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, <1>0 be the principal indecompos able character of 0, 0° respectively which corresponds to <po If Y is a p '-element in 0 then
<1>(y) =
, 'i; i;(?)' , <1> 0(yO) = / Co (y ) n D / <1>0(/) .
In particular CP (1) / D / <1>°(1). =
PROOF. Every irreducible Brauer character of 0 has D in its kernel by (III.2. 13). Thus if {
{I
o
if x is conjugate to y - I , otherwise
4]
CHARACTERS IN BLOCKS
163
. >,PROOF. Let <1>i, <1>? be the principal indecomposa ble character of 0, 0° respectively which corresponds to
b
(�)
If S is an integral domain let Hn,d (S) be the space of homogeneou s polynomials of degree d in n indeterminat es X I , . . . , Xn with coefficients in S. The group GLn (S), and hence any subgroup, acts on Hn,d (S) in a natural w ay. The next result is due to E. Cline. LEMMA 4.28. Suppose that F is a field and char F = P > O. Let I be the ideal of F[x I, . . . , xn ] generated by all x P, where x E F[xI, . . . , x,, ] and the constant term of x is O. Then I n H n,(p-l)n (F) has codimension 1 in H n,(p-l)n (F) and is mapped into itself by GLn (F). PROOF. Clearly I, and hence I n H n,(p :":l)n (F), are mapped into themselves by GLn (F). Since char F = p, I is the k space spanned by all monomials x :n with max {aJ � p. The result follows since Hn.(p-l)n (F) contains X �I only one such monomial which does not satisfy the inequality, namely x f - 1 X ;'- I . 0 •
•
•
•
•
•
THEOREM 4.29 (Thompson [1981 D. Let 0 be a finite group and let V be an n-dimensional C[ 0] module. Let d ( 0) be the smallest integer d such that H", d (C) contains a one dimensional invariant subspace. Then d ( 0 ) � (p - l)n� where p is any prz)ne which does not divide 1 0 1 ·
then 1
<1>(y) / 0 : Co (y) / = (<1>, a )o = (<1>0, a ) /0/ =/
00
�o / <1>0(y O) / 0° : Coo (yO) / .
Consequently the first equation i s proved. The second equation follows from the fact that Coo (yO) = Co (y )/Co (y) n D as D is a p -group and y is a p '-element. 0 COROLLARY 4.27. Let D be a p-group with 0 = D Co (D ). Let {
PROOF. There exists an algebraic number field Fa and an Fo[O] module Va such that Va 0Fo C = V. It may be assumed that F � K. Let VI = Va 0Fo K. Since p ,r 1 0 / it follows from (III.3 .3) that up to isomorphis m there exists a unique R -free R [ 0 ] module W with W 0R K = VI . Furthermo re Hn,(p - l)n (R ) = H n,(p -l)n (R ) is completely reducible. Thus by (4.28) Hn,(p -l)n (R ) has a one dimensiona l subspace which is preserved by O. Thus by (lII.3.3), H n,( p -l)n (R ) has a pure R [ 0 ] submodule of rank 1. This implies the result. 0 The following consequenc e of (4.29) which is due to Thompson can be proved by making use of the main result of Feit and Thompson [1961] .
164
CHAPTER IV
COROLLARY 4.30. Let G, V, n, d (G) 4n 2 •
be
[4
defined as in (4.29). Then d (G) �
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) � 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 p3 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 0 be a subgroup of G. Suppose that for a fixed i, (Xu )o is irreducible for all u with dUi I- O. Then ('Pi )O is an irred�cible Brauer character. PROOF. Let �s be the Brauer character afforded by the irreducible R [0] module is. Let Us be the projective R [0] module corresponding to Ls and let cPs be the character afforded by Us. Suppose that is is isomorphic to a submodule of (Li ) 0 . Then is is isomorphic to a submodule of ( Vi )o . Hence Us I ( U )o . Thus
(4.32)
where aus denotes the appropriate decomposition number of O. Let eis be the multiplicity of is as a composition factor of (Li )o . Then
165
duieis for all u. Hence (4.32) implies that eis = 1 and cPs (1) = cJ>i (1). Hence Us ( Vi ) This implies in particular that the socIe of (Li )o is irreducible and isomorphic to is. Apply the same argument to 'P 'f . Thus (cJ> 'f ) o = cP � and i � is isomor phic to the socIe of (L no . Hence th �re exists an exact se quence (Li )O is � O. As eis = 1 this implies that Ls I (Li )o and so (Li ) = Ls. 0 ?=
=
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 I G : H I = k > 1 . Let n = km. By induction there exists v E Hm,do (C) with do � 4m 2 , v I- 0 such that the one dimensional space spanned by v is fixed by H. If { Xi } is a cross section of H in G and w = TIi (VXi ) E H mk,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 HI- G. This implies that any normal abelian subgroup of G is central. By Bertrand's postulate there exists a prime p with 2n + 1 < P < 2(2n + 1). Thus p � 4n + 1 and so p - 1 � 4n. Let P be a Sp -group of G. Then P is abelian and P <J G. See Feit and Thompson [1961]. Thus P � Z(G) and so G = P x Go by Burnside's transfer theorem. Clearly . d (G) = d ( Go). By (4.29) d(Go) � (p - 1)n � 4n 2 • 0
&s (1) = Lu cLsXu ( 1 ) � Lu duiXu (1) = cJ>i (1),
SOME OPEN PROBLEMS
0.
�
O
COROLLARY 4.33. Let 0 be a subgroup of G. Let B be a block of G. Suppose that (Xu )o is irreducible for every Xu in B. Then ('Pi )O is an irreducible Brauer character for every 'Pi 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 Sp -subgroup of G and let N = No (P). Let B be the principal block of G. The following are equivalent. (i) G has p- length 1 . (ii) If Xu is in B then (Xu )N is irreducible. (iii) If 'Pi is in B then ('Pi )N is irreducible. PROOF. (i) =? (ii). Since G = Op ',p,p' (G) the Frattini argument implies that G = Op , (G) N. If Xu E B then 0r, (G) is in the kernel of Xu by (4. 12) and so (Xu )N . is irreducible. (ii) =? (iii). This follows from (4.33) . . (iii) =? (i). P = Op (N) is in the kernel of irreducible Brauer characters of N and so P is in the kernel of 'Pi for every 'Pi in B. Thus P � Op',p (G) by (4.12). 0 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.
1 66
CHAPTER IV
167
SOME OPEN PROBLEMS
(I) If I C I 'Pj (xi )1 'Pj (1) E R is it true that I C I 'Pj (xi )1 'Pj (1) = A ( C ), 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
I C I 'Pj (xdl'Pj (1) need not be in R. As an example let p = 2 and let G = II , the smallest lanka group. If Xi is an element of order 3 and 'Pj is chosen suitably then 'Pj (l) = 56 and 'Pj (Xi ) = - 1 . Thus I C I 'Pj (xi )I'Pj (l) - 209/2 � R. See Fang [1974] .
=
(II) Is I G I /'Pj (l) E R for all j ? Willems [1981] has shown that the answer to (II) is yes if and only if I e ,
{'Pi ( I )}?
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.
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 - I � P - {I} for x, y E C. Let ' P 1 = pa and let I Co (x ) I p = P b for x E C. Then there exists an irreducible character X of G with p a - b I x (l). If furthermore P is a Sp -group of G then X is 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 0/
=
2: X (x )X (x - I ) X, x
x (1)
where X ranges over all the irreducible characters of G. By assumption tfJ(z ) = 0 for z E P - {I}. Since 0/(1) = I Co (x ) ' it follows that
'" X ( x )X(x - I ) (X 1 ) p. f' X (1 ) As p a - b is the exact power of p dividing the denominator of the left hand side and X (x ), X(x - I ), (Xp, ! p )p are algebraic integers, there exists X with p a -b I x (1). Suppose that P is a Sp -subgroup of G. Then I C o (x) I = ( 1 ) IPI '+' ,/, P,
P
P
P,
=
P
'" I G : Co (x ) I X (x )X (x - I ) (X 1 ) = I G . P I X (l) As I G : P I � O (mod p ) and I G : Co (x ) l x (x )l x (l) is an algebraic integer there exists X with I G : Co (x ) l x (x) � O , x (l)
f'
P,
P
P
•
•
(V) Can the number of blocks be characterized in terms of the structure
for 1T a prime in a suitably large p -adic field . Let B be the block containing D x. By (4.4) D is a defect group of B and X is of height 0 in B.
B y using the first main theorem on blocks, (lII.9.7), i t i s 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.
In case G is solvable Ito [1951a], [195 1b] 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 [1978] .
of G ?
(VI) What are some necesary and sufficient conditions for the existence of blocks of defect O?
CHAPTER IV
1 68
[5
(VII) Is there a function of p and d which bounds Cij for 'Pi, 'Pj in a p-block of defect d ? If G has a p -complement, in particular if G i s p -solvable, then i t follows from (4. 15) that Cij � p a, where p a = / G /p. Fong [ 1 961] has shown that for G p -solvable Cij � pd. See (XA.6). At one time it had been conjectured that Cij � p d for all G. This conjecture was shown to be false by Landrock [1973] for the group Suz(8) and p = 2, where pa = 64 and C l l = 160. The following results proved by Chastkofsky and Feit [1978] , [1980a] , [1980b] show that Landrock's result is not an isolated example. Let p = 2, let Pm be a Sz-group of Gm and let c IT) be the Cart an invariant of Gm corresponding to the principal Brauer character. If Gm = Suz(2 m ) or Sp4(2 m } then lim c \T)// Pm / 3/2 = 1 . m ---+oo
m ---+OO For information about the decomposition numbers of Suz(2 m ) and SU3 (2 m ) 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 Cart an invariants of simple groups even when the character table is known. For instance the Ree groups 2 G2 (32 11 + 1 ) all have a Sz-group which is elementary abelian of order 8. Yet the Cart an 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 inde composable R [ G ] modules for Chevalley groups in characteristic p can be found in J.E. Humphreys [1 973b], [1976] . However the groups SL2 (p " ) constitute the only class o f Chevalley groups for which much i s known about Cart an invariants. For instance every Cart an invariant is a sum of at most two powers of 2 and c 1 1 = 2" for SL2 (p " ) in characteristic p. See J .E. Humphreys [ 1973a], Jeyakumar [1 974], 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 'P be the Brauer character afforded by the natural 2-dimensional representation of SL2 (2" ) over the field F of 2" elements. Let u be the Frobenius automorphism. Let S be the group of integers modulo n. For i E S define 'Pi = 'P a' and for I � S let 'PI = I1i E I 'Pi. Then every irreducible Brauer character of SL2 (2" ) is of the form 'PI for some subset I of S, and conversely these are all the irreducible Brauer characters of SL2 (2" ).
THEOREM. C u = 0 unless for each i E S with i E I n J and i + 1 � I n J we have i + 1 � I and i + 1 � J, in which case Cu r - i I UJ i . =
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 / G / p "go with (p, go) = 1 . Suppose that G contains exactly go p '-elements. If C I I � pa then G has a normal p -complement. =
PROOF. By the Cauchy-Schwartz inequality and (3 .3) p 2a = ( CPI, 11 ] )2 � ( CP], CP] ) ( 11 1, 11 1 ) = C l IP " � p 2". Thus 11 1 and CP I are proportional. Hence CP I ( Z ) = CP I (l) if and only if Z is a p '-element in G. Thus the set of all p '-elements in G is the kernel of C/) I and this kernel is a normal p -complement. 0 COROLLARY 5.3 (Brauer an d Nesbitt [1941]). Let / G / = paq brC where p, q, r are distinct primes. If G contains exactly q brC p '-elements then G has a normal p-complement. PROOF. By (4. 15) (iv) C I I � p a. The result follows from (5 .2).
0
(VIII) Is d�i � p d whenever Xu, 'Pi lie in a block of defect d? B y (XA.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 ?
1 70
CHAPTER IV
[5
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 Fang [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). Knorr [ 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 R [ 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 chatacters in the block B. Let
ko(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 d ---+oc ko(d) = oo? (ii) Is k (d ) � p d ? 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) � p d generalizing the known fact that this inequality holds for d � 2. See (VII. 10. 17). P. Fang 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 H � (1). Then H has at most k (d ) - 1 conjugate classes. PROOF. Let P be an elementary abelian p -group of order p d. 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
COROLLARY 5.5. Suppose that H satisfies the same conditions as in (5.4). Then H has at most j p 2d conjugate classes. PROOF. Clear by (4. 18) and (5.4).
0
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 B be the block of NG (D) with B G = B. Does the number of irreducible characters of height 0 in B equal the corresponding number for B ? (XII) Let mp ( G ) denote the number of irreducible characters of G whose degree is not divisible by p. Is mp ( G ) = mp (NG (P)) for a Sp -group P of G ? It i s clear that an affirmative answer to (XI) implies that (XII) has an affirmative answer. McKay [1972] fi rst 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 n (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 n (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 {'P n be the set of all irreducible Brauer characters of CG (y ). If ? is an irreducible character of CG (y) then
�(yx) = c�(x ) for all x E Cc (y ) where c is a p a th root of unity since y is the center of Cc (y). Thus XS (yx ) = L; d �; 'P r (x ) for all p '-elements x in Cc (y ) where each d�; is an algebraic integer in the field of p a th roots of unity over the rationals. The linear independence of {'P n implies that d;; is well defined. The numbers d {; 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 U for some u E G. Then Cc (z ) = Cc (y t. Further more after a suitable rearrangement 'P � (x U ) = 'P r (x) for all p '-elements x E Cc (y). Since XS (ZX U ) = Xs ((yx t ) = Xs ( Yx) for E C c (y ) i t follows that Xs ( zx U ) = L d ;; 'P � (X U ). Hence d ;; = d :i. Thus the set of higher decomposition numbers with respect to y depends only on the conjugate class containing y. x
If rr is an automorphism of the field of p a th roots of unity over the rationals then it follows directly from the definition that rr 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 d i; I: 0 for some t, i. Let B be the block of c Cc (y ) containing 'P r . Then B is defined and XI E B c . Furthermore y is conjugate to an element of the defect group of B G. PROOF. Let P = ( y > and let H = Cc (P) = Cc (y ). Let e be the centrally primitive idempotent in R [H] with B = B (e). By (III.9.4) B O is defined. Let XI 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 XI. Let s be the Brauer mapping with respect to ( G, P, H) and let 0 be the character afforded by V s ( e ). By (IIL4. 1 1) and Nagao's theorem (IIL7.5), XI (xy ) = O (xy ) for all p '-elements x in H. Since d i; I: 0 this implies that 'P r is afforded by an irreducible constituent of V s ( e ) . Hence s ( e ) e = e. Therefore B C = B by (IIL9.4). The last statement follows from (2.4). D
HO
HO
1 73
HIGHER DECOMPOSITION NUMBERS
CHAPTER IV
1 72
o
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
be a complete set of representa tives of the p '-classes in Cc (y). It is easily seen that {yx n is a complete set of representat ives of the conjugate classes in G which lie in the section containing y. Let CY = (c n denote the Cart an matrix of C c (y ). 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 ali i, j
LEMMA
Ls (d;i )* d :j XLsEB ( d ;i ) * d :j = O. (ii) Ls ( d ;i ) * d ;j = c� for all i, j. =
(iii) LxsEB (d ;;) * d ;j = 0 unless 'P r and 'P J lie in the same block B of Cc (y) and B C = B. In that case LxsEB (d ;; ) * d ;j = c � . By definition
('P r(x J )) * ' (d ;; )* ' (d :; ) ('P f (x j )) = (Xs (yx J )) * ' (Xs (zx j )) =
( � Xs (yx r)*Xs (zx j )).
If y i� . not conjugate to z then Ls xs (yx r)* Xs (zx j ) = 0 for all i, j. By (3.6) ('P r(x J )) is nonsingular. This implies one of the equations in (i). Since
Ls Xs (yx r)*Xs (yx J )
=
I Cc (YX r ) I Oij = I C Cdy) (x r) l oij
it follows from (3.8) that
( � Xs (yx r)*Xs (YXn ) = ('P r(x n) * '( c �) ('P i (x J )).
Hence the non singularity of (cp i(x J )) implies (ii). Fix i, j. Let B" B 2 respectiv ely be the block of Cc (y ), Cc (z ) respectively with 'P r E BI and 'P J E B 2 • If ( d ;i ) * d :j I: 0 for some s then by (6. 1) B � B ? and XS E B �. Thus
=
L
X,EB
(d�i) * d:j = L (d;i)*d:j =0
otherwise.
This completes the proof of (i), and by (ii)
L
XsEB
(d;i)*d ;j = c � if B � = B ? = B =0
otherwise.
CHAPTER IV
1 74
Since c � = 0 if B I -I B 2 , (iii) follows:
k
0
The next result is a refinement of (3. 14). See Osima [1960b] . LEMMA 6.3. Let B be a block. If () = Ls bsXs with bs E K define (} B = Lxs E B bsXs. (i) If () vanishes on a p -section of G then (} B vanishes on that p-section. (ii) If Z I and Z 2 are in different p-sections of G then
L XS (ZI) * Xs (Z 2) = o.
Xs E B
PROOF. (i) Suppose that () vanishes on the p -section of G which contains the p -element y. Then LS,i bsd ;i 'P r (x ) = (} (xy ) = 0 for all p '-elements x in C G (y ). Thus Ls bsd �i = 0 for all i. Hence by (6. 1) L bsd �i = 0 for all i and Li bsd ;i 'P r (x ) = 0 for every p I-element x in C G (y ) as so () (xy ) = L B
xs E
B
xs E B
required. (ii) Let (} = Ls Xs (Z I)*Xs. Then (} (z ) = O for every element Z in the p -section containing Z 2 . The result follows from (i). 0 (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 L Xs (x )Xs (y ) = 0 where Xs 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. Xs E B (q ) Xs (x )Xs. By assumption no element of G is in the p -section containing y and also in the q -section containing X - I . Thus by (6.3) (ii), (} (z ) = 0 for every element Z in the p -section containing y. Hence B by (6.3) (i) (} (p )(y ) = O. (ii) Immediate by (i). 0
PROOF. (i) Let () = L
The following notation will be used in the rest of this section . B I , B 2 , are all the blocks of R [ G ] . {Yi } i s a complete set of representatives o f the conjugate classes i n G consisting of p -elements. Thus {YiX ;i} is a complete set of representatives of the conjugate classes in G. •
• •
1 75
HIGHER DECOMPOSITION NUMBERS
[6
is the number of conjugate classes in G.
k(Bm ) is the number of irreducible characters in Bm. I (Yi ) is the number of conjugate classes in C G (Yi ) consisting of p '_
elements.
I (y;, Bm ) is the number of irreducible Brauer characters of C G (Yi ) which lie in blocks B of C G (Yi ) with B G = Bm. X = (Xs (YiX r)) is a character tab Ie of G. If y is a p -element in G then 'PY = ('P i(xI)) is a table of irreducible �rauer characters of C G (y ). 'P is the direct sum of the matrices 'PYi• D = (d�;) where s is the row index and (i, j) is the column index. Dm is the submatrix of D where s ranges over all values with Xs E Bm and for each i, j ranges over values for which 'P;i is in a block B of C G (Yi ) with = Bm. Thus D m is a k (Bm ) x ( L; l(y;, Bm )) matrix. If T is a ring of algebraic numbers in K and S is a subset of G define
fJ G
{ a I a = � asXs, as E T, (z ) = 0 for Z S } . ChT (S, Bm ) = { a I a = L asXs, as E T, a (z ) = 0 for Z s}.
, ChT (S) =
a
g
g
Xs E Bm
In case T is the ring of rational integers write Ch(S ) = ChT (S) and Ch(S, Bm ) = ChT (S, Bm ). LEMMA 6.5. (i) I(y; ) = Lm I(y;, Bm ) for all i. (ii) k = Lm k (Bm ) = L; l(y; ). (iii) X , 'P, D are all k x k matrices and X = D'P if the rows and columns are
suitably arranged.
PROOF. (1II.2.8) and (6. 1) irnply (i). (ii) and (iii) are clear by definition. 0 LEMMA 6.6. (i) D is the direct sum of the matrices Dm after a suitable
arrangement of rows and columns. (ii) Let T be a field of algebraic numbers in K. Then for all m dimT (ChT ( G, Bm )) = k (Bm ) =
L l(y;, Bm ) i
(iii) Dm is a nonsingular k (Bm ) x k (Bm ) matrix and ± (det Dm ? is a power of p for each m. (iv) Let y be a p-element. There exists a block B of C G (y ) with B G = Bm if and only if y is conjugate to an element of the defect group of Bm.
1 76
CHAPTER IV
PROOF. (i) Immediate by (6. 1). By definition dimT (ChT ( G, Bm )) k (Bm ). By (6.5) D is a square matrix. As D is nonsingular each Dm is a square matrix. Thus in view of (i) both (ii) and (iii) will follow once it is shown that ± (det Dm )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 Dm to D!. Thus (det Dm )2 ± det(D!'Dm ). By (6.2) D!' Dm is the direct sum of Cartan matrices. Hence ± (det Dm ? is a power of p by (3.9). (iv) Suppose that Y is not conjugate to any element of the defect group P of Bm. By (2.4) d �i = 0 for all XS E Bm. Since Dm is nonsingular, this implies that Bm -I jj 0 for any block jj of Co (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 Co (P) such that for XS E Bm, � -I 0, where x is in the conjugate class C. Choose XS of height 0 in Bm. Thus XS (x ) -1 0. Since x E Co (P) it follows that x E Co (y) and XS (xy ) -1 0, and so Xs (xy ) -1 0. Hence d �i -1 0 for some i. The result follows from the second main theorem on blocks (6. 1). 0
'rationals, where h ranges over all the integers with 1 � h � I G I and (h, I G I ) = 1 and ah is the image of a under the automorphism which sends , Zn be a complete set of a I G I th root of unity onto its h th power. Let z representatives of the conjugate classes in S. For i = 1, . . . , n define (}i = Ls asiXs where asi = Lh ahXs (z �h). Thus each asi is a rational integer. If Z E G and Z is not conjugate to z ;n for any m with (m, I G I ) = 1 then (}i (Z ) = Lh ah Ls Xs (Z � h )Xs (Z ) = O and so Oi E Ch(S). Suppose that Lj bjOj = 0 for rational integers bj• Then for each i
==
1
1,
=
•
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 Ip )th root of unity. (ii) If Z E S then Z h E S for all integers h with (h, I G I ) = 1. Then dimT (ChT (S, Bm )) = L Y; ES I( Yi, Bm ) for all m. PROOF. In view of (6.8) dimT (ChT (S)) = L Y;ES I (y; ) under either assump tion. Thus if the result if false then for some m ', dimT (ChT (S, Bm, )) < 2: I( Yi, Bm, ) y;ES by (6.7) (ii). Hence by (6.7)
l(y;, Bm ) dimT (ChT (S)) = 2: m dimT (ChT (S, Bm )) < 2: m 2:
0
PROOF. Clearly rank(Ch(S)) � n. Thus it suffices to exhibit n linearly independent elements in Ch(S). Let {ah } be a normal basis of the field of I G Ith roots of unity over the
.
The following refinement of (6.6) (ii) is due to W. Wong [1966] .
PROOF. (i) Since G - S is a union of p -sections this is clear by (6.3) (i). (ii) Suppose that Ls asXs E ChT (S, Bm ). Then Ls,j asd �j 'P J(x ) = 0 for all Y = Yi g S and all p '-elements x E Co (y ). Hence Ls asd �j = 0 for all Y = Yi g S and all j such that 'P ] is in a block jj of C o (y ) with jj O = Bm. Thus the vector (as ) satisfies L Y;ES I ( Yi' Bm ) homogeneous linear equations. By (6.6) (iii) these equations are linearly independent. Therefore by (6.6) (ii)
LEMMA 6.8 (Suzuki [1959]). Let S be a union of n conjugate classes in G. Assume that if Z E S then z h E S for all integers h such that (h, I G I ) = 1 . Then rank(Ch(S)) = n.
•
where in the last sum U, h ) ranges over all pairs with Zi conjugate to Z 7Since {ah } is a basis of the field of I G I th roots of unity over the rationals this implies that bi = O. Thus {(}i } is a linearly independent set. 0
LEMMA 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) ChT (S) = Eft ChT (S, Bm ). (ii) If T is a field then dimT (ChT (S, Bm )) � LY;ES I ( Yi, Bm ) for all m.
dimT (ChT (S, Bm )) � k (Bm ) - 2: I( Yi, Bm ) = 2: / ( Yi, Bm ). y;ES y;ES
177
HIGHER DECOMPOSITION NUMBERS
y;ES
= 2: l(y; ) = dim; (ChT (S)). y;ES -
This contradiction establishes the result.
0
Xs is a p-conjugate of XI if XI = X � for some automorphism of the field of I G Ith roots of unity over the rationals which leaves the p'-th roots of unity fixed. Two p-conjugate characters have the same irreducible Brauer (J"
1 78
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CENTRAL IDEMPOTENTS AND CHARACTERS
characters as constituents and so lie in the same block. XS 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 I G I 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 Bm into families of conjugates under the action of �\1. Let ni be the number of irreducible characters in the ith family. Then the following hold. (i) The number of orbits of the columns of Dm under the action of M is equal to the number of families in Bm. (ii) If P � 2 or if the 4th roots of unity are in the fixed field of M then after suitable relabeling the ith orbit of columns of Dm consists of ni columns.
a
PROOF. If P � 2 or if the 4th roots of unity are in the fixed field of M then M is cyclic. Since Dm 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). 0 COROLLARY 6. 1 1 . Suppose that p � 2. The number of p-rational irreducible characters in Bm is at least as great as the number 1(1, Bm ) of irreducible Brauer characters in Bm.
1 79
Xs � ( Z ) Z . es - I G(l)I Z� G XS 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 -1
_
PROOF. Xs E B if and only if ws (e ) � O. Thus by (7. 1) e=
X�B es = I � I Z�G { X�B Xs (l )Xs ( ) } Z-
l
Z.
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 is a p '-element then z
PROOF. Since the ordinary decomposition numbers are rational Dm has at least 1(1, Bm ) rational columns. The result follows from (6. 10). 0 Some results concerning the fields generated by the entries of X, 'P 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 es be the centrally primitive idempotent of K[ G] corre sponding to Ws . Then
This proves the second equation. D LEMMA 7.3. Let e be a centrally primitive idempotent in R [G]. Let B be the block corresponding to e and let A be the central character of R [ G ] corresponding to B. For each j let G denote the conjugate class of G with Xj · E G. Then e = Lj aj¢ with aj E R. If iijA ( G ) � 0 then B and G have a common defect group.
PROOF. By (7.2) e = Lj aj¢ with aj E R. Let D be a defect group of B and let � be a defect group of G. If iij � 0 then Pj � G D by (III.6.5) since e E ZD (G ; G). If A (G) � 0 then D � G Pj by (III. 6. 10). D The next result is due to Osima [1966] in case character.
a
is the principal Brauer
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1 80
THEOREM 7.4. Let e be a centrally primitive idempotent in R [ G ] and let B be the block corresponding to e. Let Xr be a character of height 0 in B. Let a be a rational integral linear combination of {'Pi I 'Pi E B } with v(a (l)) v(Xr (l)). Let 11« be defined as in (1.2) and let 11a = Ls bsX,,· Define =
1 &ill " 11« ( Z ) z. br z�O Then fa E R [ G] and e - la E J(Z(G : G)). f J«
-1
- IG1
_ _
8]
SOME NATURAL MAPPINGS
181
. where X; E C, Xj E �, Xs E Cs. If j G lp ;:= P Q then 1 C I I � I 0 (mod p Q ) but I G 1 X (l) � 0 (mod p2Q ). Hence C�e = O. Therefore {M(l - eo)}2 = O. Since MeM(l - e) = 0 this implies that M 2 = (Meot Since ei c�rresponds to a block Bi of defect 0, Bi contains a unique irreducible R [ G ] module and Ze; = Rei. Hence Me; = Zei = Rei and so Mea = E9�= Rei. Therefore ==
==
1
R e- 2 = ffi R-e-·,. 0 M2 = ffi W ; =1 i=1 , W n
n
PROOF. By (4.7) fa E R [ G ] . By (4.1) and (7. 1) Jfa
8. Some natural mappings
- &.ill " � &D1 " s ( Z - I ) Z z� x - b, x�B XS (1) 1 G 1 O
� LB &ill br Xs ( 1 ) es X;E where es is the centrally primitive idempotent in K[ G] corresponding to, OJs• Thus if XS � B then OJs (e - f:, ) = OJ" (e ) - OJs (fa ) = O. If XS E B then =
� OJs ( e Jfa ) -- OJs ( e ) OJs If V a ) -- 1 X'b(1) , XS (1) ' Hence by (4.7) OJ" (e - fa ) 0 (mod ) Thus Ws (e - fa ) 0 (mod 1T) for all s and so A (e - 1) = 0 for every central character of R [ G ] . 0 _
_
_
==
1T .
==
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 R [ 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 dimR M2 . PROOF. By (111.6.2) M is an ideal of Z. Let eI, . . . , en be all the centrally primitive idempotents of R [G] corresponding to blocks of defect O. By (7.3) e; E M for i = 1, . . . , n. Let eo = L�= I ei. Then M = Meo EB M(l - eo). We next show that M2 = (Meot Let 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) Xi ) (Xj)X (xs) c" C C".C.e = L L 1 I I � I X ( X G1 ( ) S
XE B
1
X l
To a large extent this section consists of results which are reformulations of some results proved earlier in this chapter . We will follow Broli(� [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 ChT (S, B ) we will write ChT (G : S, B ) to emphasize the dependence on G. We will also write ChT (G : G) = ChT (G). Greg denotes the set of all p i-elements in G. The Grothendieck algebra A �(K[ G D is isomorphic to Ch( G) in the obvious way. By (3. 12) Ch(G : Greg ) is isomorphic to the Grothendieck algebra A � (R [ G D. Let Ch T,proj (G) denote the ring of all T-linear combi nations of { 4\ }. Let ChT,proj (G : B ) = ChT,proj (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 = b O, c = c o, d = d O as follows: where b ( CPi ) = Lu du;Xu, b : ChT,proj (G) � ChT (G), c : ChT,proj ( G) � ChT (G : Greg ), where C ( CPi ) = Lj Cii'Ph where d (Xu ) = Li dui'Pi. d : ChT (G) � ChT ( G : Greg ), By (3. 1) the following diagram is commutative. Ch T,proj (G)
Y
� d
ChT ( G) � ChT (G : Greg )
It is clear from the definitions that b, c, d respect blocks. In other words
. (d(O)) B = d ( O B ) for 0 E ChT (G). Similarly (b (O)) B = b ( O B ) and (c (O)t = c( 0 B ) for 0 E Ch T,proj ( Q). Furthermore if X E Ch T ( G) and cP E Ch T,proj ( G)
then by (3.4)
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1 82
(8. 1) ( d (X ), <1» = (X, b ( <1» ). This situation will now be generalized. Let y be a p -element in G. Assume that T contains a primitive I (y > Ith root of 1. If ?; is an irreducible character of Cc (y ) then there exists a I (y > Ith root of 1 , (y ) such that ?;( yx ) = (y )?;(x ) for x E Cc (y ). Define the linear map Wy of ChT (C c (y )) by Wy (?;) = ( y )?; for any irreducible character ?; of C c (y ). Then for any ?; E ChT (C o (y )) and any E Cc (y ) it follows that (wy (?;)) (x ) = ?;( yx ). For any subgroup H of G let Res� : ChT (G) � ChT (H) be defined by Res�(e ) = eH• Let Ind� : ChT (H) � ChT (G) be defined by Ind�(e) = e C. Let c Y' c denote the map c = c CG( Y ) for the group C c (y ). Define Wi:
Wi:
Wi:
x
LEMMA 8.2. (i) The following diagram is commutative. ChT,proj (Cc (y ))
bY:/
ChT (G)
--�)
�
ChT (Cc (y) : C c ( Y ) reg)
(ii) If Y/ E Ch1�proj (Cc (y)) then b y,C ( Y/ ) vanishes outside the p -section which contains y and it is a class function such that (b Y' c ( y/ )) (yx ) = y/ ( x ) for x E C c ( Y )rcg . (iii) If e E ChT (G) then d Y' c ( e ) is an R -valued function on Cc (y ) defined by (d y, C (e )) (x ) = e (yx ) for x E Cc ( Y ) rcg . PROOF. Statements (ii) and (iii) are direct consequences of the definitions. Then (i) follows from (ii) and (iii). D LEMMA 8.3. If e E ChT (G) and Y/ E ChT,proj (Cc (y )), then ( d Y C (e ), Y/ )cG( Y ) = (e, b Y' c (y )) c . '
PROOF. By (8. 1) and the Frobenius reciprocity theorem (III.2.5) (d Y' C (0 ), Y/ )cdY ) = (Resgo (Y ) ( 0 ) , (wy f l o b Co( Y ) (y ))Cd Y ) = ( 0, b Y'C (y/ )) c . D
SOME NATURAL MAPPINGS
1 83
Let B be a block of G and let 13 1 , 132, • • • be all the blocks of C c (y ) with B F = B. If ?; is a class function defined on Cc (y ) let ?;B = Li ?;B;. The next result is equivalent to the second main theorem on blocks (6. 1). THEOREM 8.4. Let 0 E ChT (G) and let Y/ lowing hold. (i) b Y' o (
E ChT,proj (Cc (y )).
Then the fol
.PROOF. By (8.3) (i) and (ii) are equivalent. Thus it suffices to prove (ii). Since Xu (yx ) = Li d �i 'Pi (x ) for x E Cc (Y ) reg it follows from (8.2) (iii) that d y,C (Xu ) = Li d �i 'Pi. The result is now seen to be a reformulation of (6. 1). D Let ZT (G) denote the center of T[ G]. If S is any subset of G let ZT (G : S) denote the set of all a = LXE C axx E ZT (G) with ax = 0 for
X � S.
In the remainder of this section we will only be concerned with the case K. Observe that ChK,proj (G) = ChK (G : Greg ).
T=
By definition ChK (Q) 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 e E ChK ( G) then e B = e e, where (8 e ) (a ) 0 (ea ) for a E ZK ( G). Let S be a p -section of G. For e E ChK (G) define ds (e ) by 0
(ds ( e )) (x ) =
{
=
0
e ( x ) if x E S, o
if x � S.
For a = LXE C aXx E ZK (G) let Os (a) = LXES aXx. Then ds is the transpose of Os. In other words e ( os (a )) = (ds (e)) (a ) for a E ZK ( G), e 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 b y,C d y,C = ds. (ii) If e E ChK (G) then (ds (e )t = ds ( e B ) . (iii) If a E ZK (G) then Os (ea ) = eos (a ). 0
PROOF. Since T = K, (i) follows from (8.2)(ii) and (iii). Then (8.4) implies (ii). Since Os is the transpose of ds, (ii) implies (iii). 0 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 = Lx axx E ZK ( G) then ae = Lx bxx. 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 b Y = b Y o, d Y = d Y o . If y/ E ChK (Co (y ) : Co (Y )reg) then by (8.2) (ii) b Y (y/ ) vanishes outside the p -section which contains y, and it is a class function such that (bY (y/ ») (yx ) = y/ (x ) for x E Co (Y )reg. If 8 E ChK (G) then by (8.2) (iii) (d Y (8 ») (x ) = 8 (yx ) for x E Co (Y )reg. Define (3 Y, o Y analogously as follows. For S a p-section of G which contains y, ES
SCHUR INDICES OVER Qp
[8
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1 84
ES
'
'
(3 Y : ZK (Co (y ) : Co (y )reg) � ZK (G : S ), where
1 85
Qp Let 'P be an irreducible Brauer character of G. Let F be a finite field such that 'P (x ) E F for all x E G. By (1. 19.3) there exists an irreducible F[ G] module which affords 'P. 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(X) = F(X (x ), x E G). Then there exists a positive integer mF (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 mF (X) / n. The integer mF (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 1 1 . Schur indices over
LEMMA 9. 1 . Let F be a field of characteristic O . Let X be a n irreducible character of G. Then the following hold. (i) mF (X) = mF (x ) (X ). (ii) If 8 is afforded by an F[ G] module then mF (X) / (l�, X ). (iii) If F c;;, L then mF (x ) / m (x ) [ L (X) : F(X)]. (iv) There exists an extension L of F (x ) with [L : F (x)] = mF (x) and mL (X) = 1 . L
and
z
runs over a cross section of Co (y ) in G.
LEMMA 8.6. (i) If a E ZK (Co (y ) : Co (Y )reg ) and 8 E ChK (G) then (d Y (8 ») (a ) = I G : Co (Y ) I -1 8 « (3 Y (a » . (ii) If Y/ E ChK (Co (y ) : Co (Y )reg) and a E ZK (G) then (b Y ( y/ ») (a ) = I G : Co (Y ) I y/ (o Y (a » . PROOF. Direct consequence of the definitions.
0
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 ), Co (y » . Let so(e ) be the central idempotent in R [Co (y )] such that so(e ) = s(e). Then the following hold. (i) (3 Y (so(e )a ) = e(3 Y (a ) for a E ZK (CO (y ) : Co (Y )reg ) . (ii) o Y (ea ) = so (e )o Y (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 - I and so it follows from (8.4). 0
Suppose that F(X) = F. Let V b e 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] = mF (X f 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, Qp is well known. Schur indices over Qp are studied for instance by Yamada [1970]. We state the following without proof. THEOREM 9.2. Suppose that F is a finite extension of Qp. Let X be an irreducible character of G. If L is any finite extension of F(X) with mF (X) / [L F(X)] then mL (X) = 1 . In particular (9.2) implies that L can be chosen t o be unramified or totally ramified. On occasion one or the other of these choices is conve nient. The next result is due to Brauer [1945]. See also Brauer [1947], Gow [1975]. :
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1 86
THEOREM 9.3. Suppose that dUi -1 0. Then m Qp (Xu ) I dUi [Qp (Xu, 'Pi ) : Qp (Xu )] . PROOF. Let Zp be the ring of p-adic integers in Qp. By (1. 19.3) there exists an irreducible Zp (
l
COROLLARY 9.4. If Xu is irreducible as a Brauer character then m Qp (Xu ) = 1 . PROOF. Immediate from (9.3). COROLLARY 9.5. If P m Qp (Xu ) = 1 .
t
PROOF. Clear b y (9.4).
D
I G I and X is an irreducible character of G then D
For related results concerning Schur indices see Brow§ [1977]. 10. The ring
A HR [ G])
1 87
Before proving (10.1) we deduce some consequences. COROLLARY 10.2. Suppose that P is a p -group with P
D
COROLLARY 10.3. Let P be a Sp -group of G. Suppose that P
A O ( G) = A � (R [ G]) is the ring of all integral linear combinations of generalized Brauer characters of G. The ideal A �1) (G) of A O( 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 �1) (G), and hence in particular a criterion for A �I)(G) to be a principal ideal. The argument is from Feit [1976]. See also Alperin [1976e] . THEOREM 10. 1 . Let X I , . . . , Xm be a complete set of representatives of all the conjugate classes of G consisting of p '-elements. Let v( I C G (xi ) l ) = ai for 1 � i � m. Let YJ E A �o< G). The following are equivalent. (i) For 1 � i � m, YJ (Xi ) = P aiUi, for some unit Ui in some algebraic number field. (ii) rr �= I YJ (Xi ) = ± rr�= l p ai. (iii) For all E A �l) (G) there exists 'Ya E A O ( G) with YJ'Ya = Furthermore if YJ satisfies (i), (ii), (iii) then 'Ya is uniquely determined by ll'
THE RING A�(1� [ G])
10]
ll' .
ll'.
:=
LEMMA 10.4. Let 0 be a complex valued class function on G. Let y , Ym be a set of pairwise nonconjugate elements of G such that O(Yi ) -I 0 for 1 � 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 Yi. Let L ( 0 ) be the linear transformation on X defined by L ( O)f = Of. Then dim X = m, the characteristic roots of L (O) are O (YI), . . . , O(Ym ) and det L (O ) = �;: 1 O (Yi). 1,
•
•
•
PROOF. Define fi E X by f; (Yi ) = Oii. Then {fi } is a basis of X and so dim X = m. Furthermore L (O)f; = O(Yi )f; for all i. Thus L (O) has the required characteristic values and so the required determinant. D
1 88
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[11
PROOF OF (10 . 1) . Let Pi be a Sp -group of C G (Yi ). Since yt E A 71) (G) it E A 71) « Yi > X Pi ). Thus yt (Yi ) = p Q;Ui for some algebraic follows that yt integer Ui . Therefore n�= 1 yt (Yi ) = n�= 1 Ui n�= l pQ;. Clearly n;: 1 yt (Yi ) is rational valued and so n�= 1 Ui is a rational integer. The equivalence of (i) and (ii) is now clear. Condition (iii) holds if and only if ytA O( G) = A 71) (G). Since ytA o( G) � A 71) ( G), this holds if and only if I A O( G) : ytA O( G) I = I A O( G) : A 71) ( G) I. By (10.4) yt (Yi ) = ± I1 Ui IT p Q;. I A O(G) : ytA O(G) I = I det L (y ) I = ± I1 i=1 i=1 =1 By definition I A O( G) : A 71) (G) I = det C, where C is the Cart an matrix of G. By (3. 1 1) det C = n�= 1 p Q;. Thus condition (iii) holds if and only if n�= 1 Ui = ± 1 , which is equivalent to condition (i). If yt satisfies (ii) then yt (Yi ) -I- 0 for all i. Thus Ya (Yi ) = a ( Yi )yt (Yi t l and so Yo is uniquely determined by a. D (y; ) x P;
i
1 1 . 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 alternat ing 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. Fang [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 = V Q9 V = V @ V* by (II.2.8). S = {X E E l X = X'}, T = {X + X' I X E E}. Clearly T � S and S, T are both F[ G] modules. Vo(G) is the F[G] module with dimF Vo(G) = 1 and InvG ( Vo (G» = Vo (G).
SELF D UAL MODULES IN CHARACTERISTIC 2
11]
1 89
THEOREM 1 1 . 1. If V;;6 Vo ( G) then V is of symplectic type. PROOF. Since V = V* there exists M E E with M- 1A (x )M = A (x - I ), for all x E G. Thus M'A (x )'(M't l = A (x - 1 ) and so A (X - I), = (M't I A (x )M' for all x E G. Hence M'M- I A (x ) = A (x )M'M - 1 for all E G. Thus by Schur's lemma M' = cM for some c E F. Thus M = cM' = c 2 M. Hence c 2 = 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 dimF V = 1 and V = Vo (G). If W = V then M defines a nondegenerate symplectic form on V. 0 x
COROLLARY 1 1 .2. -Let 'P be a real valued irreducible Brauer character of G in characteristic 2. Then either 'P = 1 G or 'P (1) is even. PROOF. Let U be an F[ G] module afforded by 'P. If U is of symplectic type then 'P (1) = dimF U is even. The result follows from ( 1 1 . 1) with U = V. 0 For i < j let X;j denote the matrix in E with 1 in the (i, j) and U, i ) entries and 0 elsewhere. Let Xii denote the matrix with 1 in the (i, i ) entry and 0 elsewhere. Then {Xij I i � j } is an F-basis of S and {Xij I i < j } is an F-basis of T. By (1.19.3) there exists a finite subfield Fo of F and an Fo[G] module Vo such that V = ( VO)F. Let V�2) denote the Fo[ G] module derived from V by applying the automorphism of Fo which sends a to a 2. Let V(2) = ( V�2 ) )F. LEMMA 1 1 .3. (i) There exists a submodule Eo of E with EIEo = Vo(G). (ii) EIS = T. (iii) S IT = V(2) .
PROOF. (i) Let E 1 denote the space of all matrices on V where X � A (X - l )XA (x ) for x E G. Then as F[G] modules EI V Q9 V* = V Q9 V = E. Let E2 be the set of all scalar matrices in E. Then E2 = Vo( G) is a submodule of E 1 • The result follows as El = E f. (ii) Define f : E � T by f(X) = X + X'. Then f is an F[ G] homomorphism of E onto T with kernel S. (iii) Without loss of generality F may be replaced by its algebraic closure. �
1'
I' I i.
190
[11
CHAPTER IV
Thus the map sending a to a 2 is art automorphism of F. Let x E G and let A (x ) = ( aij ) . Then A (x )' XssA (x ) = ( aSiasj ) = L a ;i Xii + X 0
Let M = ( mij ) be an upper triangular matrix in E, i.e. Define the quadratic form OM on V by
OM (v) = L mijXiXj where i,j
v =
mij =
0 for i > j.
(xt, . . . , Xn ) .
Define fM E HomF (S, F) by
fM « Xij )) = L mijXij. i,j
LEMMA 1 1 .4. Let M be an upper triangular matrix in E. Then OM is G-invariant if and only if fM E HomF[ G ] (S, F) where F = Vo(G).
PROOF. Let v = (X l , . . . , Xn ) E V and let A
= ( aiJ E E.
Then
OM (vA ) = L mijaisajtXsxt. i,j,s,t Thus OM (vA ) = OM (v ) if and only if the following two equations are satisfied for all s < t. L mijaisajs = msS,
(1 1 .5)
i,j
L mij ( aiS ajt + ajS ait ) = i,j
SELF DUAL MODULES IN CHARACTERISTIC 2
191
THEOREM 1 1 .7. V is of quadratic type if and only if there exists a submodule So of S with S/So = Vo(G). PROOF. Every quadratic form on V is of the form OM for some upper triangular II).atrix M. Every map in HomF (S, F) is of the form fM for some upper triangular matrix M. Hence the result follows from (11 .4). 0
i
for some X E T.
1 1J
COROLLARY 1 1 .8. If V is not of quadratic type then there exists a nonsplit exact sequence O � Vo (G) � W � V � O. 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 O � To � T � Vo ( G ) � O. Thus by (1 1 .3) (iii) there exists an exact sequence O � Vo (G)� S/To � V (2) � 0. As Vo (G) is not homomorphic image of S it follows that the sequence is not split. Hence there also exists a nonsplit sequence of the required form. 0 a
COROLLARY 1 1 .9. If V is not in the principal 2-block then V is of quadratic type. PROOF. Clear by (1 1 .8). 0 COROLLARY 1 1 . 10. If G is solvable then V is of quadratic type.
mst .
(11 .6)
For all s < t
fM (A I XssA ) = L mijaisajs, i,j
fM (A ' XstA ) = L mij ( aiS ajt + aitajs ) . i,j Thus fM (A ' XA ) = fM (X) for all X E S if and only if (1 1 .5) and (11.6) are satisfied. Consequently fM (A IXA ) = fM (X) for all X E S if and only if OM is A -invariant. 0
PROOF. Induction on ' G ' . If ' G ' = 1 the result is clear. Suppose that ' G ' > 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 O2',2 (G) is in the kernel of V by (4. 12). As O2',2 (G) I (1) the result follows by induction. 0
1]
193
SOME ELEMENTARY RESULTS
PROOF. Let {L; } be the set of all conjugate classes of H and let E L;. Let nx ) = 0 if x E G - H. If C is a conjugate class of G with E C the definition of induced characters implies that Z;
z
CHAPTER V
Thus
=
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 R [ G ] corresponding to B. v is the exponential valuation defined on K with v (p ) = 1 . Xl, X2, .. are all the irreducible characters of G. 'P I, 'P2, . . . are all the irreducible Brauer characters of G. WS is the central character of K[ G ] corresponding to XS. If H is a subgroup of G and h : Z(K; H : H�K is linear, define G h : Z(K ; G : G�K by � h G (C) = 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 R are splitting fields for every subgroup of G. See for instance Brow§ [1972] , [1973] ; Hubbart [ 1972] ; Reynolds [197 1 ] . A
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 character of H. Let � G = �s asXs and let w be the central character of K[H] corresponding to �. Then
L asXs (l)ws s
= �G (l) w G .
I G : H I � ( I ) L� c w (L; ) = � G (I) w G ( C). . �
0
LEMMA 1 .2. Let H be a subgroup of G and let Bo be a block of H. Assume that Bo contains an irreducible character � such that �G is irreducible. Then B� is defined and �G E B � .
PROOF. Let �G = XS and let w be the central character of K[H] corre sponding to �. By (1 . 1) Ws = w G . Thus w G = Ws is a central character of R [ G] . 0 LEMMA 1 .3. Let Bo be a block of the subgroup H of G for which B� is defined. Let � be an irreducible character in Bo and let �G = �s as Xs . If B is a
block of G then
v( � a X (1) ) > ( � a X (1)) X
v
X
B
s s
V (� G (1»
B
s s
= v ( �G (1»
if B -I- B�, if B
=
Bf
PROOF. Let w be the central character of K[H] corresponding to �. Let e be the central idempotent in R [ G ] corresponding to B. Then Ws (e ) = 1 if XS E B, and Ws (e ) = 0 if XS g B. Also w G (e) 0 if B -I- B � and w G (e) = 1 if B = B �. The result now follows from ( 1 . 1). 0
=
COROLLARY 1 .4. Let Bo be a block of the subgroup H of G for which B � is defined. Assume that Bo and B � have the same defect. If � is an irreducible character of height O in Bo then some character XS of height 0 in B � occurs as a constituent of �G with multiplicity as ¢ 0 (mod p ). PROOF. Clear by (1.3) .
