argument due to Dade. In his paper Janusz also shows that given any tree whatsoever, there exists a split symmetric algebra having only finite number of indecomposable modules up to isomorphism such that these indecomposable modules can be described in terms of the given tree. In particular this result indicates that the question of what trees can occur as Brauer trees may be very difficult. If some tree does not occur it must be possible to decide that some symmetric algebra with only finitely many indecomposable modules up to isomorphism is not a group algebra. (See the remark on p. 306.) denotes the Brauer tree. Vertices of T will be In this section TT denoted by P, P, or Q. The exceptional vertex, if it exists, is denoted by Pe „ . E or E, will denote either an edge of T or O. If L, corresponds to the edge E, write L, <-> E. If E = 0 let (0) E. Set t = (ID I— 1)/e. By (2.13) and (2.19) t is the multiplicity of any L, in ,?0, for do, 0, do, O. If du, 0 let X,,, be defined as before (2.20). If d / 0 then the statement dual to (2.20) implies that T(,)--- L, for some j with d„,/ O. Let P be a vertex of T incident to the edge E. Let L E and let x„ correspond to P. Define V(E, P) = X„, where T(,?„,)----- L. By (2.25) V(E, P) is determined up to isomorphism by E and P. By (2.20) V (E, P) is serial. Let E be an edge of T and let P, and P, be the vertices incident to E. If E, is incident to P, we will define modules V(E,, E, E,: n) for suitable integers n. We will allow the case that E, = 0 or E, = E for exactly one of i = 1 or 2 if P, = Pex . Several cases will be considered. We begin with some preliminary definitions. Let V, (E)= Rad( V(E, P,)) for i = 1, 2. Let U be the indecomposable projective R [G] module corresponding to L where L E and let Mo(E) = U/S ( U). Then the dual of (2.20) implies that Rad(Mo (E)) = V 1 (E) ED V,(E). Suppose that E,E,, E, are distinct with E 0 such that

324

[12

CHAPTER VII

PI

P2

0

E,

E

E2

is contained in T. Define modules M, (n) for i =1,2 for suitable integers n as follows. If E. 0, M, (1) = V, (E). If E,/ 0 and P Pe. let M (1) be the submodule of V, (E) such that S (V, (E)I M,(1))<- E,. Since V, (E) is serial M, (1) is uniquely determined by (2.13). Suppose that E,/ 0 and P, = Pe„. Let 1 n t. Let M, (n) be the submodule of V, (E) such that S( V, (E)IM,(n))=-- L where L <-* E, and L, occurs with multiplicity n as a constitutent of V, (E)IM,(n). Since V, (E) is serial M1 (n) is uniquely determined by (2.13) and (2.19). Define .

V(E,, E, E2 1) = M0(E)/(MI( 1

If P, = Pe„ and 1 n

)

ED M2( 1 )).

t define

V(E 1 , E, E2 n) = Mo(E )/(M, (n)

MI ( 1 )),

where {i, = {1, 2 } . Suppose that E l E = E2 with E/ 0 such that P2 = Pe „ and P, E,

E P. E2

is contained in T. If E, = 0 the corresponding edge is missing. Define M1 (1) as above. If L E then L occurs with multiplicity t — 1 as a constitutent of V2(E). If 1 s. n t — 1 let M2(n ) be the submodule of V2(E) with S(V2(E)/M2 (n)) .--- L such that L occurs with multiplicity n as a constituent of V2(E)/M2(n). Define V(E l , E, E2 n) = V (E 2 , E, E : n) = Mo(E )/(M, (1) ED M2 (n)).

In all cases V = V(E l , E, E2 n) V (E 2, E, E, : n) is indecomposable as T(V) is irreducible. If furthermore E, =0 then V is serial as Mo(E)/M, (1) V (E, P,) for i j and so is serial. LEMMA 12.1. Let V= V(E,,E, E2 n) and V'= n') be defined as above where E 0, E' 0, E, E2 and EE. Let L -* E and E'. Suppose that L,/ L,. Then V and V' cannot have two nonisomor-

121

THE INDECOMPOSABLE R[G] MODULES IN B

325

phic composition factors in common unless they are L, and L, and E and E'

have a common vertex.