0
1 94
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[1
THEQREM 1 .5. Let B be a block of G with defect group D. Suppose that A B (C) -I 0 for a conjugate class C of G. Then there exists z E C with z E CG (D ) such that the p -factor y of z is in Z(D). PROOF. By (III.9.7) there exists a block Bo 2f NG (D) with defect group D such that HR = B. Let Ao = A o Since A R( C}f 0 there exists a conjugate class Co of NG (D) with Co � C and Ao(Co) -1 0. Choose z E Co. By (III.6.10) z E CG (D). If ( is an irreducible character in Bo then ( z ) -I O. Hence by (IV.2.4) y is conjugate in NG (D) to an element of D and so y E D. Since D � CG (z ) � CG (y ) it follows that y E Z(D). D B
PROOF. (i) Let Xr be a character of height 0 in B R . By (III.6.10) and (IV.4.8) there exists a p '-element x such that D is a Sp -group of CG (x ) and X' (x )wr (x - I ) 1'= 0 (mod ) If y E Z(D) then D is a Sp -group of CG (xy ) and Xr (xy ) Xr (x ) 1'= 0 (mod ) Thus v (wr (xy » = v (wr (x » = O. Hence Wr (xy ) 1'= 0 (mod ) for all y E Z(D ). Let ( be an irreducible character in Bo and let W be the central character of K[H] corresponding to (. Then G wr = W . Hence if y E Z(D) there exists a conjugate z of xy in H with W (z ) -I o. By (1.5) the p -factor of z is conjugate to an element of Z(Do). (ii) and (iii) are immediate by (i). (iv) If H = (I) then B R has defect 0 by (iii) and so contains exactly one irreducible character Xt . Hence if ( is the unique character in Bo then G ( = LXs (l)Xs and so by (1.3) G v (I G I) = v «( (1» = V (Xt (1)2) = 2 v (I G I). Thus v (I G I) = 0 contrary to assumption. D 7T .
7T
7T .
LEMMA 1 .7. Let B be a block of G and let be an automorphism of G such that X � = XS for all Xs E B. If y is an element in a defect group of B then y O- is conjugate to y in G. (J'
PROOF. By (IV.4.8) there exists a p '-element x and a character Xr of height
195
in B such that Xr (x )wr (x - I ) 1'= O (mod 7T) and a Sp -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 ) Xr(X ) 1'= 0 (mod ) If y and y O- are not conjugate in G then by (IV.6.3) ==
L
Xs E B
'
THEOREM 1 .6. Let Bo be a block of the subgroup H of G for which B R is defined. Let Do be a defect group of Bo and let D be a defect group of B R . Then (i) Every element in Z(D) is conjugate to an element in Z(Do). (ii) The exponent of Z(D) is at most equal to the exponent of Z(Do). (iii) If B� has defect 0 then B R has defect O. ' (iv) If p i I G I then H-I (I) .
==
INERTIA GROUPS
I Xs ( YX ) 1 2 =
7T
L
Xs E B
.
xs « yx )" )Xs (yx ) = 0
Since I Xs (yx )I � 0 for all s this implies that Xs (yx ) = 0 for all XS E B. Hence in particular X' (yx ) = 0 contrary to the previous paragraph. D 2.
Inertia groups
Throughout this section G is a fixed normal subgroup of G. In general a , tilde sign will be attached to the quantities associated with G. For instance X\, X 2 , . . . are all the irreducible characters of G. WS is the central character of K[ G ] corresponding to Xs . Let A be a subset of G which is a union of conjugate classes of G. Let 0 be a complex valued function defined on A which is constant on the conjugate classes of G which lie in A. If z E G define OZ by OZ (x ) = O(x z-) for all x E A . Then O Z is defined on A and is constant on the conjugate classes of G which lie in A. The inertia group T( 0) of 0 in G is defined by T(8) = {z I z E G, OZ = O} . Clearly G � T(O) . If W is an irreducible R [ G ] module which affords cPi then it is easily seen that T ( W) = T (cPi )' Similarly if V is an R -free R [ G ] module which affords Xs then it is easily seen that T( VK ) = T(Xs ). However T( V) need not be equal to T( VK )' Let B be a block of G. The inertia group T(B ) of 13 in G is defined by
I
T(13 ) = {z z E G, 13 Z = B} .
LEMMA 2. 1 . Let 13 be a block of G. If Xs, cPi E 13 then T(cPi ) � T(13) and T (Xs ) � T(13 ) .
PROOF. Clear by (IV.4.9).
D
THEOREM 2.2. For any group H with G � H � G let �s (H) be the set of all irreducible characters of H whose restriction to G have Xs as a constituent. Let S)l?(H) be the set of all irreducible Brauer characters of H whose restriction to G have 'Pi as a constituent. (i) If T(cPi ) � H then the map sending 8 to 8 G defines a one to one correspondence between �?(H) and S)l?(G) .
1 96
[2
CHAPTER V
2]
INERTIA GROUPS
197
(ii) If V is an irreducible R [G] module which affords a Brauer character in 2I? (G) then a vertex of V is contained in T (iPd. (iii) If T(Xs ) � H then the map sending () to () G defines a one to one correspondence between 2Is (H) and 2I s (G). (iv) If Xt E 2Is (G) then there exists an R -free R [G] module which affords Xt and whose vertex is contained in T(Xs ) .
PROOF. If such a chain exists then by (2.3) consecutive characters belong to blocks which cover a fixed block of G. Suppose conversely that XS and Xt are in blocks which cover a fixed block of G. Let Xm be an irreducible constituent of (Xs )o . By (2.3) there exists XT in the same block as Xt such that Xm is a constituent of (XT )O . The chain xs, XT, Xt satisfies the required conditions. 0
PROOF. (i) It suffices to prove the result i n case T(iPi ) = H. Let 1/1 E 91?(H). By (11.2.9) (I/I G ) H = 1/1 + () where iPi is not an irreducible constituent of (}o . Let YJ be an irreducible constituent of I/I G with YJH = 1/1 + (} 1 . By (111.2.12) 1/10 = e iP for some integer e > O and YJ (I) = I G : H I I/I(1) = I/I G (I). Hence I/I G = YJ E 2I?( G). Furthermore if 1/11, 1/12 E 2I?(H) and I/I ? = I/Ii' then ( I/I ?)H (1/Ii')H and so 1/1 1 1/12 ' Therefore the map sending () to (} G is one to one. It remains to show that it is onto. Let 'Pj E 2I?( G). Let V be an R [ G] module which affords 'Pj and let � be an R [0] module which affords iPi . By (111.2.12) Va is completely reducible and so I( Vo, W; ) � O. Thus by (111.2.5) I( VH' W�) � 0 and so I ( VH W) � O for some irreducible constituent W of W�. Since ( W:-I)o = IH : 0 I W; it follows that W; is a constituent of W Thus if 1/1 is the Brauer character afforded by W then 1/1 E 2I? (H). Since I ( V, W G ) = I( VH' W) � 0 the irreducibility of V and WG implies that V = W G and hence 'Pj = 1/1 G as required. (ii) Clear by (i). (iii) If p is replaced by a prime not dividing I G I this follows from (i). (iv) Clear by (iii). 0
THEOREM 2.5 (Fang [1961], Reynolds [1963]). Let 13 be a block of O. The map sending B to B G defines a one to one correspondence between the set of all blocks of T(13 ) which cover 13 and the set of all blocks of G which cover 13. Furthermore if B is a block of T (13 ) which covers 13 the following hold. (i) The map sending () to () G defines a one to one correspondence between the sets of all irre_ducible characters, irreducible Brauer characters respec tively, in BA and BA O . (ii) With respect to the correspondence defined in (i) B and B 0 have the same decomposition matrix and the same Cartan matrix. (iii) B and B G have a defect group in common. (iv) 13 is the unique block of G covered by B.
i
=
=
'
H.
block B of G covers a block 13 of G if there exists XS E B and Xt E B such that Xt is a constituent of (Xs )o . A
LEMMA 2.3. The blocks of G covered· by the block B of G form a family of blocks conjugate in G. If 13 is covered by B and XS E B then some constituent of (Xs )o belongs to 13. If Xt E 13 there exists XS E B such that Xt is a constituent of (Xs )O . PROOF. This is a reformulation of (IVA . I 0).
0
LEMMA 204. XS and Xt belong to blocks which cover the same block of 0 if and only if there exists a chain Xh Xs, Xh , . . . , Xj" = Xt such that for any m either Xjm and Xim +' are in the same block of G or (Xim )O and (Xim -l-J G have a common irreducible constituent. =
PROOF. Let B be a block of T(13 ) which covers 13 and let � be an irreducible character in B. By (1.2), (2.1) and (2.2) (iii) B O is defined, ( 0 E B 0 and B 0 covers 13. Suppose that B is a block of G which covers 13. Let Xt E B. By (2.3) (Xt )o has an irreducible constituent Xs E B. By (2.1) and (2.2) (iii) Xt = � G for some irreducible character � of T(B ) such that Xs is an irreducible constituent of (a . Let B be the block of T(13 ) with � E B. Then B covers B and B 0 = B by (1.2). Let 1/11 , 1/12 be irreducible Brauer characters of T (B ) in blocks which cover 13. If iP E 13 and iP is an irreducible constituent of (1/1 ?)o then iP is a constituent of (l/Ii )O for i = 1 , 2. Thus if I/I ? = I/I? then (1/11)0 and (1/12)0 have a common irreducible constituent which is in 13, and so 1/11 = 1/12 by (2.2). Therefore the map sending () to () G defines a one to one correspondence between the sets of all irreducible characters, irreducible Brauer characters respectively, of T(13 ) and G which lie in blocks that cover 13. 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 () of T (B) is in B if and only if () 0 E B o . Thus the map sendIng B to B O is one to one as required. Furthermore (i) and (ii) are proved. (iii) Let D be a defect group of B. By (111.9.6) D � D for some defect group D of B o . By (i) B and B 0 have the same defect. Thus D = D.
[3
CHAPTER V
198
(iv) Since R Z
=
R for all
z
E T(R) this is clear by (2.3).
PROOF. By (2.5) B = fj G for some block fj of T(R) which covers R. The result follows from (2.5) (iii). 0
�E� 3. 1 . _Let a � Z (R ; G : G ) n R [O]. If A is a central character of R [ G] then A (a ) A G (a ) . =
0
LEMMA 3.2. Let {Rim } be the set of all blocks of 0 where Rim = Bjn for some
E G if and only if i = j. Let eim be the centrally primitive idempotent of R [ 0] with Rim B (eim ). Let ei Lm eim . Then {ei } is the set of all primitive idempotents in Z(R ; G : G) n R [0] . =
PROOF. Any idempotent in Z (R ; G : G) n R [ 0] is a sum of centrally primitive idempotents in R [0] and is invariant under conjugation by , elements of G. The result follows. 0
LEMMA 3.3 (Fang [1961]). Let Rim and eim be defined as in (3.2). For each i let Bil , Bi2 ' . . . be all the blocks of G which cover some Rim and let eim be the centrally primitive idempotent in R [ G ] with Bim B ( eim ). Then ei Lm eim = L m eim .
for all
Aa (a ) = AB (a )
PROOF. By (3.2) and (3.3) B covers R if and only if Aa (e) = AB (e) for every idempotent in Z(R ; G : G) n R [0]. By (1.16.1) R is a splitting field of Z(R ; G : G) n R [0]. The result follows since two central characters of a commutative algebra are equal if they agree on all idempotents. 0 unique block B of G which covers R.
The notation of the previous section will be used in this section. For the results in this section see Brauer [1967a], [1968] ; Fang [1961] ; Passman [1969].
=
Then B covers R if and only if E Z(R ; G : G) n R [O] .
LEMMA 3.5. Suppose that GIO is a p -group. If R is a block of 0 there is a
3. Blocks and normal subgroups
z
a
199
BLOCKS AND NORMAL S UBGROUPS
O.
0
COROLLARY � .6. Let B be a block of G and let R be a block of 0 covered by B. Then T(B ) contains a defect group of B.
PROOF. Clear by definition.
3]
=
=
PROOF. By (3.2) {ei } is a s�t of pairwise orthogonal central idempotents in R [G]. Thus it suffices to show that eimei � 0 for all i, m. Let V be a nonzero R [G] module in Bim . By (2.3) ( VG )ei � O. Hence Vei = Veim ei � O and so eim ei � O. 0
LEMMA 3.4 (Passman [1969]). Let B be a block of G and let R be a block of
PROOF. Suppose that BI and B2 are blocks of G which cover 13. Let Ai = AB. for i 1 , 2. If C is a conjugate class of G consisting of p '-elements the� C E Z(R ; G : G)n R [0]. Thus by (3.4) A 1 ( C ) = A13 (C) A ( C ) Hence by (lV.4.2), (IV.4.3) and (IV.4.8) (iii) B1 = B2 • 0 =
=
2
,
A block B of G is regular with respect to 0 if AB (C) = 0 for all conjugate classes C of G which are not contained in O. B is weakly regular with resp'?..c t to 0 if there exists a conjugate class C of G with C .� 0 such that AB ( C) � 0 and the defect of B is equal to the defect of C. In case 0 is determined by the context the phrase "with respect to 0 " will be omit�ed. By (111.6.10) a regular block is weakly regular. Further more if AB (C) � 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 = R G for every block R of 0 which is covered by B. (iii) B = 13 G for some block R of O.
PROOF. (i) =? (ii). Let 13 be a block of 0 which is covere<.! by B. Let C be a conjugate cla.§s of G. !:f cg 0 then (A13 )G ( C) = 0 AB ( C). If C � G then by (3.4) A13 ( C) AB ( C) . Hence by (3. 1) (A13 )G = AB . (ii) =? (iii). Clear. (iii) � (i). If C is a conjugate class of G with cg 0 then AB (C) = (A13 )G (C) = 0 and so B is regular. 0 =
=
LEMMA 3.7. Let B be a block of G which is regular and let 13 be a block of O. Then B covers R if and only if B
=
13 G.
.
CHAPTER V
200
201
BLOCKS AND NORMAL SUBGROUPS
PROOF. If B covers B then B = B G by (3.6). If B = B G then by (3.1) , A8 (a ) = (AB ) G (a ) for all a E Z(R ; G : G) n R [G]. Thus by (3.4) B covers B. D
,
(iv) Let e = LacC where C ranges over all conjugate classes of G and E R. There exis.!s a conjugate class C consisting of p I-elements with C � G, iic � 0, AB (C) � 0 such that the defect of B is equal to the defect of C.
ac
LEMMA 3.8. Let B be a block of G. Then there exists a block of G which is regular and covers B if and only if B G is defined. In that case E G is the unique block of G which is regular and covers E.
PROQF. (i) :::;> (ii). There exists a conjugate class C of G with C � G and AB (C) � 0 su�h that B_ and C have a common defect group D. By (3.4) AB; (C) = A8 (C) = AB (C) � 0 and so by (III.6.10) a defect group of Bi is a
PROOF. If E G is defined then E G is regular by (3.6) and E G covers E by (3.7). The converse follows from (3.7). The last sentence is clear. D
. subgroup of some defect group of B. (ii) � (iii). Trivial. _ (iii) :::;> (iv). Since AB (e) = 1 it follows that iicAB ( C) � 0 for s2me C. By (IY.7.2) C consists of p '-elements. By (3.3) C � G. If ei = Lai,cC for each i then for some i ii i, C � O. Hence by (IV.7.3) the defect of C is equalJ:o the defect of Bi and so is at most equal to the defect of B. Since A (C) � 0, (III.6.10) now implies that B and C have the same defect. (iv) :::;> (i). Trivial. D
LEMMA 3.9. Let B be a block of G with defect group D. If C G (D) � G then B is regular. PROOF. If AB (C) � 0 for some conjugate class of G then by (III.6.10) D � C G (z ) for some E C. Hence z E C G (D) � G and so C � G. 0 z
LEMMA 3. 10. Suppose that P is a p -group with P <J G and C G (P) � G. Then every block B of G is regular with respect to G. For every block E of G, the block E G is defined and is the unique block covering E.
PROOF. Let B be a block of G and let D be the defect group of B. By (III.6.9) P � D. Thus C G (D) � C G (P) � G �nd B is regular by (3.9). Let B be a block of G. By (3.8) E G is defined and is the unique block covering B. D COROLLARY 3.11 . Suppose that P is a p-group with C G (P) � P <J G. Then the principal block is the unique block of G. PROOF. By (3. 10) with G = P, E G is the unique block of G where E is the unique block of G. Hence G has exactly one block which must necessarily be the principal block. D
THEOREM 3.12 (Fang [1961]). Let E be a block of G and let BJ, B 2 , be all the blocks of G which cover E and let B = Bj for some j. For each i let ei be the centrally primitive idempotent of R [ G] with Bi = B ( ei ) and let e = Lei . The following are equivalent. (i) B is weakly regular. (ii) A defect group of any block Bi is conjugate to a subgroup of a defect group of B. (iii) The defect of any Bi is at most equal to the defect of B. • • •
COROLLARY 3. 13. Let E be a block of G. The following hold. (i) There exists a weakly regular block B of G which covers B. (ii) Let 13 be a block of T(E) which covers E. Then 13 is weakly regular if and only if 13 G is weakly regular. PROOF. (i) Choose a block B of maximum defect among those which cover B. By (3.12) B is weakly regular. (ii) By (2.5) 13 and 13 G have the same defect. Thus E has maximum defect among all blocks 13i of T(B ) which cover E if and only if E G has maximum defect among all blocks E f'. The result follows from (3.12). D THEOREM 3.14 (Fang [1961]). Let E be a block of G and let B be a block of G which covers B and is weakly regular. There exists a defect group D of B with D � T(E). For any defect group D of B with D � T(E) I T(B ) : DG I � O (mod p ) and D n G is a defect group of B. PROOF. By (2.6) T(B ) contains a defect group D of B. Thus by (2.5) and (3. 13) (ii) it may be assumed that G = T(E). Let e be the central idempotent o f R [G] with E = B (e). Since G = T(B ), e = LacC 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 � G, iic � 0, AB ( C) � 0 such that D is a defect group of C. Let E C with D � C G (z ) . Let L be the conjugate class of G with E L. Thus C = U L where x ranges over a cross section of N G (L) in G. By (3.4) z
x
x
z
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CHAPTER V
A B (C) = AB (C) =
2:X AB (L X ) = 2:X A�
Since T(13 ) = G this implies that
BLOCKS AND QUOTIENT GROUPS
-1
(L) .
be an R [ G ] module in a natural way. Similarly every character or Brauer character of GO is a character or Brauer character of G.
to
I T(13 ) : NG (L ) I AB (L ) = AB ( C ) � O .
Thus I T(13 ) : NG (L )I � O (mod p ) and AB (i ) � o. Since NG (L ) = CG (z)G and D is a Sp group of CG (z ) it follows that I T(13 ) : Da I � 0 (mod p ). . GeAB (L Slllce ) 0 it follows from (IV.7.3) that 13 and L have a common defect group. Clearly D n a is a Sp -group of Co (z ) and so is a defect group of L. D -
A
=
LEMMA 3.15. Let B be a weakly regular block of G and let 13 be a block of G covered by B. Let XS be of height 0 in B and let Xt be a constituent of (Xs )a with Xt E B._ Then the r�mification index of Xs is not divisible by p, Xt has height 0 in B_ and I T(B ) : T(Xt )1 � O (mod p ) . PROOF. By (2.5) and (3. 13) it may be assumed that G = T(13 ). Let D b� a defect group o! B. Let d = v (I D /) and let d = v (I D n a l). By (3. 14) d . the defect of B. By (III.2.12) (Xs )6 = e LzX : where { z } is a cross sect�on <:f T(Xt ) In G and e is the ramification index of XS. Since GDIG = DIG n D it follows from (3.14) that v(1 G I) - d = v ( 1 aD : D I) = v(1 a : a n D /) v(1 a I) - d. Thus v (Xs (l» = v(l a l) - d. Since v (Xt (1) � v (l a l) - d and Xs (l) = e I G : _T(Xt ) IXt (1) it follows that e I G : T(Xt ) 1 � 0 (mod p ) and Xt is of height o in B. D IS
=
COROLLARY 3.16. Let 13 be a block of a. There exists Xt E 13 of height 0 such that T(Xt ) contains a Sp-group of T(13 ). PROOF. By (3.8) there exists a block B of G which covers 13 and is weakly reg�lar. Let XS be of height 0 in B. By (2.3) some constituent Xt of (Xs )6 is in B. The result follows from (3. 15). D
4.
203
Blocks and quotient groups
The notation of the two previous sections will be used in this section. Furthermore GO = G l a and in general a superscript ° will be attached to the quantities associated with GO. Every R [GO] module may be considered
LEMMA 4. 1 . Let B O be a block of GO. Then there exists a block of G which contatns B ° and covers the principal block of a. Conversely if B is a block of G which covers the principal block of a then B contains a block of G O . PROOF. Two R [ GO] modules in B O are linked by a chain of R [G O] modules. Hence they are also linked as R [G] modules and so lie in the same block B of G. Thus B O � B. If XS E B O � B then the principal character of a is a constituent of (Xs )6 . Thus by (2.3) B covers the principal block of a. Conversely suppose that B covers the principal block of a. By (2.3) there exists XS ,E B such that (Xs )6 contains the principal character of a as a constituent. Thus by (III.2.12) a is in the kernel of XS . Let B O be the block of G O with XS E B O . Then B O � B by the previous paragraph. D LEMMA 4.2. Let B O be a block of GO and let B be a block of G with B O � B. Let D be a defect group of B. Then D contains a Sp -group of a and D O = Dala contains a defect group of B O . In particular if d is the defect of B and d O is the defect of B O then d O � d - v(1 G I). Furthermore if B O contains an irreducible character or an irreducible Brauer character of height 0 in B then D O = Dala is a defect group of B O . 'PROOF. Let P be a Sp -group of a and let P be a p -group with P � P such that pO = pala is a defect group of B O . Let V be an R -free R [G O] module which affords an irreducible character of height 0 in B O . Thus p O 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 � G D. If v (ep; (1» or v (Xs ( 1 » equals v ( I G I ) - d then v ( I G I ) - d � v ( I G : G I) - d () and so d () � d - v (1 G I). Thus d O = d - v (1 G I) by the first part of the statement. D LEMMA 4.3. 5uppose that a is a p '_group. Let BJ , . . . , Bm be all the blocks of G which have a in their kernel and let Dj be a defect group of Bj for j = 1 , . . . , m. LetB? , . . . , B� be all the blocks of G O . Then m = n and after a suitable rearrangement Bj = B � for all j. Furthermore D 7 = Dj a I a is a defect group of B j , PROOF. Since a is a p I-group the principal character is the only irreducible character in the principal block of a. Hence a block of G covers the principal block of a if and only if it has a in its kernel. Thus by (IV.4. 11)
any two modules in Bj are linked by a chain of R [ G O] modules and so B. a block of GO. There�ore by (4. 1) m = n and Bj = BJ after a suita�le rearrangement. Since G is a p i_group Bj and BJ have the same defect and so by (4.2) DJ is a defect group of BY - 0
character in B has D in its kernel, 'Pi is the unique irreducible Brauer character in B. Then XS has height 0 in B by (4.5). Since (Cii ) is the Cart an matrix of B, (IV.4.I6) (i) implies that Cii I D I proving the last statement. 0
PROOF. Le! d = v (/ D /). Choose 'Pi E B with V ('Pi (1)) = v (/ G /) - d. By (IIL2.I3) G is in the kernel of 'Pi . Let BO be the block of GO with 'Pi E BO. By (4.1) B o � B. Let dO be the defect of BO. The result follows from (4.2). 0
THEOREM 4.7. Suppose that G = DC G (D) for some p-group D. Let B be a block of G with defect group D and let X = Xs be the canonical character of B. Let ( , ( be all the irreducible characters of D. Define (}j as follows : if y � D, Odz ) = 0 = (i (y ) x (x ) if y E D where y is the p -part of z and x is the p I-part of z. Then {{}j U = 1 , . . . , n } is the set of all irreducible characters in B.
204
CHAPTER V
LEMMA 4.4. Suppose that Cj is a p-group. Let B be a block of G and let D be a �efect group Of B. Th: G � D and there exists a block BO of GO such that B � B and D = D IG �IS a defect group of B O.
LEMMA 4.5. Suppose that {; is a p-group which centralizes all p i-elements in G. The inclusion BO � B defines a one to one correspondence from the set of all blocks of G o ont the se,f ot ll blocks of G. If D is a defect group of the block B of G then D� = DG I G �IS a defect group of the block B ° of GO where BO � B.
PROOF. In view of (4. 1) and (4.4) it suffices to show that if B is a block of G and B ? , B g are blocks of GO with B� � B and BO2 C B then BO = BO . Let XS E B � and XI E B g . If .t is a p i-elemen t in G then {; � C 2 (x). Thus C G (x )! G = C G o (XO) where XO is the image of x in GO. Since Xs, XGI E B this implies that o I GO I � = ws (x ) - I (x ) I G l XI x ) = w = / C G o (XO) / XI (( l ) (mod 1T ) . / C G o (XO) / Xs (1) for all p i-elements x in G. Hence by (IV.4.3) and (IV.4.8) B? = i3 g . 0 _
1
_
COROLLARY 4.6. Suppose that D is a p -group and G = D C G (D). Let B be a block of G with defect group D. Then B contains exactly one irreducible character XS which has D in its kernel and B contains exactly one irreducible Brauer character 'Pi . Furthermore Xs (x ) 'Pi (x) for all p i-elements x in G, Xs . 0 and
PROOF. By (4.5) there exists exactly one block BO of GID with BO C B. " �urthe�more BO is a block of defect O. Let Xs, 'Pi respectively, be the unique IrreducIble character, irreducible Brauer character respectively, in BO• Thus Xs (x) = 'Pi (x) for all p i-elemen ts in G. Since every irreducible Brauer
205
BLOCKS AND QUOTIENT GROUPS
=
If G = DC G (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.
1 , . . .
n
PROOF. If z E G then n = (j for all j. Thus if XI E B then by (IIL2.I2) (XI )D = e(j for some j and some integer e > O. Let z E G, let y be the p -part of z and let x be the p i-part of z. If y � D then XI (z ) = 0 by (IIL6.8) and (IV.2.4). Suppose that y E D and let H = D x <x > . Then (XI )H = (ja for some character a of (x > . Thus XI (xy ) = (j (y )a (x ) and XI (xy ) - £LW (j (1) {}j ( xy ) (j (1) dli X (x ) - � (j (1) XI (x ) - £LW
where 'Pi E B. Consequently XI = [dti l(j (1)] {}j . Let A be the set of all p I-elements in G and let A be the image of A in GO = GID. Since X is a character of GO which vanishes on all p -singular elements we see that -
-
°
I II} I I' = I b I ,�) �} (y ) 1 ' '�A I X (x l I ' = I%l .�" I X(x ) I '
Thus Hence dli
=
I (z ) 2 = 1 . I G 0 1 zE2:G O I x I
i 112 d;i . 1 = I XI 1 12 = (j d; (1 )2 I {}j = (j (1 )2
=
(j (1) and so XI
=
{}j.
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CHAPTER V
[5
Therefore there exists a subset S of {I, . . . , n } such that { OJ /j E S} is the set of all irreducible characters in B. If XI = OJ then dti = ?;j (1) . Thus
I D 1 = Cii = L d;i = L ?;d1t I jES This implies that S = {I , . . . , n } .
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 Brow§ [1979], Broue [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.
LEMMA 5 . 1 . Let P be a p -group, p e G. Let N = NG (P) and let H <J N with CG (P) C H If Bo is a block of H then B t? is defined. PROOF. By (3. 10) Bi;' is defined. Thus by (III.9.2) and (III.9.4) Bf? = (B �f is defined. 0
LEMMA 5.2. Let D be a p-group, D � G. Let N = NG (D) and let H
map sending Bo to B i;' defines a one to one correspondence between blocks B of N and representatives Bo of N -conjugate classes of blocks of H. If B = B(? has defect group D tht:n by (3. 14) D n H is a defect group of Bo and v (I DH I) = v (I T(Bo) I) . Suppose that Bo satisfies (i) and (ii). Let DI be a defect group of B f? By (3. 14) and (i) D n H and DI n H are both defect groups of Bo . Thus I D n H I = I DI n H I · By (3. 14) and (ii) v (I DH I) � v (I T(Bo) /) =
5]
PROPERTIES OF THE BRAUER CORRESPONDENCE
v(1 DIH I). Thus I D I � I DI I · Since D
<J
207
N, D � Dj • Therefore D = DI.
D
Let D be a p -group, D � G. (D, B ) is a block pair in G if B is a block of (D ) with defect gr04P D. If (D, B ) is a block pair in G then by (III.9.4) B A is defined for every group A with D CG (D) � A C G. Let (D, B ) and (D *, B *) be block pairs in G. (D *, B *) weakly extends (D, B ) if D <J D *, D * n CG (D ) � D and B Y = B for all y E D * . (D *, B *) extends (D, B ) if D <J D * and B H = B *H where H = D CG
D *Co (D ). (D *, B *) properly extends (D, B ) if (D *, B *) extends (D, B ) and D r£ D *.
LEMMA 5.3. Suppose that (D *, B *) is a block pair which extends (D, B ). Let H = D *Co (D). Then (D * , B *) we'akly extends (D, B ) and D * is a defect group of B H = B *H. be a defect group of B *H with D * � Do . By (3. 14) Do Ca (D ) = H and Do n DCo (D) = D. Thus Do n D Co (D) = D * n D Co (D) and Do/D = H/D Co (D ) = D */D. Hence I D * I = I Do l and so D * = Do . By (3. 1 0) 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 Bo be a block of Go with defect group D. If Co (D ) � Go there exists a block pair (D, B ) in G with B Oo = Bo . o . (ii) Let (D, B ) be a block pair in G. Let Do be a defect group of B . Then I D I :-;;; I Do I · Furthermore I D I < I Do I if and only if there exists a block pair in G which properly extends (D, B ) . (iii) Let (D * , B *) be a block pair in G which weakly extends the block pair (D, B ) in G. Let ?;, ?; * respectively, be the canonical character in B, B * respectively. Then ?;� G (D *) is irreducible. Furthermore (D *, B *) extends (D, B) if and only if the multiplicity of ?; � G(D *) as a constituent of ?; CG (D * ) is not divisible by p. PROOF. (i) Since D = D n D Ca (D) this follows from (5.2). (ii) By (III.9.6) I D I :-;;; I Do I ·
208
CHAPTER V
[5
Suppose that (D * , B *) is a block pair which properly extends (D, B ). Let H = D *Co (D ). Then B O = (B H )O = (B * H )O and so I D I < I D * I ::s:: I Do i
6 . Blocks and their germs
Let B be a block of G. Suppose that B = B 5' 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 [1 967] . The results in the rest of the section are related to work of Reynolds [1970]. Let B be a block with defect group D. By (III.9.7) there exists a unique block B I of No (D) such that B ? = B. The block B I is called the germ of B with respect to D. If D is determined by the context BJ is called the germ of B.
by (lII.9.6). Assume conversely that I D I < I D o I . Let T be the inertia group of B in No (D), let P be a Sp -group of T and let H = PCo (D). Let D * be a defect group of B H . By (3. 14) H = D *Co (D) and D * n D Co (D) = D. By (i) there exists a block pair (D *, B *) in G with B * H = B H . Thus (D *, B *) extends (D, B ). By (5 .2) v( I H I) = v (I T I) > v(I D Co (D) I). Thus D � D * and so (D * , B * ) properly extends (D, B ). (iii) Let A be a group with Co (D *) C A <J D *Co (D *). Then D *Co (D *)/D * = A /A n D * . Since D * is in the kernel of � * this implies that �! is irreducible. Hence in particular � � G (o') is irreducible. Let 8 be an irreducible character of A which has D * n A ·· in-its kernel such that £ �G(D*) is a constituent of 8 CG (0*) . Then A = Co (D *)(D * n A ). Thus �! is a constituent of 8 as D * n A is in the kernel of both 8 and � ! . 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 Co (D *). Since D is in the kernel of � it follows that D is in the kernel of every irreducible constituent of � OCG (D*) . Thus if a is the multiplicity of � �G(D *) as a constituent of � CG (D *) (III.2.S) implies that DC G (D
= (�, {� b CG (D .)} » . Since D *Co (D *) n D Co (D) = D Co (D *), (II.2.9) implies that a = (�, { � bcG (O · )}D CG (D » = (�, {� * H } D CG ( D » where H = D *Co (D). Let � * H = 8 + 80 where 8 E B H and 80 is a sum of irreducible characters none of which is in B H . Thus « 80)DCG (D), �) = O. Since D is in the kernel of 8 it follows that � is the unique irreducible character of D Co (D) which is a constituent of 8 OCG (D) . Hence
a = (�, {� * H } OCG ( D » = (�, 8 D CG (D » and so 8 OCG (O) = a{ Therefore 8 (1) = a�(1). However v(�(1» = v(I DCo (D) : D I) = v( I H : D * I) = v(I H : D *Co (D *) I � *(1» = v� * H (1» . Consequently v(8 (1» = v(a ) + v (� * H (1» . Thus (1 .3) implies that B H = B * H if and only if v (a ) = O. 0
209
BLOCKS AND THEIR GERMS
6]
LEMMA 6. 1 . Let Bo be a block of a subgroup Go of G and let Do be a defect group of Bo . Suppose that Co (Do) C Go . Then B = B 5' is defined and
Z(D) C Z(Do) C Do C D for some defect group D of B. In particular if B has an abelian defect group then Do is a defect group of B.
I
PROOF. By (5.4) (i) B = B 5' is defined. Let d be the defect of B and let do = v (I Do I) be the defect of Bo . The proof is by induction on d do . If d do = 0 then by (III.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 Co (Do) . Suppose that d do > O. By (5 .4) (ii) there exists a block pair (D *, B *) in G which properly extends (Do , Bo). Since B *0 = B 5' = B induction yields the existence of a defect group D of B such that -
-
-
Z(D) C Z(D *) C C o * (Do) = Z(Do) C Do C D * C D . This proves the first statement. The second statement is an immediate consequence of the first. 0 A more direct proof of the next result can be found in Brauer [1964a] Theorem 3. See also Brauer [ 1 971 a] , [1971b] ; Osima [1964] ; Hamernik and Michler [1973] ; Hamern*- [1973] . THEOREM 6.2. Let Bo be a block of a subgroup Go of G and let Do be a defect group of Bo. Assume that Co (Do) C Go. Then Bo is the principal block of Go if and only if B 5' is the principal block of G.
CHAPTER V
210
[6
6]
BLOCKS AND THEIR GERMS
211
PROOF. By the third main theorem On blocks (SA) (i) Bo = B fo for some , · block B I of Do Co (Do). Thus it suffices to prove the result in case
isomorphism Go HIH = Gol Go n H. Furthermore E f is defined and B f = E f . Also I H : Go n H I == 1 (mod p ) .
For any subgroup A of G let B I (A) be the principal block of A and let (; 1 (A ) be the principal character of A. Let D be a defect group of B f and let P be a Sp -group of G with D � P. Let do = v(1 Do I) and let d = v(1 D I). The proof is by induction on d do . Suppose that d do = O. By (III.9.6) it may be assumed that D = Do . By (3. 10) Bo = Bl (Do Co (Do» if and only if B� = Bl (N) where N = No (Do) . Let V be the R -free R [G] module which affords (; r (G). Let f be the Green correspondence with respect to (G, P, No (P» . Thus f( V) affords (; 1 (No (P» . Thus by (III.7.7) and (III.9.7) if B is a block of No (P) then BJ (G) = B o if and only if B = Br (No (P» . Suppose first that Bf = B I (G) then Do = LJ = P and so B� = Bl (N). Assume conversely that B � = Bl (N). Thus Do is a Sp -group of N and so Do is a Sp -group of G. Hence Do = D = P and B f = (B�f = Bl (G). Suppose that d - do > O. By (SA) (ii) there exists a block pair (D * , B *) in G which properly extends (Do, Bo) . Assume that B f = B 1 (G). Thus B * 0 = B 1 (G) and so by induction B * = BJ (D *Co (D *» . Thus by induction Bf( = B * H = B 1 (H) where H = D *Co (D ). By (3.7) Bf( covers Bo and so Bo = Bl (Do Co (Do» . Assume conversely that Bo = BJ (Do Co (Do» . Hence (; 1 (DoCo (Do» is the canonical character of Bo . Thus by (SA) (iii) (;1 (D *Co (D *» is the canonical character of D *Co (D *). Therefore B * = B l (D * Co (D *» and o so by induction B f = B * = B J (G). D
PROOF. The isomorphism Gol Go n H = Go HIH defines a one t o one map (; � t from the set of all irreducible characters of Go with Go n H in their kernel to the set of all irreducible characters of GoH with H in their kernel. Thus too = (;. 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 f. Let (; be an irreducible character in Bo . By (1 .3) there exists an irreducible character X in B f with X � (; 0. Thus (; � X o by Frobenius reciprocity (III.2.S). Since Go n H is in O the kernel of XOv ' it is in the kernel of (;. Since (; was arbitrary in Bo this implies that Go n H is in the kernel of B o . Suppose conversely that Go n H is in the kernel of Bo . By Frobenius reciprocity (111.2.5) t � (; Oo H . Since H <J G a cross section of GoH in G is a complete set of representatives of the (Go, H) double cosets in G. Thus the Mackey decomposition (II.2.9) implies that
Go = DoCo (Do) .
-
-
COROLLARY 6.3. Suppose that y is a p -element in G such that Co (y) has a normal p -complement. If X = Xi is in the principal block of G then X (yx ) =
X (y )
for every p'-element x in Co (y ) .
PROOF. By (IV.6. 1) and (6.2) X (yx) = Lj d � 'PI ( x ) where 'P I ranges over the irreducible Brauer characters in the principal block of Co (y). By (lVA. 12) (ii) 'P J (x ) = 'PJ(I) for all j and all p '-elements x in Co (y) . D
LEMMA 604. Let H be a normal p '-subgroup of G. For any subset A of G let .Ii denote the image of A in G = G I H. Let Bo be a block of a subgroup Go of G for which B f is defined. (i) H is in the kernel of B f if and only if H n Go is in the kernel of Bo . (ii) Suppose that Go n H is in the kernel of Bo . There exists a block Eo of GoH with H in its kernel such that Bo corresponds to Eo by the natural
where x ranges over a cross section of Go H in G. Therefore
o Thus t is the sum of all the irreducible constituents of (; G which have H o in their kernel. Hence «(; ° r = t . If B is a block with B � T then V « (; O )B (1» � v «(; O (1» by (1 .3). Since o G (;0 (1) = I GoH : Go I t (1) it follows that v « (; G (1» = v(t (1» . Since o «(; ° r = t this implies the existence of a block BJ � T with V « (; O )Bt (1» = v«(; G (I» . Hence B l = B f by (1 .3) and so H is in the kernel of B f . (ii) The existence o f E o follows from (4.3). Let (; b e an irreducible character in Bo . Let w, w be the central character corresponding to (;, t respectively. Thus � corresponds to Eo . Let X I , X2 , . . . be all the irreducible o characters of G and let bs = (Xs , (; G ). By (i) «(; ° r = t . Thus by (1.1) ,
o
Since B f � T it follows from (IVA.22) that « (; G (1)1 (; G (1» w 0 = w . o Evaluating at 1 we get that « (; 0 (1)1 (; 0 (1» = 1 . Therefore w 0 = w . Conse quently E f is defined and E f = B f . Furthermore
Hence
I H Go n H I :
==
1 (mod p ).
6]
[6
CHAPTER V
212
213
BLOCKS AND THEIR GERMS
PROOF. By (6.4) it may be assumed that Do is a defect group of Bo.
0
(i) Let d = v (I D I) and let do = v (I Do I). The proof is by induction on d - do. If d do = 0 then Do = D Z and by (3. 10) the germ of B with respect to D Z covers Bo. The result follows since 07T ' (C O (D Z » = 07T, (No (D Z » . Suppose that d do > O. By (5.4) (ii) there exists a block pair (D *, B *) in G which properly extends (Do, Bo). Let H = D *Co (Do). Suppose that D * C D Z . Let Ho be the kernel of Bo. By (IV.4. 1 1) Ho is a p i_group. Furthermore Ho is the kernel of B f/ = B * H and 071', (Co (D Z » C Ho . It follows from (6.4) that 07T, (Co (D Z » C Ho n D * Co (D *) and so is in the kernel of B * . The result follows by induction. (ii) Since the kernel of Bo contains all 'TT '-elements in Co (Do) the
THEOREM 6.5. Let 'TT be a set of primes with p E 'TT. Let B be a block of G
-
with defect group D. Let N = No (D) and let B l be the germ of B. Assume that 0 ,(N ) is in the kernel of BI . Suppose that Go is a subgroup of G, Bo is a block of Go with defect group Do, Co (Do) C Go and B � = B. Then 071" ( Go) is in the kernel of Bo .
-
7T
PROOF. By (111.9.6) it may be assumed that Do C D. By (5.4) Bo = B i'o for
some block Bz of Do Co (Do). Let d = v(I D I), do = v(I Do I). The proof is by induction on d - do . Suppose that d - do = O. By (111.9.2) B i' = B � = B and B i' = (B�) G. Thus B � = BI by (111.9.7). By (3. 10) Bl covers Bz. Therefore 0 ,(N) = 071',(D Co (D » is in the kernel of Bz• Thus 07T,(GO) n D Co (D ) is in the kernel of Bz and so 071',(GO) 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 (Do, Bz). Let H = D *Co (Do). By induction 071', (H) is in the kernel of B � = B * H. Since B � covers Bz it follows that 0 7T ,(H) = 07T,(DoCo (Do» is in the kernel of Bz• As 07T,(GO) n DoCo (Do) C 071',(DoCo (Do» , (6.4) implies that 07T'( Go) is in the kernel of Bo = B fio . 0
assumptions of (i) are satisfied. Thus the result follows from (i).
71'
The following example shows that the assumptions in (6.7) (ii) that
-
with defect group D. Assume that 07T,(No (D » is in the kernel of the germ of B. Suppose that y is a p-element in G such that Co (y ) has a normal S71',-subgroup. If X = Xi E B then X (yx ) = x (y ) for every 'TT ' -element x in Co (y ) . 'P I ranges over blocks of Co (y ) which have 07T (Co (y » in their kernel. Thus 'P Hx ) = 'P Hi ) for every 'TT I -element x in Co (y ). 0
071" (Co (Do» is a S7T'-group of Co (Do) cannot be dropped even in case 'TT = {p } . Let (y ) be a cyclic group of order 4. Let S = S5 be the symmetric group on 5 letters and let A = A5 �e the alternating group. Let G = (y ) S where (y ) <J G and y Z = Y for E A, Y Z = Y - I for E S - A. Let p = 2. Let t be the irreducible character of degree 4 of S which is a constituent of the permutation representation of S on 5 letters. Then tA is irreducible. Thus tA is in a block of defect O. By (3.5) the block of S containing t is the only block of S covering that of tAo Thus it is weakly regular. Hence by (3 .14) there exists an involution x E S - A such that (x ) is a defect group of the block of S containing t. 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 t 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 DoCo (Do) = (y ) A and by (3. 10) B = B � for some block Bo of Do Co (Do). Oz, (Do Co (Do» = (l) is in the kernel of Bo . lt is easily verified that No (D ) = ( x y ) W Thus W' = Oz, (No (D» . Hence by (4.3) the principal block of No (D ) is the only block which has W' in its kernel. Thus by (6 2) Oz' (No (D» is not in the kernel of the germ of B. z
COROLLARY 6.6. Let 'TT be a set of primes with p E 'TT. Let B be a block of G
PROOF. By (lV.6. 1) and (6.5) X (yx ) = Lj d � 'P Hx ) where
_
z
,
THEOREM 6 . 7 . Let 'TT be a set of primes with p E 'TT. Let B be a block of G
with defect group D. Let Do be a p -group contained in G and let Bo be a block of DoCo (Do) such that B � = B. (i) Suppose that whenever Do C D z for some E G, 071" (Co (D Z » is in the kernel of Bo. Then 07T, (No (D» is in the kernel of the germ of B. (ii) Suppose that 07T ' (C O (Do» is a S7T ' -groUP of Co (Do) and 07T ' (C O (Do» is in the kernel of Bo. Then 07T, (No (D» is in the kernel of the germ of B. z
-
0
.
.
THEOREM 6.8. Let L, M, HI, Hz be subgoups of G with LHI C MH2 and
H1 , H2 p i_groupS. Assume that [M, H2] C H2, [L, Hd C HI, L C M and H2 n HI L C HI . Let B be a block of L with defect group D such that CM (D) C L. Assume that HI n L is in the kernel of B. Let B be the block of LHI which has HI in its kernel and corresponds to B by the natural
CHAPTER V
214
[6
isomorphism from L /L n HI to LHdHI . Assume furthermore that B MH2 is defined. Then B M is defined and H2 n M is in the kernel of B M. If B M is the block of MH2 corresponding to B M by the natural isomorphism of M/M n H2 onto MH2/ H2 , then 13 MH2 = B M. The following diagram illustrates the statement. M ----
B
'
L
---
PROOF. By (111.9.5) B M is defined. Since M n H2 <J M and M n H2 n L � L n HI it follows from (6.4) (i) that M n H2 is in the kernel of B M. Since [L, H2] � H2 and L n H2 � L n HI is in the kernel of B, there exists a block B 2 of LH2 with H2 in its kernel such that B 2 corresponds to B by the isomorphism L /L n H2 = LH2/H2. By (4.3) B 2 has D as a defect group. The Frattini argument implies . that C MH21 H2 (DH2/H2) = C M (D )H2/H2 and so C MH2 (D ) � C M (D )H2 � LH2 . Thus by (I1I.9.5) B lj is defined for LH2 � H � MH2 . We will next show that (6.9) Let ? E B and let t E 13 2 with tL = r Let {Xs } be the set of all irreducible characters in B rH2 and let as = (XS, t MH2)MH2 • By (1 . 3 ) v «(MH2( 1)) =
(LasXs (1)). As ( is the unique irreducible character of LHI with ? � tL and HI is its kernel and since HI is in the kernel of each X" it follows from Frobenius 1J
reciprocity (I1I.2.5) that
as
= « XS ) L H J , () U-l J = «XS ) L , ?) L = «XS ) 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 {(XS )M } is the set of all irreducible characters in B '. Hence 13 ' = 13 MH2 . However
v ( L as (Xs ) M ( l) ) = v «(MH2 (1)) = v (I MH2 : LHl l x ( 1 )) = v (I M : L I ? (1)) = V (? M (1)) . Hence B ' = B M by (1 .3), and (6.9) is proved. Since H2 n HI L � HI it follows that HI H2 n LHI = HI . Consider the natural isomorphisms
7]
ISOMETRIES
215
L /L n HI = LHJHI = LHI/LHI n HI H2 = LHI H2/ HI H2 = LH2/ LH2 n HI H2 . Let B3 be the block of LHI H2 \vith HI H2 in its kernel which corresponds to B. Let B � be the block of LH2 with LH2 n HI H2 in its kernel which corresponds to B3 • Let ? be an irreducible character in B, let t be the unique irreducible character in B 2 with (L = ? and H2 in its kernel. Let () be the unique irreducible character of LH, H2 with HI H2 in its kernel such that ()L = r Then () E 133 by definition. Furthermore ()LH2 E B � . Since « ()LHJL = ? = (L it follows that B � = 13 2 . Thus 13 2 corresponds to 133• Since B iHJH2 is defined, (6.4) implies that B i H J H2 = B3 and so by (6.9) � B H2 = 13 rH2 = B M. By (6.4) B �H2 = 13 MH2 . Therefore 13 MH2 = B M . 0 In (6.8) it is essential to assume that B MH2 is defined. This does not follow from the other conditions. For instance let M = L = HI = (1) . Let l/J be the central character of (I) , let C� {I} be a conjugacy class of H2 and let C' = {x - I I x E C} . Then l/J H2 (C)l/J H2 (C' ) = 0 but l/J H2 (CC' ) = 1 C I � o .
7. lsometries
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 5 of blocks of N let V(N, A : 5) = 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(N, N) = V(N). M(N, A : 5 ) = Let M(N, A : X) = V(N, A : X) n M(N). Let V(N, A : 5 ) n M(N) where. 5 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 = LasXs with complex as let a Y = LXs E yaSXS ' If Y is the set of all irreducible characters in a union 5 of blocks of G let a s = a Y. It is evident that if a is a generalized
character of G then so is a Y. ' 'TT is a set of primes and 'TT is the complementary set of primes.
216
[7
CHAPTER V
If p E 7T then Sp ( 7T, N) is the union of all p -blocks B of N such that if D is a defect group of B then 07 T , (NN (D)) is in the kernel of the germ of B. If Z E G then Z-rr denotes the 7T -part of z. The 7T-section of G containing Z is the set of all elements in G whose 7T-part is conjugate to Z7r ' V7r (N, A : X) = {a I a E V(N, A : X), a (yx ) = a (y ) for any element y E A and any element x E 07r , (CN (y ))} .
M7r (N, A : X) = V7r (N, A : X) n M(N) . Similarly the subscript 7T applied to any of the sets defined above means that it is intersected with V7r (N, A : X). In this section we will be concerned with the following two hypotheses.
7]
21 7
ISOMETRIES
PROOF.
Let
C = CG (y). Since y, X-rr have relatively prime orders,
C c (X7r ) = CG (YX7r ) = CG « yx )7r ) � C . By (7.3)
(xxoY = xu for some Z E C,
u E 07r ' ( C) n C c (X7r ) = 07T' ( C) n CG « yx) 7r ) � 07T, (CG (yx ) ,, ) .
Therefore
{3 (yxxo) = (3 « yxx oY ) = {3 (yxu ) = {3 (yx )
by the definition of
V7T ( G).
0
7.5. (i) a E V7r (N, A : X) if a E V(N, A : X) and for y E A, is a linear combination of irreducible characters of CN (Y7T ) all of which have 07T (CN ( Y )) in their kernel. (ii) Suppose that (7.2) is satisfied. Then a E V7r (N, A : X) if and only if a E V(N, A X) and a is constant on 7T-sections in N. (iii) Suppose that (7. 1) is satisfied. If Z E G and Z7r E A then Z7r' = z [ Z2 with Zl E CN (Z7T ) and Z2 E 07r, (CG (Z7r )). If furthermore z : E A for some x E G and z :, = z � z � with z ; E CN (z :) and z � E O-rr, (CG (z :)) then a (z7T zl) = a (z :z O for all a E V7T (N, A ).