PROOF. Suppose that L. L, are both composition factors of V and V'. Then c, c„, c, c, are all not zero. If i, j, s, t are pairwise distinct then there exist m, n, t), w such that d„„d,„s / 0, d„,d„,/ 0, ds„ds,s / 0 and d,„,d„„/O. This implies that T contains the following subgraph.

snt

where the notation is the obvious one. This contradicts the fact that T is a tree. Suppose that { i, j, s, t} is a set of 3 distinct elements, say i,j, t with s Then an argument similar to that in the previous paragraph implies that T has the subgraph

contrary to the fact that T is a tree. If {i, /}= {s, t) then clearly E

is a subgraph of

T.

'

0

It will be necessary to consider the following two types of subgraphs of T. Let k be any integer with k 1.

(I)

Po P1 0-0-0 E„

E,

Pk Pk 0

o

+1

[12

CHAPTER VII

326

Po Ea P1 E E,

Pe„ 0

Q

# PX

E,+„ Pk-El Pk 0-0-0

E,

In type (I) Ek are distinct edges. In type (II) Eh-F, and E114 24 1-1 correspond to the same edge for i = 1, s. Possibly one or both of Po, Pk±i may equal Q. Given a graph of type (I) or (II) let A denote either the set of all even integers i with 0 i k or the set of all odd integers i with 0 i k. Given a graph of type (I) or (II) let V, (n,) = V(E,_,, E„ E, 4 1 : n,) where n, 1 and E_, = Ek±l = 0. Let L, -*E1 . Thus in case of type (II) 1, 11-Fi "-j s. Observe that L1# L1 if i/ j but i j (mod 2). The Lh+2,+1-1 for 0 definition of V, (n,) implies that S(Vo(no ))------ L i , S(V k (nk ))---- Lk-I and for 0 < i < k, L,-, L,+1. Choose A. Define X = ED,e.3 V, (n,). Then if OELI, k = 2m,

2L,; _, i=1

e 2L21

if 0E 4 k = 2m +1,

_,)

J-=‘ i

Loe,

E

,

if 0zi, k = 2m,

2L21)eL2rn

(

if 0sZ4k=2m+1.

2L2,)

Lo e J=.

If 2/4 I S(X) there exist submodules L., C S(Vi_ 1 (n)) and /42 C S(V1±1(n1±1)) with L1 1 = L,2=L. Let g, : L1 > 1.12 be an isomorphism. Let Y be the submodule of X consisting of all elements of the form Ei (y1 — gi (vi)) where i ranges over integers with 2L S(X) and y, EL,. Define M = X/Y. In effect M is obtained from X by identifying the submodules /.41 , L12 whenever 2/4 S(X). The module M is the primary object of interest in this section. The module V(E,_„ E, E±, n,) with E_ I #0 and • ± I #0 may be represented by the cone —

:

L

,

0

121

THE INDECOMPOSABLE

ITZ[G]

MODULES IN

B

327

Then X is represented by the direct sum of cones. 14 +2

0

•••

0

0 14+1

0 14 +,

L1+3

Intuitively M is constructed from X by identifying isomorphic modules at the feet of adjacent cones and may be visualized as the following connected graph, where the dotted lines may not exist.

o \o/\,0 0/ L-1

The next three results are concerned with showing that this picture is

indeed accurate. Let X° = M and let ° denote the image of any element or submodule of X in X° = M. We will also write V, (n1 ) = VAn,). Since y (no n Y = (0) for all i, V,(n,). If furthermore T is a subset of A such that V?(Ili). whenever i ET then i + 2 T then {ERE,- (ni)} 0 =

ED,„ Li .