LEMMA a C N ( Y7T )
7T
HYPOTHESIS 7. 1 . (i) N is a subgroup of G. A is a subset of N which is a union of 7T-sections of N such that any two 7T-elements in A which are conjugate in G are conjugate in N. (ii) If y is a 7T-element in A then CG (y ) = CN (y )07r, (CG (y)). HYPOTHESIS (ii)
7.2. (i) (7. 1)
is statisfied.
07r, (CG (y)) is a S7T'- groUP of CG (y) for all 7T-elements y in A.
We first prove some elementary results.
LEMMA 7.3. Let H <J G, H a 7T'-subgroup. Let x E H, Y E G. Then there exists u E CG (Y7r ) n H and Z E H such that yx = (yu y . PROOF.
:
PROOF.
(i) and (ii) are immediate by definition. (iii) The first statement follows directly from (7. 1). If z : E A then by (7. 1) (i) there exists y E N such that z ; = Z7r ' Then z ? z ? = z ;' = (z ;Y (z �Y Since xy E CG (Z7r ), z ? and (z �Y are in 07T, (CG (Z7T ))' Let xy = uv with u E N and v E 07r' (CG (Z7T )). Then
Define T
= {(Y7r y t l z ranges over a cross section of CH ( Y7r ) in H, t E C H (Y7r Y } '
T � y7rH and ' T ' = ' H ' . Thus T = Y7T H. Since H <J G it follows that (yx )7r E Y7rH and so (yx ) 7r = (Y7r Y t for some Z E H and some 7T '-element t E CG (y �). Thus t = 1 and (yx )7r = y :. As H <J G, (yx )7r' = Y7r ' XO for some Xo E H. Hence (yx ):�I = (Y7T, xoy-1 = Y7r' u for some u E H. Since Y7T' u centralizes (yx ) :-- 1 = Y7r it follows that u E CG (Y7r) n H. Therefore (yx y-I = Y7T « yx ):�) = Y7r Y7r' U = yu. 0 LEMMA 7.4. Suppose that (3 E V7r ( G ). Let y be a 7T-element in G. Let x E CG (y ) such that x7r, y have relatively prime orders and let Xo E 07r ' (CG (y )). Then (3 (yx ) = (3 (yxxo) .
Then
.
==
Since
u, y, z : z ;
and
(Z7TZ IYv
==
(z7rz [Y
(mod
07T, (CG (Z7T ))) '
Z7T Zl are all in N it follows that
(z : z ;Y = (Z7TZ1Y ZO for some Zo E 07T, (CG (Z7T )) n N � 07T , (CN (Z7T )). Therefore
a (z7T zJ) = a «z7TzlYZO) == a «z :z ;Y ) = a (z : z O. 0 Suppose that (7. 1) is satisfied. If a E V7T (N, A ) define a T E V7T ( G ) as follows. If Z7r is not conjugate to an element of A let a T (z ) = O. If z : E A for some x E G then ZX E CG (z :) and so z :, = Zl Z2 where zl E CN (z :) and z2 E 07r, (CG (z :)). Let a T (z ) = a (z :z J). By (7.S) (iii) a T is well defined. Thus T is a linear mapping from V7T (N, A ) to V7T ( G ).
CHAPTER V
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[7
Suppose that (7.1) is satisfied and A is a union of p -sections for some p E If y is a p-element in A, Xl a p'-element in CN (y ), X2 E 07T ' (C O (y » and X = XI X2 then a T (yx ) = a (YXI)'
LEMMA 7.6.
77.
PROOF. By (7.4) aT (yx ) = aT implies that a T (yx 1 ) = a (yx 1 ) '
(YXI)' Since yXI E A the definition of 0
LEMMA 7.7. Suppose that (7. 1 ) is satisfied. (i) If a E V7T (N, A ) and {3 E V7T (G) then (aT, {3 )o (ii) If a I , a2 E V7T (N, A ) then (aJ , a2)N = (a � , a ;)o .
T
= (a, {3N)N.
PROOF. (i) If Y is a 17 -element in G let {y};', {y }� respectively, be the 17-section of G, N respectively, containing y. Let ')' = a T (3 * . Then ,), (z ) = 0
if z E {y };' and y is a 17 -element of G which is not conjugate to any element of A. If y is a 17-element in A then Co (y ) = CN (y )07T' (Co (y » . Thus there exists a cross section {Xi } of CN (y ) � Co (y) with {xJ � 07T' (C O (y » such that every 17 '-element in Co (y) is of the form ZXi for some i and some 17 '-element Z E Co (y). Since ,), E V7T (G) this implies that 1
�
(z ) L.J ,), I G I zE{y}� -
=
1
1
�'
�
I C o (y ) I x ECG (y) ')' (yx ) = I CN (y ) I x ECN (y) ')' (yx ) '
L..J
L.J
= � L ')' (z ) I N I z E{y};;:
where 2:' means that X ranges only over 17 '-elements. Thus if y ranges over 17-elements in A which form a complete set of representatives of all 17-sections in G which meet A then
(a T, {3 )o = (ii) Let
I
b i Ly T
L ')' (Z ) =
z
E {y}�
I
�I L
{3 = a; in (i). Then
I I Z�N al (z )a2 (z )*
(a �, a D o = (aJ , (a ;)N )N =
�
Y
L ')' (Z ) = (a, {3N )N .
Z
E {y};;'
aI (z )a ; (z )*
� I z� = I�I Z�N aI (z )a2 (z )* = (aJ , a2)N .
=
I
0
The existence of T is a very useful tool for the investigation of certain questions concerning the structure of finite groups provided M7T (N, A Y �
7]
ISOMETRIES
219
M(G) or equivalently T maps generalized characters in V7T (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 17 is the set of all primes it is easily seen that a T = a 0 for all a E V7T (N, A ) and so M7T (N, A Y � 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 M7T (N, A Y � M(G) whenever hypothesis (7. 1) is satisfied. However Reynolds [1963] and Leonard and McKelvey [ 1 967] have shown that M7T (N, A Y � M(G) provided hypothesis (7.2) is satisfied. This result and its proof is a generalization of an earlier theorem of D ade [1964] who proved the same conclusion under the stronger hypothesis that (7.2) holds and CN (y) is a 17:-group for every 17-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 [ 1 963], section 9. All these results are of course generalizations of the case that 17 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 [ 1 964a] , Gorenstein and Walter [1962] , section 1 1 , 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 17 = {p} for some prime p. In this case a proof that M7T (N, A Y � M(G) will be given below.
THEOREM 7.8. Let p E 17. Assume that (7. 1) holds and A is a union of p-sections of N. Let Bo be a p -block in Sp ( 17, N) such that V(N, A Bo) 1= (0). Let D be a defect group of Bo and let Bo be the germ of Bo with respect to D. Then (i) NN (D)C o (D) = NN (D) 07T ' (CO (D » . (ii) 01T, (NN (D» is in the kernel of Eo and there exists a unique block Eo of NN (D ) Co (D) with 07T' (C O (D» in its kernel such that Eo and Eo coincide as blocks of NN (D) NN (D) Co (D) 07T. (NN (D»" 07T' (C O (D» N (iii) Let y E A n D and let B (y ) be a block of CN (y) with B (y ) = Bo . Then 07T' (CN (y» is in the kernel of B (y ). If E (y ) is a block of Co (y) with 07T' (C O (y» in its kernel such that B (y ) and E (y ) coincide as blocks of :
[7
CHAPTER V
220
CN (y) �� 07T , (CN (y » - 0 7T , (CG (y» G then B (y ) = B f? -
PROOF. Let y E A n D and let B (y ) b e a block o f CN (y ) with B (y t = Bo . Let Dl be a defect group of B (y ). By (5.4) there exist block pairs (Di, Bi ) in N for i = 1 , . . . , n such that (D i+l , B i+1) properly extends (Di ' Bi ) for i = I , . . . , n - l, B 7N (y) = B (y ) and Dn is a defect group of Bo . The following assertion will be proved by induction. and Bi is in For i = 1 , . . . , n CG (Di ) = CN (Di ) 07T, (CG (Di »
Sp (7T, Di CN (Di » .
Assume that the statement has been proved for i < m. If m = 1 let P = (y ) . If m > 1 let P = D m - l . Hence by (7 . 1 ) or induction CG (P) = CN (P) 07T' (CG (P» , Let Z E CG (Dm ) � CG (P), Thus Z = ZI Z2 with ZI E CN (P) and Z2 E 07T, (CG (P» . Since Dm � NN (P) it follows that Dm {07T' (CG (P» n N} is a group and [Dm' Z rl = [Dm, Z ;-1] is contained in 07T , (CG (P» n N. As Dm is a Sp-group of Dm {07T, (CG (P» n N} there exists l Z3 E 07T. (CG (p» n N with ZI Z ;- E NN (Dm ). Thus
Z3 Z2 E NG (Dm ) n 07T, (CG (P»
�
CG (Dm ) n 07T, (CG (P»
C 07T, (CG (Dm »
.
I
Since Z = (Z I Z ;- ) (Z3 Z2) this implies that CG (Dm ) = CN (Dm ) 07T, (CG (Dm » as the opposite inclusion is trivial. By (6.5) Bi is in Sp (7T, Di CN (Di » . The statement follows by induction. (i) Since V ( N, A : Bo) I:- (0), (IV.6.2) implies the existence of y E A n D and a block B (y) of CN (y ) with B (y t = Bo. Since Dn is conjugate to D in N the result follows from the previous paragraph. (ii) Immediate by (i). (iii) Let Bi be a block of Di CG (Di ) such that 07T' (Di CG (Di » is in the kernel of Bi and Bi coincides with Bi as blocks of
Then Di is a defect group of Bi and so (Di Bi ) is a block pair in G. Suppose that B f" = Bo for i = 1 , . . . , n. It will first be proved by induction on n - i that B f' = B f? Suppose that n - m = O. Replacing Dm b y a conjugate i n N i t may be assumed that D = Dm . By the first main theorem on blocks (1II . 9 . 7 ) '
7]
ISOMETRIES
221
B �N(D) = 13o . Thus B �N (D)CG(D) = Bo by (6. 8) with M = NN (D ), L = D CN (D ), HI = H2 = OTr, (CG (D» , and so B � = B f? Suppose that n - m > o and the result has been proved for m < i � n. Thus B �+1 = B f? Since (D m+l , B m+l) extends (Dm, Bm ) it follows from (6. 8) that (D m+l , B m+l) extends «Dm , Bm ). Therefore B � = B �+1 = B f? This proves the statement in the previous paragraph. Since B 7N (y) = B (y ) it follows from (6.8) applied to CN (y ) and CG (y ) y G that B 7° ( ) = B (y ) . Thus B (y ) = B 7 = B f? D
E 7T. 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 7T-part is conjugate to an element of A. For any p -block Bo of N such that V(N, A : Bo) � (0) and Bo � Sp (7T, N) let Bo be defined as in (7.8) (ii). Let B be a p -block of G with B � Sp (7T, G) and let BO l , B02 , . . . be all the p -blocks of N such that V (N, A : BOi ) � (0), BOi � Sp ( 7T, N) and B &; = B. Then the following hold. (i) If a E V7T (N, A : UiBoi ) then a T E V ( G, {A }� : B ) . (ii) V7T (N, A : U i Boi r � VTr ( G, {A }� : B ). G (iii) If a E VTr (N, A : U i BOi ) then a T = (a )B . (iv) If B � Sp ( 7T, G ) then MTr (N, A : U i Boi r � M7T (G, {A }� : B ) . THEOREM 7.9. Let p
Tr
PROOF. If Y is a p -element in G and (3 is a function defined on CG (y ) let (3y be the function on the set of all p '-elements x in CG (y ) defined by {3y (x ) = (3 (yx ). (i) Let aT = (a T )B + B . It suffices to show that By = 0 for all p -elements y in G. Let y be a p -element in G. Let SI be the union of all p -blocks 13 of CG (y) with B G = B and let S2 be the union of the remaining p -blocks of CG (y ). For t = 1 , 2 let V'(CG (y) : S/ ) be the space of all complex linear combinations of irreducible Brauer characters in St . By (IV.6.2) (a T ): E V' (CG (y ) : SI) and By E V' (CG (y ) : S2)' it suffices to show that a ; E V' (CG (y ) : Sl) because then
By = a ; - (a T ): E V' (CG (y ) : SI) n V' (CG (y ) : S2) = (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 = L ai where ai is a linear
combination of irreducible characters in BOi. If Y is not conjugate to an .element of a defect group of BOi then (ai )Y =10 by (IV.2.4). If x is a p '-element in CG (y ) then X = Xl X2 where Xl E CN (y ) and X2 E 07T, (CG (y» . By (7.6) a ;(x ) = ay (Xl ) ' Thus a ; = ay is a linear combination of irreducible
222
CHAPTER V
(7
Brauer characters of Co (y )!07T' (C O (y » C N (y )/07T, (CN (y » . Hence (4.3) and (7.8) (iii) imply that a � E V' (Co (y ) : SI) 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 T and a O vanish on the p -section of G which contains y. Hence by (IV.6.3) (a O )B 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 T and (a O )B agree on the p -section of G which contains y. Let C = Co (y ). By the Mackey decomposition (11.2.9) a 0 (yx ) = 2: {a �z n c f (yx ) for every p I-element x in C, where NzC ranges over all the (N, C) double cosets of G. If yx E c A Z then y E c A Z and so by (7. 1) (i) there exists Z l E N with Z I Z E C. Thus NzC = NC. Hence { a ° } Y = { (a Nnct}y . Therefore by (IV.6.1) { ( a O )B } y = ({ (a Nnct} S ) y where S is the union of all blocks Bi of C with B ? = B . B y (7.4) ( a Nn c) y i s a linear combination o f Brauer characters o f N n C all of which have 07T, (N n C) in their kernel. Since C/07T, ( C) = N n C/07T{N n C) there exists a linear combination f3 of Brauer charac ters of C all of which have 07T'( C) in their kernel such that (a Nne) y and f3 agree as functions on the set of p i-elements in C/07T,(C). Since y E Z(C) it follows that { ( a Nne)C}y = { ( a Nnc)y } c. Thus if cP is the character afforded by an indecomposable proj ective module of C which has 07T'( C) in its kernel then =
z
({ a Nnct}y, CP)� = « a Nne) , cP Nn c)J.,nc = ( f3 Nn C, cP Nnc)J.,nc = ( f3, CP )� where the prime indicates that the summation in the inner product ranges over p i-elements. Thus by (IV.3.3) { ( a Nnct}y f3 is a linear combination of irreducible Brauer characters of C none of which have 07T' ( C) in their s. c kernel. Hence by (6.5) ({ (a Nn c) y ) y f3 By the second main theorem on blocks (IV.6.1) (a N nc) y is a linear combination of Brauer characters in blocks B ( y ) of N n C with B (y t � UiBoi . Thus f3 is the corresponding linear combination of Brauer charac ters in blocks B'(Y ) of C. Since 13 g = B it follows from (7.8) (iii) that f3 S = f3. It remains to show that f3 = (aT ) Y ' Let x be a p i-element in C. Then x = X I X2 with X I a p i-element in N n C and X2 E 07T' ( C). Therefore since B is in Sp ( Tr, G) it follows from (7.6) that -
=
Thus f3 = (a T )y as required. (iv) Clear by (iii). 0
223
ISOMETRIES
7]
LEMMA 7 . 10 Let p E Tr. 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 YeN, A : Bo) ;;6 (0) and Bo � Sp ( Tr, N). Let 130 be defined as in (7.8) (ii). Then 13 � � Sp ( Tr, 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 Bo . Thus D is a defect group of 130 • By (IV.2.4) it may be assumed that y E D. Since 07T' (Co (y » is a S7T' - group of Co (y) it follows that 07T, (D Co (D » is a S7T'-group of D Co (D ). There exists a block pair (D, B2) in G with B �N(D)CG(D) = 130• Since 130 covers B2 it follows that 07T, (D Co (D » is in the kernel of B2• By (6.7) (ii) B � = B ? � Sp ( Tr, G). 0 THEOREM 7 . 1 1 . Let p E Tr. 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 Tr -part is conjugate to an element of A. Let B 1 , • • • , Bn be all the p -blocks of G in Sp ( Tr, G). For each i let Bi l , Bi2 , . . . be all the p -blocks of N in Sp ( Tr, N) such that V ( N, A : Bij ) ;;6 (0) and 13 g = Bi where 13 ij is defined as in (7. 8) (ii). For each i let Si = U jBij . Then the following hold. (i) V(N, A : B ij) = V7T (N, A : B ij) for all i, j. (ii) V(G, {A };' : Bi ) = V7T ( G, {A };' : Bi ) � r all i. (iii) V7T (N, A ) = V7T (N, A : lJ �= 1 Si ) = EB�=1 V7T (N, A : Si ). (iv) V7T ( G, {A };') = V7T ( G, {A };' : U �=I Bi ) = EB�= l V7T (G, {A };' : Bi ) . (v) V7T (N, A : Si r = V7T (G, {A };' : Bi ) for i = 1 , . . . , n. (vi) If {A };' n N = A then M7T (N, A : Si r M7T (G, {A };' : B; ) for i = 1 , . . . , n. (vii) Suppose that for every p -element y in A and every integer h with (h, I G i) 1, y h E A. The for i = 1 , . . . , n =
=
rank M7T (N, A : Si r = rank M7T ( G, {A };' : Bi ) = dim V7T (G, {A };' : Bi ) . PROOF. (i) and (ii). Since 07T'(C O (Z7T » � 07T'(C O (zp » for all z E {A };' these results follow from (6.5). (iii) The second equality follows from (IV.6.3) and (i). Clearly
V7T (N, A : .CJ = 1 Si ) � V7T (N, A ) .
Suppose that a E V7T (N, A ). For y a p -element in A , let B b e a block of N with (a B ) y ;;6 O. If B 1 ;;6 B2 then by the second main theorem (IV.6. 1) no irreducible Brauer character of C N (y) is a component of both (a Bt) y and (a B2) y . By (7.4) (a B ) y is a linear combination of irreducible Brauer
224
[7
CHAPTER V
characters of eN (y ) with O rr , ( CN (y» in their kernels. Hence B � Sp (71", N) by (6.7) and (7.2). Thus 13 0 � Sp (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 V7T (N, A ). Then m = dim V� (G, {A };'). Since V7T (N, A : Si r � V7T ( G, {A };' : Bi ) by (7.9) (i), both (iv) and (v) follow from (iii). (vi) By (7.9) (i v )
7]
a E M7T (G, {A };' : Bi ). Let {3 = aN. Since {A };' n N = A it follows that {3 E V7T (N, A ) and a = {3 T. By (iii) {3 = L{3j with {3j E V7T (N, A : Sj ). Thus by (iv) and (v) {3j = 0 for i � j. Hence {3 E V7T (N, A : Si ). Since (3 E M(N) the result follows. (vii) Since A is a union of p -sections it follows that if z E {A };' and (h, 1 G I) = 1 then Z h E {A }�. Thus by (IV.6.9) rank M7T (N, A : S; ) =
dim V7T (N, A : Si ) and rank M7T (G, {A };' : Bi ) = dim V7T (G, {A };' : Bi ). The result follows from (v). 0
COROLLARY
7.12. Suppose that (7.2) holds and
then a T is a generalized character of G.
PROOF.
Clear by (7.9) (iii) and (7. 1 1) (iii).
71"
= {p}. If a E M7T (N, A )
0
Suppose that (7.2) is satisfied. Let X be a set of characters of N/07T, (N) with M(N, A : X) � M7T (N, A ). The set X is coherent if T can be extended to a linear isometry from M(N : X) into M( G). If � is an irreducible character in X and T is coherent theri " C W = 1 . Thus ± C is an irreducible character of G. The next result yields some information concerning the values assumed by C 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. 1 3) below was omitted in the statement.
THEOREM 7.13. Assume that (7.2) is satisfied and 07T, (N) is a S7T'-groUP of N. Let X be a set of irreducible characters of N/07T, (N). Suppose that the
following assumptions are satisfied. (i) If P E 71" then the set of p -singular elements in A is a union of p -sections of N. (ii) X is coherent. (iii) V7T (N, A : X) = V7T (N, N 0 7T ' (N) : X). Then the following conclusions hold. (i) If � E X then (C, a T)o = ( � ;. , a )N for all a E V7T (N, A : X). -
225
(ii) Let X = {�s } and let g = L£'s (l)£,s . For each t there exists YI E M(N) such that (YI, £,s ) = 0 for all s and an integer al such that
Furthermore there exists Y E V (N) and a rational number
a
such that
£, ; (z ) = £,r ( z ) + £'t (l) {ag (z ) + y (z )} for all z E A.
M7T (N, A : S r � M7T (G, {A };' : Bi ).
Let
ISOMETRIES
(i) Let 71"1 be the set of all primes p E 71" such that if a E V7T (N, A : X) then a vanishes on all p -singular elements in N. Let 71"z = 71" - 71"1 ' For each p E 71"z let S (P ) be the union of Sp ( 71", G) and all p -blocks of G of defect O. We will first show that if � E X and p E 71"z then e£,T E S (P ) for e = 1 or - 1. Choose p E 71"z . Let £' E X and let e = 1 or - 1 such that e£,T is a character of G. Let Ap be the set of all p -singular elements in A. By assumption (i) Ap is a union of p -sections and (7.2) is satisfied if A is replaced by Ap . Furthermore V7T (N, Ap ) � Vrr (N, A ) and T defined on V7T (N, Ap ) is the restriction to V7T (N, Ap ) of T defined on V7T (N, A ). Let X = {£,s } with £' = £'1 . For each s let as = £,s (I)£, - £,(l)£,s . By assump tion (iii) {as } is a basis of Vrr (N, A : X). Since p E 71"z there exists t such that at does not vanish on Ap . Let at = {31 + {3z where {31 vanishes on N Ap and {3z vanishes on Ap . Thus {31 � 0 and {3 � = ({3 fY)S (P) by (7.9) (iii) and (7. 1 1) (iii) applied to (31 E V7T (N, Ap ) . Therefore by (IV.3.4) a ; = () + 11 where () E V (G : S (p » and 11 is a complex linear combination of charac ters afforded by projective R [GJ modules not in S (P ). Furthermore () � 0 and a ; = £'t (l)£,T - £,(1)£, ; . Thus either e£,T E S (P ) or �t ( l )C vanishes on all p -singular elements of G. In the latter case C vanishes on all p -singular elements of G and so e£,T is in a p -block of defect O. Hence in any case e£,T E S (P ). Let £' E X. Define 1/1 as follows :
PROOF.
-
1/1 (z ) = £,T (z ) =0
if z is p -singular for some p E 71"z and the 71"-section of z in G meets A, otherwise.
Since e£,T E U p E 7T2 S (P ) it follows from (6.5) that C is constant on p -singular 71"-sections which meet A for all p E 71"z . Thus 1/1 E V7T (G). Hence if a E V7T (N, A : X) (7.7) implies that
(£,T, a T)o = ( l/I, a T )o = ( I/IN, a )N = (£';. , a )N '
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CHAPTER V
[8
(ii) Let ( ? ;)N = ?r + '22 ari ?i + ')' where (?h ')'1 ) = 0 for all j and implies that for all i, j, t ,
t.
£&2 ati - (1) a l ?, l
for all
.
]
and
t.
at/ .
{ ��l) g (z ) ?,
+ ')'t
(z )
}
= ?, ( 1)
{(
s g ( z ) + ')'s (z ) ?s l)
s
}
for all Z E A. Thus the second equation in the statement of (ii) follows by setting and
8. 7r-heights
In this section we will use the notation of Chapter IV, section 8. Let
v ( I G I) = a .
Let e be a centrally primitive idempotent in R [ G ] . Let B = B (e ) be the block corresponding to e. L et D = D (B ) be the defect group of B and let d = d (B ) denote the defect. Thus v (1 G : D (B ) i) = a - d (B ). In this section we will primarily be concerned with ChR (G, B ) and ZR (G )e. Observe that ChK ( G, B ) is the dual space of ZK (G )e. See BroU(� [1976b] , [1978b], [1979] , [1980] for the results in this section and further related results.
LEMMA
If a E ZR (G)e and e E ChR (G, B) then v ( e, (a)) � v ( 1 G : D (B ) I ) .
81 .
.
By (111.6.6) e = Tr ( ) for some E R [ G ] . Since a E ZR (G) it follows that ae = Let { Xi } be a cross section of D in G. Since e is a class function on G it follows that
PROOF.
gc
Trg (ac).
c
acxi) = I G : D I e (ac ).
0
For a E ZR (G)e let s (a ) be the largest integer such that 1Ts(a ) 1 O (a ) in R for all e E ChR (G, B ) . Define the 1T -height of a, hIT (a ) by ( 1T h,, (a » = (1T·s (a ) I G : D I - I) ,
The first equation in the statement of (ii) follows by setting at = lf Z E A then ? ( l) C (Z ) - ?I (l) ? ;(S) = ?s (l) ?I (Z ) - ?, ( l)?S (Z ) for all s and t. Thus the first equation in the statement of (ii) implies that ?s (1)
O(a ) = e (ae ) = e ('22x � I This implies the result.
(l )a'i = ?i (l)ati for all i, j and t. Hence in particular
227
1T-HEIGHTS
Then (i)
?i (l)a'i - ?i (l)a'i = «C)N' ?i (l) ?i - ?i (l) ?i )N - ( ?,' bi (l) ?i - ?i (l) ?i )N = ( C' ?i ( l) C - ?i (l) C) o - {?i ( 1)8it - ?i (1)8jl } = O .
Therefore ?i
8]
where the parentheses denote a fractional ideal of K. For e E ChR (G, B ) let t ( e ) be the largest integer such that e I (Ii) I e (a ) in R for all a E ZR (G )e. Define the 1T height of e, h7T ( e ) by -
(1T h " (Ii) = ( 1T ' (IJ) I G : D I- I ) . o
By (8. 1)
h7T (a ) and h7T ( 0 ) are nonnegative integers for a E ZR (G)e
E ChR (G, B ).
and
Xu E B. By (IV.7. 1) v (Xu (1 » = v (Xu (e ) . For a E ZR (G) XU (a )JXu ( 1) = Wu (a ) E R. Therefore (Xu ( 1» = (Xu (e » � (1T h,, (xu ) I G : D i) � (Xu (1 » . Thus if (1T m ) = (p) it follows that h7T (Xu)lm is the height of Xu. In particular Xu has height 0 if and only if it has 1T-height O. This argument also shows Suppose that
that h7T (e ) = O. Since e is a primitive idempotent in ZR (G) it follows that ZR (G)e is a local ring. Thus ZR (G)eJJ (ZR (G)e) = R as K was chosen large enough. Let A be the central character of ZR (G )e. By (IV.4.2) A = for any
Wu
Xu E B.
LEMMA 8.2. Let a E ZR (G)e. The following are equivalent. (i) For every e E Ch (G, B ) with h7T ( e ) = o, v ( e (a » = v ( I G : D I ). (ii) There exists e E ChR (G, B) with v ( e (a » = v(1 G : D I). (iii) h7T (a ) = O. (iv) a � J (ZR (G)e). R
(i) � (ii) and (ii) � (iii) are immediate. (iii) � (iv). Suppose that a E J (ZR ( G ) e ) . Then (a (1) = (a ) = o for all E B. Hence e (a ) == o (mod 1Tp v{/ O : O i l) for all e E ChR (G, B ) and so h7T (a ) rf O. (iv) � (i). Let e E ChR (G, B) with v ( e (a » > p v, where v = v (1 G : D I). Let 0 = Then
PROOF.
Xu )JXu
Xu
'22 cuXu.
O (� ) =
p
L Cu � p = L cu (� p ) wu ( a).
Wu
228
CHAPTER V
[8
By assumption ( 8 (a )jp V ) = O. Since a � J (ZR (G)e ) it follows that Wu (a ) = A (a ) I: 0 for all u. Therefore
I cu
(X�P» )
==
0
(mod 1T )
corresponding to B. L et X = Xu be it character of height 0 in B. L et Co be the conjugate class of G containing xy. Since D is a Sp-group of Cc (xy ) and X (xy ) == X (x ) � O (mod 1T ) it follows that v (x (xy » = v (X (x » = O and so Therefore A (Co e ) = A (Co) = Wu (Co) I: O . Thus Co e � J(ZR (G)e ) but Co e E ZR (G : S )e. 0
Thus h-rr ( 8 ) I: O.
8.6. Define a on G by a (x ) = 1 if x is a p -element and a (x ) = 0 otherwise. Then a B E ChR (G, B ) and h 7T ( a B ) = O .
LEMMA
0
8.3. Let 8 E ChR (G, B ). The following are equivalent. (i) For every a E ZR (G)e with h-rr (a ) = 0, v ( 8 ( a » = v (1 G : D /). (ii) There exists a E ZR (G)e with v ( 8 (a » = v (1 G : D /) . (iii) h7T ( 8 ) = O.
LEMMA
(i) � (ii) and (ii) � (iii) are immediate. (iii) � (i). This follows from the fact that (8.2) (iii) implies (8.2) (i).
PROOF.
0
Let 8 E ChR (G, B ). Then h7T ( 8 ) = 0 if and only if v ( 8 (1» =
If v ( 8 (1» = v(1 G : D I) then clearly h7T (8) = O. Suppose that h7T ( 8 ) = O. Since h7T (e ) = 0 and v ( 8 ( e » = v ( 8 (1» , the result follows from (8.3). 0
PROOF.
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 ZR (G : S )e = (0). (ii) If Y is not conjugate to an element of Z(D ) then ZR (G : S ) e � J(ZR (G)e ). (iii) If Y is conjugate to an element of Z(D ) then ZR (G : S ) e g;, J(ZR (G)e ).
LEMMA
(i) By (IV.2.4) 8 (x ) = 0 for all x E S and all 8 E ChR (G,B ). Since ChK (G, B) is the dual space of ZK (G)e this implies the result. (ii) This follows from ( 1 .5). (iii) It' may be assumed that y E Z(D ). By (111.6. 101 there exists a p ' -element x such that D is a Sp -group of Cc (x ) and A ( C) I: 0 if C is the conjugate class containing x and A is the central character of R [GJ
PROOF.
22 9
7T'-HEIGHTS
v (Wu ( Co» = v (1 G : D 1) - v (X (1» = v(wu ( C» = O .
and so
LEMMA 8.4 . v (1 G : D /).
8]
PROOF. Let {qi } be the set of all primes distinct from p which divide 1 G I · Thus q � I E R for all i. Hence by (IV . 1 .3) a E ChR(G) and so a B E ChR (G, B ). Let 8 E ChR (G, B ) with h-rr ( 8 ) = O. Let a = L 8 (x -l)x. Let S = Greg and let 8s be defined as in Chapter IV, section 8. Then 8s (a )e E ZR (G)e and by (IV.8.S) a B (8s (a )e ) = a ( 8s (a )e ) = a ( 8s (ae » = a ( 8s (a » = 8 (1) .
Thus h7T ( a B ) = 0 by (8.3).
0
8.7 (Broue [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 = LXES ax x. Then v(ay ) ;?: v (1 Cc (y ) : D I ). Furthermore v ( ay ) = v(1 C c (y ) : D I) if and only if a � J(ZR (G)e ).
THEOREM
Let a be defined as in (8.6). By (8. 1) v(a B (a » ;?: v ( 1 G : D I). By (8.2) equality holds if and only if a = ae � J (ZR (G)e ). Since
PROOF.
a B (a ) = a (ae ) = a (a ) = ay 1 G : Cc (y ) I , equality holds if and only if
COROLLARY Then
v (ay ) = v(1 Cc (y ) : D I).
0
8.8 (Brauer [ 1 976a]). Let B be a block with defect group D.
where Xu, 'Pi ranges over all irreducible, irreducible Brauer characters respec tively in B.
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230
PROOF. Let e be the centrally primitive idempotent of R [ G ] correspond ing to B. By (IV.7. 1) and ( IV . 7.2 ) e E ZR (G : Greg) and e = 2": ax x with I G I a l = 2": Xu (1)2 = 2": 'Pi (1) <1\ (1) . Since e E J(ZR (G)e ) the result follows from (S.7). D
9]
SUBSECTIONS
23 1
By (IV.S.6) and (IV.S.7) m (y, B ) o ()
= () 0 f.1- (y, B)
for () E ChK ( G ) .
The definitions of {3 Y and 0 Y imply that if a E ZK (C G (y ) : C G o Y (3 Y (a ) = a. Thus m (y, B ) and f.1- (y,B ) are idempotent. Define
(y ) ) then reg
°
9. Subsections
The results in this section are mostly due to Brauer [196S] , [ 1 97 1 a] though the presentation here is based to some extent on Broue [ 1 97Sb] . The notation is the same as in the previous sectiGn and in Chapter IV, section S. Let y be a p -element in G and let B be a block of G. If Xu E B and d �j � 0 then by the second main theorem on blocks (IV .6. 1), 'P r belongs to a block B of C G (y ) with 13 G = B. For fixed y and j, the column of higher decomposition numbers d �j is said to belong to the p -section S which contains y and is said to be associated with the block B. Let B be a block of C G (y ) with B G = B. The set of columns d �j , as 'P r ranges over the irreducible Brauer characters in 13 is a subsection as sociated to B or simply a subsection. It clearly depends on y and 13. Denote it by S (y, B ). A subsection S (y, B ) is a major subsection if 13 and B = 13 G have the same defect.
If S (y, B ) is a major subsection then y E ZeD) where V is a defect group of B. Furthermore V is a defect group of B = 13 G
LEMMA 9. 1 .
PROOF. Clear by (6. 1).
D
Let y be a p -element in G and let linear map m (y, 8 ) : ChK ( G ) � ChK (G where
13 be a block of C G (y ). : S) by
m (y· B ) « () ) = b Y ({dY « ())}B ), for () E ChK ( G ) , b Y and d Y are defined in Chapter IV, section S. f.1- (y. B) : ZK ( G ) � Zk ( G : S)
Define the
Define
by f.1- (y, B ) (a ) = f r ( eo Y
(a ))
for
a E ZK (G) ,
where (3 Y and 0 Y are defined in Chapter IV, section S and e is the centrally primitive idempotent of R [C G (y )] corresponding to 13 .
Thus that
X S", B ) = Xu ° f.1- (y, B )
1 w u(y, B ) = _- X (uy, B ) . Xu (1) and W S", B) = wu ° f.1- (y, B ) . This implies in particular
and if a E ZR ( G ) then w S", B) (a ) E R. Furthermore if Xu and X are in the same block then v
w S", B) (a )
==
w �, B ) (a ) (mod 1T )
a E ZR ( G ) . Let D be a defect group of the
for
THEOREM 9.2 (Brauer [ 1 96S] , (4A), (4C)). 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 13 G = B.
then
If Xu E B
V (X S", B ) (y )) = v(I C G (y ) : D 1) + h (Xu ) , where h (Xu ) is the height of Xu ' PROOF. (i) Let e be the centrally primitive idempotent in R [ G ] corre sponding to B. Let {13i} be the set of all blocks of C G (y ) with 13 � = B and l� t ei be the centrally primitive idempotent in R [C G (y )] corresponding to Bi • Let S be the p -section containing y. By (S.S) there exists a E ZR (G : S)e with a E J(ZR (G : S)e). By (S.7) v(ay ) = v(I C G (y) : D I) , where a = 2": ax x . By (IV.S.7) o Y (a ) = 8Y (ea ) = 2":iei o Y (a ). The coefficient of 1 in o Y (a ) is ay . Let iii denote the coefficient of 1 in eio Y (a ). Then a = 2": iii . Thus there exists j with y v(aj ) � v(ay ) = v(1 C G (y ) : D I ) . Let Vj be a defect group of Bj • By (6. 1) it may be assumed that Vj c;;, D. Thus (S.7) applied to the group C G (y) with y replaced by 1, implies that
v(1 C G (y) : D I) � v(1 C G (y ) : Vj I ) � v(aj ) � v(1 C G (y ) : D I) ·
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Hence equality must hold and so D = Vj as required. (ii) Let a be the sum of all elements in G which are conjug �te to y. Thus a E ZR (G) and t F (a ) = 1 . Hence the coefficient of y in J.L (y, B ) (a ) = f3 Y (e) is the coefficient c of 1 in e. Since e � J(ZR (C G (y ))e ), it follows from (8.7) applied to C G (y ) with y replaced by 1 that v(c) = v (1 C G (y ) : D I)· Thus J.L ( y, B ) (a ) � J(ZR (G)e ) by (8.7). Hence if Xu E B then
w �, B ) (a ) == wu (J.L ( y, B ) (a )) � O (mod 1T ) which i s equivalent to t h e assertion to be proved. Let y be a p -element in G, let B be a block of irreducible characters in B G . Define
SUBSECfIONS
233
xv (1) m (Y' B) P d m (uvy, B) == p dXU (1)(ly ww (mod 1T p hu ) . XW Furthermore v (m � B )) � hu - d and equality can hold only if (y, B ) is a major subsection and hv = O. PROOF. (i) The first equality follows from the definition. Thus , 1 ( m ( y, B ) (Xu ) , Xv )G = 2: 2: d �i 'P r x ) (d
I C G ( y ) I 'Pl, 'PJ E B xE Co (y reg
0
C G (y ) and
9]
let
Xu, Xv
be
By (IV.6.2) this yields the second equality . Since implies that p d ( Y �) has integral entries. Thus intege r. (ii) By (i)
)
B
(
�j)* 'Pj (x -I ) .
has defect d, (IV.4. 16)
d p m �� B ) is an algebraic
m �.;, B) = (m ( y· B)(Xv ) , m ( y,B )(Xw )) G where * denotes complex conjugation. In case y = 1 and P is a Sp-group of G then 1 p i m �l� B ) = a uv, where a uv is defined in Chapter IV, section 4. Thus these numbers are a generalization to arbitrary sections of the n�rr.tbers , , · th en ( m (uvy, B ))U - m (uvy , B ) for a uv ' Observe that if (T is a field automorphIsm suitable y ' and B ' which do not depend on u or v . LEMMA 9.3.
Fix u and
v.
Then �(y, B ) m �� B ) = 8 uv .
PROOF. Clear by definition.
0
Let y be a p-element in G. Let B be a block of C G (y ) and let G . Let a, d be the defect of B, B = B G respectively. E B Xu, Xv, Xw (i) p d m ��B ) is an algebraic integer and
Then
�u m �� B ) (m �� B))* = m<;;;."B ) .
_
Let I (B) denote the number of irreducible Brauer characters in B.
B) . 2: u m � B) (m �� )*
m :;;; " ) = where y and
j"J? L w" (C)x" (yx )*,
C ranges over the conjugate classes i n the p -section which contains yx E C. As Xu and XW are in the same block this implies that
, � IGI , IGI y, B) (mod 1T ) uv == m (wv Xu (1) XW (1) d since p m �� B ) is an algebraic integer by (i). As Xw has height 0 this implies
that
'1\ , P d m (uvy, B ) == p d� m (wvy, B ) ,
v
XW (1)
2: m ��B) = I (B) .
Suppose that XW has height O . Let hu , hv denote the height of Xu, Xv respectively. If hu � hv then (iv)
=
� m ( y, B)
where ( Y �rl is the Cartan matrix of B and * denotes complex conjugation. (ii) (iii)
(m (y,B ) (Xv ), Xu ) G (m (y, B ) (Xw ) , Xu ) t; 2: u
(iii) By definition m �� B) = (m � B )) *. Thus if M = M ( y,B) = (m �� B)) it follows from (ii) that M2 = M. Hence the trace of M is the rank of M. By definition �u m �� B ) is the trace of M. By (i) the rank of M equals the rank Qf the Cart an matrix of B which is I (B). (iv) By definiti on
THEOREM 9.4.
2: d � i Y H d �j)*, 'P¥, cpfEB
=
Similarly
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CHAPTER V
[9
The required congruence now follows. Since v (p d m �;., B ) � 0 ,
(
)
(1)X (1) m (Y B ) >- h " + h + d - v p Xu Xw d ww (1)2 Thus the congruence implies that v (m <,:� B ) � h" - d and if equality holds then hu + hv + d - d = h" - d. The result follows as d � d. D d
v
'
�
U
v
•
9.5 (Brauer [196S] (5H» . Let B be a block of G with defect group D. Let y E Z(D). Let S (y, 13 ) be a major subsection associated to B. Suppose that Xu, Xw E B and Xw has height O. Let Xu have height h". Then v (m <':.;, B ) = h" - v(1 D I) .
THEOREM
PROOF. For ()
E ChK ( G )
let
Thus 8 E ZK ( G ) and () 1 ( 82) = «()I,
()2) . By (9.4)
m <,:.;, B ) = (m ( y, B ) (Xu ), Xw )o = m (y, B ) (Xu ) (Xw ) = Xu (p. ( Y, B )(Xw » = X<,:, B ) (Xw ) Xu (1)W � , B ) (Xw ) =
= =
Since
Xw
1"61) Wu (p. (y, B ) (L Xw (X - 1)x) ) 1u61) Wu (L X�, B ) (X - I )X ) .
(9.2) that = v (X�' B ) (y » = v (1 Co (y ) : D I) . (S.7) LX�, B ) (X - l )X � J(ZR (G» . Consequently has height 0 it follows from
v (X�' B ) (y - l »
Thus by
( (L X�, B ) (X - I )X )) = O.
v Wu Therefore
v(m <,:.;, B ) = v(Xu (1» - v(1 G 12 = hu - v(1 D I). D 9.6. Let B be a block with defect group D. Let 13 be a block of D Co (D ) with 13 0 = B and let T(13 ) be the inertia group of 13 in No (D). Let y � Z(D) be a complete set of representations of conjugacy classes of G
THEOREM
9]
SUBSECTIONS
235
which meet Z(D). For y E Y let My denote a complete set of ( T(13 ), No (D) n Co (y» double coset representatives in No (D ). Then each subsection Sy, x = S (y, (13 X )Co (Y ) with y E Y and x E My is a major subsec tion. Conversely every major subsection associated to B is conjugate to exactly one subsection S y,X ' PROOF. If Y E Z(D), then D Co (D ) � Co follows from (111.9.2) that
(y ). Thus if y E Y and x E My
it
« 13 X )C o (Y ) O = (13 X ) O = 13 0 = B .
Since D <] D Co (D) i t follows that D i s contained in the defect group o f 13 and so by (1II.9.6) S y, x is a major subsection. Suppose conversely that S (y, B ) is a major subsection associated to B. After y is replaced by a conjugate it may be assumed by (9. 1) that y E Y. Thus D Co (D) � Co (y ). By (5.2) B = 13 � o (Y ) for some block 130 of D Co (D) and 130 = 13 X for some x E No (D). It is clear that x may be chosen to lie in My . Suppose finally that S y,x is conjugate to Sy ', x' in G, where y, y ' E Y, x E My and x ' E My " Thus there exists z E G such that
/
(9.7) This implies in particular that z Co (y '). Thus
Co (y Y
=
Co (y ') or equivalently Co (y )z =
(13 XZ ) C o (Y ') = (13 x,) co (Y ') .
The block of Co (y ') on the left has defect group D xz = D \ while that on the right has defect group D X ' = D. Thus D is conjugate to D Z in Co (y '). Thus in (9.7) z may be chosen to be in No (D ). Hence the first equation in (9.7) implies that y = y ' since y, y ' E Y. Hence z E No (D ) n Co (Y ). Now (9.7) becomes (13 XZ)C o (Y ) = (13 x ')c o
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236
Let B be a block with defect group D. Let B be a block of G D C G (D ) with B = B and let T(B) be the inertia group of B in N G (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 ', B ') with y ' E Z(D) n C. Let nB denote the number of conjugate classes of major subsections associated to B. Then n � is the number of T(B ) conjugate classes of Z(D) n C. Furthermore
COROLLARY 9.8.
PROOF. By (9.8) and (9.9) I Z(D ) I /e :<s pd + n + l . Hence p h 1 Z(D) 1 :<S pd + n . 0
and My for y E Y be defined as in (9.6). By (9.6) n � = This implies the first statement. Each (T (B ), N G (D) n � Y E ynC C G (y » double coset contains at most e cosets of N G (D ) n C G (y ). Thus
Y
Y
Let B be a block of defect d and let h be the maximum height of an irreducible character in B. Let nB be the number of conjugate classes of major subsections associated to B. Then nB :<S p d - h and if h > 0 then nB < pd - h .
LEMMA 9. 9.
PROOF. Let Xu be an irreducible character in B of height h. Let S (y, B ) be a major subsection associated to B. By (9.2) m �� B ) '" O. By (9.4) (i) (iv) p d - h m �� B ) is a totally positive algebraic integer. By (9.3) �p d- h m �� B ) :<s pd - h , where the sum ranges over all major subsections associated to B. Thus the arithmetic-geometric inequality implies that
Thus nB :<S p d - h . If equality holds then m �� B ) p -( d - h) for all major B subsections and in particular m �l�B ) = P -( d - h) and so v(m �1� » = h - d. By (9.4) (iv) h = O . 0 =
Let B be a block with defect group D. Let 1 D 1 p d and let n be the largest integer such that p n :<S e, the inertial index of B. Let h be the height of an irreducible character in B. Then h :<s n + d v(I Z (D ) I) .
COROLLARY 9. 10.
=
-
Thus
p h I Z(D ) I :<s p de <
Suppose that the notation is as in (9. 10). Let p ' be the index of the Frattini subgroup of D in D. If d > 0 then h < d + �r ( r + 1) - v ( I Z(D) I).
PROOF. Let
1 My I :<s 1 NG (D) : NG (D) n C G (y ) I :<s e 1 My I . By (9.6) nB = � EY I M I and I Z (D ) I = � Y EY I N G (D ) : N G (D ) n C G (Y ) I . Thus nB :<s I Z (D) I :<s en B ' 0
nB :<s p d - h .
COROLLARY 9. 1 1 .
PROOF. By (5.2) Thus
Y 1 My I ·
237
SUBSECfrONS
9]
as
r > O.
T(D)/D C G (D ) is isomorphic to a p '-subgroup of GL, (P ).
; =1
The result follows from (9. 10) . . D
Suppose that B has an abelian defect group of rank r > O. Then the height of an irreducible character in B is less than �r(r + 1) .
COROLLARY 9.12.
PROOF. Clear by (9. 1 1).
D
The estimate in (9. 1 1) is extremely crude and obvious improvements can be made in (9. 1 1) and (9. 1 2). However these methods do not appear strong enough to answer Question (IX) in Chapter IV.
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 ) = p d, every irreducible character in B has height 0, every decomposition number is 1 and the unique Cartan invariant is p d.
THEOREM 9 . 1 3 =
PROOF. By (9.8) there are at least pd subsections associated to B. Thus by (IV.6.S) k (B ) � p d. Hence by (lVA.21) it suffices to show that B contains exactly one irreducible Brauer character. This will be done by induction on
IGI·
If D � Z(G) the result follows from (4.6). Suppose that D g Z(G). Choose y E D, y � Z(G). Let S (y, B ) be a subsection associated to B. Since D is abelian it is a major subsection. Let {Xu } be the set of all ir:educible characters in B. As D is a defect group of 13, the inertial index of B is 1 . As y � Z( G), induction may be applied to C G (y) '" G. Thus 13 �as a unique irreducible Brauer character 'P r . Hence the Cart an matrix of B is (p d ). By (9.5) d � ; '" 0 for all u. The arithmetic-geometric inequality implies that
238
CHAPTER V
[9
Therefore
Since there are p d subsections associated to B, (IV.6.5) implies that E has a unique irreducible Brauer character for each such subsection S (y, E ). This is the case in particular for S (1, B ). 0 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 Broue and Puig [1980] . It is a good deal more complicated than (9. 1 3). We next prove a technical preliminary result. LEMMA 9. 1 4 .
Let (gij ) 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 s be a primitive p nth root of 1 for some n and let tr denote the trace from Q( s ) to Q. If a is a nonzero vector whose entries are algebraic integers in Q(s ) then tr(Q (a, a )) � [Q(s ) : Q] q = p n - I (p - 1)q. i
PROOF. Let t = p n-l (p - 1). Let 'L::� ei s , where each ei is a vector with i integer coordinates. Then Q ( a, a ) = 'L:,j:o Q (ei' ej ) . Since tr(1) = t, tr(s ) = _ p n - 1 if i == 0 (mod p n -1) but i � O (mod p n ) and tr(s i ) = O otherwise, it follows that tr( Q ( a, a )) = p n - I
1-1
2: As ,
s =o
(9. 15)
where
i,j with i,j ranging over all values congruent to s mod p n - I and 0 � i,j < can be rewritten as As
with
i,j
= 2: i<j Q (ei - ej, ei - ej ) + 2:i Q (ei' ei )
as above.
t. This (9. 16)
9]
23 9
SUBSECTIONS
For a fixed s let No = No be the number of i with O :!S i < t, i == s (mod p n- I ) and ei = O. Let Nl = N� be the number of i with 0 � i < t, i == s (mod p n - I) and ei nonzero. Then No + NJ = P - 1. If No = p - 1 then As = O. If No = 0 then the second term in (9. 16) is at least (p - 1)q and so As � (p - 1)q. If 0 < No < p - 1 then the first term in (9. 16) is at least No NI q � No q and the second term is at least NJ q and so As � (p - 1)q. Since a -1 0, some ei is nonzero . Thus As � 0 for all s and As � (p - 1)q for at least one value of s. The result follows from (9. 1 5). 0
Let . B be a block of defect d. Let S (y, E) be a major subsection asspciated to B. Let k (B) denote the number of irreducible characters in B and let I (E) denote the number of irreducible Brauer characters in E. (i) Let (; be the Cartan matrix of E and let 6 be the quadratic form corresponding to p d {;- I . Let q (E ) be the minimum value assumed by 6 on the set of all nonzero vectors with integral coordinates. Then q (E)k (B ) � p d I (E) . (ii) If B contains no irreducible characters of positive height then k (B )2 � P 2d I (E ) .
THEOREM 9.17.
PROOF. Let {Xu } be the set of all irreducible characters in B. Choose n so that if s is a primitive p n th root of 1 then m �� B ) E Q( s ) for all u, v. (i) By (9.5) m ��13 ) -1 0 if XW has height O. Thus by (9.4) (ii)
m �� 13 ) = 2: I m �� 13 ) 1 2 > 0 v
for all
u.
By (9.4) (i) pdm �� 13) = 6 ( a , a ) , where a = ( d �i) ' Hence p n - l (p - 1)q (E) � tr (p d m �� 13 ») by (9. 14). If this is summed over all u then (9. 14) (iii) implies that p n - l (p _ 1)q (E)k (B) � tr(p d I (E )) = p n - 1 (p - 1) p d I (B ) . The result follows. (ii) By (9.5) pd m �� B ) is a nonzero algebraic integer for all u, v. By (9.4) (ii) 'Lu l m � 13) 1 2 = m � 13 ) . If k = k (B ) and t = p n - l (p - 1) the arithmetic-geometric inequality implies that
= kt1 2: p 2d m ( y, B )a _
a
vv
,
240
CHAPTER V
[10
where (T ranges over the Galois group of Q(£ ) over over v, (9.4) (iii) implies that
Q. If this is summed
k � k1t L L p2d m <';'; B = k1t tp2d l (B ) = -l p2d l (B ) . k ) (J"
(J"
U
0
10. Lower defect groups
This section contains results of Brauer [1969a] and reformulations of some of these results by Iizuka [ 1 972] . Related results can be found in lizuka 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 Bro U(� [1979] . See also Olsson [1980] . The notation introduced in Chapter IV, section 8 will be used in this section. Furthermore e l , e2 , . . . are all the centrally primitive idempotents in R [ G ] . For each t, BI is the block corresponding to el and k (BI ) denotes the number of irreducible characters in BI • For a subset S of G, S = LxES X E R [ G ] and S is the image of S in R [ G ] . Observe that ZR (G) = Efj I ZR (G)el and Zi? (G) = Efj I Zi? (G)el• Furthermore Zi? ( G )el = ZR (G ) el i s a n ideal o f Zi? (G) which i s a local ring and ZR (G)el is an ideal of ZR (G) which is a local ring.
BI it is possible to associate k (BI ) conjugate classes {Cj 1 1 � j � k (BI )} such that the following conditions are satisfied. (i) Every conjugate class of G is associated with exactly one block. (ii) {Cjer } is an R -basis of Zi? (G)el•
LEMMA 10. 1 . With each block
PROOF. Let { G } be all the conjugate classes of Zi? (G)el • Then
G. Let {a f} be an R -basis of
l {a i} 1 = dimi? Zi? (G)et = rankR ZR (G) el = k (BI ) . l lj Let Ci = LI Lj a l , ) a j and let A denote the matrix (a l ,n), where (t, j) is the row index and i is the column index. Since A is nonsingular it is possible to arrange the classes {G} so that A=
*), C ' A,
1 0]
LOWER DEFECT GROUPS
where AI is a nonsingular indexed by (t,j) for each
k (BI ) x k (BI ) matrix and the columns of AI t. The result follows since a jel = 8sIa j . 0
241
are
Let �(G) be a complete system of representatives of the conjugate classes of p -groups in G. For P E �(q) let Vp (G) be the linear subspace of Zi? (G) which is spanned by all ¢ where P is a defect group of G . For P E �(G) let
�(P : G) = { Q I Q � o P, Q E �(G)} . Define U(P : G) = LOE'{l(P: O) Vo (G) and V(P : G) = Vp (G)EB U(P : G). Let V, (P : G) = V(P : G) n Zi? (G)et • By (111.6.3) V(P : G) is an ideal of Zi? (G). Therefore VI (P : G) = VI (P : G)et and if
U, (P : G) = U(p : G) n VI (P : G)
then
V(P : G) = Efj v, (P : G), (10.2) V(P : G)/ U (P : G) = EB VI (P : G)/ UI (P : G). LEMMA 10.3. Let P E �(G), let B = BI and let { Cf} be defined as in (10.1). Let C �, • . • , C�, mB (P) be the subset of { Cf} consisti�g of those classes which have P as a defect group. Then the image of {C�,j el 1 1 � j � mB (P)} in VI (P : G)J UI (P : G) is an R -basis of this space. The number mB (P) = mB (P : G) depends only on G, B and the conjugate class of P. 1 ,
PROOF. It is clear that {C�,j el 1 1 � j � mp ( Q ), Q � oP} is a basis of VI (P : G) for all t and P by (10.2). Thus {C�,j el 1 1 � j � mB (P)} is a basis of V, (P : G)J UI (P : G). The last statement is an immediate conse quence. 0 If mB (P) i' 0 then multiplicity mB (P).
P
is a
lower defect group of B. It is said to occur with
P E �(G) and let B = BI . Then mB (P : G) = � mB (P : No (P)), where B ranges over all the blocks of No (P) with J3° = B.
LEMMA 10.4. Let B
PROOF. Let
N = No (P) and let s be the Brauer homomorphism with
242
[10
CHAPTER V
respect to (G, P, N). Let e be the primitive idempotent of Zft. (G) corre sponding to B and let s(e) = L e, where l ranges - over all the primitive idempotents in Zft. (N ) such that B G = B for B corresponding to e. By (III.9.8) s is an R -isomorphism from Vp (G) onto Vp ( N ). Thus by (10.3) s induces an R -isomorphism from V, ( P : G )/ U, (P : G ) onto V(P : N)s (e)J U (P : N )s(e). By (10.2)
EB V (P : N) l / U(P : N ) l, where l ranges over all the primitive idempotents in Zft. (N ) that occur in the sum s (e) = L e. Therefore mB (P : G) = dimft. ( V, (P : G )/ Ut (P : G)) V(P : N )s(e)/ U(P : N )s(e) =
= L dimft. (P : N)l/ U (P : N ) l ) = L m13 (P N ). D :
10]
LOWER DEFECT GROUPS
LEMMA 10.7. Let B = B, . Let P E �(G) and let y E �o (G). Let { Cn be defined as in (10.1) and let C�, P, I , , C�, p, mS (P) be the subset of { Cj} consisting of those classes which have P as a defect group and are contained in S (y ). Then the image of { C�,P,j i, / 1 � j � m � (B ) } in W" AP : G) / T" y (P : G) is an R - basis of this space. The number m �(P) = m � (P : G) depends only on G, B, the conjugate class of y and the conjugate class of P. .
.
•
PROOF. The first statement follows from (10.3) and (10.6). The last state ment 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 m �(P).
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 P � G D.
LEMMA 10 8. (i) LY E!J3o( G ) m �(P) = mB (P) for all P E �(G). (ii) LP E!J3 ( G) mB (P) = k (B ).
By (III.6. 10) D is a lower defect group of B as e second statement follows from (III.9.6) and ( lOA). D
PROOF.
PROOF.
E Zft. (G)e. The
Let �o (G) be a complete set of representatives of the conjugate classes of p -elements in G. If y E �o (G) let S (y ) denote the p -section which contains y. Let W(y : G) = W( y ) be the subspace of Zft. (G) which is spanned by all G with Cj � S (y ). Then Zft. (G) = EB y W(y ), where y runs over �o (G). By (IV.8.5) (iii) W (y )e, � W (y ) for all y and t. Thus if W, (y ) = W( y ) n Zft. ( G ) then W, (y ) = W( y )e, and W(y ) = EB , w, (y ). Furthermore Zft. (G)et = EBY E \l30 ( 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; e, with C; � S (y ). Thus the number of such classes depends only on G, B and the conjugate class of y. Define
W" A P : G) = V, (P : G) n W(y ) and
Tt,y (P : G) = U, ( p : G ) n W(y ). Then v,
(P : G) = E� W " y (P : G), Y 't;fG )
VI (P : G)J UI (P : G) = EB W';, y (P : G)/T" y (P : G) . yE��()( G )
( 10.6)
243
.
Immediate from the definitions.
D
LEMMA 10.9. Let y be a p -element in Z(G). Then m �(P) = m 1 (P) for all blocks B and P E �(G). PROOF. The map which sends a to ay defines an R -isomorphism on Zft. (G). It clearly preserves the ideal V, (P : G) for all t, P and sends W(l) onto W(y ). Thus it sends Wt, l (P : G )J TI, l (P : G) onto W y ( P : G)J ,£,y (P : G) and the result follows from (10.7). D "
THEOREM 10.10. Let y E �o ( G), P E �(G). Then m � (P : G ) = L L m 1 (Q : C G (y )) , Q
13 where Q ranges over the elements in �(CG (y )) which are conjugate to P in G and jj ranges over all the blocks of C G (y ) with jj G = B . PROOF. Let C b e a conjugate class o f G with defect group P contained in S(y ). Let So be the p -section of C G (y ) containing y. Then Co = C n So is easily seen to be a conjugate class of C G (y ) which has a defect group Q . that is conjugate to P in G. Conversely every such conjugate class Co of C G (y ) is of the form C n So for suitable C. Let s be the Brauer homomorphism with respect to ( G, (y ) , C G (y )). If C is a conjugate class of
244
[10
CHAPTER V
}n S {y) �ith defect group P then the definition of s implies that s( C ) = Co + C', where C' is a union of conjugate classes of CG (y), none of
G
which is in So . Let B = Bt and let e be the primitive idempotent of ZR ( G ) corresponding to B. Let s{e) = L m lm , where each em is a primjtive idempotent of ZR (CG (G» . By the previous paragraph s induces an R -isomorphism from W ( y : G) n Vp (G) onto Lo W(y : cG (y » n Vo (CG (y » , where Q ranges over all elements of � (G) which are conjugate to P in G. Thus by (10.6) and (10.7) s induces an R -isomorphism from Wt, y (P : G)/Tt, y (P : G) onto E9 0 Lt \iV;, y (Q : CG (y » /Tt, y (Q : CG (y » . The result now follows from (10.9). 0 If Y 1= 1 then (10. 10) implies that m � (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 k (P).
THEOREM 10. 1 1 . Let S = S (l) be the section of G consisting of p i-elements. For a E R [ G ] let Tr(a ) = LxE G X - 1 ax. Let C �,P,j be defined as in (10.7) and for each t, P, j choose x �,p,j E C�,p,j . Then for B = Bt, {C�,p,j l p E � (G) , 1 � j � m k (P)}
is an
R -basis
of ZR ( G : S )et and {I CG (x �,p,J 1 CLp,j et I P E � ( G ) , 1 � j � m k ( P )} is an R -basis of Tr(R [S ])et . Furthermore ZR ( G : S )et/Tr(R [ S Det ChR ( G : G reg, Bt )/ChR, proj (G : G reg, Bt ) =
as
R
modules.
For a E ChR (G : Greg , Bt ) define a * = LXES a (x -1)x E ZR ( G : S). Then the map sending a to a * is R -linear. By (IV. 3 . 1 1 ) it is onto ZR ( G : S). By (IV.2.3) ChR, proj ( G : Greg)* � Tr(R [ S ]). For a conjugat: class C let D (C) be a defect group of C. If x E C then Tr(x ) = I CG (x ) I C. Thus { D (C)C I C � S } is a basis of the R module Tr(R [ S ]). Therefore ZR ( G : S )/Tr(R [ S ]) is a torsion R module whose invariants are { I D ( C) I I C � S } . By (III.3 . 1 1 ) this is the set of elementary divisors of the Cart an matrix of G and so ChR, proj ( G : Greg) * = Tr(R [s]). By (IV.7.2) (a Bt ) * = a * et . Hence Ch R ( G : Greg, Bt )* = ZR ( G : S)et and Ch R,proJ ( G ·. G reg, Bt )* = Tr(R [ S ])et • This establishes the required isomorphism.
PROOF.
1 1]
245
GROUPS WITH A GIVEN DEFICIENCY CLASS
It follows from ( 10.7) and the fact that R is a local ring that has the required basis. Let T, be the R module with basis
ZR ( G
: S )et
{ I CG (x �,p,j) 1 C Lp,j et I p E �(G) , 1 � j � m k (P)} , where
B
=
Bt .
Then T, � Tr(R [ S ]) et . Clearly
{ I D ( C �,p,J l l p E � (G) , 1 � j � m k ( P )} is the set of invariants of the R module ZR
EBt ZR ( G : S ) et / Tt and so T,
=
(G : S ) et / T, .
Hence
=
ZR (G : S )/Tr (R [S])
=
EBt ZR ( G : S )er /Tr(R [ S ])er
Tr( R [S ])et for each
t.
0
COROLLARY 10. 12. Let B = Bt and let m � O be an integer. Then the multiplicity of p m as an elementary divisor of the Cartan matrix of B is given by Lm k (P), where P ranges over all groups in � (G) of order p m. By (10. 1 1) Lm k ( P ) is the multiplicity of p m as an invariant of ZR ( G : S )et /Tr(R [S ])et . The result follows from the isomorphism in (10 . 1 1). 0
PROOF.
COROLLARY 10. 1 3 . For y E �o (G) let l (y,B ) denote the number of irreduc ible Brauer characters in blocks 13 of CG (y ) with 13 G = B. Then 2: m h ( P ) = l (y, B ) .
PE'l5( G )
PROOF. By (10. 10) it suffices to prove the result in case y = 1 . In this case it follows from (10.12) as I (1, B) is the rank of the Cart an matrix of B. 0 COROLLARY 10. 14. Let D be a defect group of B. Let 13 be a block of D CG (D ) with 13 G = B and let T(B) be the inertia group of 13 in NG (D). Then m �(D ) is equal to the number of T(J3 ) conjugate classes in Z(D ) n S (y ). By (IV.4. 16) and (10. 12) m k (D ) = 1 . The result now follows from (9.8) and (10. 10). 0
PROOF.
1 1 . Groups with a given deficiency class
.
The results in this section are due to Brauer [ 1 964a] .
246
CHAPTER V
[11
Let r be an integer. The group G has deficiency class r for the prime p if G has no non principal 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 1 1 . 1 .
If p'
{' I G I
then
G
has deficiency class r for the prime p.
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 1 1 .2. G
LEMMA 1 1 .3 . r � n then G
Suppose that P <J G and I p 1 = p ". If G is of deficiency class is of deficiency class O.
PROOF. By (4.4) every block of G has defect at least block is the only block of G. 0
n. Thus the principal
LEMMA 1 1 .4. Let P be a p -group with I P I = p " and let G = PCo (P). Then G has deficiency class r if and only if G /P is of deficiency class r - n. PROOF. Clear by (4.5).
0
THEOREM 1 1 .5 . If G has deficiency class r then for every p -element y of G, Co (y ) is of deficiency class r. Conversely if r > 0 and Co (y) has deficiency class r for every element y of order p in G then G has deficiency class r. PROOF. Let y be a p -element, let Bo be a nonprincipal block of Co (y ) and let d be the defect of Bo . By the third main theorem on blocks (5.4), B&' = 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 Co (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)ti) there exists a nonprincipal block of Co (y ) of defect d and so Co (y) is not of deficiency class r. 0 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. 1 8) implies the next result.
11]
GROUPS WITH A GIVEN DEFICIENCY CLASS
THEOREM 1 1 .6.
247
There exist only finitely many groups G of deficiency class whose Sp -group has a given order.
In case ( 1 1 .6).
p = 2, Brauer [ 1 97 4c]
0
has proved the following generalization of
Let G be a group of deficiency class r � 0 for the prime 2. Suppose that a Sz-group P of G contains an elementary abelian subgroup of order 2'+1 . Let I P 1 = 2 a • Then there exists a bound f(a ) depending only on a such that I G I � f(a ). THEOREM 1 1 .7.
The proof of ( 1 1 .7) is not very difficult but depen ds on specia l prope rties of involu tions. It is not known wheth er an analog ous result is true for odd primes . . If y is a p -elem ent in G it is possib le to apply (1 1 .4) to Co (y). Thus ( 1 1 .5) can be used to give an induct ive characteriza tion of group s of positiv e deficiency class. Such a characterization does not seem to be possib le for group s of deficiency class 0 since the secon d statem ent of ( 1 1 .5) is false in this case. For instan ce let q be a prime with q == 1 (mod p) and let G be a Frobe nius group of order pq. Then G does not have deficiency class o for p but I Co (y) I = p for every eleme nt y of order p in G and so Co (y ) has deficie ncy class 0 for every eleme nt y of o�der p in G. This construction shows the existe nce of infinit ely many groups of deficie ncy class 1 with a Sp -group of order p. It follows from ( 1 1 . 6) that an analog ous situati on canno t occur for deficie ncy class O. It has been shown that a simple group of Lie type in charac teristic p has deficiency class 1, Dagge r [ 1971], 1. E. Humphreys [ 1 97 1 ] . This also shows the existen ce of infinit ely many group s with deficiency class 1 . Amon g these there are only finitel y many with a given Sp -group . A related result of a slightly different sort can be found in Tsushima [1977] .
irreducible R [G] module up to isomorphism. Then irreducible R [ G ] module up to isomorphism. 2]
RADICALS AND NORMAL SUBGROUPS
PROOF. Clear by (1. 19.4).
249
B
contains a unique
0
LEMMA 1 . 3 . Let H <J G and let V be a projective indecomposable R [GJ module. Then VH = E9 U; where each U; is a projective indecomposable R [H] module and for each i there exists x E G with Vi = V�.
CHAPTER VI
PROOF. Immediate by (1. 13 .6) and Clifford' s theorem (111.2. 12). 1 . Blocks and extensions o f
R
will be used The notatio n introd uced at the beginning of Chapt er III on will also throug hout this chapte r. The follow ing assum ptions and notati be used. K is a finite extension of Qp , the field of p -adic numb ers. A R is the ring of integers in K. i n K of K By (111.9. 10) and (1. 19.4) there exists aAfinite unramified. extens � . condItIons are such that if R is the ring of intege rs in K then the follow mg
A satisfie d. K is a G. of H up subgro every for [H] R of field g splittin a is R (I) splittin g field for every p '-subgr oup H of G. (u) be th� (II) Let B be a block of R [G] with defect group D.A Let such that If [G] R of B block a Galois group of K over K. Then there exists ' V is an irreducible R [ G] module in B then _
I
V Q9 R R
=
r-1
L yO"j
j =O
with Y E 13.
13 i If furthermore 13 U) = u is the block of is a defect group of B U) by (111.9. 10).
R [ G]
which contai ns v ui then
R [G] module in B is R [D]-projective and there exists an irreducible R [G] module in B with vertex D.
LEMMA 1 .2. Let B
0
be a block of R [G]. Suppose that B contains a unique 248
R [G] covers
the block
b of R [H]
LEMMA 1 .4. Suppose that H <J 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 D CG (D) � H. Then the following hold. (i) B = b G is the unique block of R [GJ which covers b. (ii) D is a defect group of b. PROOF. (i) Choose the notation so that B covers G, where G is defined as in (II). by (V.3.9) B is regular and so by (V.3.7) 13 = G G is the unique block which covers G. Then 13 U) is the unique block which covers G U). 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 G and so by (II) D is a defect group of b. 0
D
Then !!. very LEMMA 1 . 1 . Let B be a block of R [ G] with defect group D.
PROOF. Clear by (111.4. 14).
S uppose th at H <J G. The block B of A A if B covers b U) for some j.
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 Knorr [ 1 976] . Some related results have been proved by Ward [1968], Huppert and Willems [ 1 975], Willems [ 1 975] .
LEMMA 2. 1 . Let V be an R [H]-projective R [ G] module and let M = V/ VJ(R [G]). Then M is R [H]-projective if and only if VJ(R [G]) = VJ(R [H]).
250
[2
CHAPTER VI
PROOF. Let I = leR [ G D and let 10 =J(R [HD. As loR [ G ] = R [G]lo, Vlo
is an R [ G ] module. Suppose that VI = Vlo. Let A = R [ G ] , B = R [H] and S (l.4. 1 1) implies that M is R [H]-projective. Suppose conversely that M is R [H]-proj ective. Then _
=
10•
Then
O � ( Vl/ Vlo)H � ( V/ Vlo)H � MH � O is a split exact sequence as ( V / Vlo)H is completely reducible. Thus O � Vl/ Vlo � V/ Vlo � M � O is a split exact sequence as M is R [H]-proj ective. Hence V / Vlo = M E9 Vl/ Vlo and so VI/ V10 = ( V / VJo)l Ml + ( VI / V10)1 = ( VI / V10)1. Thus Vl/ Vlo = (0) by Nakayama's lemma (I.9.8) and so VI = Vlo. D =
COROLLARY 2.2. Let V be a projective indecomp�sable R [ G ] module and let M = V/ VI ( R [ G D. Then H contains a vertex of M if and only if Vl ( R [ GD = Vl( R [ il D· PROOF. Clear by (2. 1).
D
2J
As U;1 (R [H]) � U;1 ( R [ G D 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 = V / Vl(R [ G D for some projective indecomposable module V, the result follows from (2.2). D
COROLLARY 2.4. Let V be an irreducible R [H] module. Assume that every component of vG lies in a block with a defect group which is contained in H. Then VG is completely reducible. PROOF. By assumption there exist centrally primitive idempotents {ei } such that a defect group of each ei lies in H and such that VGe = VG, where e = L ei . By (2.3)
(
be the block corresponding to e. The following are equivalent. (i) H contains a defect group of B. (ii) Vl( R [ G D n = Vl( R [HD n for all R [ G ] modules V in B and all integers n ;?; O. (iii) l(R [ G De = l( R [HD R [ G ] e. (iv) Vl(R [ G D = Vl(R [HD for all projective indecomposable R [ G ] modules V i n B.
)
VGJ(R [GD = V Gl(R [ G De = V _® R [ G ] l( R [ G De R[ H ) = V ® l (R [ G D e R[ H ) = V ® l ( R [HD R [ G ] e = VJ (R [ H] ) ® R [ G ] e R[H) R[H) = (0).
Hence VG is completely reducible.
THEOREM 2.3. Let e be a centrally primitive idempotent in R [ G] and let B
25 1
RADICALS AND NORMAL SUBGROUPS
D
COROLLARY 2.5. Suppose that p ,{ 1 G : H I . Let V be an irreducible R [H] module. Then VG is completely reducible.
_
PROOF. (i) � (ii). By ( 1 . 1) every module in B is R [H]-proj ective. Thus (ii) follows from (2. 1) by induction on n. (ii) � (iii). This is clear 'as eR [ G ] = R [ G] e is a module in B. (iii) � (iv). Let R [ G ] e = EB Vi , where each Vi is indecomposable. Each proj ective indecomposable R [ G ] module in B is isomorphic to some Vi . By the hypothesis EB VJ( R [ G D = R [ G ] el( R [ G D = R [ G ] el(R [HD = EB VJ( R [HD·
PROOF. Since H contains a defect group of every block the result follows from (2.4). D The next result is quite old. See e.g. Green and Stonehewer [1969] , Villamayor [ 1 959] .
COROLLARY 2.6. l(R [G]) = J(R [HD R [ G ] if and only if p
,{
I G : HI .
PROOF. If l(R [ G]) = l(R [HD R [ G ] then R [G]/l(R [GD is R [H]
projective by (2. 1). Thus every irreducible R [ G ] module is R [H] projective and so p ,{ I G : H I by ( 1 . 1). If P ,{ 1 G : HI then every R [ G ] module is R [H]-proj ective. Thus (2. 1) implies the result . D If W is an indecomposable R [ G ] module let w ( W) = WH ( W) denote
[2
CHAPTER VI
252
the integer so that WH is a direct sum of w ( W) nonzero indecomposable R [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 proj ective R [ G ] module then w ( U) = w ( U / UJ( R [ G ]) = w ( U / UJ( R [H])
by Clifford's theorem (III.2. 1 2) and (I. 1 3 .7). THEOREM 2.7. Let B be a block of R [ G] which covers the block b of R [H] .
3]
SERIAL MODULES AND NORMAL SUBGROUPS
253
UJ ( R [ G ] )" / UJ( R [ G ])n+ l � (0). Let UH = ��= 1 Pi , where each Pi is a projective indecomposable R [ H ] module. By (V.2.3) and (1 .3) each Pi is
serial. By (2.3)
VH = ( UJ( R [ H] )" / UJ( R [ H] ) n + l )H
w
=
P;l ( R [H])" / PiJ ( R [ H] )n + 1 E9 i= l
and so w ( V) ::s; 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. 0
Assume that H contains a defect group of B. Then the following statements are equivalent. (i) Every indecomposable projective R [ G ] module in B is serial. (ii) Every indecomposable projective R [H] module in b is serial. If furthermore (i) or (ii) is satisfied then all irreducible modules in B have the same width.
It should be mentioned that in general not all irreducible R [ G ] modules in V will have the same width. For instance suppose that p = 2, H = SL2(8), G is the semi direct product of H with a cyclic group of order 3 and R contains the field of 8 elements. Then the principal block of G contains an irreducible R [ G] module of degree 6 and width 3, and of course it also contains the one dimensional module of width 1 .
PROOF. Suppose that (i) is satisfied. Let U be a proj�ctive indecomposable R [ G ] module in B of maximum width. Let UH = E9 �= 1 Pi , where each Pi is a projective indecomposable R [H] module. Then w = w ( U ) =
COROLLARY 2.8. Let B be a block of R [ G ] which covers the block b of
w ( U / UJ( R [H]) .
Let n be a positive integer such that PJ ( R [H] )" � (0). By (1 .3) PiJ( R [H] )" � (0) for i = 1, . . . , w. Let V = UJ( R [H])" / UJ( R [H] )"+I . Then w
VH = E9 P;l(R [H] )" / PiJ( R [H] ) ,, +
1
R [H] . Assume that H contains a defect group of B. Then the following statements are equivalent. (i) Every indecomposable projective R [ G ] module in B is serial. (ii) Every indecomposable R [ G ] module in 13 is serial. (iii) Every indecomposable projective R [H] module in b is serial. (iv) Every indecomposable R [ H] module in B is serial.
PROOF. Clear by (I. 16. 14) and (2.7).
i=1
and s o w ( V) � w . By (2.3) V i s irreducible since U i s serial. Let Uo b e the projective indecomposable R [ G ] module which corresponds to V. Then w ( Uo) = w ( V) � w ( U) . Hence the choice of U implies that w ( Uo) = w. The indecomposability of the Cartan matrix (I. 16.7) now implies that every projective indecomposable R [ G ] module in B has width w and so every irreducible R [ G ] module in B has width w. Thus (i) implies the last statement. Furthermore PiJ( R [H] )" / PiJ( R [H] ) ,, +1 is irreducible as w ( V) = w. Hence each Pi is serial. Since w is the width of every projective indecomposable R [ G ] module in B, the previous paragraph implies that if U is any projective indecom posable R [ 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 proj ective indecomposable R [ G] modules in B. Let n be a positive integer such that V = _
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
posable R [H] module in b is serial. (v) Let T(b) denote the inertia group of b in G. Then T(b)/H is cyclic.
LEMMA 3.2. (i) B = b O is the unique block of R [G] which covers b. (ii) b contains a unique irreducible R [H] module V up to isomorphism. CHAPTER VI
254
(i) Clear by (3. 1) (ii) and (1 .4). (ii) This follows from (3.1) (iii) and (1. 19.4).
[3
PROOF.
0
By (V.3.14) T(b)/H is a p '-group. Let a be an irreducible R [G/H] module with aT(b)/H faithful. By (3. 1) (i) dimR a = 1. Let V be defined as in (3.2) (ii). If M is an R [G] module we will write Ma = M @ a. In particular ai is defined by induction as ai = a i - I @ a.
3.3. (i) There exist integers s, e > 0 with e I I T(b) : H I and an irreducible R [G] module W such that W, Wa, . . . , Wa e - I are pairwise nonisomorphic modules in B, W = Wa e and every irreducible' R [ G] module in B is isomorQhic to some Wa i. Furthermore V = S (EB;=l Wa i ). (ii) If R = R then s = 1 and e = I T(b) : H I .
THEOREM
Suppose first that R = R. 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)/H. Let s = 1 and let e = I T(b ) : H I . 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 M - eA (z )Me = A ( x - ezx e ) = A ( x -e )A (z )A (x e ) for all z E H. Thus by Schur's Lemma Me = cA (x e ) for some c E R. Let F be a finite extension field of R with Co E F such that c � = c. Define A (x ) = CalM. Thus A extends to a representation of T(b) over F. This representation is irreducible as its restriction to H is irreducible. Since R = R is a splitting field of T(b) there exists an R [ T(b)] module X such that XH = V. Consequently Xa i is irreducible for all j and (Xa i )H = V. By (3.2) (i) Xa } is in b T(b) for all j. Let (3 denote the trivial R [H] module. By (11.2.3) e-I (b ) = ( XH @ (3 r (b ) = X @ (3 T (b ) = Xa i.
PROOF.
VT
2:o
If Y is an irreducible R [ T(b)] module in b T(b ) then V I YH by (1.4) and so IR ( VT(b) , Y ) � 0 by (111.2.5). Thus Y = Xa i for some j. If Xa = Xa] for some 0 � i < j < e then by (111.2.5) i
}=
1 < lR ( VT(b ), Xa i ) = I( V, V) = 1 . Let W Xc. Then Wa i ( Xa i ) O by (V.2.S). The result is proved in case R R. We will now consider the case of general fl. The previous part of the proof imp.l ies the existence of an irreducible R [G] module W in B so that yO = EB;=l Wa i, where e = I T(b ) : H I . Furthermore Wa i = Wa i if and only if i j (mod e) and every irreducible module in B is isomorphic to Wai for some j. Let (u) be the Galois group of R over R. By (1. 19.4) V @R R = EB�=�I Y O"i , where m is the smallest integer with y u m = y. Let W be an irreducible R [G] module with W I W @R R. Let e, t be the smallest positive integer such that Wa e = W, ( yO t' = yO respectively. Then e I I T(b) .:. H I . Also t i m as ( yN t = ( yO" )N. Let U be the direct sum of all distinct R [G] modules of the form WUiai. Then 3J
SERIAL MODULES AND NORMAL SUBGROUPS
=
255
=
=
==
therefore
(4 ) � R = u = E!1 ( YO ti• J =l Wa i
R
v G � R = V l? R
(
,=1
r = ; ( $, ( VG t') = ; (ffl ) 1? R
By (2.4) and (3.2) VO is completely reducible. Thus (1. 19.4) implies the result. 0 Waf
If M is an R [G] module then I (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" - l (M)) for n ?; 2.
3.4. Suppose that V @R R is the direct sum of m irreducible R [ G] modules. Then the following hold. (i) If M is any irreducible R [G ] module in !} then M @R R is the direct sum of m algebraically conjugate irreducible R [ G] modules. (ii) If U is a projective indecomposable R [G] module in B then U @R R is the direct sum of m algebraically conjugate projective indecomposable R [ G] modules. (iii) If M is an indecomposable R [G] module in B then I(M @R R ) = ml (M) . G.0ROLLARY
_
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). 0
[4
3.5. (i) If U is a nonzero indecomposable projective R [G] module in B then U is serial and I ( U) = 1 D I . (ii) Let M be an indecomposable R [ G] module in B. Then M is serial and Socn (MH ) = (Socn (M» H for all n. CHAPTER VI
256
THEOREM
Let P be a projective indecomposable R [H] module which corresponds to V. Let P be a projective indecomposable R module which corresponds to V. By (IV.4. 16) the unique Cartan invariant in 6 is 1 D I . Thus I (P) = I (P) = 1 D 1 by (3.4). By (2.8) U is serial. Thus I ( U) is the smallest integer n with UJ(R [G])n = (0). Hence by (2.3) I( U) is the smallest integer n with UJ(R [H])n = (0). Since P I UH it follows from (1.3) that l( U) is the smallest integer n with PJ ( R [H]) n = (0). Thus I ( U) = I(P) = I D I as P is serial. This proves (i). Furthermore { UJ(R [G])n }H = { UJ(R [ G])H Y and so Socn ( UH ) = {SoCn ( U)}H for all n. By (2.8) M is serial. Hence M = SOC k ( U) for some projective indecom posable R [ G] module in B and some integer k. Therefore Socn (MH ) Socn +k ( UH ) = {Socn+ k ( U)}H = {Socn (M)}H
PROOF.
for all n.
=
0
COROLLARY 3.6. Let M be an indecomposable R [G] module in B. Then Ai @R R = EB �r, where 'T ranges over a factor group of the Galois group of R over R and each IV!" is indecomposable with I (M) = 1 (MT ) for all 'T.
Furthermore Soc(M) @R R
EB Soc(My·
By (3.5) M = Socn ( U) for some integer n �nd some principal indecomposable R [ G] module in B. Since U @R R EB (r with V a principal indecomposable R [G] module, the result follows from (3.4) (ii). 0 =
PROOF.
=
4. The radical of
R [ G]
This section contains various results concerning J(R [G]). The material up to (4.8) is due to Tsushima [1971a] . See also Tsushima [1978a]. The following notation will be used in this section. VI, V2, • • • is � complete set of representatives of the isomorphism classes of irreducible R [G] modules. For each i, U; is a projective indecompos able R [G] module corresponding to �. Let 'Pi, $i be the Brauer character
4]
afforded by Vi, U; respectively. Choose the notation so that 'PI 1 0 , the principal irreducible Brauer character. Let ti be the trace function afforded by V; . The following hold. THE RADICAL OF R [ G ]
257
=
'Pi (x ) for x a p '-element in G.
(4. 1) ti (x ) = ti (xp) where Xp ' is the p '-part of x in G. (4.2) Let F; R (ti (x ), x E G) and let T; be the trace from F; to R. Choose the notation so that T1(t1 ), T2(t2), • • • are all the distinct elements of the set { T; (ti )}. By (1.19.3) and (1. 19.4) { T; (ti )} is the set of all trace functions affC?rded by irreducible R [G] modules. Furthermore J (1� [G]) = J(R [G] @R R ). ti (x ) =
=
A
4.3. Let { Vi I j E S } be a set of pairwise nonisomorphic irreducible R [ G] modules. Let I = Is be the intersections of the £!nnihilators of all �, j E S. Let { aj I j E S} be a set of nonzero elements in R. Then Ann(I) = R [G]c, where c = LX E O { LjES ajtj (x - l )}x.
�EMMA
Let s = Lj ES aA . Define t E HomR (R [G], R ) by ( LxEo bxx )t = bl • By (1. 14.9) t is symme�ric and ponsingular. Clearly s is symmetric and nonsingular in HomR (R [G]II, R ). Thus PROOF.
( yELO byy) ct { x,yELO by jLES ajtj (x - 1 )xy } t = yELO by jESL ajtj (y ) =
The result follows from (1. 16.17).
0
COROLLARY 4.4. (i) For all i, LXE O T; (ti (x -l» x E Ann(J(R [G]) . _ (ii) Let a l , ·a 2 , . . . be nonzero elements of R. Then Ann (J ( R [G]) = R [G]c, where c = LX E O {Li aa� (ti (x -I» }x.
It suffices to prove the result in case R = R. Hence T; ( ti ) ti for all i. Then (i) f<2}lows from (4.3) with 1 = Ann( V; ) and (ii) follows from (4.3) with 1 = J(R [G]). 0 PROOF.
=
4.5. Suppose that R = R. Let p n be the order of a Sp -group of G. Let I be the annihilator of EBjES Vi, where S is the set of all j with p n + I ,r $j (1). Then J (R [ G ]) C I and Ann(I) = R [ G]c, where c is the sum of all p-elements in G. LEMMA
CHAPTER VI
258
[4
PROOF. Clearly J(R [ G ]) C 1. Let Wi (1) p nhi for each i. Thus hi is an integer for all i. by (lV.3.8) Li hi'Pi (x ) = 0 if x � 1 and Li hi 'Pi (1) = p - n I G I . Thus if a = p - n I G I then a � O. Clearly iii � 0 if and only if i E S. Thus (4. 1) and (4.2) imply that for x E G if x is not a p -element, =
if x is a p -element. Therefore LXE G {a -[ (4.3). 0
L i E S iii (ti (x - I ))} X =
c. The result follows from
Hence i=
P300F. (i) I f Y E G then iy c n X E G R [ G ] iXY = L Thus i is an ideal of R [G ] . Theref�)f e i m : 1 c i m ( R [ G ]I) = i m I for any integer m � 1 . Hence by induction I m + 2 C F 1 m for any integer m � O. Thus i is nilpotent. R [ G ]r = IG (rHX (r ) R [ G]) = IG ({rH (I)YR [ G ]).
that
R [ G ]I =
C�o ( rH (I)YR [G l ) .
(ii) By (1. 16. 16) the left and right annihilators of an ideal in F[ 0 ] or F[ H] coincide. Thus by (ii)
( 2: ( 2:
) ( 2: )
i = IG rH (IYR [ G ] = rG xE G IH (IYR [ G ] xE G R [ G ] IH (IY = rG x EG
)
x EG
xE G
�EMMA 4.7. X Let H1 be a subgroup of G and let I be a nilpotent left ideal of R [H] . �et I = x - Ix and let i = n X E G R [ G ] r. Then the following hold. (i) I is a nilpotent ideal of R [ G ] . (ii) rG ( i ) = LX E G rH (IYR [ G ] and i = IG {L x E G rH (IYR [ G ]}, where r, I denote right and left annihilators respectively. (iii) If I is an ideal of R [H] then i = n x E G rR [ G ] .
implies
10 ({rH (I)Y R [ G ] ) = 10
rG (i) = rG { lG ( xE2:G rH (IYR [ G ] )} = xE2:G rH (IYR [ G ] .
THEOREM 4.6. Ann(J( R [ G])) is the Reynolds ideal of R [ G ] .
(ii) By_ (1. 16. 15) 1 = IH (rH (I)). This IG (rH (I)R [ G]). Therefore if x E G then
Do
Thus by (1. 16. 15)
If x is a p '-element of G let Cx E R [ G ] be the sum of all the elements in G whose p I-part is con j �gate to x. The ideal of R [ G ] generated by all Cx is the Reynolds ideal of R [ G J . See O'Reilly [ 1 974] for related concepts.
PROOF. Since Cx E R [ G ] and_ J(R [G]) = J( R [ G]) Q9R R i t suffices to prove the result in case R = R. By (lV.3. 10) there exists for each p '-element x of G an R -linear combination La;!i of t l , t2 , . . . such that La;!i (x ) = 1 and La;!i (z ) = 0 for all p '-elements z not conjugate to x. If y is a p -element in CG (x ) then ti (xy ) = ti (x ). Thus (4.4) (i) implies that Cx E Ann(J(R [ G ])). For each i, L Y E G ti (y - I)y = Lti (x - l )cx. Thus by (4.4) (ii) Ann(J(R [ G])) is generated by all ex- 0
259
THE RADlCAL OF R [ G ]
4]
THEOREM 4.8. For a Sp -group P of G let Ep = LXEP X. Then Ann(J( R [ G] )) C L p E pR [ G ] , where P ranges over all the Sp -groups of G. PROOF. Let P be a Sp -group of G. Then J(R [P]) = {Laxx I Lax = O} and so
Ann(J( R [P])) = EpR [P] . Apply (4.7) with I = J(F[P]) . By (4.7) (i) ....... 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 LXE G axx E J(R [ G]) then for all x E G
2: axy 2: ayx = O.
yES
=
yES
L (ii) Suppose that for every Sp -subgroup P of G, L YEP aXY = Y EP ayX = 0 for R L all x E G. Then XE G axx E J ( [ G]). = PROOF. Let c be the sum of all the p -e1ements in G. Then L is the coefficient of z in ( L x E G axx )c = c ( Lx G axx ). Thus (i) L y E S ay z
E
y E S azy
follows from (4.6). Let Ep be defined as in (4.8). Apply the previou s paragraph to P. Then (ii) follows from (4.8) and (1. 16. 15). 0 In general neither of the conditio ns in (4.9) is both necessary and sufficient for an element of R [ G ] to be in J( R [ GJ). This can be seen as follows.
260
CHAPTER VI
[4
Let G be a group with a Sp -group of order p n such that p n +1 l lPi (1) for some i. For instance G = SL2 (4) = As and p = 2. Then lPi (1) = 8 for some i. Let I be defined as in (4.5). Then J( R [ GJ) � I but every element in I satisfies (4.9) (i). Let G = S�2 (4) = As and let p = 2. Let P be a Srgroup, let Ep = LX EP X and let E = Li = 1 (Ep )\ where p X l , . . . , p Xs are all the Srgroups of G. Thus E is the sum in R [ G ] of all �lements 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 [ GJ). If z is an involution with z e P then (zy )2 = 1 for y E P if and only if y = 1. Thus if E = Laxx then L Y E P a ZY = 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 dimR J( R [ G J). THEOREM 4.10 (Wallace [1958], [1961 J). Let p n be the order of a Sp -group of G. The following hold. (i) p n - 1 � dimR (J( R [ GJ)) � I G 1 (1 - 1 /lP 1 (1)). (ii) p n - 1 = dimR (J( R [ G J)) 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 lPI is rational valued. Then dimR (J(R [ GJ)) = 1 G I (1 - 1/lP1(1)) if and only if G has a normal Sp -group. PROOF. It may be assumed that R R. Since VJ( R [ GJ) C J( R [ GJ) and dimR VI = 1 + dimR Vtf( R [ GJ) it follows that =
(4. 1 1) If G is a p -group then clearly equality holds in (4. 1 1). Suppose that G is a Frobenius group with Frobenius complement P and Frobenius kernel H, where P is a Sp -group of G. Then any irreducible character of G which does not have H in its kernel is of defect 0 and 'PI is the only irreducible Brauer character of G which has H in its kernel. Therefore dimR (J (R [ G J)) = I G 1 - 2:
'Pi
(1)2 = p " - 1 .
Suppose co�versely that dimR (J( R [ G J)) = p n - 1 . Then (4. 1 1) implies that dimR (J(R [ GJ)) = lP, (l) - 1 . Therefore U;J( R [ GJ) = (0) for i � 1 and so 'Pi is in a block of defect 0 for i � 1. Thus in particular 'PI is the only Brauer character in the principal block. Hence G = O p' ,p (G) by (IV.4. 12). This proves (ii).
THE RADICAL OF R [ G ]
4]
26 1
By (IV.4. 15) lPi (1) � lP I (l) 'Pi (1) for all i. Thus
1 G 1 = 2: lPi (l)'Pi (1) � lPI(l) 2: 'Pi (1)2 (4. 1 2)
= lPI(l) { 1 G 1 - dimR J( R [ GJ)}.
This shows that the second inequali ty in (i) holds and complet es the proof of (i). Suppose equality holds in (4. 12). Then lPi (1) = lP1(1)'Pi (1) for all i. Therefore lPi = lP1 'Pi for all i. Since lP, is rational valued (IV.10.1 ) implies a that lPI(Yj ) = ± p ai for any p '-elemen t Yh where p j i s the order o f a (y Sp -group of Co (Yj ). For any p '-elemen t y of G let c ) be the number of element s whose p I-part is y. Then c (y ) � I lPl (y ) 1 for every p I-elemen t y in G. Hence
IGI=
y
2:
a
p '-eJement
c (Y ) � 2: l lP , (y ) I � 2: c1> I (y ) = I G I ( c1>" 'P l)= I G I .
Y
y
Therefore in particular c (1) = lPl (1) = P n is the order of a Sp -group P of G. Hence P contains all the p -elements in G and so P <1 G. Suppose conversely that P <1 G. Then the set of irreducible Brauer characters of G and of G / P coincide. Thus Hence dimR (J( R [ GJ)) = 1 G 1 (l - 1/p ). As G has a p -complement H it follows that lPl = (lH ) 0 . Thus c1> 1 (1) = p n . 0 "
The assumption in (4. 10) (iii) that lPl is rational valued is probably unneces sary, but this remains an open question . In case G has a p complem ent H, lP l = (lH)O i s rational valued. Thus i n particul ar c1> 1 is rational valued if G is p -solvabl e. Other propert ies of the radical of a group algebra can be found in Clarke [1969] ; Hamern ik [ 1975b] ; Koshitan i [1977] , [ 1 978] , [1979] ; Morita [195 1 ] ; Motose [1974c] , [1977] , [1980] ; Motose and Ninomi ya [1975a], [1980] ; Spiegel [ 1974] ; Tsushim a [1967] , [1968] , [ 1 978b] , [1 979] ; Wallace [1962a] , [1962b], [1965] , [1968] . Properti es of the radical of the group ring of a p-group have be� n studied by Hill [ 1 970] ; Holvoet [1968] ; Jennings [194 1] . I f G i s p-solvab le o f p-Iength 1 see Schwartz [ 1979] .
262
CHAPTER VI
5. The radical of
[5
R [ G]
The notation o f Chapter I V section 8 will be used i n this section. Thus it is assumed that R R. For () E ChK (G) define =
()�
= !Q2 IGI
�
x�G
() ( X - I )x.
lu�i)
=
Most o f the results i n this section are due t o BroUt� [ 1 976b], [1978b] . LEMMA 5 . 1 . Let B be a block and let e be the corresponding centrally
primitive idempotent in R [ G]. Then =
Xv E B, wv (Xu ) = ouv . Thus for a E ZK (G)e, a = �XuEB Wu (a )Xu . (i) If a E ZR (G)e then Wu (a ) E R for all This implies the result. (ii) It is easily seen that J(�XuEB RXu ) = 1T (�XuEB RXu ). Furthermore J(�XuEB RXu ) consists of all E �XuEB RXu with Wu (a ) == 0 (mod 1T) for all Xu E B. The result follows as J (ZR (G)e ) consists of all a E ZR (G)e with wu (a ) == O (mod 7T) for all Xu E B. 0 u.
(
)
LEMMA 5.3. Let I = �u R « I G I /Xu (l))Xu ). Let e be a centrally primitive
idempotent in R [ G] and let B be the corresponding block. (i) Ie �XuEB R « I G 1 / Xu (l))Xu ). (ii) Ie is an ideal in �XuEB RXu and Ie is an ideal in ZR (G )e. (iii) For every positive integer n, Ie {J(ZR ( G ) ) e r � 7T "le. =
PROOF. (i) Clear by definition. (ii) By (5.1) Ie � ZR (G)e. Thus it suffices to prove the first statement.
This follows from the fact that XuXv = ouvXu and so
1TR [G]e + LEB R [G] XuI G(ll ) Xu . Xu
PROOF. It suffices to show that
A = 1TR [G] + � R [G]
(iii) Since Ie is an ideal in � XuEB RXu it suffices to show that for n � 0
(lu��) Xu)
or equivalenty that
If Xp' is the p '-part an element x in G then Xu (x ) == Xu ( xp ) (mod 1T ). Thus (lV.3.12), (4. 1) and (4.2) imply that
L R [G]
u
LEMMA 5.2. Let e be a centrally primitive idempotent in R [G] and let B be , the corresponding block. (i) ZR (G)e � �XuEB RXu . (ii) J(ZR (G)e ) = 1T(� XuEB RXu ) n ZR (G)e.
a
E ZR (G).
Since 1TR [G] � J(R [G]) it follows that J(R [G]) = J(R [ G]). ' In fact J(R [G]) is the inverse image in R [G] of J(R [G]). Let A denote the inverse image of Ann (J(R [G])) in R [G]. 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 ] .
Ae
263
PROOF. For
Thus 0 E ZK (G). By (lV.7. 1) (y, J is the set of centrally primitive idempo tents in K[ G]. Furthermore
Xu
THE RADICAL OF R [ G ]
5]
- I )X) , (XuI G(ll ) Xu) L' R [G] , ( xELG ti (X =
where {ti } is the set of all trace functions of irreducible R [ G] modules. The result now follows from (4.4). 0
J(ZR (G)e y � 7T " L RXu .
Xu EB
This follows from (5 .2) (ii). 0 THEOREM 5.4. Let e, B, I be defined as in (�.3). Let d be the defect of B. (i) J(ZR (G)e ) = (1Tp - dle ) n ZR (G)e. (ii) For every integer n � 1 , J(ZR (G)e y � (7T "p-dIe ) n ZR (G)e.
PROOF. (i) Clearly (7Tp -dle ) n ZR (G )e is an ideal of ZR (G)e. By the definition of I, XU E p dle for Xu E B. Thus �X"EBRXu � p - dle. Hence by -
(5.2) (ii)
J (ZR ( G )e ) 7T =
( x�B RXu ) n ZR ( G)e � 1Tp -dIe n ZR ( G )e.
264
CHAPTER VI
[5
Let Xv be a character of height 0 in B. Then
Therefore Wv (1Tp - dle ) � 1TR and so 1Tp - dIe n ZR (G)e is annihilated by the cen tral character Wv of R [ G ] e. Hence 1Tp - dIe n ZR ( G)e � J (ZR ( 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 ) "-l � ( 1T
2: R Xu ) 1T n - 1 p - dle
Xu E B
The result follows as Ie is an ideal in L.xuE B RXu by (5.3) (ii).
By (5. 1 ) and (5.7) Wu ( co )Xu (1)1 1 G 1 E R for all u. Let Bo be the union of all blocks of defect O. Since {! G 1 / Xu (I)} 1 0 if and only if Xu is in Bo. This implies that "" -- ;:c - = C o = L.J Wu ( Co) Xu ' ?