LEMMA 12.2. (i) (ii)

PROOF. (i) By definition Y C S(X)C

Rad( V, (0)C Rad X.

S(V,(0)g_ i Ed

iEd

Therefore T(V, (n,))

L,.

icd

(ii) Clearly S(X)° &cm. and S (X)° C S(X ° )= S(M). Thus it suffices to show that any irreducible submodule of M is in S(X) ° . Suppose the result is false. There exists an indecomposable submdodule Z of S 2(X) which is not irreducible such that S(Z)C Y and z° irreducible. Let T be the set of all i such that z° is isomorphic to a

328

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CHAPTER VII

composition factor of V, (n, ). Z ma/ be replaced by Z n fe,E, V. (it )1. Thus it may be assumed that Z C 610,Er (n.). Since Z is indecomposable S(Z)C Rad(Z). Thus S(Z)= Rad(Z) as S 2(Z)= Z. The dual of (2.20) implies that 1(Z) = 2 or 3. Let E be the edge corresponding Z ° . If Z ° is isomorphic to a constituent of S (Z) then (2.23) implies that E is incident with P. and furthermore E is the only edge of T which is incident with Pe„. Assume that j, j + 2 E F for some j. Then L, + , is a composition factor of S (V, (n,)) and S(V, ±2(n1 ,2)). Thus Z ° and L,,, are both isomorphic to composition factors of V, (n, ) and Vh±2(n,±2). Suppose first that E, and E, +2 do not have a common vertex for any i + 2 E F. By (12.1) L,±1 Z ° . Thus F = {j, J + 2 } or the following configuration occurs in T with s > 2

where possibly E, = In this case F C {j, j +2, j + 2s}. Therefore in (n,) C L,,, Z ° . Since Y n (o) it follows that any case Y n Y n Z Z ° and so S(Z) — Z ° which is not the case since both vertices incident with E,,, are incident with other edges in T. Suppose next that E, and E1±2 have a common vertex. By (12.1) Z ° or L, ±1 . Up to symmetry the following configuration must occur in T.

ED,

p cx o

'

„

If Z° L, +, then F = {j, j +2}. Let Y0 = Y n V,+2(n,+2)). Then (i) Z° . Thus /(Z)- /(Z1+ /(Y1= 2 and so S(Z) Z° . Therefore Yo, Z C Z, ED Z2 where Z, C (1), z2 C V,+2(n,± 2). Both Z1 and Z2 are indecomposable, all composition factors of Z, (1)4 are isomorphic to Z ° and / (Z 1 ) = 1, /(Z2)= 2. Therefore Rad(Z) C Rad(Z, z2) = Rad(Z2) C V,+2(n1 +2). Thus Yo n Rad(Z)= (0) and so Y n z = (0). Hence Z Z ° contrary to the reducibility of Z. If Z(' L, then F C {j, J +2, j + 4} and if j +4E T then L,,, L, otherwise z° L, is not a constituent of V, ±4 (n1±4). Hence

121

THE INDECOMPOSABLE R

[G] MODULES

IN

B

329

Since Pe „ is not incident with E, it follows that z° S(Z). As S(Z) fl (o) it follows that S(Z) .--- L,,,. Thus z n = (0) and so Z is isomorphic to submodule Z' of V, (1)6) V,+2(n 1 + 2 ) with S(Z') n Y Since L, occurs with multiplicity 1 in V,,(n1 ± 2) and L, C S(V,±2(n1 +2)) it follows that Z' C y (1) IED Z 1 , where z, C V,±2(n,±2) and Z 1 Thus Rad(Z') C y (1). As Rad(Z') n Y = S(Z') n Y/ (0) this contradicts the fact that V, (1) n Y = (0). Therefore if i E F then i + 2 F. Consequently {ED, er (n, )}° — (1), Er V?(n,). Hence Z = Z° contrary to the fact that Z is reducible. LI 12.3. Let i E 4. (i) If E. is not incident with Pe„ then L. occurs as a composition factor o f M with multiplicity at most 2. If the multiplicity is 2 then L I S(M). (ii) If E. is incident with Pe, then the multiplicity of L. as a composition factor in M is equal to the multiplicity of L 1 as a composition factor of V, (n1 ). This is n, in case (I) and n, + 1 in case (II). LEMMA