D
COROLLARY 5.5. Suppose that R1T m = Rp. Let I, B, e, d be defined as in (5.4). (i) J (ZR (G)e ) md +n � 1T nIe for all integers n � 1 . (ii) J (ZR ( G )e )md + l � 1TZR (G)e.
-?
Xu E Bo
If Xu is in a block of defect 0 then Xu (y ) = 0 for y E S, Y 1 1 . Thus Wu ( co) = 1 . Hence c 2 = L.xuE Bo XU ' 0 6.
n ZR ( G)e.
265
p-RADICAL GROUPS
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 [ 1 969], and Motose [ 1 974a] , [1974b]. Much of this has been strengthened by Knorr [ 1 977] . Related results can be found in Brow� [ 1 978a], Isaacs and Scott [1972], Khatri [1974] . Define the following sets of subgroups H of G. Q:( G ) = {H I VG is completely reducible · for every irreducible R [H] module V}. :B(G) = {H l p ,f I G : H I }·
PROOF. (i) This follows from (5.4) (ii). (ii) This follows from (i) as Ie � ZR ( G ). D
ffi(G) = {H I J ( R [ G ] ) � J ( R [H] ) R [ G J}.
Br o w� has pointed out that the next result, due to Tsushima [1971b] , can be proved by these methods . See also Reynolds [1972] .
Since J (F[ G ] ) = J ( R [ G ] ) 0 R F for an extension field F of R it follows that ffi( G) depends only on G and p.
THEOREM 5.6 (Tsushima [1971b]). Let c E R [ G ] be the sum of all the p-elements in G. Then c 2 is the sum of all the centrally primitive idempotents in R [ G ] which correspond to blocks of defect O.
LEMMA 6 . 1 . ffi(G) = Q:( G).
PROOF. Let S be the set of all p -elements in G. Let Co = LyE S Y E R [G] . Thus Co = c. Since c is in the Reynolds ideal it follows from (4.6) that , c E Ann(J ( R [ G ])). Thus Co E A. As Wu (Xv ) = Duv it follows that (5.7) Since XuXv = DuvXu this implies that
PROOF. Let H E ffi( G) and let V be an irreducible R [ H ] module. Then VGJ ( R [ G ] ) = ( V 0 R [ G ] ) J ( R [ G ] ) = V 0 J ( R [ G ] ) R [H]
R [H] _
_
� V 0 J ( R [H ] ) R [ G ] R [H]
= VJ ( R [H ] ) 0 R [ G] = (0). R [H]
Thus VG is completely reducible and so H E CI( G). Therefore ffi( G) � Q:( G) . Suppose that H E Q:( G). Let {Xi } be a cross section of H in G. Th�n an arbitrary element a E J ( R [ G ] ) is of the form a = L.i aiXi with ai E R [ H] .
CHAPTER VI
266
[6
If V is an irreducible R [H] module then V G a = 0 as V G is completely reducible. Therefore 0 = ( V 0 1)a Li Vai 0 Xi . Hence Vai = 0 for all i. Hence each ai annihilates every irreducible R [H] module and so ai E J(R [H]) for all i. Thus a E J (R [H] ) R [G] . Hence H E fft(G) and so Gr(G) � fft(G). 0 =
LEMMA 6.2. fft(G) � :B(G). PROOF. Let H E fft(G). Let V = Inv H ( V) be a one dimensional R [H] module . By (6.1) H E Gr(G) and so V G is completely reducible. Thus w i V G, where W = Inv G ( W) and dimR W = 1 . If Q is a Sp -group of H then Q is a vertex of V. Let P be a Sp -group of G with Q � P. Similarly W has P as a vertex . Since W I V G, we must have P = Q by (III.4.6). 0
A group G is a p-radical group if m(G) = :B(G) or equivalently Gr(G) = :B(G). 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 R [ G ] to
be in the radical. THEOREM 6.3. The following conditions are equivalent. (i) G is p-radical. (ii) m(G) contains a Sp -group of G. (iii) J(R [ G]) = n xE G J(R [PX ])R [G], where P is a Sp -group of G. (iv) Ann J ( R [G] ) = Lp EpR [G] where P ranges over all Sp -groups of G
and Ep = LX EPX. (v) LxE G axx E J(F[G]) if and only if L Y EP a Y = L Y EP a y = 0 for all X X X E G and every Sp -group of G.
PROOF. (i) � (ii). Clear by definition. (ii) � (iii). Since P E m(G), J(R [ G]) � J(R [PX ])R [G] for all x E G. Thus J(R [G]) C n X E G J(R [PX ])R [G]. The opposite inclusion follows from (4.7) (i). (iii) � (i). By (6.2) it suffices to show that :B( G) � fft( G). Let H E :B(G) and let P be a Sp -group of G with P � H. By (4.7) (i} n xE H J(R [PX ])R [H] � J(R [H]). Therefore
n J(R [PX ])R [G] � n J(R [PX ])R [HX ]R [ G ] xE G xE G xE G
6]
267
p -RADICAL GROUPS
By (4 .7) (i) n E G J(R [HX ])R [G] � J(R [G]). By assumption J(R [ G]) = X nxE G J(R [PX ])R [G]. Hence all these sets are equal and so
J(R [G]) = n J(R [HX ])R [G] C J(R [H])R [G]. xE G Therefore H E m(G) and so :B(G) � m(G). (iii) G (iv). Since J(R [P]) = EpR [P] each of the statements follows
from the other by taking annihilators. (iv) ¢:> (v). Statement (v) is equivalent to the fact that a in R [G] is in J(R [G]) if and only if aEp = Epa 0 for all Sp -groups P of G. In other words J(R [G]) = Ann(Lp EpR [ G ]). Thus (iv) and (v) can be derived from each other by taking annihilators. 0 =
COROLLARY 6.4. Let P be a Sp -group of G and let V Invp V be an R [G] module with dimR V = 1 . Then G is p-radical if and only if V G is completely reducible. =
PROOF. By (6. 1) and (6.3) G is p -radical if and only if P E Gr(G). The r� sult follows as up to isomorphism V is the unique irreducible R [P] module. 0 THEOREM 6.5 (Khatri [ 1 973]). Suppose that H
hold.
(i) (ii) (iii) (iv)
H E :B(G) if and only if H E fft(G). If G is P -radical then G IH is P -radical. If H is a p-group then G is p-radical if and only if GIH is p-radical. If H E :B( G) then G is p -radical if and only if H is p-radical.
PROOF. Let P be a Sp -group of G. (i) If H E :B( G) then H E Gr( G) by (2.5). Thus H E m( G) by (6. 1). The converse follows from (6.2). (ii) Let V be an irreducible R [PHIH] module. V G is completely reducible as an R [ G ] module if and only if V G is completely reducible as an R [ G IH] 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 R [P] module. Since H C P, P is an _ irreducible R [PIH] module and so V G is a completely reducible R [ G IH] module. Thus V G 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 reducible then by Clifford's theorem (III.2. 12) ( V G ) H is completely reduc-
268
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[6
ible. Thus by the Mackey decomposi tion (II.2.9), VH is completely reduc ible. If VH 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 and G2 are p-radical.
=
G1
X
G2 • Then G is p-radical if and only if G1
CHAPTER VII
PROOF. If G is p -radical then G1 and G2 are p -radical by (6.5) (ii). Suppose that G1 �nd G2 are p -radi �a1. Let Pi be a Sp -group of Gi for i = 1 , 2. By (6.5) J( R [ G; ]) C J( R [P; ]) R [ G; ] . Since J( R [ G ]) J( R [ G d) R [ G2] + J( R [ G2]) R [ G d as R [G] = R [Gd 0R R [G2 ] modules it follows that =
J( R [ G ]) C J( R [Pd) R [ G ] + J( R [ P2]) R [ G ] . Let p = p\ x P2 . �hen P is a _Sp -grc:up o f G and J(R [ P; ]) C J(R [P ]) for i = 1, 2. Thus J( R [ G ]) C J( R [ P ]) R [ G ] . Thus P E ffi ( G ) and so G is p -radical by (6.3). 0
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 Qp , the field of p -adic numbers. R is the ring of integers in K. D is a cyclic subgroup of G.
D = (y ). For 0 � i � a, Di is the unique subgroup of D with I D : Di I = p i . C
=
C G (Di )'
Therefore C <J N; and 269
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[1
B is a block of R [ G ] 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) R is a splitting field of R [H] for every subgroup H of G. K is a splitting field of K [H] for every p i-subgroup H of G. (ii) There exist blocks 13 = 13 (0) , . . . , 13 (r - I ) of R [ G ] such that D is a defect group of each B Ul. Furthermore if V is any irreducible R [ G ] module then V i s i n B i f and only if V 0R R = L.;:� yU) with yU) i n B U). (iii) There exists an element a in the Galois group of K over K such . that after a possible rearrangement B crj = 13 U) for j = 0, . . . , r - 1 and B CT' = B. Choose the notation so that B CTj = B U). For any integer j' define B U') = B Ul where 0 � j � r - 1 and j' == j (mod r ) . Since N = No � N; for 0 � i � 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 Bi of R [N; ] with 13 ;' = 13 = Ba . Thus also B f'k = 13k for o � i � k � a. By (111.9.3) and condition (iii) above, this yields that B �)Nk = B � ) 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 B �). Let bk be a block of R [ Ck ] which is covered by 13k for 0 � k � a. For j = 0, . . . , r - 1 define b � ) = br. Then B �) covers b�). Observe however that be need not be equal to bk • By (V.2.3) the blocks covered by 13k are conjugate under the action of Nk. Thus b r = b � for some z E Nk . By (V.3.9) 13k is regular with respect to Ck . Thus by (V.3.6), b r;:k = Bk . By (V.3. 14) D is a defect group of bk • The group Nk /Ck is a group of automorphisms of the cyclic p -group Dk. 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 I I (P _ l)p a - l . Therefore by (V.3. 14) I T(bk ) : Ck I I (P - 1) where T(bk ) is the inertia group of bk in Nk. Furthermore T(bk )/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 [ Cd blocks {b �j} as in (111.9.101. Let Bk be the block of R [Nd which corresponds to the set of all R [Nd blocks { B �j} as in (111.9.10). Thus Bk covers b � for all x E Nk . If H is a group and V is an R [H] module define: I ( V) = number of factors in a composition series of V. S ( V) = S I ( V ) = Soc( V) is the socle of V.
1J
BLOCKS WITH A CYCLIC DEFFECT GROUP
27 1
S " ( V) is the inverse image in V of S ( V/S " - I ( V» for n ;?; 2. T( V) = V/Rad V. Rad I ( V) = Rad( V) and Rad" ( V) = Rad(Rad" - I ( V» for n � 2. Suppose that JL is a one dimensional representation o f R [H] . We will write VJL for the tensor product of V with the underlying module which affords JL. Thus in effect we are identifying JL with its underlying module. If VI, V2 are R [H] modules for any group H then 1R ( VI, V2) denotes the intertwining number of VI and V2• If V is an R [H] module then VK = V 0R K. LEMMA 1 . 1 . If 0 � k � a - I then (i) Co = Ck n T(bo), (ii) T(bd = T(bo)Ck , (iii) T(fh )/Ck = T(bo)/Co.
PROOF. (iii) is an immediate consequence of (i) and (ii). (i) Clearly Co � Ck n T(bo). Suppose that x E Ck n T(bo). Then x E No and x d E Co for some integer d with d I (p - 1). Since x E Ck � Ca - I this implies that x E Co. (ii) Let Ck � H � Nk such that H/ Ck is a Hall p i-subgroup of Nk / Ck. Thus H/Ck is cyclic and T(bk ), T(bo)Ck are both contained in H. Hence it suffices to show that I T(bk ) : Ck I = I T(bo)Ck : Ck I · Let t be the number of blocks of R [ Ck ] which are covered by b :1. Thus t = I H : T(bk ) I . By (V.3.9) and (V.3. 10) br:nNo covers a block b of R [ Co] if and only if b HnNo = br:nNo• By the First Main Theorem on blocks (111.9.7) this is the case if and only if b H = (b Ck y-l = b r:, or equivalently b Ck is covered by br:. Furthermore t is the number of blocks of R [ Co] which are covered by br:nNo• Thus t = I H n No : T(bo) I . Since I H : Ck I is the number of elements ()f Da - I which are conjugate to -l y pa in G is follows that ' H : Ck , = ' H n No : Co l . Therefore Hence
t I T(bk ) : Ck I = I H : Ck I = I H n No : Co l = t I T(bo) : Co l ·
I T(bo)Ck : Ck I = I T(bo) : T(bo) n Ck I = I T(bo) : Co l = I T( bd : Ck I .
0
For x E Na - I define a (x ) by x - I y pa - l x = y pa - l a (x ) . Thus a is a one dimensional representation of R [Na -t] whose kernel is Ca - t • By abuse of notation we will also use a to denote the restriction of a to any subgroup of Na I • We will also denote the underlying one dimensional R [Na -a module by lY.
272
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In view of (1.1) { a j 1 0 � j � I T(bo): Co /} is a set of distinct representa is the trivial tions of R [ T ( bk )] for 0 � k � a - 1 . Furthermore iT(bo):Coi one dimensional representation of every subgroup of T(bk ). a
=
a°
LEMMA 1 . 2 . Suppos� that for some k with 0 � k � a - I, bk contains a
unique irreducible R [ Ck ] module up to isomorphism. Then bk contains a unique irreducible R [ Ck ] module up to isomorphism. Furthermore up to isomorphism bo contains a unique irreducible R [ Co] module.
PROOF. By (V.4.6) bo contains a unique irreducible
isomorphism. The result now follows from (VI.1.2).
R [Co] module up to 0
THEOREM 1 .3. Fix k with 0 � k � a 1 . Assume that up to isomorphism bk contains a 'inique irreducible R [ Ck ] module V and bk contains a unique irreducible R [ Ck ] module V. (i) There exist integers s Sk , e ek, with e I I T(bd : Ck I , and hence e I (P - 1), and an irreducible R [ Nk ] module W such that W, Wa, . . . , Wa e- 1 are pairwise nonisomorphic modules in Bk . W e = W and every irreducible R [Nd module in Bk is isomorphic to some Wa j• Furthermore VNk = S ( E9;= 1 Wa ; ). (ii) If R = R then s = 1 and e I T(bo) : Co l . -
=
=
a
=
PROOF. This is a direct consequence of (1.2), (L I6J3) and (VI.3 .3).
D
The index of inertia of B or the inertial index of B is the maximum number of pairwise nonisomorphic irreducible R [No] modules in Bo. By (1.2) the assumptions of ( 1 .3) are satisfied for k O. Thus the index of inertia of B is the integer eo defined in (1.3). In case R = R it is equal to I T(bo) : Co l and so coincides with the index of inertia defined in Chapter V, section 9. Since N � Na - 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 nonprojec tive R -free R [D]-projective R [Na - 1 ] modules. Such a correspondence is also defined for R [G] modules and R [Na- 1 ] modules. The following notation will be used 13 = Ba - 1• C; Na - 1 , If V is a nonprojective R -free R [D ]-projective R [G] module then V is the R [ d ] module which corresponds to V by the Green correspondence. =
2]
STATEMENTS OF RESULTS
273
If V is a nonprojective R [D ]-projective R [G] module, V is defined similarly. If V is projective define V = (0). By (III.7.7� V is a �onprojective R -free R [G] or R [G] module in B if . and only if V is in B. LEMMA 1 .4. Let .x(D; , d), ID(D;, d) .be defined as in Chapter III, section S for 0 � i � a 1. Then the following hold. (i) .x(D;, d) = {(I )} for 0 � i � a 1 . (ii) If A E ID(D;, d) then A n D = ( I ) for 0 � i � a 1 . (iii) If �, W are R [D ]-projective R -free R [G] modules or R [D ] projective R [ G ] modules then (a) HO( G, (I), V ) = HO ( d, (I), V), (b) HO (G, ( I ), HomR ( V, W » = HO( d, ( I ), HomR ( V, W» . -
-
-
PROOF. (i) (ii) Suppose that A E ID(Di, d). Thus A � d n D 7 for some
x E G - G._ Since Da - 1 n D � - l = (I) it follows that A n Da - 1 C Da - 1 n p f = ( I ). T�us A n D = (I) as D is cyclic. This proves (ii). Sinc� .x (Di' G ) � ID(D;, G) and every element of .x (Di' d) is a subset of D, (i) is an immediate consequence. (iii) This now follows directly from (III.S . I 0). 0
�EMMA l oS. (i) Let V be a nonprojective R -free R [G] or a nonprojective
R [ G] module in B. Then Vo = V EB A l EB A 2 where A is projective, A2 is a sum of indecomposable modules in blocks other than 13 and (A 2) D is projective. (ii) Let ? be a n� nprojective R -free R [d] or a nonprojective R [ d ] module in B. Then VO = V EB A for some projective module A. (iii) Let V, W be nonprojective R -free R [G] or nonprojective R [G] and HO( G, (I), V) = HO( d, (I), V) Then B. in modules O O H (G, (I), HomR ( V, W » = H ( C;, ( I), HomR ( V, W» . I
PROOF. (i) This follows from (II.S.3) and (1.4)(i) (ii).
(ii) This follows from (III.S.4) and (L4) (i). (iii) This is a special case of (L4) (iii). 0
2. Statements of results
=
This section contains only statements of results. The proofs of these statements are given in the rest of this chapter.
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The material in this chapter is an outgrowth of the work of Brauer [1942a] . In that paper he proved (2. 1 1}-(2.19), (2.22) and (2.23) in case K = K and I 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 I > p. In particular one of his results, (11 .2) below, is in general not true for I D I > P as is shown by the examples at the end of section 1 1 . A quarter of a century later Thompson [1967b] proved results analogous to those of Brauer for the case that D is a Sp -group of G and Cc (x ) = D for all x E D, x "1 1. In doing this he used the Green correspondence and proved (1. 17.12) as well as a version of (S.6) below which is of critical importance for the whole development of this material. Almost im mediately after this, Dade [1966] was able to combine Thompson ' s methods with the theory of blocks to prove (2.1 1}-(2.19) in general for the case that K = K. Then Janusz [1969a], [1969b] using Dade ' s work as a starting point gave a complete description of all the indecomposable R [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 R [G] modules which lift to R [G] modules were determined by Michler [197S]. In case I D 1 = 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 = R in case D is a Sp -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 correspon dence and in particular by using (II.S.10). Also (2.S), (2.7) for I D 1 = 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 13 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 argu ments 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 general ized earlier work of Alperin and Janusz [1973]. His results are contained in section 10 generalized to the case of arbitrary K. Peacock [197Sa] , [197Sb],
2]
STATEMENTS OF RESULTS
275
[1977] pushed these methods further to get an alternative proof of (2.20) and some refinements for R = R by an argument �hich was independent of the results for K[ G] modules as opposed to R [G] modules. In case I D 1 = 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 S, 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, e denotes the index of inertia of 13. 2.1. B contains exactly e irreducible R [G] modules up to isomorphism.
THEOREM
2.2. B contains exactly I D i e 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 R [0] module in B is serial. There exists an irreducible R [ 0] module W such that W, Wa, . . . , Wa e - I is a complete set of representatives of the isomorphism classes of irreducible R [ 0] modules in B.
THEOREM
2.S. Let L, M be irreducible R [G]. modules in B. The following are equivalent. (i) L = M (ii) S (i ) = S (M). (iii) T(i ) = T(M).
THEOREM
.
THEOREM 1 ( V) ::::; p a
2.6. Let V be an indecomposable R [G] module in B. Then and V is R [Dk ]-projective if and only if I ( V) 0 (mod p k ) . ==
[2
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276
THEOREM 2.7. Let L be an irreducible R [ G ] module in 0 < l(i ) � e or pQ - e � l (i ) < p a .
B
then either
The next result is due to Blau and is a strengthening of an earlier result. THEOREM 2.8. Let 0 � k � a - I and let V be an indecomposable R [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 R [Ni ] in Bi is absolutely irreducible for 0 � i � a and any indecomposable R [Ni ] module in Bi is absolutely indecomposable. Let LJ, . . . , Le be a comple te set of representatives of the isomorphism classes of irreducible R [ G ] module s in B. Let 'Pi be the Brauer character afforded by Li for 1 � i � e. Let W be the module defined in (2.4) and let 1/1 be the Brauer character afforded by W. THEOREM 2 . 10. Suppose that R = R. Then every 'Pi is of height O. (1/ I G D 1 )1/1(1) E R and (II I G" : D I )1/1(1) � O. Furthermore there exist inte gers Ci for 1 � i � e with 0 < I Ci I � e such that 1 1 ( 1 (mod p Q ) . I G D I 'Pi ( 1 ) == Ci I G D 1 1/1 ) :
:
:
For i = 1 , . . . , e let Vi be the princip al indecomposa ble R [G] modul e which corres ponds to Li. Let cJ>i be the charac ter afforded by Vi. Let A be the set of all characters of No which are afforded by irreducible . K [No] modul es in Bo and which do not have D in their kernel
STATEMENTS OF RESULTS
2]
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 Qp for which ( ) is satisfied. It should be noted that ( ) need not be satisfied for the extension of Qp 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 am denote a primitive m th root of l over Q3 and let Ko = Q3(a3). Let x be an element of order 15 in G. Then there exists a unique irreducible character ( of G with ( x) = a S + a JS1 . Thus ( x ) � Ko and A = {( + ('T} where a fs = a is. Hence ( ) is not satisfied for the block B of Ko[ G] which contains (. In this case ( ) is satisfied for any nonprincipal block of K [ G ] where K = Qp (L�=o a fD. *
*
*
*
l
*
THEOREM 2.12. B contains exactly e + I A I characters which are afforded by irreducible K [ G] modules. If I A 1 = 1 these will be denoted by Xo, . . . , Xe. If / A I � 1 these characters are divided into two families X I , . . . , Xc and {XA I A E A }. In the latter case let XO = LA E A XA. If x is any p'-element in G then XA ( x ) = XI-'- (x) for A, f.L E A. The characters XA, A E A defined in (2.12) are called the exceptional characters in B. By (2. 12) XA and XI-'- agree as Brauer characters for A, f.L E A. In particular they have the same decomposition numbers. Let dO i denote the XA, 'Pi decomposition number for A E A. THEOREM 2.13. dUi = 0 or 1 for 1 � i � e and 1 � u � e. If K = K then d Oi = 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 Xu or XA and X is indecomposable. =
a THEOREM 2 . 1 1 . If K = K then I A I = (p - l)le. In particular I A 1 = 1 if and only if I D 1 = p arid I T(bo) : Co l = p - l. Throughout the rest of this chapter it will be assumed that the followi ng condition is satisfied.
THEOREM 2.14. Suppose that K = K. Let 0 � u � e and let A E A. Let Xu, XA denote an R -free R [ G ] module such that Xu !.XA is indecomposable, (Xu )K affords Xu and (XA )K affords XA. Then I (Xu ) = 1 or p a _ I and I(XA ) = e or pQ - e. The value of I (Xu ) or I (XA ) is independent of the choice of Jhe moq,ule Xu or XA and depends only on Xu or XA. Furthermore I(XI-'- ) = [ (XA ) for A, f.L E A.
Condition ( ) is a hypothesis that K must satisfy. By (2. 1 1) it holds in case K = K. By (2. 1 1) no two distinct elements of A are conjugate by a
If K = K define 8u = ± 1 for u = 0, . . , e by 8u I (Xu ) (mod p Q ) where Xu is defined as in (2. 14). If p Q = 2 choose 80 � 8 1 at random.
*
.
==
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278
[2
THEOREM 2.15. Suppose that K" = K. (i) Let 1 � i � e. If dUi I- 0, dVi 1- 0 and u I- v then ou + ov = O. (ii) Let 1 � i � e. There exist u, v with ou + ov = 0 such that cJ>i = Xu + Xv' (iii) Let X = XO if I A I = 1 and let X = XA for some A E A if I A I I- 1. Then
ooX (x ) + u! ouXu (x ) = O =1 for every p '-element x in G. (iv) Suppose that G = Na - 1 . If I A I I- 1 then 01 = . . . = De = 1 and D O = - 1 . If I A I = 1 then the notation can be chosen so that 01 = . . . = De = 1 and DO = - 1. THEOREM 2. 16. Suppose that K = K. Then every irreducible character in B is of height O. Furthermore Xu (1) Dut/l (1) (mod p a ) for u = 0, . . . , e. ==
XA (1)
==
- ooet/l(l)
(mod p a ) 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 I G D I gets larger. If for instance I D 1 2 / 1 G I they contain no information. Suppose that K = K. For 0 � k � a - I let t/lk 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 Bk • Thus {t/I %} is the collection of irreducible Brauer characters in blocks covered by Bk • If x E Dk - Dk+ 1 then CG (x ) = Ck . Let d (u, x, t/I %) or d CA, x, t/lt) denote the corresponding higher decomposition number for Xu or XA respectively where u = 1, . . . , e and A E A or possibly u = 0 in case I A 1 = 1. :
THEOREM 2.17. Suppose that K = K. For 0 � k � a - 1 there exist E k = such that for x E Dk - Dk + 1 the following hold. d (u, x, t/I %) = EkOu for 1 � u � e or for u = 0 if I A I = 1 .
- EkOO d (A, x, t/l k) = 1 Ck I z
� L.J
w E T(bd
±1
'WZ (x ) for A E A,
where , is an irreducible constituent of the restriction to D of an irreducible constituent of ACK in bk. Furthermore if 0 � j � a - I and E :), . . . , E j are the signs for the group G then there exists y = ± 1 such that E ; = yEi for 0 � i � j. If G = G for some j with O � j � a - 1 then Ej = . . . = Ea - l = 1 unless
2]
STATEMENTS OF RESULTS
279
= 2 and j = 0, in which case the notation may be chosen so that Ei = 1 for O � i � a - 1.
p
It follows from (2.17) that if I A I I- 1 and A E A then XA has a higher decomposition number not equal to ± 1 unless p = 2, e = 1 and A 2 = 1 . In the latter case if XI is the nonexceptional character in B and is the irreducible character of D with kernel equal to DI, we can define X � = XAa for A E A U {I} and note that {X a satisfies (2. 1 1}-(2. 17) with Do, ot, E I , . . . , Ea - I unchanged but Eo replaced by - Eo. Then Xa = X ; is the 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(bo)-conjugate classes of nonprinci pal irreducible characters of D. ll'
THEOREM 2.18. Let r be the number of blocks of R [G] which are algebrai cally conjugate to B. Then the Schur index mo of XA over K is dOi for some i and is independent of A E A, and II XA W = m� r is independent of A E A. By (2.9) m = It�. (Li' Li ) 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 (Xu ) = Qp (4\ ), mQp (Xu ) = 1 and IIXu W = m. (ii) Given i with 1 � i � e then either there exists u, v with 1 � u, v � e such that cJ>i = Xu + Xv or there exists u with 1 � u � e and cJ>i = Xu + (e/emo)Xo. If A = 1 and e I- e then replacing K by a purely ramified\ extension it may be assumed that = 1 and so Xo can be recognized from the decomposition of the cJ>i . In this case it is called the exceptional character in B. Thus B contains an exceptional character unless I D 1 - 1 = e. Let aui = dUi for u I- 0 and let a Oi = e /(ema) if d Oi I- O. Thus if 1 � i � e, cJ>i = auiXu + aviXv for suitable u, v. It follows from (2.19) that I (e/e ). Since e /e = [ Qp (Xo,
rna
rna
* ,
=
280
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SOME PRELIMINARY RESULTS
281
Observe that (2.19) implies that each irreducible constituent of Xo; occurs with multiplicity ( I D 1 - l )/ e for do; rf O.
THEOREM 2.26. Let V be an indecomposable R [ G] module in B. Then I I(S( V» - 1( T( V» I :-::; 1 .
THEOREM 2.20. If dUi rf 0 then XUi is serial. If furthermore dVi rf 0 for some v rf u then Rad 0; = Xu; + Xv; and S ( U; ) = Xu; n Xvi .
3. Some preliminary results
COROLLARY 2.21. Every indecomposable R [ G] module in B is serial if and only if one of the following conditions is satisfied. (i) 1 D 1 - 1 = e and after a possible rearrangement XUi is irreducible for all u with 1 :-::; u :-::; e and all i with dUi rf O. (ii) 1 D 1 - 1 rf e. XU i is irreducible for u with 1 :-::; u :-::; e and all i with dUi rf 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 vert(!x, if its exists, is at the center. THEOREM 2.23. Let 0 :-::; u :-::; e. There is an ordering
.
THEOREM 2.24. If Xu is real valued then there exist at most two real valued irreducible Brauer characters which are constituents of XU , THEOREM 2.25. Let 0 :-::; u :-::; e and let X be an R -free R [G] module such that XK affords auiXu, where dUi rf 0 and X is indecomposable. Then there exi§ts j with_ dUj rf 0 such that X XUj ' Furthermore I eX) = 1 or p a _ I and I (Xuj ) = I(Xui ) if dui , duj rf O. =
In view of (2.19) and (2.25), 8u = _± 1 can now be defined for K in general as follows. If dUi rf 0 .then 8u == I (Xu; ) (mod p a ). If p a = 2 choose 80 rf 8 1 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.
For 0 � k :-::; a - I; bk contains a unique irreducible R [ Ck ] module up to isomorphism. Throughout this section it will be assumed that
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 G holds for Nk for 0 :-::; k :-::; a - I since if Dk <J G then {; = Nk • LEMMA 3.1 . For 0 :-::; k :-::; a - I, bk contains a unique irreducible R [Cd module up to isomorphism. PROOF. Clear by (1.2). 0 Let e = ea -l be defined as in (1.3). Let W be the R [ 0] module defined in (1.3). LEMMA 3.2. W, Wa, . . . , Wa e - 1 is a complete set of representatives of the isomorphism classes of irreducible R [0] modules in B. PROOF. Clear by (1.3).
0
LEMMA 3.3. Let Ca -I � H � G for some subgroup H. Let M be an R [H] module in the block which covers ba - I • Then (i) (S" (M» ca_1 � S" (Mca_J for all integers n, (ii) (Rad" (M»)ca _l d Rad" (Mca _J PROOF. Let Ca -I = 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 " - I (M»)c is completely reducible, induction implies that Rad(Mc ) � (S " - I (M»)c � S " - I (Mc). Thus Mc � S " (Mc) as required. (ii) By induction Rad" - I (Mc) � (Radn - I (M»)c. Thus Rad" (Mc) � Rad(Radn - I(M)c) � (Rad" (M» c by induction. 0
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LEMMA 3.4. Let Ca - I C Fr� 0 for some subgroup H. Let V be an irre!J.ucible R [H] module in the block which covers ba - I. Suppose that V 0 R R is the direct sum of m indecomposable modules. Then the following hold. (i) If M i! any irreducible module of R [H] in the block which cover� ba - I then M 0R R is the direct sum of m algebraically conjugate irreducible R [H] modules. (ii) If P is a nonzero indec�mposable projective R [H] module in the block which covers ba - J then P 0R_R is the direct sum of m algebraically conjugate indecomposable projective R [H] modules. (iii) If M is an indecomposable R [H] module in the block which covers ba - I then
)
I M � R = ml(M).
(
PROOF. This follows from (VI.3.4). 0 THEOREM 3.5. (i) If P is a nonzero indecomposable projective R [0] module in B then P is serial and l (p) = p a . (ii) Let M be an indecomposable R [0] module in B. Then M is serial and S " ( Mca _ J = (S " ( M»ca_1 for all n. PROOF. This follows from (VI.3.5).
0
COROLLARY 3.6. Let M be an indecomposable R [ G] module in B. Then M @R R = EB A1'" where ranges over a quotient group of the Galois group of R over R and each tr is indecomposable with I (M) = I CtVr ) for all Furthermore S CM ) 0 R = EB S ( Mf· T
T.
R
PROOF. This follows from (VI.3.6).
0
THEOREM 3.7. Let M be an indecomposable R [0] module in B. Then for o � k � a, M is R [Dk ]-projective if and only if I (M) == 0 (mod p k ).
-
PROOF. If k = a this is a consequence of (3.5). Suppose that 0 � k � a 1. Assume first that R = R. By (1.3) e = I T (bj ) : � I for 0 � j � a and there are exactly e irreducible Brauer characters t/li in B. All of these have the same degree and hence are of height O. Thus t/li (1)/ 1 0 : D I is a unit in R. Therefore if M is an indecomposable R [Dk ]-projective module in B then I (M) 0 (mod p k ) by (IV.2.2). ==
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SOME PRELIMINARY RESULTS
By (3.5) B contains exactly ep a - k pairwise nonisomorphic indecompos able modules M with I (M) 0 (mod p k ). Thus B contains at most ep a - k pairwise nonisomorphic indecomposable R [Dd-projective modules and it a k �uffices to show �hat B contains at least ep - pairwise nonisomorphic In.decomposable _R [Dd-projective modules. We will show that for each j aj aj l :V1th k � j � a, B contains at least e (p - - p - - ) pairwise nonisomorphic Indecomposable modules with vertex Dj• This clearly implies the desired result. Fix j with k � j � a. Let V be the unique irreducible module in bj (up to isomorphism) and let Wa i, 1 � i � e be the modules in Bj defined in (3.2). For 1 � s � p a -] let Xs be the indecomposable R [Dj ] module of R dimension s. Then (X�j)Dj = , � : Dj , Xs . Thus X�j and X�j have no common indecomposable component if s � t. Furthermore if s rE 0 (mod p) then e�ery indecomposable component of X�j has vertex Dj• Since V � Xl j it follows from (1 .3) that EB�= I Wa i � X�j � X�j. By (3.5) the indecomposable components of X�j in Bj are all serial and must contain some Wa i as a submodule. Thus Bj contains at least e (p a -j p a -j - l ) pairwise nonisomorphic indecomposable modules with vertex Dj• The Green correspondence between � and Na - 1 = 0 now implies that B contains at least e (p a -j - p a -j - l ) pairwise nonisomorphic indecomposable modules with vertex Dj• This completes the proof in case R = R. Suppos : n �w that R is in the general case. Let M be an indecomposable nonzero R [ 9 ] module in B. Let M_be an indecompos.?ble nonzero R [ G] module in B such that M I M 0R R. By (3.6) !vi 0R R_ EB MT where ranges over a quotient group of the Galois group of R over R. Further more I(M) I(M) by (3.4) (iii). Thus it suffices to show that M is R [Dd-projective if and only if M is R [Dd-projective. This is true by (111.4.14). 0 ==
_
=
T
=
LEMMA 3.8. e = e. _ B contains exactly e irreducible R [0] modules up to isomorphism and B contains exactly e ' D I nonzero indecomposable R [0] modules up to isomorphism. PROOF. By (3.2) B contains exactly e irreducible R [G] modules up to isomorphism. By (3.5) each indecomposable R [0] module V in B is uniquely determined by S ( V) and I ( V). Since I ( V) may take any value between 1 and p a _= '_D I , it follows that B contains exactly e I D , nonzero indecomposable R [G] modules up to isomorphism. It remains to show that e = e. By (3.7) B contains exactly (p a - p a - l )e indecomposable R [0] modules
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with vertex D up to isomorphism. Furthermore Bo contains exactly (p a - p a - I )e indecomposable R [No] modules with vertex D up to isomorphism. By (III.7.8) B and Bo contain the same number of indecom posable modules with vertex D up to isomorphism. Thus = e. 0 e
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 CM), S (M)a - l , S (M)a - 2, . . . .
PROOF. Let V be an indecomposable R [Nk ] module in Bk with W = S ( V) and I ( V) = 2. By (3.2) T( V) = Wa d for some integer d. Let P be the principal indecomposable R [Nk ] module corresponding to W. Then V = S\P). By (3 .2) and (3.8), {Pa i I 0 � i � e - 1} is a complete set of represen tatives of the isomorphism classes of principal indecomposable modules in Bk. Thus if Vo is an indecomposable R [Nd module in Bk with I ( Vo) = 2 then Vo = Va i for some i and so T( Vo) = S ( Vo)a d• As every indecompos able 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 R = R. Let z = y pa - l . Then (z ) = Da - 1 <1 Nk• Since •
•
•
•
==
p(z ) =
p a - J (dimR W)R [(z )] R [(Z )]
it follows that dimR {Inv(z ) (P)} = p a - I dimR W = dimR {spa -l (p)}. As (z ) <1 Nk, Inv(z )(P) is a submodule of P. Thus Inv(z ) (P) = s pa - l (p) since P is serial. Let U = S pa - l+,(p). Then U(z - 1) = S (P) and T( U) = Spa -l + ' (p)/ Spa -l (p). Thus T( U) = S (P)a pa -1 d • Let x be a p '-element in Nk• Let {Ui } be a set of characteristic vectors for x in U such that their images form an R -basis of T( U). Then UiZ = Ui + Wi, where {wd is an R-basis of S (P) as U(z - 1) = S (P). Let UiX = CiUi . Then for all i
Thus WiX = Cia (x )Wi for all i. Consequently { Wi } is a set of characteristic vectors for x and t/lo(X ) = t/I (x ) a (X ), where t/lo is the Brauer character afforded by S (P). Therefore
3]
SOME PRELIMINARY RESULTS
S (P)a pa - 1 d
and so
=
285
T ( V) = S (P)a - 1
- 1 == p a - 'd == d (mod e ).
0
LEMMA 3. 10. Let M, and M2 be indecomposable R [O] modules in B such that I (M,) + I (M2) � p a . Then HO (O, (I), HomR (M! , M2» = HomR [ G ] (M M2)' "
PROOF. Let g E Tr � ) (HomR (M" M2» ' By (II.3. 13) there exists an inde composable projective R [G] module P and an exact sequence with g = fh. Clearly P is in B. Thus by (3.5) h (MJ ) � S I(M) (P) and s pa-I(M2)(P) is in the kernel of f. Since [ (M,) + I (M2) � pa it follows from (3.5) that S I(M])(P) � s pa-I(M2)(p). Consequently g = fh = O. This implies the result. 0 LEMMA 3. 1 1 . Let V" V2 be irreducible R [ G] modules in B. The following are equivalent. (i) VI = V2. (ii) S ( V, ) = S ( V2)' (iii) T( VI) = T( V2)' PROOF. Clearly (i) implies (ii) and (iii). The modules V j , Vi 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 ( VI) = S ( V2) and VI :/=: V2. Then HomR[ G ] ( V; , Vj ) = (0) and so HO(G, (1), HomR ( V; , Vj » = (0) for {i, j } = { 1 , 2}. Thus by (1.5) (iii) HO( G, ( l ) , Hom R ( V; , V} » = (O) for {i,. j} = { 1 , 2}. Choose the notation so that I ( VI) � I ( V2)' Suppose first that I ( VI) + I ( V2) � p a . Hence Hom R [ G ) ( VI , V2) = (0) by (3.10). By (3.9) V, = s n ( V2) for some n contradicting the previous sen tence. Suppose next that I ( VI) + I ( V2) > p a. By (3.5) 1 ( V; ) � p a for i = 1, 2. Thus there exists a principal indecomposable module P and exact se quences
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By (III.S.12) and (3. 10) HomR [ G ] (Xl, X2) = (0). However (3.9) implies that X2 = XII S n (Xl ) for some n contrary to the previous sentence. 0 T(X2 )'
LEMMA 3.12. B contains exactly e irreducible R [G] modules up to isomorphism and B contains exactly e I D I nonzero indecomposable R [G] modules up to isomorphism. PROOF. The map sending V to V is one to one from the isomorphism classes of nonprojective indecomposable modules in B to the isomorphism classes of nonprojective indecomposable modules in B. Thus by (3.8) it suffices to show that B contains exactly e irreducible pairwise nonisomor phic modules. By (3.8) and (3. 1 1 ) B contains at most e pairwise nonisomorphic irreducible modules. For i = 0, . . . , e - 1 let V; be an R [G] module such that Vi = Wa i. There exists an irreducible R [G] module Li with HomR [ G ] ( V;, L; ) I: (0). By (III.S.13) applied to G and 0 and (1.S) (iii) HomR [ ] ( Wa i, L ) I: (0). Thus Wa i = S (L ). Hence by (3. 1 1) {L I O ::<S i ::<s e - l} is a set of pairwise nonisomorphic irreducible modules in B. 0 G
THEOREM 3.13. (i) Let F be a finite extension field of R. Let 0 � k � a and let V be a nonzero indecomposable R [Nk ] module in Bk• Suppose that VF is the direct sum of m algebraically conjugate F[Nk ] modules. Then for any j with 0 � j � a and for every indecomposable R [� ] module U in Bj, UF is the direct sum of m algebraically conjugate absolutely indecomposable F[ � ] modules. (ii) Suppose that for some k with 0 � k � a, Bk contains an absolutely indecomposable nonzero R [Nk ] module. Then for all i, every indecomposa ble R [N; ] module in Bi is absolutely indecomposable. PROOF. Clearly (i) implies (ii) so that only (i) needs to be proved. By (3.6) WF 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 - I, Bj contains a nonzero indecomposable R [�] module U such that UF is the direct sum of m algebraically conjugate indecomposable F[ � ] modules. By (3.6) S ( U) has the same property. By (3.2) every irreducible R [�] module in Bj has the required property and so by (3.6) every indecomposable R [�] module in Bj has the required property. Thus the result holds for j = k = a - 1 . Suppose that 0 � j � a - 1 . Let Vi b e a n indecomposable R [ G ] module -
3]
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SOME PRELIMINARY RESULTS
in Bj with vertex Dj and let U be the R [ 0] module which corresponds to Vi under the Green correspondence. By (III.S.7) (iii) UF 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 (III.S.7) (iii) the result holds for every nonprojective indecomposable R [G] module in B as it holds for every indecomposable R [OJ module in B. It also holds for the indecomposable projective R [GJ modules in B as it holds for the irreducible R [GJ modules in B. 0 -
LEMMA 3. 14. Suppose that h ( W, W) = m. Let 0 � k � a and let VI , V2 be indecomposable R [Nd modules in Bk• Then lEd VI , V2) 0 (mod m ). ==
PROOF. By (1. 19.1) W Q9R R EB;n= 1 W�i where is an element in the Galois group of R over R, WO is absolutely irreducible, W�i = w�j if and only if i j (mod m ). By (3.6) and (3.13) Vs Q9ii R = EB�= I ( V, )�i for s = 1, 2 and some absolutely indecomposable modules ( Vs)o. Thus =
T
==
lE� ( V I , V2 ) =
Ii?
(
VI
� R, V2 � R) = mIft. ( VI � R, ( V2)O ) .
0
LEMMA 3.1S. Let M, ;;£: M2 be indecomposable R [OJ modules in B. Let m = Iii ( W, W). (i) If I (M, ) � e then Iii (M;, �) � m for {i, j} = {I, 2}. (ii) If Iii (M" M1 ) = m then I (M,) ::<s e. PROOF. Both statements are immediate consequences of (3.9) and (3. 14). 0 THEOREM 3. 16. Let V be an irreducible R [GJ module in B. Then either I ( V) ::<S e or I ( V) � P Q - e. PROOF. Let m = h ( W, W). By (ilLS. 13), (1.S) (iii) and (3.13) dimii HO ( 0, (I), Homii ( V, V» = dimii HO( G, (I), Homii ( V, V» = h ( V, V) = m.
Suppose that I ( V) � ! p Q. Thus by (3. 10) h ( V, V) = m. Hence I ( V) � e by (3. 1S). Suppose that I ( V) > !pQ. There exists a principal indecomposable R [GJ module P and an exact sequence O � V � P � X � O.
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[3
Thus I(X) = I(P) - /( V) < 1pQ. By (III.5. 12) and (3. 10) h (X, X) = dimR HO ( 0, (1), HomR (X, X» = dimR HO(O, (1), HomR ( V, V» = m. By (3. 15) l eX) � e and so l e V) � pQ - e. 0 LEMMA 3. 17. Let V be an indecomposable R [G] module in B. (i) S ( V) is the direct sum of pairwise nonisomorphic irreducible modules. (ii) T( V) is the direct sum ofpairwise nonisomorphic irreducible modules. PROOF. The module V* is in a block with defect group D and T( V*) = S ( V)*.
Thus (ii) follows from (i) by duality. It remains to prove (i). If V is projective the result follows from (1. 14.8) and (1. 16.8). Suppose that V is not projective. Let L be an irreducible R [G] module in B. It will be shown that L occurs in S ( V) with multiplicity at most 1 . Suppose that l(i ) + / ( V) � p Q. B y (III.5. 13), (1.5) (iii) and (3.10) HomR[ G ] (L, V) = HomR [cn(i, V). By (3. 16) either I(i ) � e or l e V) � e. In either case the result follows from (3. 13) and (3. 15). Suppose that I(i ) + le V) > p Q. There exist principal indecomposable R [OJ modules PI, P1 and exact sequences
Thus l eX) + I ( Y) < p Q. By (3. 16) either I (X) � e or l eX) � p Q - e and so I( Y) � e. By (III.5.12), (3. 14) and (3.15) dimR HO(O, ( 1 ), HomR (L, V»
=
dimR HO(O, (1), HomR (X, Y»
= h e W, W) . Thus by (III.5.13) and (1.5) (iii) IR (L, V) = h ( 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 MI and Ml 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, VI, V1 are irreducible. Assume that VI � V1 and I ( VI) � / ( Vl)' Then I ( Vl) + l e V) < pQ < le VI) + le V).
PROOFS OF (2. 1 }-(2 . 1 O)
4]
289
PROOF. By (1.5) (i) there exist exact sequences for i = 1 , 2
O � V EB A I EB A1 � M; EB A ; EB A �� Vi EB A ': EB A � � O, where A I , A ;, A '{ are projective and A1, A �, A � are sums of indecompos able modules in blocks other than B. If this sequence is multiplied by the central idempotent corresponding to the block B then (1. 15.6) implies the existence of projective R [0] modules Pi such that O � V � M; EB Pi � V; � O is exact for i = 1, 2. Suppose that I ( VI) + I( V) � p Q. Then O � V � M; � Vi � O is exact for i = 1, 2. Thus S (MI) = S (Ml) and so S (MI / V) = S (Mz/ V) by (3.9). Hence S ( VI) = S ( Vz). Thus by (3.1 1) VI = V2 contrary to assump tion. Suppose that I ( V2 ) + I ( V) � p a. Since M is not projective I (M; ) -I p Q. Then there exist principal indecomposable modules PI, P2 in B such that -
[
-
�
-
O � V � M EB Pi � V; � O is exact for i = 1, 2. Since Pi is not in the kernel of gi, T( Vi ) = T(Pi ) = S (Pi ) for i = 1 , 2. Since f; ( V) n Pi -I (0) it follows that S ( V) = S (Pi ) for i = 1, 2. Thus S ( V) = T( V; ) for i = 1, 2 and so T( VI) = T( V2)' Hence by (3. 1 1) VI = Vz contrary to assumption. 0 LEMMA 3. 19. Let M be an indecomposable R [G] module with S CM) = Rad(M) such that S (M) is irreducible but T(M) is reducible. Then T(M) is the direct sum of 2 nonisomorphic irreducible modules. Furthermore . Z(S(M» -1 1 or p Q - 1. PROOF. By (3. 17) T(M) is the direct sum of pairwise nonisomorphic irreducible modules. By (3.18) I(T(M» = 2. For any irreducible con stituent V of T(M) 1 � l e V) � p Q - 1. Thus if I (S(M» = 1 or p Q - 1 , (3.18) implies that I(T(M» = 1 contrary to assumption. 0
4. Proofs of (2.1)-(2.10)
The proofs of (2. 1) and (2.2) will be given simultaneously by induction on / G I .
CHAPTER VII
290
[5
Suppose that G = Co. Then e 1 . By (VA.6) Go 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 Dk -I- D. Let GO = G /Dk. By (VA.5) there is a unique block 13° of GO which is contained in 13 and D / Dk = DO is the defect group of 13 °. By (1 .3) the index of inertia of 13 ° is 1 . Thus by induction 13 ° contains a unique irreducible Brauer character. Since every irreducible Brauer character of G has DK in its kernel it follows that 13 contains a unique irreducible Brauer character. Since the inertial index of Bi is 1 for all i it follows by induction that Bi has a unique irreducible Brauer character for o � i � k. As C = Ca-1 for k � i � a - I we see that the hypothesis of section 3 is satisfied. The result follows from (3. 12). Suppose finally that G -I- Ca -1• Hence by induction Gk 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). D =
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). (204) follows from (3.2) and (3.5). (2.5) follows from (3. 1 1). (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 1 1 G : D 1 d·I mR V == 1 G : D 1 d·ImR. V (rna d p a ) for any R [GJ module in B. Thus (2. 10) is a direct consequence of (2.7). D 5. Proofs of (2.11)-(2.17) in case
0 1 = 1, 00 = - 1.
G = Co, statements (2. 1 1 )-(2.17) are true with e
PROOFS OF (2. 1 1 )-(2. 1 7) IN CASE K
=
1,
=
K
291
PROOF. This is a direct consequence of (VA.7). D LEMMA 5.2. Statement (2. 1 1) is true. PROOF. By (5. 1), T(bo)/Co 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) : Co], b(;·(bo) has (p " - l)/e irreduc ible characters which do not have D in their kernel. Thus 1 A 1 = (p a - 1)/e by (V.2.5) (i). Since e I (P - 1), 1 A 1 = 1 if and only if e = p - 1 and I D 1 = p. D =
LEMMA 5.3.
B contains exactly e + 1 A 1 irreducible characters.
PROOF. Let y S -I- 1 . The number of conjugates of y S in D is 1 N : C 1 where yS
E Di - Di+1• The number of blocks of C covered by Bi is ( lI e ) 1 Ni C I . Therefore the number of irreducible Brauer characters of CG (y S ) = C in blocks which are mapped to B by the Brauer correspondence is 1/ e times the number of conjugates of y S in D. By (2. 1) there are exactly e irreducible Brauer characters in B. Hence by (IV.6.6) (ii) B contains exactly e + 1 A 1 irreducible characters. D LEMMA SA. In case
:
G = Ca-1 the decomposition numbers in B are all 1 .
PROOF. B y (5.3) there are p a irreducible characters i n
B = Ba-1 • B y (2.1) there is a unique irreducible Brauer character in B. The result follows from (VA.6). D LEMMA 5.5. Let X be an R -free R [Ca-1] module in
ba-1 such that X is indecomposable. Then XK has no composition factor with multiplicity greater than 1 .
PROOF. B y (204) X i s serial. Hence X/Rad(X) i s irreducible. Thus there
exists an indecomposable projective R [Ca -1] module P and a commutative diagram P
Yl
K=K
Throughout this section it is assumed that K = K. LEMMA 5 . 1 . In case
5]
X � X/Rad(X) � O. Since 7TX C Rad(X) it follows that X/Rad(X) X /Rad(X) and Rad(X) is the unique maximal submodule of X. Therefore g is an epimorphism and the result follows from (SA). D =
292
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The next result was first proved by Thompson [1967b] in a critical special case. LEMMA 5.6. Let 11 be a character of G such that cJ>i = 11 + 11 ' for some i where 11 ' = 0 or 11 ' is a character. Let eJ, . . . , ep a be all the irreducible characters of Ca - l in ba - l . Then I (11ca - " es ) - (11ca - " et ) I � 1 for all s, t. If 11 rf 0 and 11 rf cJ>i then (11ca - " es ) rf (11ca - p et ) for some s, t. PROOF. By (1. 17. 12) there exists an R -free R [G] module Y such that YK affords 11 and Y is indecomposable. If Y is projective then 11 = cJ> and the
result is trivial since YCa -1 is projective. Suppose that Y is not projective. By (1 .5) (i) Yo = Y EB A I EB A2 where A l is projective and A2 is a sum of modules in blocks other than B. By (111.5.8) and (1.4) (i) Y = Y is indecomposable. If 11 1 and 112 are the characters afforded by (A 1)K and (A2)K respectively then «112)ca _ p eS ) = 0 for all s and «11 1)ca _l, es ) is independent of s. Hence it may be assumed that G = 0 and Y = Y. It follows from (3.5) that Y is serial and YCa -1 = EB Ux, where U is an indecomposable R [Ca - d module in ba - 1 and x runs over a cross section of T(ba -1) in Na-I• Thus YCa _ 1 = Yo EB A, where Yo = U and A is a sum of modules in blocks other than ba - 1 • Hence it may be assumed that G = Ca - 1 • Now (5.5) implies that (11ca - p es ) � 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 cJ>i = 2X + 11 ' for some i, where 11 ' = 0 or 11 ' is a character. If e is any irreducible character of Ca - I then (2Xca_p e) is even contrary to
(5.6).
0
LEMMA 5.8. Let X be an R -free R [ G] module in B such that XK is irreducible and X is indecomposable. Then one of the following holds. (i) XK is irreducible. (ii) There exists a principal indecomposable R [0] module U and an exact sequence
5]
PROOFS OF (2. 1 1 )-(2 . 1 7) IN CASE K
K
293
PROOF. IK (XK, XK) = 1 . Since X is not projective. this implies that
dimR HO( G, (I), HOffiR (X, X» = 1 . Thus b y (1.5) (iii) dimR HO( G, (I), HomR (X, X » = 1 . Since X = X by (111.5.8) and (1.4) (i), it is serial. Thus. there exists a principal indecomposable R [ 0] . module U and exact sequences
O � Y � U � X � o. o � v � U � X � O, As X is R -free, Y is a pure submodule of U and so Y = V. By (2.6) I ( U) = p a . Hence I (X) � p a /2 or I ( Y) � p a /2. Let_ Yo = X if I (X) � p a /2 and let Yo = Y if I ( Y) � p a /2. Thus in any case I ( Yo) � p a /2. By (111.5. 12) HO (G, (1), HomR ( Yo, Yo» is a nonzero cyclic R module. By (3 . 10) 0 = Tr&>(HomR ( Yo, Yo» = Tr&>(HomR ·( Yo, Yo» . Thus by (111.5.15) HomR[ G ] ( Yo, Yo) = HO(G, (I), HomR ( Yo, Yo» = R. Hence rankR HomR[ G ] ( Yo, Yo) = 1 and so IK « Yo)K, ( YO)K ) = 1. Therefore Yo is irreducible. 0 LEMMA 5.9. Let 0 � k � a - 1. Suppose that G = Cko Let G O = Ck /Da - I • Define � for 0 � j � a - 1 by Da - 1 � � and � /Da - l = CGo(DdDa - l). Then G
�
z
�
z
(iv) Let X = X O be an irreducible character of G which has Da - I in its kernel and let X O be in the block b� of G O . If x E Dj - Dj + 1 for 0 � j � a - I and E � then d (X o , xDa - l/ Da - h ( ljI J) Hj )O = d (X, x, ljIJ). z
with YK irreducible.
=
294
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PROOF. The first statement, (i) and (ii) follows from (V.2.S) and (V.3. 14).
The remaining statements are consequences of the Second Main Theorem on blocks (IV.6.1). 0 At this point (2.1 1) and (2. 13) have been proved except for the fact that dAi is independent of A. This will follow once (2. 12) is proved. Observe that (2. 1S) (i) is an immediate consequence of (2. 1S) (ii). Also (2. 1S) (iii) is an immediate consequence of (2. 12) and (2. 1S) (ii) since for 1 � i � e, 'Pi occurs with multiplicity ou + ov = 0 for suitable in the generalized character o oX + L: = l ouXu . The proofs of (2. 12), (2. 14), (2. 1S) (ii), (2. 1S) (iv), (2. 1 6) and (2. 17) will be given simultaneously by induction on / G / . If G = Co these results are true by (S. l). Suppose that G -I Co. Assume first that G = Ck for some k with 0 � k � a - I . Without loss of generality k may be chosen so that G -I Ck- 1 • Since G -I Co, / D / > p. Thus by induction all the results are true for Ck-1 and also for G O = GIDa-l• Let A be the subset of A consisting of all those characters in A which have Da-1 in their kernel. By (VA.S) there exists a unique block B O of G O with B O � B. By induction there exist irreducible characters Xl , XI., A E i1 ° in B and these are precisely all the irreducible characters in B which have Da -1 in their kernel. Let () J , ()A, A E A be the irreducible characters of Ck-1 in bk- l • For any element x in G let Xp, Xp' denote the p -part, p '-part of x respectively. Let co, . . . , Ck be the signs defined in (2. 17) for the group G O . Let c b, . . . , d-l be the signs defined in (2. 17) for the group Ck-1• By induction and (2.1S) the values of ou, U = 0, . . . , e are determined in all smaller cases. Let � be defined as in (S.9). By induction there exists 1" = ± 1 such that the signs for b� multiplied by 1" yield the signs for (b:�\lt Thus if a is the nonexceptional character of Hk-dDa-1 in (b :.".\ 1 )0 = (b � _l) Hk - I /Dk - l , then the generalized decomposition number of X l at xDa-1/Da-1 for x E D - Dk is 1" 0 times the generalized decomposition number of a, where 0 = 01 for b� . If x E D then (S.9) (iv) implies that the sign c which equals the generalized decomposition number of a in (b :�\It at xDa-11Da- 1 must equal the generalized decomposition number of a in b :.".-;1 at x. By (S.9) (ii) H a = {3 k -l , for {3 E bk-1 and so by (2. 17) applied to Ck-I, the decomposition number of {3 at x is equal to c. By the observation following (2. 17) we may assume that {3 = () Hence there . exists l' = ± 1 such that c j = YCj for 0 � j � k - 1 . If A E A U {I} then induction applied t o Ck-1 yields that if � i s an u, v
°
l.
5]
PROOFS OF (2. 1 1 )-(2. 1 7) IN CASE K
=
K
295
irreducible constitutent of the restriction to D of an irreducible constituent of ACk _1 in bk-1 then o
()A (X )
=
�(Xp )t{!k -l(XP)
/�; / ZEN�Ck_1
C (xp )t{!f(xp)
if Xp E Di - Di+1, 0 � i � k - 2.
As CG (x ) � Ck-1 for x E D - Dk this implies that
if Xp EGD,
If xp E Dj - Dj+1 for some j � k then by induction and (S.9) (iv) XA (x ) = - CjOo�(xp )t{!k (xp ) , where DO is defined by induction on G O . If k < a - I then since G O = CGo(Dk /Da-1), induction implies that DO = 1 and Ci = 1 for k � i < a - 1. If Xp E Da-1 then Xp is in the kernel of XA and so XA (x ) = XA (xp ) = �(xp )t{!dxp). Thus for A E A a U {I} o if Xp EGD, �(xp )t{!k (xp ) (S.10) XA (x ) = - ci oa '" yz (xp ) 'f' i ( Xp')
� Z�i �
./. z
If p a = 4 let DO = - 1 , 0 1 = 1 . In general 0 1 can be defined by induction and 01 + DO = O. Since c ; = CiY for 0 � i � k - 1, comparison of the last two equations yields that for A E A 0 U {I} �(Xp ) [t{!d Xp ) - Y0 1 t{! f- 1 (Xp)] if Xp EGDk, G XA (x ) - 1'0 1 () A (x ) o otherwise. _
For
fL
{
{
E A define 71>,-
(x ) =
g (xp ) [X1 (Xp) - 1'0 1 () f'(xp)] o
otherwise, where g is an irreducible constituent of the restriction to D of an
[5
CHAPTER VII
2 96
irreducible constituent of j.LCk - 1 in bk- t � The previous equation implies that if A E A 0 U {I} then XA = 71A + ),01 () ? For j.L E A define XIL = 711L + ),01 () �. It follows from the characterization of characters (IV . 1 . 1) that 711-'- is a generalized character. Thus also XIL is a generalized character. Direct computation shows that XIL satisfies equation (5. 10) with A replaced by j.L and ( replaced by g. In particular XIL (1) > O. We next compute Il xlL W . Let y S E Di - Di+1 and let A be the set of all elements in G whose p-part is y S. If i � k then LA I XIL (X ) 12 = LA I X 1 (X ) 12 by (5. 10). Suppose that i < k then
� 1 x. (x ) 1' =
<
;, "p�omJ 1 � 1 <�, C (y ' ) "':(x ) I' in Cj
If Z and Z l are in different cosets of blocks. Hence
C
then 1/1 ; and 1/1 ; 1 are in different
� 1 XIL (x ) 12 = 1 �i I Z�i 1 gz (y S ) 1211 1/1 ;1I'2 = � 1 X l (X ) 12 .
x
x
This implies that I I xlL W = II xl ll2 = 1 . Hence XIL is an irreducible character for all j.L E A U {I}. Equation (5. 10) shows that each XIL is in B, the XIL are all distinct and XIL (x ) = XA (x ) for all p '-elements x and A, fL E A. Thus (2.12) holds . . Furthermore (5. 10) implies that (2.17) holds for G with signs, say e '; for i � k and e '; = ei for 0 � i � k - 1 . (Recall that for G = Ck, 01 = 1 and 00 = - 1.) Since XIL (1) = X l (1) is the degree of I/Ik it follows that XIL is irreducible as a Brauer character. This implies that (2.14) and (2.16) hold. since LA u{ 1 } XA 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 � - 1 . Assume next that G = Na -t • Let A o b e the set o f irreducible characters of Co in bo which do not have D in thei! kernel. Let 0 1, OA, A E Ao be the irreducible characters in ba- t • It follows directly from (2.17) applied to Ca- 1 that T(ba-1)/Ca-l permutes the set {()d in (p a - I )1 e orbits each of length e. Thus if fL = A No E A let XIL = () ? Then by (V.2.5) there are exactly 1 A 1 = (P Q - 1)/e irreducible characters in B whose restriction to Ca-l has an exceptional character as a component. Since B has e + I A I irreducible characters by (5.3), there must be e distinct constituents of () 7. Call these Xl , . · · , Xe. Since ()l is irreducible as a Brauer character, (1.3) implies that () 7 = L:=l Xu and each Xu is irreducible as a Brauer character. Furthermore Ell.
a
5]
PROOFS OF (2. 1 1 )-(2. 17) IN CASE K
=
K
297
Xu (X ) = 'Pu (X ) for all p '-elements x for = 1, . . . , e after a suitable rearrangement. If 1 A I Jf. 1 th�n the characters XA, A E A are the exceptional characters. It is clear that 1 (Xu ) = 1 for 1 � � e. Thus Ou = 1 for 1 � � e. If A E A then XA (x ) = L;=l 'Pi (x ) for all p '-elements x, since () l = ()A on the set of p '-elements and so () 7(x ) = () ?(x ) for all p '-elements x. Thus
u
u
E ll.
==
-
IL
T
cr
T
298
CHAPTER VII
[5
and
{ ( OA - OjL ) G } (X ) = (OA - OjL) (x ) for x E (; with Xp conjugate in (; to an element of D. A standard argument (see e.g. Feit [1967b] section 23) shows that there exists a sign 0 = ± 1 and irreducible characters XA, A E A such that (OA - OjL ) G = O (XA - XjL ). Since (XA - XIL ) (X ) = O (OA - 0IL ) (X ) for x E (; with Xp conjugate in (; to an element of D, it follows that the higher decomposition numbers for XA and I/Ik for some k with 0 � k � a - I are not zero. Hence by the second main theorem on blocks (IV.6.1) XA is in B for all A E A. Furthermore (XA )O = OOA + C LILEA OIL + r where c is some integer and (r, OIL ) = 0 for all JL E A. Possibly c and r depend on A. Let XA be an R -free R [O] module such that (XA )K affords XA and XA is indecomposable. Let YA be defined as in (5.8). Then (XA )K affords OA if o = 1 and ( YA )K affords (h if 0 = - 1 . (If ' A , = 2 then 0 is not uniquely defined. Thus 0 can now be defined so that 0 = 1 if and only if I (XA ) � p Q /2.) Therefore (5.1 1) in any case. Since XA and XIL hav_e the sa�e irreducible Brauer constituents, it follows from (1.5) (i) that I (XA ) == l (XIL ) (mod p Q ) and so (5.12) for A, JL E A as 1 � l CXA ), l (XIL ) � pa - 1. Suppose that X is an exceptional character in B with X I: XA for all A E A. Let X be an R -free R [0] module such that XK affords X and X is indecomposable. Define Y as in (5.8) .. Define YA similarly corre'sponding to XA. Then HomR [ G ] (X, XA ) = (0) for all A E A. Thus by (1.5) (iii) and (III.5.12) Let YOA = XA if l (XA ) = e and let YOA = YA if I (XA ) = p Q - e. Then in any case I ( YoA ) = e. Let Yo = X if YOA = XA and let Yo = Y if YOA = YA. Therefore I ( Yo) = e or p a - e. Thus I ( Yo) + l ( YoA ) � p a. Hence by (3. 10) and (111.5.15) HomR [o ] ( Yo, YOA ) = (0) and so IK « Yo)K, ( YOA )K ) = (0) for all A E A. This is impossible as ( YO)K must involve some constituent which affords an exceptional character. We have shown that B contains exactly ' A ' exceptional characters if I A I I: 1 and they all have the same degree. Define 00 = - o. By (5.3) B contains exactly e nonexceptional irreducible characters if I A , I: 1 and
6]
THE B RAUER TREE
299
e + 1 nonexceptional irreducible characters if ' A I = 1 . If Xo is an R -free R [0] module which affords LA E A XA, where ' A I I: 1 and Xo is indecompos able then (1.5) (i) implies that l (Xo) == L l (XA ) == 00
(mod p Q ). A (2.12), (2.14), (2. 16) and (2.17) are now immediate by (5.8), (5.1 1) 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 � u < � e such that ou = ov and lPi = Xu + Xv + 1] for some character 1] Let OJ, . . . , Opa be all the irreducible characters of Ca - 1 jn ba - I where 01 is the nonexceptional character in ba -I • If Ou = 1 then l(Xu ) = 1 . Thus 0 1 is the only constituent in bQ -1 of the character afforded by « XU )K ) ca - I and it occurs with multiplicity one. If ou = - 1 then 1 (Xu ) = p a I and so OJ for 2 � j � P Q are the constituents in ba - I of the character afforded by « XU )K )ca l and each occurs with mUltiplic ity one. Thus by (1.5) (i) « Xu ) ca - p 01 - OJ ) = Ou for j � 2. Consequently « Xu + Xv )ca 01 - ( 2) = 20u contrary to (5.6). This contradiction completes the proof of (2. 15). D v
I
I.
_
_I'
6. The Brauer tree
LEMMA 6. 1 . For 1 � i � e, 1 � u � e, Qp (Xu ) � Qp (�i ). If I A I I: 1 then also Qp (Xo) � Qp (�i ) . PROOF. If X is a p -rational character then Qp (X) is in the field generated by all
PROOF. Let Xv be defined as in (2.14) and let Xu = X�. Then l (Xu ) = l (Xv ) and so ou ov (mod p a ). Since e � 2, P Q > 2 and so ou = ov . The result follows from (2. 15). D ==
LEMMA 6.3. Suppose that I A I I: 1 . Then ,\\ I: X ::- for any automorphism of Q p (Xu ), where 1 � u � e, A E A. a
300
CHAPTER VII
[6
PROOF. Suppose that XA =IX�. Since er permutes the irreducible Brauer characters of each Ck, it follows from (2. 17) that all the higher decomposi tion numbers of XA are ± 1 . Then (2. 17) implies that p = 2, e = 1 and Xu = X l , X � = X- I . Thus if x E D - D l then as 00 + 01 = 0, XI( X ) = eOOl I No : Co l 1/10(1) = - X-l( X ) � 0
and so X I (X t � X- I (X ).
0
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 dUi � o. We will later define the Brauer graphs of a block of F[ G] with cyclic defect group, where F is any finite extension of Qp, 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 dUi � 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. 0 LEMMA 6.5. Let T be a connected tree and let er, p be automorphisms of T
which fix no edge. (i) er fixes at most one vertex in T. (ii) If erp = per and (P) P, p ( Q ) = Q for vertices P and Q then P = Q. (J
=
PROOF. (i) Suppose that fixes vertices P and Q of T with P � Q. Then fixes the unique path from P to Q and so fixes some edge contrary to assumption. (ii) Since erp (P) = per(P) = p ep) it follows from (i) that P = p cP). Thus P = Q by (i) applied to p. 0 a
a
THEOREM 6.6. If I A I � 1 then Qp (Xu ) = Qp (�i ) for 1 � u � e, 1 � i � e. If I A I = 1 then the notation can be chosen so that Qp (Xu ) = Qp (�i ) for 1 � u � e, 1 � i � e. PROOF. By (2.9) M = Qp (�i ) is independent of i. Choose K with M � K such that K is a Galois extension of Qp and M is the maximal unramified
PROOFS OF (2. 1 1 )-(2 . 1 9)
7]
301
subfield of K. Let (j be the Frobenius automorphism of K and let er denote the restriction of (j to M. Thus (er) is the Galois group of M over Qp. Let be the Brauer graph of B. By (6.4) is a tree. Let er also denote the automorphism of defined by er. Hence eri fixes no edge of if eri � 1 . B y (6.5) there i s a vertex P of such that for all j with er i � 1 , the set of vertices fixed by eri is either empty or consists of P. If I A I � 1 then P corresponds to u = 0 by (6.3). If I A I = 1 choose the notation so that P corresponds to u = O. Thus x �j � Xu for 1 � u � e. Hence Qp (Xu ) = M for 1 � u � e by (6. 1). 0 T
T
T
T
T
COROLLARY 6.7. If the notation is chosen suitably in case I A 1 = 1 then in any case for 1 � u � e the Schur index mQp (Xu ) = 1 and for 1 � u � e Xu = LXu;, where {Xui } is a set of pairwise distinct algebraically conjugate characters. PROOF. By (IV.9.3), (2. 13) and (6.6) mQp (Xu ) = 1 . The second statement is an immediate consequence. 0 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 (2.1 1)-(2.19)
By assumption (2. 1 1) 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 dUi = 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. PROOF OF (2. 18). By (6.3) and condition ( ) of section 2, XA is not conjugate to any other character of 13 for A E A. Let rnA denote the Schur index of XA over K. Let 13 = 13 ( 1 ), . . . , B (r) be all the [G] blocks conjugate to B. Then for A E A, XA = rnA L�=l X �) where X �) is an irreducible character in B U). *
R
302
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[8
There exists i with 1 � i � e and dOi 1=. O. By (2. 13) dOi = rnA. Thus " XA W = m � r = d �J. The result follows as dO i is independent of A E A. 0
PROOF OF (2. 1 9). (i) This follows directly from (6.6) and (6.7). (ii) Let lPi = 2:.: duiXu . By (1. 17.8) dUi = Xum dUi if 1 � u � e, =0
"
W
dOi = I XAm 1 1 dOi . 2 B y (2. 13) and (6.7) dUi = 0 or 1 for 1 � u � e. B y (2. 18) dOi = (m/m �r)dO i and so dO i = 0 or m /m o r. Since c[Ji is the sum of m algebraically conjugate principal indecomposable characters it follows that m = re /e. Thus dO i = 0 or e / emo. The result follows from (2. 15). 0 8. Proofs of (2.20)-(2.25)
LEMMA 8. 1 . Suppose that dUi 1= 0, dVi 1= 0 for some i, u 1= v. Then Rad 0; XUi + Xvi and S ( O; ) = Xui n Xvi .
303
PROOFS OF (2.20}-(2.2S)
submodule XAj of XOj such that (XAj )K affords XA. Thus Lj � XAj � XOj and so S 2(XAj ) � S2(XOj ). Since S 2(XUj )lS (XUj ) 1= (0), S 2(XOj ) has a composition series with composition factors Lj, Lj. Thus S 2(XAj ) � S (XOj ) by (2. 13). Hence XAj = S (XAj ) = Lj and the second case of the lemma holds. Suppose that XVj = S (XVj ) = Lj . By (2. 1 9) either v 1= 0 or v = 0 and e = e = I D 1 - 1 . In either case Xv is the sum of m algebraically conjugate characters X �, each of which is irreducible as a Brauer character. Let ij denote the irreducible R [G] module which affords Xv as a Brauer character, then ij affords X �. By (2. 14) [( i n = l« ij t ) = 1 or p a - 1 . Since Lj 0ft R = E9 ij by (2.9) it follows that I (Lj ) = 1 or p a - 1 contrary to (3. 1 9). 0
PROOF OF (2.20). In view of (8. 1) it only remains to show that XUi is serial. Suppose it is not. Let n ;?= 1 be the smallest integer such that s n +l(Xui )/s n (XUi ) is reducible. Then Xu; /s n - I(XUi ) has a submodule M such that S (M) is irreducible and M/ S (M) is completely reducible but not irreducible. Let S (M) = Lj. By (2. 1 9) M/S (M) Ls EB Lt for some s -I t. Since dUj 1= 0, dus 1= 0, dut 1= 0 (8.2) yields a contradiction as M is isomorphic to a submodule of C0. =
=
PROOF. 1 ( 0; ) = I (Xui ) + I (XVi ) as lPi = auiXu + aviXv ' The existence of the Brauer tree implies that L is the only common irreducible constituent of XUi and Xvi . Thus if S ( Ui ) � XU i n Xvi then Li occurs with multiplicity at least 2 in both XUi and XVi and so u = v = 0 by (2. 13). This contradicts the fact that u 1= v. Thus XUi n XVi = S ( O; ). Hence I (Xui + XVi ) = 1 ( 0; ) - 1 and so Rad 0; = XUi + Xvi . 0 LEMMA 8.2. Let s 1= t with dus 1= 0, dut 1= 0 for some u. If dUj 1= 0 then S 2( C0 )/S ( C0 ) ¥: Ls EB Lt unless j = s or t, d Oj 1= 0 and dOi = 0 for all i 1= j. ( The last condition is equivalent to the fact that XA (x ) = 'Pj (x ) for all p '-elements x in G.)
PROOF. Assume that S2( C0 )/S ( C0 ) = Ls EB Lt . By (2. 15) and (6.2) there exists v 1= u with dVj 1= O. By (3. 1 9) and (8. 1) Ls EB Lt = S 2(XUj )/S (XUj ) EB S2(XVj )/S (XVj ).
Suppose that XVj 1= S (XVj ). Then Lt say, is a constituent of both XVj and XUj and so j = t by (2. 19). Hence Lj is a constituent of XVj with multiplicity at least 2. Thus = 0 by (2. 13). Let A E A. By (1. 17. 12) there exists a pure v
8]
PROOF OF (2.21) AND (2.22). Clearly (2.22) is a reformulation of (2.21). Thus it suffices to prove (2.22). By (2.20) Ui is serial for 1 � i � e if and only if there exists u with dUi 1= 0 and XUi 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). 0 PROOF OF (2.23). If the result is false then by (2.20) there exist i, i ', such that XUi has a factor modul � M with S (M) = Lj, S 2(M)/S (M) = Ls and Xui ' has . a factor module M' with S (M') = Lj, S 2(M')/S (M) = Lt for s 1= t. Then dus 1= 0 and dUj 1= O. Furthermore S \ C0 )/ S ( C0 ) = Ls EB Lt by (3. 1 8) contrary to (8.2). 0 PROOF OF (2.24). Let dUi 1= O. Since Xu is real valued (X�;)K affords Xu and X�i is indecomposable. Thus by (2.20) X � i = XUj for some j with dUj 1= O. Let 'P (ll, , 'P (s ) be the ordering of the irreducible Brauer constituents of XUj defined in (2.23). Then by (2.23) 'P (s )*, . . . , 'P ( I )" is a cyclic permutation of 'P (ll, , 'P (s ). Thus if 'P (n )* = 'P (l) then 'P (k + l) = 'P ( n -k)* . Hence 'P (n -t ) = 'P (n -t ) * if and only if n - t == t + 1 (mod s ) or 2 t == n - 1 (mod s ). There are at most two solutions to this congruence which implies the result. 0 •
•
.
.
.
.
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304
LEMMA S.3. Let V, W be R [G] modules with S ( V) = Lj. If Lj occurs as a composition factor of W with multipicity n then IR ( W, V) � nIR (Lj, Lj ). PROOF. It may be assumed that V � �. Thus by (1.16.4) h ( W, V) � IR ( W, � ) = h ( [J*; , W*) = nIR (L *; , L D = nIR (Lj, Lj ). D
-
_
Hence I ( Y) = 1 . By (III.5.12), HO(O, (1), HomR ( Yo, Y)) � (O) and so HomR ( Yo, Y) � (0). Thus Y = T( Yo). By (2.S) T( Yo) is determineg by S (Lj ). Thus by (2.5), T( Yo) = Y is deterrr:ined by Lj• By (2.S) S (X) is determined by Y, and hence by Lj• Thus X is determined up to isomor phism. Consequently X is determined up to isomorphism. Thus in particular X is serial. It remains to show that if X = XUj then X = XUj. By definition the multiplicity of S (X) as a constituent of X is 1 if u � 0 and ( I D 1 - 1)/e if = O. Hence (S.3) implies that if u � 0, m = II auiXu W = rankR HomR[ G ](X, Xuj ). h (X, X) � m ( I D 1 - 1) if u = 0, e This implies that IR (X, X) = rankR HomR[ G ] (X, Xuj ). Thus there exists g E HomR[ G ] (X, XUj ) such that g induces an isomorphism g from X to XUj. As XUj is serial, Rad Xuj is the unique maximal submodule of Xuj. Hence g
\
u
)
SOME PROPERTIES OF THE BRAUER TREE
305
is an epimorphism. Therefore XUj = X /Z for some module Z. As both Xuj and X are R -free this implies that rankRZ = rankRX - rankRXUj = O. Consequently Z = (0) and so Xuj = X. D 9. Some properties of the Brauer tree
PROOF OF (2.25). By (2.9) and (2.19) 1 (�Ui ) = 1 (.�Ui ) where Xui is an R -free R [G] module which affords XU. Thus l (X) = I (Xud is l or p a - 1 by (2. 14). There exists j such that Lj is isomorphic to a submodule of X. We will first show that X is determined up to isomorphism by I (X) and j. If G = Na - 1 this is clear by (2.4). Thus it may be assumed that G � O = Na - l . By (III.5.13) HO(G, (I), HomR (Lj, X)) � (O). Thus by (1.5) (iii) HO( 0, (I), IJomR (Lj, X)) � (0). Suppose thaj I (X) = 1 . Then HomR (Lj, X) � (0). Hence X = T(Lj ). Thus by (2.5) X is uniquely determined up to isomorphism. Hence also X is uniquely determi!.led up to isomorphism. Suppose that l(X) = p a 1 . Let P, Po be principal indecomposable R [OJ modules such that the following sequences are exact O� Y�P�X�O _
9]
Let Ko be an arbitrary finite extension field of Qp. Let K be a field which satisfies condition ( ) of section 2 such that Ko � K and K is a totally ramified extension of Ko, i.e. Ko and K have the same residue class field. Let Bo be the block of Ko[ G] which corresponds to B. Then the Brauer tree of Bo is defined to be the Brauer tree of B. Thus every block of Ko[ G] with a cyclic defect group has a Brauer tree for any field Ko with [Ko Qp J finite. It is sometimes convenient to associate the irreducible Ko[ 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 Xu, Xv correspond to distinct vertices on the same edge then by (2.19) and (2.25) Du + Dv = O. It follows from the results of section 2 that if G = Na - 1 then the Brauer tree is a star with the exceptional vertex, if any, at the center. *
:
(
:1< 0
0
\
°
ex
.......
°
G = Na - 1
°
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(S) has the following tree. The degrees are written by the vertices.
+4 014
1
°
°
35 ex
A more spectacular example is given by the principal 13-block of the automorphism group of Suz(8). 014 01 0 14 0 14 64 105 1 64 0 0 14 0 14 01
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If P is any odd prime then the tree for the principal p -block of PSL2(P ) looks as follows. There are �(p - 1) edges. �(p + 1) P-1 1 p-1 p+1 0 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 PSL2(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 S5 has the following Brauer tree . o _--!,-- o __-!o- o __� o
___
o
' "
==
1
4
4
6
1
The principal 5-block of Suz(8) has the following tree. 4 64 01 --°
f 91
10 14
The next two results were proved by Tuan [1944] in case a Yang [1977] .
9]
1. See also 307
SOME PROPERTIES OF THE BRAUER TREE
=
THEOREM 9. 1 . Suppose that G has a cyclic Sp -group. Then every [ G] module in the principal p -block is equivalent to an Fp [G] module where Fp is the field of p elements.
R
PROOF. As the principal p-block contains the principal character, the result is a direct consequence of (2.9). D
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 st:(aight line segment. D
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 Li is denoted by Ei • The vertex corresponding to Xu is denoted by Pu. If 1 !S i !S e then T - {Ei } is the disjoint union of two trees. Let these be denoted by. To(Bi ) and Tl(Ei ) where To(Ei ) contains the vertex Po. n (Ei ) is the number of vertices in TJ(Ei ). Thus n (Bi ) is the number of vertices in T which are separated from Po by the removal of the edge Bi • d (Pu ) is the number of edges on the unique path in T which joins Pu to Po. LEMMA 9.3. (i) For O !S u !S e, ou
=
( - l)d(P.,l o o.
(ii) Let 1 � j � e and let P = Pu be the vertex on Then one of the following holds. (I) oo( - l)d ( P ) = ou = 1 ; n (Ej ) = I ( Lj ). (II) oo( - l)d (P ) = ou = - 1 ; n (Ej ) = p a - 1 (Lj ).
[ 10
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308
Ej
which is in
'T1(Ej ).
PROOF. (i) This is an immediate consequence of (2. 19) and (2.25). (ii) This is proved by induction on n (Ej ). If n (Ej ) = 1 then XUj = Lj where P = Pu . Thus I (Lj ) == ou (mod p a ) and the result follows from (i). Suppose that n (Ej ) � 1 . Let E1, • • • , Es be all the edges in TI (Ej ) which have P as a vertex. Hence E I, . . . , Es, Ej are all the edges in which have P as a vertex. Furthermore n (Ej ) = 1 + �;= I n (E; ). By (1.5) (i) 'T
ou == I (Lj ) +
By induction and (i) Therefore
ou
=
:! I (L )
;=1
I(Lj ) - ou
(mod p a ).
:! n (E; ) == I (Lj ) - ou {n (Ej ) - l}
;=1
I (Lj ) - ou n (Ej ) == 0
(mod p a ). The result now follows, as n (Ej ) � e and by (2.7) I (Lj ) � p a. D
(mod p a ).
I(Lj ) � e
or pQ - e �
LEMMA 9.4. Suppose that I D 1 - 1 i: e. Choose j with dOj i: 0 and let L1 , • • • , Ls be all the irreducible R [G] modules (up to isomorphism ) which are constituents of XOj• Then one of the following holds. (i) 00 = - 1. l (L ) = n (E; ) � e for 1 � i � s and ��= I I (L ) = e. (ii) 00 = 1. p a - 1 (L; ) = n (Ei ) � e for 1 � i � s and �;= I {p a - l (L )} = e.
PROOF. E1 , • • • , Es are precisely the edges of which have Po as a vertex. Let Pi i: Po be the other vertex on E; for 1 � i � s. Thus d (P; ) = 1 and so ( - 1 )d ( P) = - 1 for 1 � i � s. Since �:= n (Ei ) = e, the result follows from (9.3) (ii). D 'T
1
LEMMA 10. 1 . Suppose that G has a cyclic Sp -group (x ). Let I (x ) 1 = p n. 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 n - a l ( V) � d. If furthermore D <J G then p " - a l ( V) = d. 10. Some consequences
10]
PROOF. By (1 .5) (i) it may be assumed that G = G and V = V. By (3.4) it may be assumed that K = K. By (3.5) it may be assumed that G Ca - 1 Then No « x )) = Co « x )). Hence Burnside's transfer theorem implies that G = (x )H with H = Op,(G). Replacing x by a conjugate it may also be assumed that D � (x ). By (111.3.8) every indecomposable R [G] module U in B is of the form U = Uf{ for some R [DH] module Uo in a block with defect group D. Hence V = Vf{, where Vo is an R [DH] module in a block with defect group D. By the Mackey decomposition (11.2.10) ( Vc?)(x ) = {( VO) D }(x ) . Thus the minimum polynomial of y on Vo is ( Y - 1)\ where d = p n - a do. Therefore it may be assumed that V = Vo and x = y. Since H is a p i-group, VH = EB M;, where each M; is irreducible. Choose M = � with Mel (Rad V)H. Then ��,:-� M ( y - 1Y is an R [G] module which is not in Rad V. B contains a unique irreducible R [G] module W up to isomorphism by (2. 1). V is serial by (2.4). Thus V = ��,:-� M(y - It Since M I WH it follows that dimR M � dimR W. Therefore I ( V)dimR W = dimR V � d dimRM � d dimR W. Hence I ( V) � d. Suppose that D <J G. By (111.3.7) WH = M and I ( V)dimR W = dimR V = d dimR W. Hence in this case I ( V) = d. D The next result is the celebrated Theorem B of Hall and Higman [1956]. The proof given here is in the spirit of that of Thompson [1967b]. For an alternative proof in a critical special case, see Feit [1967a]. As can be seen from the proof, the integer a which occurs in the second possibility of (10.2) is related to the defect of a p-block of a quotient group of a suitable subgroup of G. A more precise formulation can be found in Knapp and Schmid [1982]. SOME CONSEQUENCES
309
=
•
THEOREM 1 0.2. Suppose that G is p-solvable with Op (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 Go = (x )Op,(G). Then one of the following occurs. (i) d p n. (ii) There exists a prime q and a positive integer a such that Go has a nonabelian Sq -group, p a _ I is a power of q and p Q - 1 1 1 Go / . Furthermore p ,,- a (pQ - 1) � d < p ". =
=
PROOF. Suppose the result is false. Let G be a counterexample of minimum order and let H = Op (G). Since Cc (H) <J G and Op , (C c (H)) � H it follows that Cc (H) � H. The minimality of G now impli : s that G = Go. Without loss of generality it may be assumed that K = K. There exists a prime q such that x pn-I does not centralize any Sq -group of H. By the Frattini argument (x ) normalizes a Sq -group Q of H. Thus the minimality of G implies that H = Q is a q -group. Let V = E9 V;, where each V; is indecomposable. Let Ai be the kernel of ( Vi ) Since V is faithful ,n Ai = (1). Hence O (G /Aj ) = 0) for some j. Hence it may be assumed that V = Vi 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 W Then B covers the principal block of A . By (V.2.3) A is i n the kernel of V as A is a p'-group. Thus A = (1) and so W is faithful as Op (G) = (1). Hence it may be assumed that V = W is irreducible. Suppose there exists an indecomposable R [(x p )H ] module VI such that V I V? By (111.3.8) V = V? Let dl be the degree of the mini�um . polynomial of xP in VI. Then d = pd l and the r� sult follows by mductlOn. Hence it may be assumed that V ,{ V? for any R [(xP )H] module VI. 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 XK absolutely irreducible. Consequently dimR V I I G I . Furthermore I( V) = 1 or p a _ l by (2.14). Clearly d � p ". Thus d < p " as G is a counterexample. He�ce Va has n � R [D]-projective summands. Thus Va = V by (1.S) (i). If I ( V) = 1 then V is irreducible and so x p n-I is in the kernel of V. This contradicts the fact that V is faithful. Therefore I( V) = p " - 1. By (2. 1) dimR V = (p " - 1) dimR W, where W is the unique irreducible R [0] module in E. Therefore p " - 1 1 1 G I and so p " - 1 is a power of q. By (10.1) d ;?; I ( V) = p " - 1. This completes the proof. 0 CHAPTER VII
310
H.
[10
p
H.
The next result includes a theorem of Peacock [1979].
THEOREM 10.3. Suppose that G has a cyclic Sp-group (x ). Let l (x ) 1 = p ". Let V be a faithful irreducible R [G] module in B and let d be the degree of the minimum polynomial of x acting on V. Assume that I D I ;?; p 2 . Then dimR V ;?; d ;?; p ,,- a (p a - e ) � p ,, - a (p a - p + 1).
10]
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 a and so Va = V by (1 .S) (i). By (10.1) I ( V) < p a - e and so by (2.7) I( V) � e. By (2.4) every indecomposable module in E is serial. Thus there exists an indecomposable projective R [0] module P and an exact sequence P � V � O. Let z = y p a - I . Then (z ) = Da - I and p(z ) is a multiple of (z ) Thus dimR (P(I - z Y /P(I - Z y+ l ) is independent of i for 0 � i � P - l . Each P(1 - Z )i is an R [0] module. As all irreducible modules in E have the same dimension, this implies by (2.6) that I(P/P(I - z )) = p a - l . Since V = Vc is a faithful R [0] module it follows that V (1 - z ) � O. Hence P/P(I - z ) = V/ V(1 - z ) as P is serial. therefore I ( V) � p a - I � P > e as a � 2 by assumption . This contradicts the previous paragraph. 0 SOME CONSEQUENCES
31 1
(z ) .
The next result strengthens a Theorem of Lindsey [1974] .
COROLLARY 10.4. Suppose that G has a cyclic Sp -group (x ) with l (x ) 1 = p " � p 2 . Assume that (x ) n (x )Z = (1) for z E G, z � Nc ( x »). Let V be a faithful irreducible R [G] module. Then dimR 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). 0
The next result is a generalization of (S.6) which generalized a result of Thompson [1967b]. In case I D 1 = p this first appeared in Feit [1967b].
THEOREM 10.S. Assume that K = K. Let Y be an R -free R [G] module in B and let Y/ be the character afforded by YK. Suppose that Y = MI EB · . . EB M", where each M; is a nonzero indecomposable R [ G] module. Let () , () be all the irreducible characters of Ca - I in ba - I . Then I ( Y/ca - P ()s ) - ( Y/ca -I, ()t ) I � n for all s, t. J, . . .
pa
PROOF. Without loss of generality it may be assumed that Y is indecom posable. If Y is projective then ( Y/ca -P ()s ) is independent of s and the result is proved. Thus by (1. 17. 1 1) no M; is a projective R [G] module. By (1 .S) (i)
10]
Yo = Y ffi A I ffi Az where A I is projective and Az is a sum of modules in blocks other than B. If 1]i is the character afforded by (Ai )K for i = 1, 2 then « 1] z)ca - P Os ) � 0 f<;?r all s and « 'T1 1 )ca - P Os ) is independent of s. By (111.5.8) and (1.4) Y Y !VII ffi · . . ffi !VI" and each M is indecomposable and nonzero. Hence it may be assumed that G = G and Y = Y. By (111.4.6) there exists an R [ T(ba - I )] module Yo such that Y� Y, ( Y�h( ba-t) = Yo ffi A where A is a sum of modules in blocks which do not cover ba - I and Yo is a sum of n indecomposable modules. Thus it may be assumed that G = T(ba - I ). By (1 .3) every irreducible R [G] module remains irreducible when restricted to Ca - I . Thus by (3.5) (ii) each (M; ) cu - t is indecomposable. Hence it may be assumed that G = Ca - I . It suffices to show that (1], Os ) !S n for all s. Suppose this is false. Thus there exists 0 = Ot for some t with ( 71, 0 ) = n l > n. Let M be a K[G] module which affords 0 and let YI = Y n n M. Then Yt is a pure submodule of Y and so S ( Yt) � S ( Y). Therefore I ( S ( YI)) !S n since every indecomposable R [Ca - t] module in ba - I is serial. Thus YI is a sum of at most n nonzero indecomposable modules. Thus it may be assumed that Y = Yt • Hence YK affords the character n l O. Consequently rankR HomR[ G ) ( Y, Y) = IK ( YK, YK) = n � . Since HomR[ G ) ( Y, Y) = Inv G HomR ( Y, Y), it is a pure submodule of HomR ( Y, Y). Hence In ( Y, Y) = dim R Homn [G) ( Y, Y) ?: IK ( YK, YK) = n �.
31 2
[ 10
CHAPTER VII
=
=
PROOF. There are exactly 2e modules XVj up to isomorphism with dvj ;i 0 since by (2.19) there exist exactly 2 values of for each j with dvj ;i O. Let @j = �VjXv + awj Xw and let X = Uj /Xvj . Then T(X) Lj is irreduc ible. Thus X is indecomposable. Since XK affords awjXw it follows from (2.2�) that X = XWj' for some j I with dwj, ;i O. Thus in particular if T(XUi ) Lj then CPj = aujXu + avjXv for some v. Therefore the following sequence is exact. v
Hence
n � !S In ( Y, Y ) = � In ( M; , Y ) !S nnl .
Thus n l !S n. This contradiction establishes the result. i=1
0
The next result is due to Green [1974a] in case K = K. It generalizes an earlier result of Alperin and Janusz [1973]. THEOREM 1 0.6. Suppose that dui ;i 0 for some u, i with O !S u !S e, 1 !S i !S e.
There exists an infinite exact sequence
=
=
(10.8) Consequ ently the exact sequence (l0.7) exists. Furthermore for all s, fs (Ps ) XVj for some v, j with dvj ;i O. As there are exactly 2e modules XVj up to isomorphism with dvj ;i 0, it suffices to prove the re�ult for any one of �hem. For convenience it may be assumed that I (XUi ) = 1 _by (2.25). Furthermore after a change of notation it may be assumed that XUi W as defined in (2.4). In view of (1. 15.6) and (1.5) (ii) the exactness of (l0.8) implies that O � XVj � if � XUi � O is exact !or some projective module if in B. By (2. 14) [ (XUi ) + I (X j ) < 2p a . Hence U is a principal indecomposable module. Thus there exist principal indecomposable modules Ps in B and an exact sequence =
=
Let W be the unique irreducible R [ G] module in B. Then In (M; , � ) = min{ I (M; ), I (� )} as every R [ G] module in B is serial. For 1 !S s !S p a , Os is irreducible as a Brauer character. Hence I ( Y) = n I . Thus for 1 !S i !S n nl .
(10.7)
=
l
I (� ) = I ( Y) =
313
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) Ps Ps +Ze and fs (Ps ) = fs +ze (Ps+Ze ) for all s. (ii) {[fs (Ps )] / 1 !S s !S 2e } = {XVj / l !S j !S e, O !S v !S e, dvj ;i O}. . (iii) {[PZS - I ] / 1 !S . s !S e } = {[Pzs ] / 1 !S s !S e } = {[ Uj ] / l !S j !S e }.
=
IR (M; , Y) = j� IR (M; , � ) !S j=t � =1
SOME CONSEQUENCES
v
/
Furthermore Is (Ps ) = l7js ) fo_r all s. By (1.4) and (111.5.8) XVj = XVi for all v, j with dvi ;i O. Thus the following sequence is exact
Choose the notation so that '\V; Wa l -i for 0 � i � e 1. Let V(i, I) denote the R [OJ module in 13 such that S ( V(i, I» = \V; and I ( V(i, I» = I. By (2.8) the following equations can be read off for all s. (10.9) Thus {t. (ps ) / 1 � s � 2e } is a set of 2e pairwise nonisomorphic modules. Hence also {js (Fs ) / 1 � s � 2e } and {fs (Ps ) / 1 � s � 2e } are each sets of 2e pairwise nonisomorphic modules. Thus every module XVj with dvj "l 0 is of the form fs (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 �2e such that S(fs (Ps ) = Lj . 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 Lj = S(fzs - I(PZS-1», By (2.5) �nd QO.9) there is a unique value of s with 1 � s � e such that HomR (ij, !zs- I(Fzs-1» "I (0). Thus by (3.10) applied to 0 and (III.5.13) applied to G and by (1.5) (iii) there is a unique value of s with HomR (Lj, f2s-I(Pzs - 1» "I (0). D [10
CHAPTER VII
314
==
-
Suppose that dui "l 0 for some u, i. Green [1974a] has defined a function on {I, . . . , e} as follows. Let Ps be defined as in (10.6). If Pzs = � then P2s + 1 = UaUl ' By (10.6) a- is a permutation on {I, . . , e }. It also follows from (10.6) that if the notation is chosen suitably then there exists an exact sequence
a
.
It is clear that � I + I (fs (Ps » (fs + I(Ps + 1» = p a for all s. The permutation a- = a-ui depends on the choice of u and i. However the definition of a-ui implies that if dvj "l 0 then a-vj = a-ui in case au = av . If au "I av then a-ui(Jvj = (1 , 2 ' . . e ). r----.-
COROLLARY 10.10. Suppose that dui "l 0 for some u, i. Let athat dvj "l 0 for some v, j with au = av . Then T(Xvj ) = LaU l '
=
a-ui .
Assume
PROOF. Clearly au = av if and only if XVj = hs + I(P2s + 1 ) for some s. By definition S(f2s + I (P2s + 1» = S (Pzs ) and T(f2s + I (Pzs + 1» = T(Pzs + 1 ) = S(P2S + I ) ' Thus if Pzs = � then S(f2S + I(Pzs + 1» = Lj and T(fzs + l(PZS + 1» = L(T (j ) ' D
10]
In Green's terminology the ordered sequence SOME CONSEQUENCES
315
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 ab stract 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 (2)
2 ---
a(l)
/
This process can be carried out for any permutation but wi1l 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. 1 1 . Let B = 13 be an R [G] block with a cyclic defect group D. Let I D I = p d Let C be the Cartan matrix of B and let 0 be the quadratic form corresponding to pdC- 1 • Let q be the minimum value assumed by 0 on the set of all nonzero vectors with integral coordinates. Then q = e. .
PROOF. By (IV.3.1 1) and (2 . 1 2) {Xu / 1 � u � e } is a basic set for B. Let D O , Co denote the decomposition matrix, Cartan matrix respectively with respect to this basic set. Let 0° be the form corresponding to p d (COr l . Then 0° is integrally equivalent to 0 and q is the minimum value assumed by 0° on the set of all . nonzero vectors with integral . coordinates. By (2.12)
[10
where t = (pd - 1)le. Let 1 denote the matrix all of whose entries are 1. Then CO = (D 0)' D 0 = I + tl. Thus p d (COr l = p dI - tl. Hence if a = (ai ) then CHAPTER VII
.3 1 6
o (a, a ) =
(10.12)
a 7 + t L (ai - aj t ! i <j i=1
Let No be the number of ai = 0 and let Nl = e - No. The last term in (10. 12) is at least NoN1 and the other term is at least N1 • Hence if a -I- 0 then Nl -I- 0 and so O (a, a ) � Nl + NoN � N I + No = e. Thus e ::s q. If ai = 1 for all i then O (a, a ) = e and so q = e. 0 1
THEOREM 10.13. Let B be a block of R [G] with an abelian defect group D of order p d. 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 k (B ) ::s p 5d /3.
PROOF. Let y be an element of maximum order p c in D. By (V.9.2) there exists a major subsection S (y, B ) associated to B. By (lVA.S) there exists a unique block BO of CG (y )/(y ) contained in B. Further more D I(Y ) is the defect group of B o . By (III.2. 13) y is i� the kernel of every irreducible Brauer character in B. Hence B and B O both have I(B ) irreducible Brauer characters. If q (B ) and q (BO) are defined as in (10. 1 1) then q (E ) = q (EO) by (IVA.27). If r ='2 then EO has a cyclic defect group and so q (B ) = [ (B ) by (10. 11). Thus k (B) ::S p d by (V.9. 17) (i). If r = 3 then [ (B O) ::s k (BO) < p d -C by the previous paragraph as B O has an abelian defect group of rank 2. Hence by (V.9.17) (i) k (B ) ::S p 2d - C. This implies the result as � � d. 0 _
-
_
c
The next result was first announced in Brauer and Feit [1959].