PROOF. If k 2 then 4 = {i} and M = V, (n,). The result is clear. Suppose that k > 2. (i) As E, is not incident with Pe„ , L, occurs with multiplicity 1 in V, (n,). If j/ i and L, is a constituent of V, (n,) then the following subgraph of T must occur (up to symmetry) with P/ Pe where E,, may not be there. „,

E,

p o

In any case this implies that L, I S (X) and so L, I S (M) and the multiplicity of L, -in S(M) is 1. (ii) As E, is incident with Pe„ , one of the following subgraphs of T must occur (up to symmetry) E, o Pe.

o

E, o

o

Pe„

330

CHAPTER VII

[12

In the first case L i occurs with multiplicity it, + 1 in V, (ni ) and multiplicity 1 in V,, 2(n1 , 2) and in the socle of both. Thus L occurs with multiplicity ni + 1 in M. In the second case L, occurs with multiplicity ni in y (n,) and does not occur in any V, (ni ) with j i, j Ez.E LEMMA 12.4. M is indecomposable.

PROOF. Suppose that M = M, ED M2. For s = 1, 2 let f, be the projection of M onto Ms . For any i E 17 (n,)C V(n, )f V(n,)f,. We will first show that VAn, ) V?(n,)f, for s =1 or 2. Suppose this is not the case. Hence V?(n,)f,/ (0) for s =1, 2. Let Ws = VAn,)f, for s = 1,2. Then 2L, J T(14/1 (i) W2) and so L, occurs with multiplicity at least 2 as a constituent of M. Suppose that E, is not incident with 1='„ . Then L, S(M) by (12.3) and so W. L, for s =1 or 2. Since L,Z S(V?(n,)) it follows that VAn,)f 17 (n1 ) for s/ t. Suppose that E, is incident with Pex . Let n be the multiplicity of L, as a composition factor of M. By (12.3)(ii) the multiplicity of L, as a composition factor of y (n,) is n —1. Let k be the smallest integer such that L, occurs with multiplicity n — 1 as a constituent of S k (Rad( V, (n,))). Since Rad( ED W2) = {Rad( V, (n, ))1 f efRad( V, (n, ))} f2 and this is a sum of at most 4 serial modules it follows that for s =1 or 2 S (Rad( 14; )) (0) and L, occurs with multiplicity n —1 as a constituent of Rad( 14/, ). Hence L, occurs with multiplicity n as a constituent of W. Therefore L, does not occur as a constituent of W, for s# t by (12.3)(ii) and so W, =0. Hence W. ---- V, (n,) for s = 1 or 2 in all cases. Suppose that i +1 E A and L,IS(M I ). Then L, S(M 2) by (12.2)(ii) and so V;),,(n,±1)f2 V, 1 (n1±1 ). Thus V f V?±I(n,,I) and

L,„ s(m,).

By changing notation if necessary it may be assumed that either 1 E A and Lo IS(M,) or 0 E A and L 1 S (M I ). Hence by iterating the result of the previous paragraph we see that L J S (M 1) for all i E {1, , k} — A. Consequently (12.2)(ii) implies that S(M)C M. Thus M2 = (0). E The definition of M shows that M depends on whether the graph is of type I or II, on the integers n,, the set A and the functions g,. It will later turn out that M is independent of the choice of functions g,. LEMMA 12.5. M determines the type of graph and the ordered set Eo, , Ek

12]

THE

INDECOMPOSABLE

R[G]

MODULES IN

B

331

up to a reversal of order. Once the order is fixed, the set A is also determined by M except in case of a graph of type II with Po = Pk+i = Q. Furthermore the integers n, for i =1, ,k are uniquely determined.