THEOREM 10.14. Let B be a block of R [GJ of defect d. Let k = k (B ) denote the number of irreducible characters in B. , Then k ::s p d for d = 0, 1, 2 and k ::s p 2 d -2 for d � 3. PROOF. Induction on d. If d = 0 the result follows from (IVA. 19). If d = 1 it follows from (2.1) and if d = 2 it follows from (10. 13). Suppose
that d � 3. If B contains an irreducible character of positive height the result follows from (IVA.18). Thus it may be assumed that every ir reducible character in B has height O. Let S (y, E) be a major subsection associated to B. By (IVA.S) there exists a unique block EO of CG (y )/(y ) contained in B. Furthermore D /(y ) is the defect group of B o . By (111.2. 13) y is in the kernel of every irreducible Brauer character in E. Hence B and B both have I (.BO) irreducible Brauer characters. Thus by (V.9. 17) (ii) k -2 ::s p 2 d[ (BO) ::s p 2dk (B O), where k (B O) is the number of irreducible characters in B o . Since BO has defect at most d - 1 , induction implies that k (B O) ::s p 2(d - I )- 2 as d = 2d - 2 for d = L Thus 1 1J
SOME EXAMPLES
317
°
1 1 . Some examples
The following result is due to Dade [1966] and shows that every possible sequence of signs Ck can occur in (2. 17).
LEMMA 1 1. 1 . For 0 � k ::s a - I let l'k = ± 1. There exists a group G and a block B of G with defect group D such that if Ck is defined as in (2.17) then Ck = l'k for O ::s k ::s a - 1 .
PROOF. For O ::s k ::s a - I let qk b e a prime with qk = - 1 (mod p a ) and 2 < q l < . . . < qa - l . Let 13k = ± 1 for O ::s k ::s a - 1. If 13k = 1 let Ok = (I). If 13k = - 1 let Ok be the nonabelian group of order q � and expo nent qk. Let D operate on Ok in such a way that Dk = Co (Od in case 13k = - 1. Define G = ( 00 x . . . x Oa - l )D. Define an irreducible charac ter Ok of Ok as follows. If 13k = 1 then Ok = (I) and Ok is the principal character. If 13k = - 1 then Ok is some irreducible character of Ok of degree qk. Let o 'be the character of 00 x . . . x Oa - l defined by 0 = rr::� Ok . Then T(O ) = G. Furthermore O G = X l + L XA where A ranges over the nonprincipal irreducible character of D and XA (z ) = A (z )X l (Z ) for z E G. {XA } is the set of exceptional characters in a block B with defect group D and X l is the nonexceptional character. It is known that if z E Dk - Dk + 1 and x is a p '-element in CG (z ) then
318
[1 1
CHAPTER VII
Thus by (2.17) '}'k
=
Ck
= n�,:-� f3i .
-
o :
D
integers in Ko. Let X be an Ro-free Ro[ G] module such that X�) affords K Then the following hold. (i) X is indecomposable. (ii) Every irreducible constitutent of X is absolutely irreducible. (iii) Let (Sij (x )) be the matrix representing x in a representation with underlying module X. Let 7ro be a prime in Ro. Then for given i ;l j there exists x such that either Sij (x ) � 0 (mod 7ro) or Sji (x ) � 0 (mod 7ro) .
If X is decomposable the matrix representation can be chosen so that S i n (x ) == Sn I (X ) 0 (mod ) for all · x E G, where n = X (l). Thus (iii) implies (i). If some irreducible constituent V of X is not absolutely irreducible then there exists an unramified extension KI of Ko such that if R is the ring of integers in KI then R is a . splitting field for V. Since V @R R I is completely reducible it follows that if Ko is replaced by KI 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 Sij (x ) Sji (x ) 0 (mod 7ro) for some i ;l j and all x E G. are in the Galois group of · Ko The Schur relations imply that if over Qp then PROOF.
7ro
I
==
==
I
u, T
L s ij (x )s fi (x - I ) = OeFT I G I l x ( 1 )
where OaT = 1 if and coincide on Qp (X ) and OaT = 0 otherwise. Let Tr denote the trace from Ko to the maximal unramified subfield KI of Ko. Thus Tr (sd x )) = Tr(sji (x )) == 0 (mod p ) for all x E G and xEG
(J
Thus in particular
T
(mod p 2) .
Since I D I = p, I K K 1 / p - 1 by (2. 17). Thus I G l l x (I) == O (mod p 2). However as X is in a block of defect 1 this is impossible. D
LEMMA 11 .2. Assume that I D I = p. Let X be an irreducible character in B. Let Ko be an unramified extension of Qp (X ) and let Ro be the ring of
==
319
SOME EXAMPLES
I G I I Ko : K l i == 0 x (l)
The f3i can be chosen so that
f3i for 0 � k � a 1. n i=k
a-I
1 1J
1
In Brauer's original treatment of blocks of defect 1, (11.2) and re lated 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 irreduc ible 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 (1 1 .5) which is concerned with the construction of examples that show that (1 1 .2) (i) is false in all other cases. (1 1 .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. Let X be the R [D ] module which has {(1 - y Y 1 1 � i � p a - I} as an R -basis. Let Y be the unique submodule of X of index p. Then {zi l l � i � p a _ l} is an R -basis of Y where z l = p (l - y ), zi = (l - y Y for 2 � i � p a 1 . Y = VI EB V2 where VI i s the vector space over R spanned by ZI and V2 is the vector space over R spanned by {Zi / 2 � i � p a - I}. -
LEMMA 1 1 .3. If a > 1 and p ;l 2 then VI and V2 are R [D ] modules. PROOF.
By definition
for 2 � i � p a - 2.
Thus it suffices to show that Zp - l ( 1 - y ) E V2• Let YI be the R module spanned by Z and let Y2 be the R module spanned by {Zi 1 2 � i � p a - I}. Thus Yj = Vi for j = 1, 2. zi (I -' y ) = zi (l - y ) = Zi + I E V2 a
I
"
zp" - .(l - y) =
"
� (� ) ( - l)Y = p�. (� ) ( - l)Y P�I a =
(� ) ( - IY [(y
- 1) + lY.
[11
CHAPTER VII
Thus
320
z,"- l (l - y )
:�1 (�") < - lY[s (y - 1) + 1] (y - 1) ',�' (�") < - 1) '5 /�1 (�" ) ( - 1)' I (y 1) == P� (�Q) ( - I ts + (1 - lya - (1 - 1) == (y - 1) :�l (�Q ) ( - I ts (mod Y2).
�
�
Let
Thus
f(x ) = (1 - x ya =
� (�Q) ( - ltx s.
Hence /,(1) = 0 and zp a_ I (I - y ) == (y - 1) [f'(I) ± p Q ] == ± p Q (y - 1) (mod Y2). Since a > 1 this implies that zp a_ I (I - y ) E pYI + Y2 and so zp a- l (1 - y ) E V2• 0
LEMMA 1 1 .4. (i) V2 is an indecomposable R [D] module. (ii) Suppose that p r£ 2 and a > 1. Let N = DE be a dihedral group with E = <x ) of order 2. Then yN = ZI EB Z2 where = WI EB W2 , WI is the trivial R [N] module, dimR W2 = p Q - 2 and W2 has a one dimensional socle which is the R [N] module whose kernel is D.
2
1
PROOF. (i) By definition X/pX is a serial indecomposable R [D] mod ule. 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 dimR A = P Q - 2. Thus the minimum polynomial of y on A has degree p a - 2 and so the minimum polynomial of y on Y/p Y has degree at least p Q - 2. This implies that V2 is indecomposable. (ii) Let fl = HI + x ), h = Hl - x ) in R [N]. Then fi yN � yN for i = 1, 2 as Y is the unique submodule of X with dimR (XI Y) = 1 and X = J(R [D]). Let Zi = t y N. Then Zi r£ 0 for i = 1, 2. As ( yN)D = Y EB Y it follows that (Zi ) D = Y for i = 1, 2. Hence ( yN )D =
llJ
VI EB VI EB V2 EB V2 where each summand is indecomposable by (i). Furthermore 21 EB 22 = yN = Vf"EB Vf". As dimR 2i = p Q - 1, dimR Vf" = 2 and dimR Vf" = - 2) it follows that 2i = W; I EB W;2 with dimR W; I = 1, dimR W; 2 = (p Q - 1) and W; 2 is indecomposable. Since V� = <x )(x ) exactly one of W; say WI is the trivial R [N] module. Let "'l = Wl j for j = 1, 2. It remains to show that S ( W2) is not the trivial R [N] module. Let 'P be the Brauer character afforded by W2 • No irreducible con stituent of ( yN)K has D in its kernel hence no irreducible constituent of (ZI)K has D in its kernel. Thus 1 + 'P (x ) = O. As 'P (x ) = - 1 it follows from (2.8) that S ( W2) is not the trivial R [N] module. 0 SOME EXAMPLES
321
2(P Q
J,
I,
THEOREM 1 1.5. Suppose that p r£ 2 and a > 1 . Let q be a prime with q + 1 == p Q (mod p Q +I ). Let G = PSL2(q). Then a Sp -group of G is cyclic of order p Q. G has an irreducible character () of degree q (the Steinberg character), which is afforded by a Q[ G] module. Furthermore there ex ists an R -free R [ G] module x such that XK affords () and X = MI EB M2 where MI is the trivial R [G] module and M2 is an irreducible R [G] module with dimR M2 = q - 1.
PROOF. The existence of () and the structure of the Sp -group of G are well known facts. Let D be a Sp -group of G. Then Co (D) = D x H for some p '-group H and I No (D) : Co (D) I = 2. In particular No (D)/H = N where N is defined in (11.4). Furthermore D n D Z = (1) for z � No (D). Thus the Green correspondence between G and No (D) is defined for modules with nontrivial vertex in D. Let X be the R [No (D)] module with kernel H such that X as an R [N] module is isomorp!Iic to ZI. Let X be the R [G] module which corresponds to X. Thus X = WI EB W2. Hence by (111.5.8) X = M, EB M2 where M = W; for i = 1, 2. Let B be the principal p -block of G. Then the index of inertia of B is 2. Since () (z ) = - 1 for z E Co (D ) - {1} it follows that () is in B, this implies that the Brauer tree for B is 0-------0--
1 q q-l Let 'P I , 'P2 be the irreducible Brauer characters in B where 'P I is the principal Brauer character. Thus M, affords 'P I . The principal block of
R [No (D)] has a unique indecomposable module, namely W2 , which has no invariants and whose dimension over R is congruent to - 2 (mod p a ). It follows from (111.5.10) and (111.5.14) that a module which affords 'P 2 corresponds to W2 . Thus M2 affords 'P 2 . Hence in particular rankR X = q. Since X has no invariants, neither does X. Thus the character afforded by XK is sum of exceptional characters and possibly e. As rankR X = e (1) and the degree of every exceptional character in B is q - 1 it follows that XK affords e. 0 322
CHAPTER VII
[12
The next result which is due to Benard and L. Scott, Benard [1976] will be used in section 1 3. In some sense it explains why it is necessary to assume that a > 1 in ( 1 1 .3).
LEMMA 1 1 .6. Let R be the ring of integers in an unramified extension of Qp. Let X be an R -free R [D] module such that XK is irreducible. Then X is indecomposable. If furthermore Y is an R -free R [D ] module with YK = XK then Y = X.
PROOF. If rankR X = 1 the result is obvious. Suppose that rankR X > 1 . Consider the map f : X X defined b y f(v) = v ( 1 - y ). Then f is an R [D ]-monomorphism and ' X : f(X) , = ± det(1 - y ). Since XK is ir reducible, the characteristic values of y are all the primitive p th roots of unity for some n. As rankR X > 1, n � O. Hence det(1 - y) = II(1 - �) = p, where � ranges over all primitive p th roots of 1 . Therefore ' X : f ( X) , = ' R , . This implies i n particular that X i s inde composable. Furthermore f(X) is the unique maximal submodule of X and so f (X) is the unique maximal submodule of X. Thus X isomorphic to its unique maximal submodule. Suppose that YK = XK• It may be assumed that YK = XK• If Y is replaced by a multiple it may be assumed that Y � X. Since ' X : Y , is finite there exists a chain of submodules Y = Xm � Xm-1 � � Xo = X such that Xi is maximal in Xi - I for 1 :::;:; i :::;:; m. By the previous paragraph Xi = Xi-1 for 1 :::;:; i :::;:; m and so Y = X. 0 �
11
11
IS
• • •
12. The indecomposable
R [ G]
modules in
B
This section contains a description of all the nonzero indecomposable R [ G] modules in B in terms of the Brauer tree of B and of its exceptional
THE INDECOMPOSABLE R [G] MODULES IN B
vertex. The results in this section �ere all proved by Janusz [1969b] and Kupisch [1968] [1969] in case R = R. Given the results of section 2 for K in general, Janusz's arguments go through without any change. The structure of the principal indecomposable R [ G] modules in B is described by (2.20). In this section ' D ' e - 2e pairwise nonisomorphic nonzero indecomposable R [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 , 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 indecompos able 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 group algebra. (See the remark on p. 306.) In this section 'T = 'TB denotes the Brauer tree. Vertices of 'T will be denoted by P, Pi or Q. The exceptional vertex, if it exists, is denoted by Pex • E or Ei will denote either an edge of 'T or O. If Li corresponds to the edge E, write Li E. If E = 0 let (0) E. Set t = ( ' D , - 1)/ e. By (2. 13) and (2.19) t is the multiplicity of any L; in XOi for dO i � 0, do; � O. If du; � 0 let Xu; be defined as before (2.20). If dUi � 0 then the statement dual to (2.20) implies that T(Xu;) = L for some j with dUj � O. Let P be a vertex of incident to the edge E. Let L E and let Xu correspond to P. Define V(E, P) = Xu; where T(Xu; ) = 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 and let PI and P2 be the vertices incident to E. If Ei is incident to Pi we will define modules V(E" E, E2 : ) for suitable integers We will allow the case that Ei 0 or Ei = E for exactly one of i = 1 or 2 if Pi = Pp . Several cases will be considered. We begin with some preliminary definitions. Let Vi (E) = Rad( V(E, Pi » 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 C U). Then the dual of (2.20) implies that Rad(Mo(E» = V1(E) EB V2 (E). Suppose that E, E1, E 2 are distinct with E � 0 such that 12]
323
a
�
�
'T
�
'T
n.
n
=
�
324
[12
CHAPTER VII
Pz
PI
-- 0 -- 0 --
is contained in Define modules Mi (n) for i = 1, 2 for suitable integers n as follows. If Ei = 0, M (1) = V; (E). If Ei -1 0 and Pi -I Pex let M (1) be the submodule of V; (E) such that S ( Vi (E)/M (1 » � Ei . Since V; (E) is serial Mi (1) is uniquely determined by (2. 13). Suppose that Ei -1 0 and Pi = Pex • Let 1 � n � t. Let M (n ) be the submodule of Vi (E) such that S ( Vi (E)/Mi (n » Li where Li � Ei and L occurs with multiplicity n as a constitutent of V; (E)/M (n). Since V; (E) is serial Mi (n) is uniquely determined by (2. 13) and (2.19). Define V(Ej, E, Ez : 1) = Mo(E)/(Ml (1) EB Mz(1» . If Pi = Pex and 1 � n � t define V(Er, E, Ez : n ) = Mo(E)/(M (n) EB � (1» , where {i, j} = {1, 2}. Suppose that El -I E = Ez with E -1 0 such that Pz = Pex and El
Ez
E
T.
12]
THE INDECOMPOSABLE R [ G ] MODULES IN B
325
phic composition factors in common unless they are Li and Lj, and E and E I have a common vertex.
PROOF. Suppose that Ls -l Lt are both composition factors of V and V'. Then Cis, Cit, Cjs, Cjt are all not zero. If i, j, s, t are pairwise distinct then there exist m, n, v, w such that dmidms -I 0, dnidnt -I 0, dvjdvs -I 0 and dwjdwt -I O. This implies that contains the following subgraph. T
=
PI
E
where the notation is the obvious one. This contradicts the fact that is a tree. Suppose that {i, j, s, t} is a set of 3 distinct elements, say i, j, t with s = i. Then an argument similar to that in the previous paragraph implies that has the subgraph T
T
w
D
Pex
-- 0 --0 -
El
E2
is contained in If El = 0 the corresponding edge is missing. Define MI (1) as above. If L � E then L occurs with multiplicity t - 1 as a constitutent of Vz(E). If 1 � n � t - 1 let Mz(n ) be the submodule of Vz(E) with S ( Vz(E)/Mz(n» = L such that L occurs with multiplicity n as a constitu ent of Vz(E)/Mz(n ). Define V(Er, E, Ez : n ) = V(Ez, E, E I : n ) = Mo(E)/(Ml (1) EB Mz(n » . In all cases V = V(Er, E, Ez : n ) V(Ez, E, E I : n ) is indecomposable as T( V) is irreducible. If furthermore Ei = 0 then V is serial as Mo(E)/M (1) = V (E, Pj ) for i -I j and so is serial. T.
=
LEMMA 12.1 . Let V = V(E I , E, Ez : n ) and V' = V(E � , E I, E� : n ') be de fined as above where E -I 0, E ' -I 0, E I -I Ez and E � -I E �. Let Li � E and Lj � E '. Suppose that Li f:. Lj . Then V and V' cannot have two nonisomor-
n
v
contrary to the fact that is a tree. If {i, j} = {s, t} then clearly T
is a subgraph of T.
E
E'
� D
It will be necessary to consider the following two types of subgraphs of Let k be any integer with k ;?; 1 . (I)
T.
[ 12
CHAPTER VII
326
12]
Then
THE INDECOMPOSABLE R [ 0 ] MODULES IN B
X is represented by the direct sum of cones. L
oLi-1/o�Li +1
(II)
0
In type (I) Eo, . . . , Ek are distinct edges. In type (II) Eh+i and Eh+2s+1-i correspond to the same edge for i = 1, . . . , s. Possibly one or both of Po, Pk+l may equal Q. Given a graph of type (I) or (II) let .1 denote either the set of all even integers i with O ::':S i ::':S k or the set of all odd integers i with O ::':S i ::':S k. Given a graph of type (I) or (II) let Vi (ni ) = V(Ei - l, Ei, Ei+1 : ni ) where ni � 1 and E-l = Ek+1 = O. Let Li � Ei. Thus in case of type (II) Lh+j = Lh +2s+l-j for O ::':S j ::':S s. Observe that Li � Lj if i I- j but i == j (mod 2). The definition of Vi (ni) implies that S(Vo(no» = L 1 , S(Vdnk » = Lk-I and for o < i < k, S ( V; (ni » = Li -I EB Li + I . Choose .1 . Define X = EBi E .1 V; (ni). Then
m 2L2j-1 EB j=1 ( � 2L' - ) EB L'rn + ' S(X) = I Lo EB ( � 2L2j) EB L2m i
'
Lo EB C� 2 L ) 2i
If 2 Li
I SeX) there exist submodules
if 0 E .1, k
= 2m,
if 0 E ,1, k
=
if 0 � Ll, k
= 2m,
2m + 1 ,
=
if 0 � .1, k = 2m + 1. �
1,
interest in this section. The module V(Ei - J, Ei, Ei + l : represented by the cone
L,.,
ni )
with
Ei - I I- 0
Li
/�o L; _ '
and
Ei+ 1 1- 0
o oLi +1/ �Li+3 Li+2
0
Intuitively M is constructed from X by identifying isomorphic modules at the feet of adj acent cones and may be visualized as the following connected graph, where the dotted lines may not exist.
The next three results are concerned with showing that this picture is indeed accurate. Let XO = M and let ° denote the image of any element or submodule of X in XO = M. We will also write V; (ni t = V�)(ni )' Since V; (ni ) n Y = (0) for all i, V?(n ) = Vi (ni). If furthermore r is a subset of .1 such that whenever i E r then i + 2 � r then {EBiEr V; (ni )}O = EBi Er V�) (ni) ' LEMMA 12.2. (i) T(M) � EBiE.1 Li. (ii) SCM) = EBi E{O" ,. ,k l-.1 Li •
Lil C S( V;-I(n-I» and Li 2 C S(Vi+l(ni+I» with Li l Li2 Li . Let gi : Lil Li2 be an isomorphism. Let Y be the submodule of X consisting of all elements of the form Li (Vi gi (Vi » where i ranges over integers with 2Li I SeX) and Vi E Lil• Define M = X/ Y. In effect M is obtained from X by identifying the submodules Li Li 2 whenever 2Li I S (X). The module M is the primary object of =
327
may be
PROOF. (i) By definition YC
Therefore
SeX) C EB i E.1 S ( V; (ni » C EB i E.1 Rad( V; (ni »
C Rad X.
T(M) = reX) = EB i E.1 T(V; (ni» = EB i E.1 Li• (ii) Clearly S(X)o = EBi E{ , ... ,kl- .1 Li and S (X)O C S(XO) = S(M). O Thus it suffices to show that any irreducible submodule of M is in S(X)o . 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 ZO i.s irreducible . Let r be the set of all i such that ZO is isomorphic to a
[12
CHAPTER VII
328
composition factor of Vi (ni ). Z may be replaced by Z n {EBiEr V; ( ni )} ' Thus it may be assumed that Z C; EBiEr V; (ni ). 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 I (Z) 2 or 3. Let E be the edge corresponding Zo o If ZO is isomorphic to a constituent of S (Z) then (2.23) implies that E is incident with Pex and furthermore E is the only edge of T which is incident with Pex • Assume that j, j + 2 E r for some j. Then Lj + 1 is a composition factor of S ( V} (nj )) and S ( V} +2(nj+2)) . Thus ZO and Lj+1 are both isomorphic to composition factors of V} (nj ) and Vh+2 (nj +2 ). Suppose first that Ei and Ei +2 do not have a common vertex for any i, i + 2 E r. By (12. 1) Lj+ 1 = Z o o Thus r {j, j + 2} or the following configu ration occurs in T with s > 2 =
=
=
=
�
o
/ �
0 --- 0
o
E; +2S - 1
E; + 2S
0
,
where possibly Ej = Ej +2s+l • In this case r C; {j, j + 2, j + 2s}. Therefore in any case Y n EBiEr V; (ni ) C; Lj+ 1 = Z o o Since Y n Z I- (0) it follows that Y n Z = ZO and so S (Z) = ZO which is not the case since both vertices incident with Ej + 1 are incident with other edges in T. Suppose next that Ej and Ej +2 have a common vertex. By (12.1) ZO = Lj or Lj +l • Up to symmetry the following configuration must occur in 'T. E;
0
E; + I
Pex
0
E;+2
Y n EB
iEr
V; (ni ) C; Lj
+
I
If ZO = Lj + 1 then r = {j, j + 2 } . Let Yo Y n ( V} (I) EB V} + 2 (nj +2)). Then o Yo = Lj + 1 = Z o o Thus I (Z) � I (ZO) + I ( YO) 2 and so S (Z) = Z o There fore Yo, Z C; z, EB Z2 where ZI C; V} (1), Z2 C; V} +2 (nj +2 ). Both ZI and Z2 are indecomposable, all composition factors of ZI EB Z2 are isomorphic to ZO and I (ZI) 1, I (Z2 ) 2. Therefore Rad(Z) C; Rad(ZI EB Z2 ) = Rad(Z2) C; V} +2 (nj +2). Thus Yo n Rad(Z) = (0) and so Y n Z (0). Hence Z = ZO contrary to the reducibility of Z. If ZO = Lj then r C; {j, j + 2, j + 4} and if j + 4 E r then Lj + 3 = Lj other wise ZO = Lj is not a constituent of V} +4(nj + 4). Hence =
EB Lj +3 = Lj + EB ZO. I
Since Pex is not incident with Ej it follows that ZO ,{ S (Z). As S (Z) n Y I- (0) it follows that S (Z) = Lj + l • Thus Z n V} +4(nj + 4) = (0) and so Z is isomorphic to submodule Z' of V} (1) EB V} + lnj +2) with S (Z') n Y = Lj + l • Since Lj occurs with multiplicity 1 in V} +2 (nj +2) and Lj C; S ( V} +2 (nj +2)) it follows that Z' C; V} (1) EB ZJ, where Z, C; V} +2 (nj +2) and ZI = Lj + l • Thus Rad(Z') C; V} (1). As Rad(Z') n Y = S (Z') n YI- (0) this contradicts the fact that V} (1) n Y = (0). o Therefore if i E r then i + 2 g r. Consequently {EBiEr V; ( ni )} = EBi Er V?{nd · Hence Z = ZO contrary to the fact that Z is reducible. 0 LEMMA 12.3. Let i E ,1. (i) If Ei is not incident with Pex then Li occurs as a composition factor of M with multiplicity at most 2. If the multiplicity is 2 then Li I S (M). (ii) If Ei is incident with Pex then the multiplicity of Li as a composition
PROOF. If k � 2 then ,1 = {i} and M V; (ni ). The result is clear. Suppose that k > 2. (i) As Ei is not incident with Pex , Li occurs with multiplicity 1 in V; (ni ). If j I- i and Li is a constituent of V} (nj ) then the following subgraph of T must occur (up to symmetry) with P I- Pex, where Ej +2 may not be there. =
Ei
P
0 ----- 0 ------ 0 ------ 0
In any case this implies that Li I S (X) and so Li I S (M) and the multiplicity of Li " in S (M) is. l . (ii) As Ei i s incident with Pex , one of the following subgraphs o f T must occur (up to symmetry) Ei
=
=
329
factor in M is equal to the multiplicity of Li as a composition factor of V; (ni ). This is ni in case (I) and ni + 1 in case (II).
E; +2s + 1
0
THE INDECOMPOSABLE R [ O J MODULES IN B
12]
0 ------- 0 --- 0
=
=
Ei
Ei + 1
0 -- 0 --0
Pex
Pex
In the first case Li occurs with multiplicity ni + 1 in � (ni ) and multiplicity 1 in Vi + 2(ni+2) and in the socle of both. Thus Li occurs with multiplicity ni + 1 in M. In the second case L occurs with multiplicity ni in Vi (ni ) and does not occur in any V; (nj ) with j � i, j E .1 . 0 CHAPTER VII
330
[12
LEMMA 12.4. M is indecomposable.
Suppose that M = MI EB M2• For s = 1, 2 Iet /' be the projection of M onto Ms. For any i E .1 , V7(ni ) C V7(ni )fl EB V7(ni )f2 . We will first show that V7(ni ) V7(ni )fs for s = 1 or 2. Suppose this is not the case. Hence V7( n )fs � (0) for s = 1, 2. Let Ws = V7(ni )fs for s = 1, 2. Then 2Li I T( WI EB W2) and so Li occurs with multiplicity at least 2 as a constituent of M. Suppose that Ei is not incident with Pex • Then Li I S CM) by (12.3) and so Ws L for s = 1 or 2. Since Li g S ( V7(ni » it follows that V7(ni )fi V7(ni ) for s � t. Suppose that Ei is incident with Pex • Let n be the multiplicity of Li as a composition factor of M. By (12.3) (ii) the multiplicity of Li as a composi tion factor of Vi (nd is n - 1 . Let k be the smallest integer such that Li occurs with multiplicity n - 1 as a constituent of S k (Rad( � (ni » ): Since Rad( WI EB W2) = {Rad( Vi (ni » } fl EB {Rad( � (n » } f2 and this is a sum of at most 4 serial modules it follows that for s = 1 or 2 S k (Rad( Ws » � (0) and L; occurs with multiplicity n - 1 as a constituent of Rad( Ws ). Hence Li occurs with multiplicity n as a constituent of W,. Therefore L; does not occur as a constituent of Wt for s � t by (12.3) (ii) and so Wt = O. Hence W, � (ni ) for s = 1 or 2 in all cases. Suppose that i + 1 E .1 and Li I S (Ml ). Then L .{' S (M2) by (12.2) (ii) and so V7+ I (ni +l)h � V7+ I (ni + I ). Thus V7+ 1 (n-rl)fl V7+ 1 (ni +l ) and Li+2 1 S (MI). By changing notation if necessary it may be assumed that either 1 E .1 and L o I S (Ml ) or 0 E .1 and L I I S (MI). Hence by iterating the result of the previous paragraph we see that Li I S (M1) for all i E {1, . . . , k } - .1 . Conse quently (12.2) (ii) implies that S (M) C MI . Thus M2 = (0). 0 PROOF.
=
=
=
=
=
The definition of M shows that M depends on whether the graph is of type I or II, on the integers ni, the set .1 and the functions gj . It will later turn out that M is independent of the choice of functions gi .
LEMMA 12.5. M determines the type of graph and the ordered set Eo, . . . , Ek
12]
THE INDECOMPOSABLE R [ G J MODULES IN B
33 1
up to a reversal of order. Once the order is fixed, the set .1 is also determined by M except in case of a graph of type II with Po = Pk + = Q. Furthermore the integers nj for i = 1 , . . . , k are uniquely determined. I
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 Eo, . . . , Ek is determined up to a reversal by S (M) and T(M) since is a tree. By (12.2) .1 is determined by T(M). If the graph is of type II, (12.2) and (12.3) imply that {Eh+l, • • • , Eh+s} is determined by the common composition factors of S (M) and T(M). Thus Q is determined as Q � Pex • Eo, . . . , Ek are determined up to a reversal of order by S (M), T(M) and the fact that is a tree by (12.2). Thus {Po, Pk+I} is also determined. If Po � Q then .1 is determined by the condition that Lo I S (M) or L o I T(M) but not both. Similarly .1 is determined if Pk+1 � Q. If Po = Pk+1 = Q then the definition of M shows that .1 may be replaced by {1, . . . , k } - .1 . Since there is at · most one i E .1 with Pex incident to Ej it follows that nj = 1 for all j except possibly for j equal to such an i. In which case ni is determined by /(M) as is known. 0 T
T
T
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 .1 and the integers nj . In view of (12.4) and (12.5) there are at least 22n ('Y) pairwise nonisomorphic, nonpro jective, reducible, indecomposable, nonzero R [G] modules in B, where 'Y ranges over all subgraphs of 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 I D i e - 2e. Once this is done it follows from (2.1) and (2.2) that every nonzero indecomposable R [G] module in B is projective, irreducible or isomorphic to one of the modules M. Case (i): 'Y is of type I. Pj � Pex for 1 � i � k. Then n ('Y) = 2 depending on the choice of .1 . Case (ii): 'Y is of type I. Pj = Pex for some i with 1 � i � k. Then n ( 'Y ) = 2 t depending on the choice of .1 and ni . Case (iii): 'Y is of type II. Po � Pk+I. Then 1 � nj � t - 1. Thus n ('Y) = 2( t - 1) depending on the choice of .1 . Case (iv): 'Y is of type II. Po = Pk+l = Q. Then 1 � nj < n - 1 . Thus n ( 'Y ) = (t - 1) as M does not depend on the choice of .1 . T
332
[12
CHAPTER VII
LEMMA 12.6. If T has no exceptional vertex then Ln (y) = I D i e - 2e. PROOF. 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 G) ways of choosing such a pair of edges. Hence
2: n (y) = 2
(; ) = e 2 - e = I D i e - 2 e
since 1 D 1 - 1 = e. 0 PROOF. In view of (12.6) it may be assumed that T has an exceptional vertex. Let PI, P2, . . . be all the vertices in T such that Pi "I Pex and Pi, Pex bound a common edge. Consider the subgraphs Tt, T2 , . . . of T such that Ti consists of Pex and all vertices and edges with the property that a path from any vertex in Ti - {PeJ to Pex goes ghrough Pi . Since T is a tree, Tl, T2, . . . yields a partition of the set of edges in T. Let ei be the number of edges in Ti . Thus Li ei = e. In Case (i) {Po, Pk+I} is any set of vertices in Ti for any i such that Po and Pk+1 do not bound an edge. The number of possible pairs of vertices in Ti is . � ei (ei + 1). The number of pairs which bound an edge is the number of edges ei . Hence there are � ei (ei - 1) possible sets {Po, Pk+1} in Ti . Therefore the number of possibilities for y is � Li (e; - ei ). Thus if y ranges over all possible graphs in Case (i), Ly n (y) = Li (e; - ei ). In Case (ii) Po may be chosen in any Ti - {Pex} and Pk+1 in any Tj - {PeJ with i "l j. Thus the number of possibilities for y is Li<j eiej ' Hence if y ranges over all possible graphs in Case (ii), Ly n ( y ) 2 t Li <j eiej . In Case (iii) 'Y lies in Ti for some i. There are �ei (ei - 1) ways of choosing {Po, Pk+1} in Ti . Thus there are � Li ei ( ei - 1) ways of choosing y in T. Thus if y ranges over all possible graphs in Case (iii), Ly n ('Y ) = (t - l)L(e ; - ei ). In Case (iv) Q may be chosen arbitrarily in T - {Pex}. Thus there are e choices for Q. Hence if y ranges over all possible graphs in Case (iv), Ly n (y) (t - l ) e. Consequently if y ranges over all possible graphs of type I or II in T =
=
2: y n (y) = 2:i (e T - ei ) + 2 t 2: i<j eiej + (t - 1) 2:i (e i - ei ) + (t - l)e
(� eiY - t � ei + (t - l)e
333
By the remarks above, ( 12.7) completes the description of all the indecomposable R [G] modules in B up to isomorphism. PROOF OF (2.26). If V is an irreducible or a principal indecomposable R [ 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 iJ
THEOREM 12.7. L n (y) = I D I e - 2e.
=t
SCHUR INDICES OF IRREDUCIBLE CHARACfERS IN B
13 ]
= te 2 - e = e (et + 1) - 2e = 1 D i e - 2e. 0
Throughout this section K K. Thus B = iJ, Xu = Xu , cPi 'Pi, XA = X" for O :!S u :!S e, 1 :!S i :!S e, A E A. Let {3 be an irreducible Brauer character in Bk for some k with O :!S k :!S a. Let M = Qp ((3 ) and let S be the ring of integers in M. By (1. 19.3) and (2.9) S is a splitting field for any irreducible S[N ] module in Bi for =
=
O :!S i :!S a.
PROOF. Clear by (2.9) and (2. 1 9). 0 If I A 1 "1 1 then XA - XI-< vanishes on all p '-elements of G for A, JL E A . Thus the maximal unramified subfield of Qp (X,, ) is the same for all A E A. Denote it by F. By (2. 1 5) F c;;;, M. This yields LEMMA 1 3.2. Suppose that I A 1 "1 1 . Let F be the maximal unramified subfield of Qp (XA ) for some A E A . Then F is the maximal unramified subfield of Qp (XI-< ) for all JL E A. Furthermore F c;;;, M and [M(XA ) : Qp (XA )] = [M : F] for all A E A. Let m (X ) = mQp (X ) denote the Schur index of the irreducible character X over Qp . The object of this section is to prove THEOREM 13.3. m (Xu ) = 1 for 1 :!S u :!S e and one of the following holds. (i) 1 A 1 = 1 and m (Xo) = [M(Xo) : Qp (Xo)] . (ii) I A 1 "1 1 . For A E A, m (X" ) = [M : F ] . It is easily seen that (13. 1}-(13.3) imply
[13
THEOREM 13.4. If X is an irreducible character in B and
CHAPTER VII
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 (Xu ) = 1 for 1 :::::; u :::::; e. (ii) If I A 1 = 1 then m (Xo) :::::; [M(Xo) : Qp (Xo)]. (iii) If I A 1 1 1 and A E A then m (x J I [M(x d : Qp (XA )].
PROOF. By (2.13) dUi and dAi = 0 or 1 for all u, A, i. Thus the result follows from (IV.9.3) and (13.1). 0
LEMMA 13.6. If I D I = p the conclusion of (13.3) holds. In particular if / A 1 = 1 the conclusion of (13.3) holds.
PROOF. If I A / = 1 then / D I = p. Thus the second statement follows from the first. Suppose that / D 1 = p. Let X = XO if I A / = 1 and let X = XA for A a fixed element of A if I A / 1 1 . By (13.5) it suffices to show that m (X) � [M(X ) : Qp (X)]. By (lV.9.2) there exists an unramified extension Ko of Qp (X) such that [Ko : Qp (X)] = m (X) and there exists a Ko[ G] module which affords X. By (11 .2) (ii) M(X) C;: Ko and so [M(X) : Qp (X)] :::::; m (x). 0
Throughout the remainder of this section it will be assumed that I A / I I 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. (13.7) m (x) � [M(X) : Qp (X)] = [M : F] . LEMMA 13.8. If D <J G then (13.7) holds.
PROOF. Let V be a Zp-free Zp [G] module such that VQp 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 <J G ,(2.15) (iv) implies that n = m (X ) [Qp (X) : Qp ] e. Since [Qp (X) : F] = (p s - p S - I )/e for some s > 0 this implies that n = m (x)(p s - p S - I) [F : Qp ].
SCHUR INDICES OF IRREDUCIBLE CHARACTERS IN B
1 3]
335
Let {3 be the character afforded by the irreducible Qp [D] module of dimension p S - p s -l over Qp. Then VD has a pure submodule X which affords (3. By (11 .6) the minimum polynomial of y on X, and hence on V, has degree at least p S - p S - I . Since D <J G, D is in the kernel of every irreducible Zp [G] module. Thus if W is a submodule of V and Wo is a submodule of W with W/ Wo irreducible it follows that W(l - y ) � Woo Hence I ( V) � p S - p S - l . Consequently VK has at least (p s - p S - l ) [M : Qp] irreducible constituents. Therefore m (x ) (p S - p S- l ) [F : Qp] = n � (p s p S - l ) [M : Qp ] . This implies that m (X ) � [M : F]. 0 _
PROOF. The Galois group of F over Qp acts transitively on the set of all blocks which are algebraically conjugate to B. Since the Brauer correspon dence commutes with field automorphisms it follows that F is the maximal unramified subfield of Qp (f.L ) . Thus if E is the extension of F generated by all p a th roots of 1 , then (2.17) implies that Qp (f.L ) = E n Qp (f.L ) = E n Qp (XIL ) = Qp (XIL ). 0
Define an equivalence relation on A as follows. For f.L l , f.L2 E A let (i be an irreducible constituent of (f.Li )D for i = 1 , 2. Then f.Ll � f.L2 if and only if (f = (�. This is clearly an equivalence relation. If ( has order p S then the equivalence class containing f.LI consists of (p s - p S - l )/e distinct elemefits, and these are all algebraically conjugate. In particular the equivalence class containing f.LI consists of one element if and only if � f = 1 and e = p - 1 . I n the proof o f (13.7) two cases will b e handled separately. Case (i). If f.L E A with f.L � A then f.L = A. Case (ii). There exists f.L E A, f.L I A with f.L � A. Let OJ, . . . , Oe be the nohexceptional characters in Bo. The Galois group of M over F acts on the set { OJ . Let T( Oi ) = La 0 f, where ranges over this Galois group. I
(Y
PROOF OF (13.7) IN CASE (i). By (2.17) Qp (X) = F. Thus there exists an irreducible F[ G] module which affords m (X )X. Hence there exists an F[No] module which affords m (X )XNo• By (13.8) and (13.9) m (f.L ) = [M : F] for all f.L E A. Thus
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[ 14
(13. 10) where 7J is a sum of characters in blocks other than Bo. Let t/J be the irreducible Brauer character in a block bo of K[ Co] which is covered by Bo. Let d = d (A, y, t/J) be the corresponding higher decomposition number. Since the multiplicity of t/J as a Brauer constituent of T( OJ )cv is divisible by [ M : F] it follows from (13. 10) that [M : F] I m (X )d. As we are in Case (i), d = ± 1 by (2. 17) and so [M : F] I m (x). 0
15]
IRREDUCIBLE MODULES WITH A CYCLIC VERTEX
337
happens that all the vertices coming from Po 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 3 is in the center of G. Let L ' = Qn and let L" be the cubic unramified extension of Qn. A nonprincipal 1 3-block of L "[G] has no exceptional characters . The tree T ' of a faithful 1 3-block B' of L '[ G] is as follows . 0 42
f.L � A, f.L � A. By (2. 17) (X - XIL )No = ) ± (A - f.L . Hence « X - XIL )No, A )No = ± 1 . Therefore either (XNo, A )No or « XIL )No, A )No is relatively prime to m (A ). As X and XIL are algebraically conjugate it follows that m (X) = m (XA ). Thus by changing notation if necessary it may be assumed that (XNo' A )No is relatively prime to m (A ).' By (13.9) m (X )XNo is afforded by a Qp (A ) [No] module. Thus by (IV.9. 1 ) m (A ) I m (X ) (XNo, A )No• Therefore m (A ) I m (x). Hence [M : F] I m (x ) by (13.8) and (13.9). 0
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 1 3-block given in section 9.
14. The Brauer tree and field extensions
15. Irreducible modules with a cyclic vertex
PROOF OF (13.7) IN CASE (ii). Choose
Let K = K and let X = ,{\ be an exceptional character in 13 if I A I � 1 and let X = XO if I A I = 1 . Let M = Qp (Xu ) for 1 � u � e. Thus M is defined as in the previous section. Suppose that L ' and L " are fields with Qp (X) � L ' � L " � M(X ) � K.
Let B" be a block of L "[ G] which splits into algebraically conjugate blocks, one of which is 13 if L " is extended to K = K. Let B ' be a block of L '[G] which splits into algebraically conjugate blocks, one o f which is B" i f L ' i s extended t o L ". Let T ' , T" b e the Brauer tree o f B ', B" respectively. According to (2.9) and (2. 1 9) 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 b y m edges i n T " and each vertex other than Po i s replaced b y m vertices. The vertex Po remains fixed. According to (13.3) the vertex Po in T ' corresponds to a character with Schur index m (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
0
192
3
105 0 Po
0 42
This section contains a proof of the following result.
THEOREM 15. 1 . Let V be an irreducible R [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 > <J G and let V be an indecomposable R [G] module with vertex P. (i) There exists an integer k with 0 � k � I P I and p l' k such that Vp is a projective R [P]/R [P] (1 - x t module. (ii) V(1 - x) = {v l v E V, v (l - x t - 1 = O}. (iii) If h E Tr� p)(HomR [(xp)l ( V, V» then h ( V) � V(l - x ). (iv) V/ V(l - x ) is an R [G] module which is R [P]-projective and which has P in its kernel. Furthermore V/ V(l - x ) is a projective R [G /P] module. (v) Let .x be a nonempty collection of proper subgroups of P. If HO ( G, .x, V) R then V / V(l - x ) is irreducible. =
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[ 15
PROOF. (i), (ii). For 0 � k � I p i , let Vk denote an indecomposable R [P] module of R -dimension k. Ry (II.2.11) ( Vf)p = I G : p i Vk• Since V is R [P]-proj ective it follows that Vp = m Vk for some integer m > O. 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(l - x t - I ) = (0). Let w = v (l - X ) k - I E V(l - x t - I . By (i) wx = w. By (ii) k = np + s for 0 < s < p and some integer n ?!= O. By assumption h = Tr� p)(g) for some g E HomR (xP)( V, V). Since (1 - x Y = 1 - x P it follows that
g(w ) = g(v (l - x Y (l - x )"P ) = g (v (l - x Y ) (l - x P ) ". Since 1 + x + . . . + x p - I = (1 - X y - l and np + p - 1 ?!= k this implies that {Tr�: � ) (g )} (w ) = g (w ) (l + x + . . . + x P - I ) = g(v (l - x Y ) (l - x P ) " (1 - x y - I = O. Thus
h (w ) = (Tr(G {Tr x P ) (g)} ) ( w ) = 0 x)
(x ) (
as required. (iv) V(l - x ) = (v(l - z ) I v E V, z E P) is an R [ G ] module as P <J G. Clearly P is in the kernel of V / V (1 - x ). Let ( Vp ) G = EB w; with W1 = V. By (i) ( W'; )p = m (R [P]/R [P] (l - x t ) for all i and some integer m depending on i. Thus ( Vp) G ( 1 - x ) = EB W; (1 - x ). Observe that ( Vp) G (1 - x ) = ( V(1 - x )p ) G since they have the same R -dimension and ( Vp ) G (1 - x ) � ( V(l - x )p ) G . Consequently
V/ V(l - x ) = WI / WI (l - x ) I ( Vp) G /( Vp) G (1 - x ) = ( Vp) G /( V(l - x )p) G = « V/ V(1 - x ))p t. Since ( V/ V(l - x ))p = Invp ( V/ V( l - x )) the previous paragraph implies that « V/ V(l - x ))p ) G is a direct sum of copies of R [ G / P ] R [ G / P ) . Thus V/ V(1 - x ) is a proj ective R [G / P] module. (v) Since every proper subgroup of P is contained in (x P ) the hypothesis implies that HO ( G, ( x P ) , V) = R. As EndR [ G ) ( V) is a local ring and Tr� p)(HomR(xp)( V, V)) is an ideal of EndR [ G )( V) by (II.3.7), it follows that if h E HomR [ G ) ( V, V) but h is not an isomorphism then h E Tr� p)(HomR[xp] ( V, V)). Thus by (iii) h ( V) � V(1 - x ). Suppose that the result is false. By (iv) V/ V(l - x ) is a projective R [G/P] module . Thus by (1.16.8) there exists an irreducible submodule M � V/ V(1 - x ) and an epimorphism f : V/ V(I - x ) � M. Let t be the natural proj ection of V onto V / V(l - x ). Thus the row in the following diagram is exact.
1 5]
IRREDUCIBLE MODULES WITH A CYCLIC VERTEX
339
V
If
v � V/ V(1 - x ) � O. Since Vp and ( V/ V(l - x ))p are R [P]/R [P] (l - x t modules and Vp is projective by (i) there exists an R [P]-homomorphism Ii such that the following diagram is commutative.
As V is R [P]-projective there exists an R [ G ]-homomorphism h such that the following diagram is commutative.
V
1' / v � V/ V(l - x ) � O. Since the result is assumed to be false f = t h is not an epimorphism. Thus h is not an epimorphism and so h ( V) � V (1 - x ). Hence f = t h = 0 contrary to assumption. 0 0
0
PROOF OF (15. 1). By (III.4. 14) it may be assumed that R is a splitting field of G and of all its subgroups. Let N = N G (P). Since P is cylic there exists G with CG (P) � G <J N such that p {' I N : G I and P centralizes all p '-elements in G. If P is the defect group of any block of G then P � P and so CN CP) � G. Thus by (V.3.10) every block of N is regular with respect to G. Let f be the Green correspondenc e with respect to ( G, P, N). Apply (15.2) to N with f( V) in place of V. By (111.5. 10) the hypothesis of (v) is satisfied. By (III.7.8) it suffices to show that if B is the block of N which contains f( V) then P is a defect group of B. Let W = f( V)/f( V) (l - x ). Then W is in B. By (15.2) W is in a block of R [G / P] of defect O. Since p {' I N : G I Clifford's theorem (111.2. 12), implies that Wo is a direct sum of irreducible modules in blocks of defect 0 of R [G/P]. By (V.4.5) each of these is in a block of R [G] with defect group P. Hence by (V.3. 14) B has defect group P. 0
TENSOR PRODUCTS OF R [NJ MODULES
By (VII.2.S) the composition factors of V('P, 1) in ascending order afford . the Brauer characters 'Pa - 1 , . . . , 'Pa . . Th e next result IS. an Immediate consequence of this fact.
2)
m 'T"
CHAPTER VIII
s of Let p be a prime . The object of this chapter is to apply the result s result the of Some p. order of Chapter VII to study groups with a Sp-group cyclic a has G that proved in this chapter can be generalized to the case ver this Sp-group and satisfies some other subsidiary conditions. Howe more general case will not be considered here. used The notation introduced at the beginning of Chapter VII will be ing follow the ion addit throughout sections 1, 2 and 3 of this chapter. In assumptions and notation will be used in these sections. K = K. P is a Sp-group of G. P = (y ). I P 1 = p. C = Co (P). N = No (P). H. By Burnside's transfer theorem C = P x H for some p i_group of D in If B is a p -block of positi ve defect of G then P plays the rolethan that Chapter VII. However the situation is a good deal simplerNa-I = G. If described in Chapter VII since C = Co = Ca-1 and N = No = le which R [N] modu M is a nonp rojec tive R [G] module let AI be the corresponds to M in the Green correspondence. e deno tes the inertial index of the principal p -bloc k of G. By (VII.2 .3) I N : C I = e. nonzero The group N has no p-blo ck of defect O. Let V be a 2.6) that (VII. and indecomposable R [N] modu le. It follows from (VII. 2.4) and V) ( S V is serial , I ( V) � P and V is determined up to isomorphism by let V ('P, 1) I ( V). If 'P is an irreducible Brauer character of N and 1 � I � p I» = l. In be an R [G] modu le such that S ( V('P, 1» affords 'P and I ( V('P,s 'P. Defin e particular V('P, 1) is an irreducible R [G] modu le which afford V('P, 0) = (0).
(I - I )
LEMMA 1 . 1. Let 'P be an irreducible Brauer character of N and let 1 � I � P. Then the following hold. (i) V('P, 1) * = V('P * a 1-\ I). (ii) For x in N let detv(
1 . Groups with a Sylow group o f prime order
340
-
34 1
2. Tensor products of
R [N]
modules
The purpose of this section is to give explicit formulas for the structure of the tensor product of two indecomposable R [N] modules. These results can be found in Feit [1966] except for (2.9) which is due to Blau [1971a]. In case N = C the re �ults o � this section are implicit in Green [1962b]. T�e results of thIS sectIOn have been generalized to the case that P is cyclIc but not necessarily of prime order. This work was begun in Green [1?62b] and continued in Sri�ivasan [1964b] and RaIley [1969]. Finally Lmdsey [1974] found an algOrIthm which gives a complete description of t�nsor products of modules in this more general case. As can be seen from Lmds�y's work, the situation in the general case is a good deal more complIcated and will not be treated here. . The principal Brauer character will be denoted by tv or simply by 1 . LEMMA 2 . 1 . Let 'P be a n irreducible Braue; character of N and let 1 � I � p. Then V('P, I) = V('P, 1) 0 V(I, I).