PROOF. By (12.2) the graph is of type I if and only if S(M) and T(M) have no common composition factors. Furthermore {Eo , , Ek } is determined by S(M) and T(M). If the graph is of type I the ordering E0 , . , Ek is determined up to a reversal by S (M) and T(M) since T is a tree. By (12.2) A is determined by T(M). If the graph is of type II, (12.2) and (12.3) imply that , Eh., s 1 is determined by the common composition factors of S(M) and T(M). Thus Q is determined as Q P. Eo, . . , E, are determined up to a reversal of order by S(M), T(M) and the fact that r is a tree by (12.2). Thus {Po, Pk* ,} is also determined. If P then A is determined by the condition that Lo J S(M) or L o l T(M) but not both. Similarly A is determined if Pk+I Q. If Po = Pk4 1 = Q then the definition of M shows that A may be replaced by {1, , k}

Since there is at most one i E A with Pex incident to E, it follows that n, = 1 for all j except possibly for j equal to such an i. In which case n, is determined by l(M) as T is known. E Let y be a graph of type I or II. Let n(y) be the number of modules M which can be associated to y depending on A and the integers n,. In view of (12.4) and (12.5) there are at least En (y) pairwise nonisomorphic, nonprojective, reducible, indecomposable, nonzero R[G] modules in B, where y ranges over all subgraphs of r of type I or II. We will now compute n(y) and then count the number of possibilities for y. This will turn out to be ID le — 2e. Once this is done it follows from (2.1) and (2.2) that every nonzero indecornposable k[G] module in B is projective, irreducible or isomorphic to one of the modules M. Case (i): y is of type I. P,/ P. for 1 Ls. i k. Then n(y) = 2 depending on the choice of A. Case (ii): y is of type I. P, = P. for some i with 1 Ls_ i k Then n(y) = 2t depending on the choice of A and n,. Case (iii): y is of type II. Po Pk +I. Then 1 n, Ls_ t- 1 . Thus n(y) = 2(t — 1) depending on the choice of A. Case (iv): y is of type II. Po = Pk +1 = Q. Then 1 n
332

LEMMA

[12

CHAPTER VII

12.6. If

has no exceptional vertex then En(y)=1D le —2e.

T

Only Case (i) can occur. Given any pair of distinct edges in T there is a unique graph of type I with these as extreme edges since T is a tree. There are () ways of choosing such a pair of edges. Hence PROOF.

E n(y)= 2 ( 2e ) = e 2 — e

=ID le —2e

since D 1-1= e. THEOREM

12.7. En(y)=1Dle —2e.

In view of (12.6) it may be assumed that T has an exceptional vertex. Let PI , P2, be all the vertices in 7 such that P # Pex and P, Pex bound a common edge. Consider the subgraphs 71, 72, ... Of 7 such that T, consists of Pe„ and all vertices and edges with the property that a path from any vertex in T, {P e,,} to P. goes ghrough P. P Since T is a tree, T1, 7- 2, - yields a partition of the set of edges in T. Let e, be the number of edges in T,. Thus PROOF.

E, e, = e. In Case (i) {Po, Pk4 is any set of vertices in T, for any i such that Po and do not bound an edge. The number of possible pairs of vertices in T, is e, (e, +1). The number of pairs which bound an edge is the number of edges e,. Hence there are e, (e, —1) possible sets flub0, Pk +1, in Therefore the number of possibilities for y is E, (e — e,). Thus if y ranges over all possible graphs in Case (i), E y n (y) = E, — In Case (ii) Po may be chosen in any T, —{Pe,,} and Pk+1 in any TI — {Pe,,} with i j. Thus the number of possibilities for y is E,,, e,e,. Hence if y ranges over all possible graphs in Case (ii), Ey n(y) = 2t E, <, e,e,. In Case (iii) y lies in T, for some i. There are e (e, —1) ways of choosing {Po, Pk +1} in T. Thus there are E, e,(e, —1) ways of choosing y in T. Thus if y ranges over all possible graphs in Case (iii), n(y) = (t —1)E(e;— e,). In Case (iv) Q may be chosen arbitrarily in T {P ex }. Thus there are e choices for Q. Hence if y ranges over all possible graphs in Case (iv), Ey n(y)= (t — 1)e. Consequently if y ranges over all possible graphs of type I or II in T Pk±i

E n(y)= E (ef

—

e,)+2tE e,e, + (t —1)

= t (E e,) — t

E (e

— e,)+ (t —1)e

E e, + (t —1)e

= te 2 — e = e(et +1)-2e =

le

—

2e.

CI

13]

SCHUR INDICES OF IRREDUCIBLE CHARACTERS IN

if?

333

By the remarks above, (12.7) completes the description of all the indecomposable P[G] modules in B up to isomorphism. PROOF OF (2.26). If V is an irreducible or a principal indecomposable 1 [G ] module in B then S( V) T(V). Thus the result follows from (12.2) and (12.3). 0

13. Schur indices of irreducible characters in ./fl Throughout this section K = k. Thus B = B, = x., 43, = i, = xA for O u e, 1 i e, À E A. Let g be an irreducible Brauer character in Bk for some k with 0 k a. Let M = Q, (0) and let S be the ring of integers in M. By (1.19.3) and (2.9) g is a splitting field for any irreducible S"[Isid module in B, for 0 i a. LEMMA 13.1. M = Qp(q);)= Q, (x.) for 1

e, 1ue.

PROOF. Clear by (2.9) and (2.19). 0 If A 1 then xA x, vanishes on all p'-elements of G for A, ti En. Thus the maximal unramified subfield of Q, (x,) is the same for all À E A. Denote it by F. By (2.15) F CM. This yields

LEMMA 13.2. Suppose that A I 1. Let F be the maximal unramified subfield of Q, (x,) for some À E A. Then F is the maximal unramified subfield of Q, (x) for all g E A. Furthermore F C M and [M(xA):Q, (x,)] = [M: F] for all À E A. Let m (x) = mop (x) denote the Schur index of the irreducible character over Q,. The object of this section is to prove

THEOREM 13.3. m (x.) = 1 for 1 ,-ç u e and one of the following holds. 1A1=1 and m(x0)=[M(X0): Q, (x.)]. (ii) A 1. For À E A, m(x,)= [M : F]. It is easily seen that (13.1)—(13.3) imply

334

[13

CHAPTER VII

13.4. If x is an irreducible character in B and ço is an irreducible Brauer character which is a constituent of A then

THEOREM

m(X)=

[Qp

(x, (p): Qp (x)J.

Theorem 13.4 is due to Benard [1976]. The proof of (13.3) given here is based partly on Benard's argument and partly on the earlier results in this chapter. The proof will be given in a series of short steps. LEMMA

13.5. (i) m(x„)= 1 for 1 u e. IA = 1 then m(x0) --5 M (x ): QP (X°)]* IA I/1 and A E A then m(XA)1[M(XA):

(ii) If (iii) If

- [

QP

(X0i•

By (2.13) d„, and d, = 0 or 1 for all u, A, i. Thus the result follows from (IV.9.3) and (13.1). D PROOF.

13.6. If IDI=p the conclusion of (13.3) holds. In particular if IA 1=1 the conclusion of (13.3) holds.

LEMMA

PROOF.

If IA J = 1 then ID I = p. Thus the second statement follows from

the first. Suppose that I D I = p. Let X = xo if I A I = 1 and let x = xA for A a fixed element of A if IA 1/1. By (13.5) it suffices to show that m (X)

[M(X): Qp (x)]. By (IV.9.2) there exists an unramified extension Ko of Q,, (x) such that [Ko:Qp(X)]= fn (X ) and there exists a K0[G] module which affords x. By (11.2)(ii) M(x) C Ko and so [M(x): Qp (x)] m (x).LI

Throughout the remainder of this section it will be assumed that IA I / 1 and x = xA for some A E A. In view of (13.5) and (13.6) the proof of (13.3) will be complete as soon as the following inequality is established. m LEMMA

[M(X):

Qp

(x)] =

(13.7)

F].

13.8. If D < G then (13.7) holds.

Let V be a Zr -free Z r [G] module such that VQ, is irreducible and x is a constituent of the character afforded by V. Let n be the number of irreducible constituents of VK. Since D < G ,(2.15)(iv) implies that n = m(x)[Q,(x): Q r ]e. Since [Q,(x): F] = (p' ps - ')/e for some s > 0 this implies that n = m (x)(ps — : Qr]. PROOF.

—

13]

SCHUR INDICES OF IRREDUCIBLE CHARACI- ERS IN B

335

Let g be the character afforded by the irreducible Q, [ID] module of dimension ps — ps' over Qp. Then V, has a pure submodule X which affords g. By (11.6) the minimum polynomial of y on X, and hence on V, has degree at least ps — p". Since D
This implies that m (x) [M : F].

(P s Ps-1 )[M Qp].

LII

LEMMA 13.9. PROOF. The Galois group of F over Q, acts transitively on the set of all blocks which are algebraically conjugate to B. Since the Brauer correspondence commutes with field automorphisms it follows that F is the maximal unramified subfield of Q, (g). Thus if E is the extension of F generated by all pa th roots of 1, then (2.17) implies that QP

)= En

=E

n QP (26, ) = (x,). Eli

Define an equivalence relation on A as follows. For 'L I , tt2 E A let be an irreducible constituent of (),, for i =1,2. Then gi if and only if = 0. This is clearly an equivalence relation. If ‘, has order ps then the equivalence class containing IL, consists of (ps — ps -1 )/e distinct elements, and these are all algebraically conjugate. In particular the equivalence class containing [L i consists of one element if and only if ‘ 1; =1 and e = p —1. In the proof of (13.7) two cases will be handled separately. Case (i). If g E A with g — À then jt = A. Case (ii). There exists E A, tt A with jt— A. Let ee be the nonexceptional characters in Bo. The Galois group of M over F acts on the set {a}. Let T(0,) = E, v, where a- ranges over this Galois group. PROOF OF (13.7) IN CASE (i). By (2.17) Q, (x) = F. Thus there exists an irreducible F[G] module which affords m(). Hence there exists an F[No] module which affords m (x)xN0 . By (13.8) and (13.9) m(p,)=[M:F] for all g E A. Thus

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[14

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m(X)XN.= [M Fi

E

tWL}

c,T(00+ i

,

(13.10)

where n is a sum of characters in blocks other than B o . Let ip be the irreducible Brauer character in a block bo of K[C0] which is covered by Bo. Let d = d(A, y, tp) be the corresponding higher decomposition number. Since the multiplicity of ip as a Brauer constituent of T(0,),, is divisible by [M: F] it follows from (13.10) that [M: m(x)d. As we are in Case (i), d = ± 1 by (2.17) and so [M : Fl m (x). D (13.7) IN CASE (ii). Choose Ix —À, A/ A. By (2.17) (x — x,) No = -± (A. — p.,). Hence ((X — Xp.)/vo, À )No = ± 1. Therefore either (xN o, À )No or ((X)No , A)N o is relatively prime to m (A ). As x and x„. are algebraically conjugate it follows that m (x) = m (x, ). Thus by changing notation if necessary it may be assumed that (xN„, À )No is relatively prime to m (À). By (13.9) m(x)x m, is afforded by a (:)(A)[No] module. Thus by (IV.9.1) m (A ) I m()(, )No . Therefore m (A ) I m (X). Hence [M: Ffi m (x ) by (13.8) and (13.9). El PROOF OF

14. The Brauer tree and field extensions

Let K = k and let x = be an exceptional character in ii if A 1/ 1 and let x = xo if IA = 1. Let M = Q, (xo ) for 1 u e. Thus M is defined as in the previous section. Suppose that L' and L" are fields with Q,(x)Ç L' C L" C M(x) C K.

Let B" be a block of L"[G] which splits into algebraically conjugate blocks, one of which is E if L" is extended to K =K. Let B' be a block of L'[G] which splits into algebraically conjugate blocks, one of which is B" if L' is extended to L". Let T', T " be the Brauer tree of B',B" respectively. According to (2.9) and (2.19) there exists an integer m and a fixed vertex Po on T', which is the exceptional vertex if there is an exceptional vertex, such that T " is an "unfolding" of T ' in the following sense. Each edge of T ' is replaced by m edges in T " and each vertex other than Po is replaced by m vertices. The vertex Po remains fixed. According to (13.3) the vertex P o in T ' corresponds to a character with Schur index in (x) while the vertex Po in T " corresponds to a character with Schur index m(x)/m. Thus if, as in section 9, the vertices of T', T " are associated to irreducible L'[G], L"[G] modules respectively rather than characters then each vertex of T ' is replaced by m vertices in T " . It so

15]

IRREDUCIBLE MODULES WITH A CYCLIC VERTEX

337

happens that all the vertices coming from P o in T ' are equal. In this way the Schur index has a natural interpretation in terms of the Brauer tree. As an example let G be the semi-direct product of Suz(8) with a cyclic group (x) of order 9, where x acts as an outer automorphism of order 3 on Suz(8) and x' is in the center of G. Let L' = (), and let L" be the cubic unramified extension of Q. A nonprincipal 13-block of L"[G] has no exceptional characters. The tree T ' of a faithful 13-block B' of L 1 [G] is as follows. o42 192

o

3

105 o

o42 where the numbers at the vertices are the degrees of the corresponding characters. The tree T " is the triple unfolding of T ' hinged at Po and is the same as the tree of the principal 13-block given in section 9. 15. Irreducible modules with a cyclic vertex

This section contains a proof of the following result. THEOREM 15.1. Let V be an irreducible I [G] module with a cyclic vertex P = (x). Then P is a defect group of the block which contains V.

This result is independent of the other material in this chapter. It is due to Erdmann [1977a] and we will give her proof here. In case G is p-solvable it had been proved earlier by Cliff [1977]. We will first prove a preliminary result. LEMMA 15.2. Let (1) P = (x)< G and let V be an indecomposable R[G] module with vertex P. (i) There exists an integer k with 0 k P and p k such that V is a projective 1 [P]/ [P (1 — x) k module. (ii) V(1 x) = {v Iy E V, v(1 x) k _ i = 0}. (iii) If h E Tr(Gx , )(HomAu. ,))( V, V)) then h (V) C V(1 — x). (iv) V/ V(1 — x) is an k[G] module which is R[P]-projective and which has Pin its kernel. Furthermore VI V (1— x) is a projective R[G I P] module. (y) Let I be a nonempty collection of proper subgroups of P. If Ir(G, I, V) ----- i then V I V(1 — x) is irreducible. —

—

338

[15

CHAPTER VII

(i), (ii). For 0 k let VI, denote an indecomposable R[P] module of k-dimension k. By (II.2.11) (1/)p --1G :PI Vk. Since V is k [11-projective it follows that Vp mV, for some integer m >0. As P is a vertex of V,p k. Now (i) and (ii) follow directly. (iii) In view of (ii) it suffices to show that h(V(1— 4-1 ) = (0). Let w = v(1 E V(1 - X . By (i) wx = w. By (ii) k = np +s for 0< s