Let U(l) = V('P, 1) 0 V(l, I ) for 1 � I � p. By (VII.2.6), V('P, p) are principal indecomposable R [N] modules. By (VII.2.4) dimR V('P, p ) = P'P (l) = dimR U(P) �nd I ( U(p» = p. � y (II.2.7) U(P) is projective. Thus U(p) is a principal mdecomposable R [G] module. As S ( U(p» affords 'P it follows that U(p) = V('P, p ). Let 1 � I � p. Then U(l) � U(P) and so S ( U(l» affords 'P. As I ( U(I» = . I It follows that U(l) = V('P, I). 0 and
V(I, p)
PROOF.
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342
If 'Ph 'P 2 are irreducible Brauer characters of � then P is in the kernel �f 'P t and 'P 2 . Thus V('Pl, 1) 0 V('P 2 , 1) is an R [NIP] modul e and S? IS med comple tely reducible . Hence V('Pl, 1) 0 V('P 2 , 1) is comple tely determ then , known is NIP by the Brauer charac ter 'P l 'P 2 . If the charac ter table of know� by (2.1) the structu re of tensor produc ts of R [N] modul es will be of thIS once modul es of the form V(l, s ) 0 V(l, t) are describ ed. The rest in the is I) V(l, that e Observ n. questio section is concer ned with the latter Thus I). (1, � of kernel the in princip al p -block for 1 � I � p and so H is ered. consId be only modul es of the Froben ius group NIH need to LEMMA 2.2. Let 1 � s � p. Then
s-l
V(l, s ) 0 V(l, p ) = EB V(a - \ p). ; =0
PROOF. As V(l, p ) is proj ective the result follows from (III.2.7).
0
The method of proof for the next result is essentially due to Green
[1962b] .
LEMMA 2.3. Suppose that
EB�= o V(A;, I; ). Then
1 � m, n � p - 1 and V(l, m ) 0 V(l, n ) =
k
EB V(A;a m - \ p ) EB ( V(l , p - m ) 0 V(l, n » ; =0
= EB V (a -j, p) EB E9 V (A;a m -Ii , p - Ii ) . n-l
k
j =O
1 =0
PROOF. There exists an exact sequence
o � V(a P - m, p - m ) � V(a P - m, p) � V(l, m ) � O. Tensor ing with V(l, n ) yields that O � V(a P - m , p - m ) 0 V(l, n ) � V(a P - m, p ) 0 V(l, n ) � V(l, m ) 0 V(l, n ) � O - -j is exact. Also V(a P - m , p) 0 V(l, n ) = EB;.:-� V(aP m , p) by (2 .2) and is exact. Thus by Schanuel's Lemma (1.4.3)
TENSOR PRODUcrS OF R [N] MODULES
2]
343
k
V(Aia P - li, p ) E9 ( V(a P - m, p - m ) 0 V(l, n » EB i =O
n-1
k
= EB V(aP - m -j, p) E9 EB V(Aia P - li, p - Ii ). i =O j =O
The result follows from tensoring this equation with V(a m - p, 1).
0
LEMMA 2.4. Suppose that 2 � t � P - 2. Then
V(l, 2) 0 V(l, t) = V(l, t + 1) E9 V(a - t , t - 1). PROOF. Without loss of generality it may be assumed that H = (I) and N is a Frobenius group. Thus it suffices to prove the result in case N is a Frobenius group of order p (p - 1) since the restriction of any indecompos able R [N] module to any subgroup of N which contains P is indecomposa ble. Thus in particular, a, a 2 , . . . , a P - 1 = 1 are p - 1 pairwise distinct irreducible Brauer characters of N. Let V = V(l, 2) 0 V(1, t). There is an exact sequence
O � V(l, t) = V(1, 1) 0 V(1, t) � V � V(a - 1, 1) 0 V(l, t) = V(a -" t) � 0. This implies that dimR S ( V ) � 2. Thus V = V(A, m ) EB V(fL, n ) for some A, fL, m, n with 0 � m � n � p. The Brauer character afforded by V is 1 + 2 L;:� a -; + a - I. Hence t - 1 � m and so n � t + 1 as m + n = 2t. This is easily seen to imply that either the result holds or V = V(l, t) E9 V(a -I, t). Thus it suffices to show that V(y - l Y � (0). Let x be an element of order p - 1 in N. Choose R -bases {Vo, VI}, {WO , . . . , WI - I } of V(1, 2), V(l, t) respectively such that ViX = a - i (x )v;, WiX == a -; (x)wi and ViY = Vi + Vi - I, WiY = Wi + Wi - I , where V - I = W - I = O. Therefore (v 0 0 w; ) (y - 1) = v0 0 Wi- I . Furthermore
(V I 0 WI - t) (Y - 1) = Vo 0 WI - l + Vo 0 WI - 2 + V I 0 WI -2 . (2.5) We will prove by induction on t that (V1 0 WI - 1 ) (Y - 1 Y = t ( Vo 0 w o).
This will complete the proof of the result. Suppose that t = 2. Then (2.5) implies that
(v J 0 WI) ( Y - 1)2 = (vo 0 WI) (Y - 1) + (vo 0 wo) (y - 1) + (v 1 0 w o) (y - 1) = (v0 0 wo) + O + (vo 0 wo) = 2(vo 0 wo).
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344
(vd:?9 Wt - I ) (Y - 1Y = (vo 0 Wt- I ) (Y - ly - 1 + (vo 0 Wt -2) (Y - lY - I (2.6) + (V I 0 Wt -2) (Y - lY - 1 = (Vo 0 w o) + 0 + (V 0 Wt -2) (Y - ly- l . I Since V I 0 Wt-2 E V(l, 2) 0 V(l, t - 1) � V it follows by induction that (V I 0 Wt -2) (Y - l Y - I = (t - l) (vo 0 w o). Thus the result follows from (2.6). 0 THEOREM 2.7. Suppose that 1 � s � t � P and s + t � p. Then the following hold. s-I (i) V(1, s ) 0 V(1, t) = EB V(a - \ s + t - 1 - 2i ). s -I (ii) V (1, s) 0 V (1, P - t) = EB V (a -e, p - t + s - 1 - 2i ). i=O (iii) V(l, p - s) 0 V(l, t) s- I = E9 V(a - I + 1 +i, p - s - t + 1 + 2i ) EB E9 V(a -i, p). i =O i =O (iv) V(l, p - s ) 0 V(l, p - t) p-I s- I + \ t - s + 1 + 2i) EB E9 l V(a i, p). V(a EB i =O
t -s - I
; =0
TENSOR PRODUCTS OF R [ N] MODULES
.
J =s+t
PROOF. (i) This is proved by induction on s. If s = 1 the result is trivial. If s = 2 the result follows from (2.4). Suppose that s � 3 . By (2.4)
V(1, s - 1) 0 V(l, 2) 0 V(l, t) = ( V(a - I , s - 2) 0 V(l, t» EB ( V(l, s) 0 V(l, t» . Thus by induction
V(l, s + t - 1) Ef) 2
(ffl V(a -\ s + t - 1 - 2i) ) Ef) V(a -(' -", t - s + 1).
The unique decomposition property now implies the result. (ii) Since s � p - t it follows that s + (p - t) � t + (p - t) � p. Thus this follows from (i) when t is replaced by p - t. (iii) By (2.3) with m = s, n = t, k = s - 1, A i = a - i� Ii = S + t - 1 - 2i and (i) it follows that
s- I
EB V(a i - I + I , p) EB ( V(l, p - s ) 0 V(l, t» i =O
I-I
s-I
J =O
i =O
= $ V(a -i, p) EB EB V(a i - t + l, p - s + t + 1 + 2i).
-
Since E9;:� V(a i - I +\ p ) = E9; = : s V(a -i, p), the result follows from the unique decom positio n proper ty. (iv) Observ e that a p - l = 1 and so
$ J
p -s - t - I
-O
.
p -s -t - I p_ V(a -"i, p) EB E9 V(a P- 1 -i, p) = E91 V(a i, p).
E9 ( V(a -\ s + t - 2 - 2i) 0 V(l, 2» ; =0
i =s+1
Thus the result follows from (iii) upon replaci ng t by p - t.
0
COROLLARY 2.8. Suppose that 1 � s � ! (p - 1). Then s- I (i) V(1, s ) 0 V(l, s)* = E9 V(a i , 2i + 1). i =O
s-l p - I -s V(l, p - s ) 0 V(l, p - s)* = EB V(a i, 2i + l) EB E9 V(a i, p). i=s PROOF. (i) by ( 1 . 1) V(1, s)* V( a S - 1 , s ) = V(a S - \ 1) 0 V(l, s). Thus by (2.7) (i) (ii)
i =O
=
V(l, s) 0 V(l, s)* = V(a ' - ', 1) 0 =
; =0
= E9 V(a - ; - I , s + t - 3 - 2i) EB ( V(1, s) 0 V(l, t» .
.
J =O
s- 2
s -J
345
Application of (2.4) yields tbat the left hand side is isomorphic to
Suppose that t > 2. Then (2.5) yields that
=
2]
This implies the result. (ii) By (1.1) and (2.7) (iv)
s- 1
{ � V(a ', 2s - 1 - 2i ) }
E9 V(a S - 1-\ 2 s - 1 - 2i). i =O
346
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v(1, p -
s) (2) V( 1, p - s ) *
= V(a ' - ' - ' , 1) ® =
3]
s- I
{ � V(a'H, 2 i + l) EB � V(a i, p) } p-I
EB V(a P - I +\ 2i + 1) EB EB V(a P - 1 -s +j, p). i=O
The result follows as a p- I = 1 . 0
j= 2s
The next result is due to Blau [197 1 a] . The proof is due to Lindsey [1974] .
THEOREM 2.9. Suppose that p -1 2. Let A be a linear character of N (i.e. A (1) 1). Let V(A, d) (2) V(A, d) A + EB A - where A + is the space of symmetric tensors and A - is the space of skew tensors. (i) If d s � Hp - 1) then =
=
GROUPS OF TYPE Lc(p )
8+
d-I
=
8 - + 2: a +2i• i=O
347
(2. 10)
Let n (a j ) ( - lY and extend n by linearity to the space of Brauer characters of NIP. If V is an R [N] module let n ( V) = n (8) where 8 is the Brauer character afforded by V. The definition and (VII.2.8 ) imply that =
if s is even, =
( - 1Y
if s is odd.
By ( 2.7) A + EB A - is the direct sum of d nonzero indecomposable modules . Furthermore each of these is odd dimensional over R . By (2. 10) n (A + ) - n (A - ) = d. This yields that A +, A - is the direct sum of all the indecomposable components of V of V(A, d ) (2) V(A, d ) with n ( V) = 1 , - 1 respectively . Now (2 .7) yields the result. 0
=
3.
A + = EB V(A 2 a I -S+\ 2i + 1), O:::;; i(mod :S;s-l ) i",s 2 i AEB l V(A 2 a I- s + , 2i + 1). O::s;i�s i""s(mod- 2) (ii) If d p - s with s � Hp - 1) then =
=
V(A 2 a S + i, p ). EB V(A 2 a s +\ 2i + 1) EB s�i�p A + = O�i�s EB l s ""s(mod- 2-) i ""s(mod- 2l ) i
Groups of type
L2(P)
A group G is of type L2(P ) if every composition factor of G is either a p -group, a p '-group � r is isomorphic to PSL2 (p ) . If P = 2 or 3 then PSL2 (p ) is solvable and so in these cases a group of type L2(P) is p -solvable. It should be emphasized that the hypotheses introduced in section 1 are assumed throughout this section. In particular it is implicitly assumed that a Sp -group of G has order p. The object of this s ection is to prove the following results.
V(A 2 a S + i, p ). EB l V(A 2 a s + i, 2i +- 1) EB s:::;; iEB A - = O�i�s l-s) �p i"'s(mod 2 i",s(mod 2)
THEOREM 3 . 1 . Suppose that G is not of type L 2 (P). Let L be a faithful R [ G ] module and let d be the degree of the minimum polynomial of y o n L . Then d � H2p - 2 ) .
PROOF. By (2. 1 ) it may be assumed that A 1 . Thus it may be assumed that H <1) and N is a Frobenius group. Hence it suffices to prove the result in case N is a Frobenius group of order p (p - 1) since the restriction of any
THEOREM 3.2. Suppose that G is not of type L 2 (p). Let L be a faithful R [ G ] module such that dimRL � p. Then Cc (P) P x Z, where Z is the center of G.
=
=
R [N] module to any subgroup of N which contains P is indecomposable. Thus in particular a, a 2 , , a P -1 1 are p - 1 pairwise distinct irreducible Brauer characters of N. Let x be an element of order p - 1 in N. Let {vo, . . . , Vd - l} be an R -basis of V( l , d) such that ViX a - i (x )Vi for 0 � i � d - 1 . Thus {Vi (2) Vj + Vj (2) Vi I 0 � i � j � d - 1 } is an R -basis of A + and {Vi (2) Vj Vj (2) Vi r 0 � i < j � d - 1 } is an R -basis of A - . Let 8+, 8 - be the Brauer character afforded by A +, A - respectively. Then it follows that •
•
•
=
=
THEOREM 3.3. Suppose that G is not of type L2(P ). Let L be a faithful indecomposable R [ G ] module and let d = dimR L. Assume that p � 1 3 . Then d � � (p - 1).
=
The proof of (3.2) and the proof of a weaker form of (3 . 1) can be found in Feit [1966] . The proof of (3.3) is due to Blau [1971a] . This last result strengthens a result in Feit [ 1 966] .
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[3
In case 3 � P � 1 1, the inequality in (3.1) is the best possible. A covering group of As, A6, A7 respectively has a faithful representation in any algebraically closed field of degree 2, 3, 4 respectively. This shows that for p = 3, 5, 7 the inequality in (3. 1) cannot be improved. The Janko group II has a faithful representation of dimension 7 over the field of 1 1 elements. Thus for p = 11 the inequality in (3. 1) cannot be improved. Suppose that p 3 13. Any transitive permutation group on p letters has a faithf� l in�ecomposable R [G] module of degree p - 2, namely Rad UI / S ( UI ), where UI is the principal indecomposable R [ G] module corresponding to the principal irreducible module. There seem to be no known examples for p 3 13 with d < P - 2 in (3. 1) or (3.3). Thus for p 3 13, the inequalities in (3. 1) and (3.3) are far from the best possible. Various improvements of the inequality in (3.3) under special conditions can be found in Blau [1974b], [1974c], [1975a] , [1975d], [1976], [1980]. For a related result see Blau [1975c]. LEMMA 3.4. Statement (3.2) follows from (3. 1). PROOF. Suppose that (3. 1) has been proved. If L is projective then Lc is projective and so Lc is indecomposable as dimR L � p. If L is not projective then MN = M and the minimum polynomial of y on L has degree d < p. Thus in any case (2.1) implies that Lc = V @ ( L�=I Yi ) where V is an indecomposable R [Pl module and YJ , . . . , Ym are irreduc ible R [H] modules which are conjugate in N. If m 3 2 then dimR V � �p and so by (3.1) G is of type L 2 (p) contrary to assumption. Thus rri = 1. Sin�e dimR V > �p it follows that dimR YI = 1. Hence H is represented by scalars on M and so H is in the center of G. 0
The next result is a preliminary lemma needed for the proof of (3. 1). LEMMA 3.5 (Brauer [1942bJ). Suppose that p 3 3. Let M be a faithful indecomposable R [G] module in the principal p-block such that M = M* and dimR M = 3. Then G is of type L2 (p ). PROOF. Without loss of generality it may be assumed that G = G'. Since
M = M*, dimR S (M) = dimR T(M). If M is reducible this implies that the irreducible constituents of M are all of dimension 1, contrary to G = G'. Thus M is irreducible. By (VII. 10.5) it may be assumed that M is an F[ G] module where F is the field of p elements. Let A denote the representation of G with underlying module M. Thus
3J
GROUPS OF TYPE L2 (P )
349
for x E G, A (x ) is a linear transformation on M. Since M = M* there exists a nonsingular linear transformation Q on M with Q - 1 A (x )Q = A '(x - I ) , where ' denotes transpose. Taking the transpose and replacing x by X - I implies that Q'A '(x - 1 )(Q - I )' = A (x ) and so A (x ) = Q'Q - I A (x )Q ( Q - I),. Thus by Schur ' s Lemma Q ' = cQ for some c E F. Since Q " = Q it follows that c 2 = 1 . As det Q = det Q ' it follows that c 3 = 1 . Thus c = 1 and Q = Q'. Since A (x ) QA '(x) = Q for all x E G this implies that G is isomorphic to a subgroup of 03(P ). The result follows as O�(p ) = P SL2 (p ) . 0 The remainder of this section is concerned with the proofs of (3. 1) and (3.3). Suppose that either of these results is false. Choose a counterexample G of minimum order; given G, choose d as small as possible. Without loss of generality it may be assumed that L is indecomposable. Since d < p, LN has no nonzero projective direct summand. Thus LN = L. By (2. 1) and (VII. I0.l) there exists an irreducible character
where
V(
Y1 �
V(I, 1) and for i > 1 ,
Y;
is in a nonprincipal block of N.
PROOF OF (3. 1). The result is trivial for p � 3. Thus it may be assumed that p 3 5.
, Suppose tirstthat d � ! (p - 1). By (2.8) LN @ L :' has no projective direct summands. Thus if M / LN @ L :' there exists an R [G] module M with MN = M. By (2.8) Yea, 3) = Yea, 3) @ YI / LN @ L :'. Let M be an R [G] module with MN = Yea, 3). Hence dimRM = 3 and M* = M as Yea, 3)* = V(a, 3). As P is not in the kernel of M, (3.5) yields a contradiction. Therefore d = p - s with s � � (p - 1). Since d < � (p - 1) it follows that 3 s - 2 > p. By (2.8) LN @ L :' has (p - 2s)m projective direct summands and for 0 � i � s - 1, LN @ L :' has m direct summands M with [(M) = 2i + 1 . Let n be the number of direct summands M of LN @ L f,; with 1 < [(M) < d. As 3s - 2 > p it follows that p - s < 2(s - 1) + 1 . Thus n = � (p - s - 3)m if s is even and n = � (p - s - 2)m if s is odd. Hence in any case n 3 � (p - S - 3)m.
350
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4]
A CHARACTERIZATION OF SOME GROUPS
35 1
Let !VI be a direct summand of. LN ® L 'tv with 1 < I (M) < d. Since 1 < I (M), P is not in the kernel of M. Let M be the R [G] module corresponding to M. The minimality of d implies that MN f:; M. Hence MN has a nonzero projective summand. This implies that LN (29 L 'tv has at least n nonzero projective summands. Consequently (p - 2s )m � ! (p - s - 3)m. Therefore p � 3s - 3. As 3 ,f p, P � 3s - 2. D
that 1 + 2(s - 3) � P - 2s and so 4s - 5 � p � 3s - 1. Thus s � 4 and so p � 3s - 1 < 13 contrary to assumption. Therefore 3s - 1 < p. By (3.7) and (3.8) mj � 1 for 1 � i � s - 1 and mj � 2 for at least s - l values of i with 1 � i � s - 1 . Hence (3.6) implies that 1 + 2(s - 2) � P - 2s. Thus 4s - 3 � P and so 4d � 3p - 3 as required. D
PROOF OF (3.3). In view of (3. 1) 'P (1) = 1. Thus m = 1 and d = do. Suppose that e � i (p - 1). By (3.1) d > e and so by (VU.2.7) d � p - e > � (p - 1) contrary to assumption. Therefore e � �(p - 1) and d = p - s with s � e. By (2.8)
4. A characterization of some groups
s -- I
p - I -s
LN (29 L :' = E9 V(a \ 2i + 1) EB E9 V(a \ p). i =O i =s Since d < � (p - 1) it follows that s > i (p + 3). D
with 1 � i � s - 1. Then
PROOF. Since P is not in the kernel of Mi, the minimality of d implies that d � 2i + 1 in case mj = O. If i rf s - 1 then d � 2(s - 2) + 1 = 2p - 2d - 3. Thus 3d � 2p - 3 < 2p - 2 contrary to (3. 1). If i = s - 1 then p - s = d � 2(s - 1) + 1 and so p � 3s - 1 . D LEMMA 3.8. There is at most one value of i with mi
=
THEOREM 4.1 . Suppose that G = G ' and a Sp -group P of G has order p. Let 1 + np be the number of Sp -groups in G. Let H = Op ( G ). Then one of the following holds. (i) p > 3, GIH = PSL2 (p). (ii) P = 2a + 1 > 3, GIH = SL2 (p - 1). (iii) There exist positive integers h, u such that n -_ hup +uu2++1 u + h This result was first proved by Brauer [1943] under the additional assumption that CG (P) = P. The proof of (4.1) given here is quite similar to that given in the above mentioned papers except that the results of section 3 are used to simplify a portion of the argument. The proof of (4. 1) requires the following result of Zassenhaus [1936] which will here be stated without proof. ,
For 1 � i � s - 1 let M; be an indecomposable R [ G] module with Mi = V ( a i, 2i + 1). Let dimR Mi = 2i + 1 + mip. We need two subsidiary results. LEMMA 3.7. Suppose that mi = 0 for some i = s - 1 and 3s - 1 � p.
The purpose of this section is to provide a proof of the following result due to Brauer and Reynolds [1958] .
1.
PROOF. Suppose that mi = 1. Since M; !Vi i t follows that M; = M; . Thus (M; )N = Ai EB V(a i, p) for some j with V(a j, p) = V(a j, p)*. Hence a 2j = 1 by (1. 1). Since by (3.6) s � j � p - s - 1 and s > i (p + 3) it follows that j = Hp - 1) in case e = p - 1 or Hp - 1). Thus the result is proved unless e = � (p - 1). If e = Hp - 1) then j = Hp - 1), Hp - 1) or Hp - 1). Since G = G', (1 . 1 ) (ii) implies that a jPa - p (p - I )/2 = 1 . Thus a i-(p - I )/2 = 1 and so j !(p - 1) (mod e ). As e = �(p - 1) it follows that j rf Mp - 1), � (p - 1). Thus j = Hp - 1) and the result is proved also in this case. D �
==
Suppose that 3s - 1 � p. By (3.7) and (3.8) mj � 1 for 1 � i � s - 2 and mi � 2 for at least s - 3 values of i with 1 � i � s - 2. Hence (3.6) Implies
THEOREM 4.2. Ld G be a triply transitive permutation group on an odd number of letters m. Assume that only the identity fixes at least 3 letters. Then m = 2 a + 1 for some a � 1 and G = SL2 (2a ). Define the function 2 F(p, u, h ) = hup + U + u + h . u+1 We first prove some arithmetical lemmas. LEMMA 4.3. Let n � 1 be an integer. Suppose that u is a nonnegative integer such that (l + up) I (p - 1)(1 + np). Then there exists an integer h � 0 such that n = F(p, u, h ) and (u + 1) I h (p - 1).
352
CHAPTER VIn
[4
PROOF. Let 1 + up = m , m 2 with m , I (p - 1) and m 2 1 (1 + np). Then p 1, up + 1 0 (mod m , ) . Hence u + 1 == 0 (mod m , ) . Also up + 1 == 0, 1 + np == 0 This implies that (n - u )p 0 (mod m 2) and so n - u 0 (mod m 2) . Thus up + 1 = m , m 2 divides (u + 1)(n - u ). Define the integer h by (u + 1)(n - u ) = h (up + 1). Hence h (up + 1) 0 (mod (u + 1)). Since u - 1 (mod (u + 1)) this implies that (u + 1) 1 h (p - 1). Furthermore n = F(p, u, h ). Suppose that h = - h ' < O. Then (u + 1) (u - n ) = h ' (up + l» O and so u 2 = U (h P + n - 1) + h ' + n > uh p. Thus u > h 'p. This contradicts the fact that (u + 1) 1 h '(p - 1). Hence h � O. 0 ==
==
==
==
==
==
I
I
LEMMA 4.4. Let n � 1 be an integer. Suppose that v is a positive integer such that (vp - 1) I (p - 1) (1 + np ). Then there exists an integer h � 0 such that u = (n - h )/v � O, n = F(p, u, h ) and (u + 1) l h (p - l). If u = 0 then v = pn - n + 1 . PROOF. Let vp - 1 = m , m2 with m , 1 (p - 1) and m 2 1 (1 + np ). Then p 1, vp - 1 == 0 (mod m ) Hence v - I == 0 (mod m I). Also vp - 1 == 0, np + 1 == 0 This implies that (n + v )p 0 (mod m2) and so n + v == 0 (mod m2) . Thus vp - 1 = m , m 2 divides (v - 1) (n + v ). Define the integer h by (4.5) (v - 1)( n + v) = h (vp - 1). Thus h � O. For fixed n, p, h let f(X) = (X - 1) (X + n) - h (Xp - 1). Thus f(v ) = O. Hence f(v ') = O where v ' = (h - n )/v and so u = - v ' is an integer. Since f(l) < 0 and v � 1 it follows that v ' < 1 and so v ' � 0 as v ' is an integer. Therefore u � O. As f( - u) = 0 it follows that (u + l) (n - u) = h (up + 1) and so n = F(p, u, h ). Furthermore h (p - l) == - h (up + l) == O (mod (u + l)). If u = 0 then n = h and (4.5) implies that v = pn - n + 1 . 0 ==
I
==
.
4]
A CHARACTERIZATION OF SOME GROUPS
353
The proof of (4. 1) will now be given. Suppose the result is false. Let G be a counter example of minimum orde!. We will show in a series of lemmas that the assumed existence of G leads to a contradiction. The notation introduced at the beginning of section 1 will be used for the remainder of this section with the following modifications. t = (p - 1)/e. B is the principal p-block of G. If t l- 1 , tl , . . . , tt are the exceptional characters in B. Xo, . . . , Xe and 0o, . . . , Oe are defined as in (VII.2.12) for the principal p -block of G where XI is the principal character of G. If t = 1 let t, = Xo. LEMMA 4.6. (i) If OJ = 1 for Xj an irreducible character then Xj (l) = 1 or 1 + np. (ii) If OJ = - 1 for Xj an irreducible character then Xj (l) = P - 1 or p (pn - n + 1) - 1. (iii) If tl- l and 00 = 1 then tti (1) 1 + np for 1 � i � t. (iv) If t I- 1 and 00 = - 1 then tti (1) = p - 1 or p (pn - n + 1) - 1 . =
PROOF. Let X be an irreducible character of G. Then 1 G c (y ) I X y ) ( X (1) I C is an algebraic integer. Since I G Cc (y ) 1 = e (1 + np ) = (p - 1)(1 + np )/t this implies that (1 + np)(p - l)X (Y ) tX (l) is an algebraic integer. Suppose that X = Xj is a nonexceptional character in the principal p-block. By (VII.2. 17) Xj (Y ) = OJ and Xj (l) = OJ + mp for some integer m. Thus x (1) I (p - 1) (1 + np). Suppose that t I- 1 and X = ti. Then by (VII.2. 17) L:=, ti (y ) = 00 and L: = I ti (1) = 00 + mp for some integer m . Thus tX (l) I (p - 1) (1 + np ). (i), (iii). Let X = Xj with OJ = 1 or let X = 2:: = 1 ti where t I- 1 and 00 = 1 . B y (4.3) n = F(p, u, h ) where x (1) = 1 + up. Since G i s a counterexample either h = 0 or u = O. If u = 0 then X(l) = 1 and so in particular t = 1 . If h = 0 then n = F(p, u, 0) = u and x (1) = 1 + np. (ii), (iv). Let X = Xj with OJ = - 1 or let X = 2:: = 1 ti where t I- 1 and 00 = - 1 . By (4.4) n = F(p, u, h ) where X (1) = vp - 1 with u = (n - h )/v. Since G is a counterexample either h = 0 or u = O. If h = 0 then :
:
354
[4
CHAPTER VIII
n = F(p, u, O) = and u = n /v. Thus v X (l) = p (pn - n + 1) - 1 by (4.4). q u
=
1 and x (1) = p - 1. If u
=0
then
The next result is a refinement of (4.6). LEMMA 4.7. (i) If 0i = 1 for Xi irreducible, j � 1 then Xi (1) = 1 + np. (ii) If 0i = - 1 for Xi irreducible then Xi (1) = p - 1 . (iii) If t � 1 and D O = 1 then t(; (l) = 1 + np for 1 ::s; i ::s; t. (iv) If t� 1 and DO = - 1 then t(j ( l ) = p - 1. PROOF. (i) Since G = G', Xj (l) > 1 for j � 1 . The result follows from (4.6) (i). (ii), (iii), (iv). There are at most e values of j with OJ = 1 since the Brauer tree is connected. By (4.6) Xi (1) ::S; 1 + np if 0i = 1 and X I (l) = 1 . Let co = 1 if Do = 1 and co = 0 if 00 = - 1 . Let c 1 = 1 - co. Then (VII.2.1S) (iii) implies that < e (1 + np ) = � (1 + np)
t 1 = t {p (pn - n + 1) - I}.
4]
A CHARACTERIZATION OF SOME GROUPS
355
Let G O = G/Op , (G). Every irreducible character in B has 01' (G) in its kernel and so may be identified with a character of G O . Suppose that there exists an irreducible nonprincipal Brauer character 'P in B with 'P (I) ::s; Hp - 1). As GO is simple, (3.1) implies that G O = PSL2 (p ) contrary to assumption. (i) Suppose that t = 1 . Let x be an element in N which maps onto an element of order p - 1 in G O . By (4.7) there exists j with Xi (l ) = P - 1. Thus (Xj )N is irreducible and so (Xi ) is the character afforded by the regular representation of (x ) . Hence the linear transformation corresponding to x in the representation which affords Xi has determinant - 1, contrary to the fact that G = G'. Thus t� 1 . If 0 0 = - 1 then by (4.7) (i (I) � Hp - 1) which i s impossible. Thus 00 = 1 . B y (4.7), t( 1 ( 1 ) = 1 + np. Suppose that ou 1 for some u with 2 ::s; u ::s; e. For any j let Pi denote the vertex on the Brauer tree corresponding to Xi' The path from Po to Pu contains a vertex Ps with Os = - 1 . Since p] is an end point of the tree, p] does not occur on this path. Thus XS has at least two nonprincipal Brauer constituents. By (4.7) xs (1) = p - 1 . Hence there exists an irreducible nonprincipal Brauer character 'P with 'P (1) ::S; Hp - 1) which is not the case. Consequently 0i = - 1 for 2 ::S; j ::s; e. The result follows from (4.7). (ii) This is a direct consequence of (i) and (VII.2.IS) (iii). (iii) Since GO is simple it follows from (3.2) that a Sp -group of GO is self centralizing. Hence (x)
C:
Thus Xi (1) � p (pn - n + 1) - 1 and t(i (1) � p (pn - n + 1) - 1. The result follows from (4.6). 0 LEMMA 4.8. (i) t� 1 , t( l ( l ) = 1 + np, XI (1) = 1 and Xi (l ) = P - 1 for 2 ::S; j ::S; e. (ii) 1 + (1 + np )/t = (p - l ) ( e - 1). (iii) G is simple and CG (y ) = (y ).
where G O has 1 + nop Sp-groups. Thus 1 + nop ::s; 1 + np and (1 + np )/t = (1(1) is an inte:ger. By (ii) p - 1 and (1 + np )/t are relatively prime. It follows that (1 + np)lt 1 (1 + nop )/t. Thus 1 + np ::s; 1 + nop and so n = no. The minimality of G now implies that G = G O . 0
PROOF. Let G] be the subgroup of G generated by all elements of order p in G. Thus G] <J G and G] has 1 + np Sp -groups. Suppose that G I � G. By induction p > 3 and G]/O p ' (G 1 ) is either isomorphic to PSL2 (p ) or p = 2Q - 1 amd G]/Op , (G]) = SL2 (p - 1). Hence G /Op' (G) has a normal subgroup A such that either A = PSL2 (p ) or p = 2Q - 1 and A SL2 (p - 1). Thus G/Op , (G) is isomorphic to a sub group of the automorphism group of A. Since G = G ' this implies that G /Op ,(G) = A contrary to assumption. Hence G = G]. This in particular implies that G/Op , (G) is simple.
Let 8 1 , 82, , be all the irreducible characters of G which are not in B. Then each 8; is in a block of defect O. Let 8; (1) = pai. Let q be a prime with q I e. Let x be an element of order q in N. •
•
LEMMA 4.9. (i) t = P - 2 - n. (ii) 2: a ; = n.
=
PROOF. (i) Since te = p - I this follows from (4.8) (ii). (ii) By (4.8) (i)
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A CHARACTERIZATION OF SOME GROUPS
357
particular this implies that (cI>, IN ) � 1 and (cI>, IN ) = 1 if and only if cI>( z ) = 1 for all z E N - P. This last condition holds if and only if cI> = cI> the principal indecomposable character which corresponds to the principal Brauer character. Hence «O; )N, IN ) � a; for all i. Thus (4.9) (ii) and (4.12) imply that «0; )N, IN ) = a; for all i. Therefore (O; )N = a;cI>o and so O; (x ) = a;. D 0,
:
= 1 + ( np 1 t + (p - 1 )2 ( e -,- 1) + P 2 L a T . Direct computation using (i) yields the result.
4]
D
LEMMA 4.10. Let z be a q -singular element in G. Then the following hold. (i) Xi (z ) = 0 for 2 � i � e. (ii) ti (Z ) = - 1 for 1 � i � t.
PROOF. (i) By (4.8) (ii) (p - 1) is relatively prime to (1 + np )/t. As I G 1 = p(p - 1)(1 + np )It, this implies that Xi is in a block of defect 0 for q, where 2 � i � e. The result follows. (ii) This follows from (i) and (VII.2. 1S) (iii). 0
LEMMA 4.13. e = 2. P = 2a + 1 for some a > 1 . Every element of even order in G is conjugate to x. A Sr group of G has order p - 1 .
PROOF. Let {Y/i } b e the set of all irreducible characters o f G. If z is· a q -singular element then (4. 10) yields that o = L y/; (I)y/; (z ) = 1 - (l + np) + p L a;O; (z ).
Thus 2: aiO; (z ) = n. Suppose that z is not conjugate to x. Then (4. 10) and (4.1 1) imply that -]
LEMMA 4.1 1 . Oi (x ) = ai for all i.
PROOF. Let T = ' + Xl - 2:;= 2 Xj . By (VII.2.1S) (iii) T(z ) = 0 for z E N, z i- y i with 1 � i � P - 1. By (VII.2. 17) T(y ; ) = e - A (Y ; ) for all i, where A is a faithful irreducible character of N. Thus I
p
L T(z ) = ep - L A (y i ) = ep. zE N Hence (TN, IN ) = 1 . Since (4.10) (i) holds for a n arbitrary prime divisor q of e i t follows that if 2 � j � e then ;=1
p-I
L X (z ) = L Xj (Y ; ) = O. z EN j Thus «Xj )N, IN ) = O. Hence also «'; )N, IN ) = 0 for 1 � i � t by the previous paragraph. Let p be the character afforded by the regular representation of G. Then (pN, IN ) = 1 + np. Hence ( 2: 0; (I) (O; )N, IN ) = np and so ; =0
(4.12) If cI> is a principal indecomposable character of N then cI> = 'P + 2: A where 'P is an irreducible character of N/P and A ranges over all the faithful irreducible characters of N. For each i, (O; )N is a sum of a; principal indecomposable characters of N. Thus cI>( z ) = 'P (z ) for z E N - P. In
0 = L Y/i (X )y/; (z ) = 1 + t + L Oi (X )Oi (z ) = 1 + t + L a;O; (z ) = 1 + t + n > O. This contradiction shows that every q -singular element in G is conjugate to x. Thus every q -singular element in G has order q. Furthermore Co (x ) is a Sq -group of G. By (4. 10) and (4. 12)
I Co (x ) 1 = L 1 Y/i (x ) 1 2 = 1 + t + L a �· = 1 + t + n.
Hence by (4.9) (i) I Co ( x ) 1 = P - 1 is even. Thus q = 2 and p = 2a + 1. If a = 1 then a Srgroup of G has order 2 contrary to the simplicity of G. Since N /P is cyclic of order e it follows that e = 2. D LEMMA 4. 14. Let S be a Srgroup of G. Then No (S) is a Frobenius group of order (p - l) (p - 2).
PROOF. By (4. 13) every element of S - (1) has order 2. Thus S is abelian. Since no element of odd order commutes with an involution, No (S) is a Frobenius group with Frobenius kernel equal to S. By (4.13) 1 S 1 = p - 1 and any two involutions in G are conjugate. Hence by a theorem of Burnside any two involutions in No (S) are conjugate. D
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PROOF OF (4. 1). By (4.13) t = 4 (p - 1 ) Thus by (4.9) (i) n = 4 (p - 3). Hence I G 1 = p(p - l) (p - 2). By (4. 14) G has a permutation repres� ntation on p letters in which No (S) is the subgroup leaving a letter fixed. Smce No (S) I. S a Frobenius group, any faithful permutation representation of No (S) on G has a triply transitive permutation representation on p letters. As I G I = p (p - 1) (p - 2) the assumptions of (4.2) are satisfied and so by (4.2) G = SL2 (p - 1). 0 .
5. Some consequences of (4.1)
The next two results are due to Brauer and Reynolds [1958] . The first of these was originally proved by Brauer under the additional assumption that Co (P) = P. Nagai [1952], [1953], [1956], [1959] has proved several results related to Brauer's original result in which analogous conclusions are reached under various assumptions about n. A slightly different sort of related result can be found in Hung [1973] . Herzog [1969], [1970], [1971] and more recently Brauer [1976b], [1979], have proved related results in case the Sp -group of G is assumed to be cyclic but not necessarily of prime order. THEOREM 5.1. Suppose that G = G' and a Sp -group P of G has order p. Let 1 + np be the number of Sp -groups in G. Let H = Op,(G). Then one of the following holds. (i) p > 3, G /H = PSL2 (P ). (ii) p = 2a + 1 > 3, G/H = SL2 (p - 1). (iii) n ?; ! (p + 3). PROOF. Suppose that neither (i) nor (ii) holds. By (4.1) n = F (p, u, h) for some positive integers h and u where 2 F(p, u, h) -_ hup +uu++1 u + h . It is easily seen that for fixed positive h, F (p, u, h ) is an increasing function of u and for fixed positive u, F(p, u, h ) is an increasing function of h. Thus n ?; F (p, 1 , 1) = ! (p + 3). 0
COROLLARY 5.2. Suppose that p is a prime factor of I G I and p] > I G I· Assume further that the following conditions are satisfied. (i) G = G'. (ii) A Sp -group of G is not normal in G.
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(iii) G has no normal subgroup of order 2. Then p > 3 and either G = PSL2 (p ) or p - 1 = 2" and G = SL2 (p - 1). PROOF. Let P be a Sp -group of G. If I p i ?; p2 then I G P 1 < p and Sylow'S theorem implies that P <J G contrary to assumption. Thus I P I = p. If (5. 1) (i) or (5.1) (ii) holds then I G : H I > 1p3 and so H = (1) and the result is proved. If (5. I) (iii) holds then I G I ?; 2P (1 + np ) ?; 2P (1 + � (p + 3)P ) = P (p + 1) (p + 2) > P 3 contrary to assumption. 0 :
The next result shows that in a very special case, groups which have a character that satisfies the condition of (IV. IO. I) can be classified. COROLLARY 5.3. Assume that a Sp -group of G has order p and G has an irreducible character YJ such that YJ vanishes on p -singular elements and YJ (x ) = ± I Co (x ) lp for every p '-element x of G. Then G is of type L2 (p). PROOF. Let H be the last term in the descending commutator series of G. If p I I G : H I then G is p -solvable and the result is proved. If p ,f I G : H I then by Clifford ' s theorem YJH is irreducible as YJ (1) = p. Thus by changing notation it may be assumed that G = G'. . Let x be a p '-element in G and let Z = Z(G). If x does not commute with a p-element other than 1 then YJ (x ) = ± 1 . If x commutes with a p-element other than 1 then YJ (x ) = ± P = ± YJ (1) and so x E Z. Thus if G has exactly 1 + np S,, -groups then G has exactly (l + np ) (p - l) I Z I p singular elefuents. Hence I G I = I G i ll YJ 1 12 = I G 1 - (1 + np ) (p - 1) I Z I - I Z I + I z i p 2 . Therefore I Z I (p2 - 1) = I Z I (1 + np ) (p - 1) and so 1 + np = 1 + p. Hence n = 1. If P = 2a + 1 > 5 then the number of Sp -groups is 1 + 1 (p - 3)p > l + p. Since SL2 (4) = PSL2 (5) the result follows from (5. 1). 0 The proof of the next result in case p = 5 or 7 depends on the classification of all finite groups which have a faithful complex representa tion of degree at most 4. This classification can be found in Blichfeldt [1917] . THEOREM 5.4 (Brauer and Tuan [1945]). Let G be a simple group with I G I = pq bm, where p, q are primes, b, m are positive integers and m < p - 1.
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Then p > 3, q = 2 and one of the following holds. . (i) p = 2a ± 1 and G = PSL2 (p), . (ii) p = 2a + 1 and G = SL2(p - 1). Conversely the groups listed in the conclusion satisfy the assumptions. PROOF. The converse follows easily from the fact that in case (i) I G 1 = !p (p - l ) (p + 1) = 2ap (p ± 1)/2 and in case (ii) I G 1 = 2 ap (p - 2). In proving the result it may clearly be assumed that q ,( m. If p = q then Sylow ' s theorem implies that a Sp -group is normal in G contrary to assumption. Thus p f: q. Hence if P is a Sp-group of G then I P 1 = p. If P = 2 or 3 then G is solvable by Burnside ' s theorem. Thus p � 5. By (VII.2.15) (iii) there exists a character ( f: Ie in the principal p -block B of G such that (d, q ) = 1 where d = ((1). Thus d I m and so d < p - 1 . Consequently ( is an exceptional character in B by (VII.2.16). Suppose that d :S ! (p + 1). If P = 5 or 7 the result follows by inspection of finite linear complex groups in dimension at most 4. If P � 1 1 then G = PSL2 (p) by (3.1). Thus (p + l)(p - 1) = � p I G 1 = 2q bm. Since (p + 1 , p - 1) = 2 it follows that q b divides either p + 1 or p - 1 . Hence either p - 1 = q b and p + 1 = 2 m or p + 1 = q b and p - 1 = 2 m . In either case q = 2 and statement (i) holds. Suppose that d > ! (p + 1). Since d < P it follows from (3.2) that Ce (P) = P. Let e = I Ne (P) : P I and let t = (p - l)le. By (VII.2. 16) d ·= p - e. Let e = q Ch where h I m. Thus d = p - q Ch and so (d, h ) = 1. Thus dh 1 m. Since ! < d this implies that h = 1 . Hence e = q C and d = p - q C. Since d < P - 1 it follows that 'c > O. These results imply that (5.5) 1 + d = 1 + P - q C = 2 + te - q C = 2 + (t - l)q c. m
If q f: 2 then (VII.2. 15) (iii) implies the existence of a nonprincipal nonex ceptional character X in B with (X (1), q) = 1. Hence X (1) I m and so X (1) < P - 1 which is impossible as X (1) ± 1 (mod p) by (VII.2.16) and X (l) > 1. Therefore q = 2. Suppose that c > 1. Then (5.5) and (VII.2.15) (iii) imply that there exists an irreducible character X in B with X(l) ¢ 0 (mod 4), X f: I e, ( Thus X(l) = n or 2n with n I m. Since X (l) :S 2 n < 2(p - 1) and x (1) = ± 1 (mod p ) it follows that x(1) = p ± 1 . Hence n = ! (p ± 1). As n I m and m < 2 n this implies that m = Hp ± 1). Consequently d :S Hp + 1) contrary ==
i.
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361
to assumption. Thus c = 1. Therefore d = p - 2 and so m = p - 2. There fore e = 2 and I G 1 = p 2b (p - 2). Since e = 2 there exists a unique irreducible character X in B with X f: I e, (i. By (VII.2.15) (iii) X (l) = P - 1. Hence (x(1), m ) = 1 and so p - 1 = r. By (IVA.23) X is not in a 2-block of G of full defect b. Since p - 2 is odd, X is the only irreducible character in B which is not in a 2-block of G of defect b. Since C e (P) = P there are no elements in G of order 2p. Thus by (IVA.24) X (l) 0 (mod 2b). Consequently X (l) = 2b and so I G 1 = p (p - l) (p - 2). Since e = 2 this implies that G contains exactly (1 + !(P - 3)p) Sp-groups. Statement (ii) now follows from (5.1). 0 ==
6. Permutation groups of prime degree
Let G be a transitive permutation group of prime degree p. Then a G has order p. Thus the material of Chapter VI is applicable to the study of G. See for instance Ito [1960a], [1960b], [1962b], [1963a], [1963b], [1963c], [1964], [1965a], [1965b], Michler [1976a], Neumann [1972a], [1972b], [1973], [1975], [1976] . Similarly the methods of Chapter VI can be applied to transitive permutation groups of degree 2p and 3p if it is assumed that p 2 does not divide the order of such a group. See for instance Ito [1962c], [1962d], L. Scott [1969], [1972]. In this section we will only prove a classical result of Burnside and a result of Neumann. See also Klemm [1975], [1977] Mortimer [1980] for related results. Sp -group of
THEOREM. 6. 1. Let 'Po be the principal Brauer character of G and let
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Suppose the result is false. Then X = 'Po + 'P I + 'P2 as a Brauer character, where 'P is an irreducible Brauer character, 'PI r£ 'Po and 'P I � Hp 1 ) Hence by (3. 1) G is of type L 2 (P). Thus P > 3 and G = P S L2 (p ) as G is simple. It is easily seen and well known that every pi-element in G is conjugate to its inverse. Thus every irreducible Brauer character of G is real valued. Thus by (VII.I0.6) the Brauer tree of G is a straight line segment. Hence X has at most two irreducible Brauer constituents. Since 'Po is a Brauer constituent of X the result follows. 0 1
-
.
THEOREM 6.2. Let G be a transitive permutation group on p letters where p is a prime. Let
61
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363
has an F-basis {VI, . . . , vn } where for Z E G, ViZ = Viz. Let l c + X be the character afforded by V. Let V 0 V = A ; EB A F where A ; is the space of symmetric tensors and A F is the space of skew tensors. Let Ho be the subgroup of G consisting of all Z in G which fix VI and V 2 . Let H be the subgroup of G consisting of all Z in G which fix the set {V I, V 2}. Thus / H : Ho / = 2 as G is doubly transitive. Let {30 = I H, {31 be the irreducible characters of H which have Ho in their kernel. LEMMA 6.5. (i) A ; affords the character {3 f? + l c + X. (ii) A F affords the character {3 ? (iii) (X, (3 ?) r£ O. PROOF. (i) {Vi 0 Vj + Vj 0 Vi / 1 � i � j � p} is an F-basis of A ; on which G acts as a permutation group. Since G is doubly transitive on {Vi } there are two orbits: {2Vi 0 Vi / 1 � i � p} and {Vi 0 Vj + Vj 0 Vi / 1 � i < j � p }. The first orbit yields an F[ G] module isomorphic to V and the second orbit has isotropy group H. Hence A ; affords the desired character. (ii), (iii) {Vi 0 Vj - Vj 0 Vi / 1 � i < j � p} is an F -basis of A F on which G acts as a monomial group. Let V;j be the one dimensional F-space spanned by Vi 0 Vj - Vj 0 Vi. Since G is doubly transitive on {Vi } it follows that G is transitive on the set of spaces V;j and H is the subgroup of G consisting of all elements which leave VI2 fixed. Since Ho is the subgroup of G consisting of all elements which leave VI 0 V2 -'--- V 2 0 VI fixed it follows that A F affords the character {3 ? This proves (ii). For 1 � i � P let Wi = 2:j i (Vi 0 Vj - Vj 0 Vi ). Let W be the F-space spanned by {Wi }. If Z E G and ViZ = Vk then WiZ = Wk . Thus W is an F[ G] module and the linear map f : V W defined by f ( Vi ) = Wi is an F[ G] homomorphism. Since 2:�=1 Wi = 0 it follows that 2:�=1 Vi spans the kernel of f. Thus W affords the character X. Hence by (ii) (X, (3 ?) r£ O. 0 �
�
From now on in this section assume that G satisfies the hypotheses of (6.4). N = Nc (P) is a Frobenius group. As N r£ G it follows that p r£ 2.· By assumption there exists an element x of order 2 in N. By (6.3) G is doubly transitive. Let H, Ho, {30, {31 be defined as above. We will use the notation introduced in section 1 . Let X b e the R -free R [ G ] module obtained b y tensoring the permuta tion module of degree p with R. By (6. 1) and (6.2) X affords the character l c + X, where X is irreducible. By (6. 1) X = 'Po + 'P as a Brauer character where 'P is an irreducible Brauer character. Let
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afforded by the principal indecomposable R [ G] module which corre sponds to 'Po , 'P respectively. Let X 0 X A + EB A - where A + is the space of symmetric tensors and A - is the space of skew tensors. The group p (x> has exactly 2 irreducible Brauer characters. let 1]0 , 1]1 be the characters afforded by the corresponding principal indecomposable R [ P ( x )] modules, where 1] 0 corresponds to the principal Brauer character. =
7]
CHARACTERS OF DEGREE LESS THAN P
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fixed is H. Thus this permutation representation affords the character f3 5'. By (6.7) and Frobenius reciprocity (( {3 5') GI , 1 GI ) = (f3 �, (l GI ) G ) = (f3 �, 1 + X ) = 2.
LEMMA 6.6. (i) A + is projective and (A +)P(x) affords the character � (p + 1)1] 0 . (ii) A - is projective and (A - ) p(;\:) affords the character Hp - 1)1] 1 .
Hence G1 has exactly 2 orbits on the set of unordered pairs {i, j}. These must necessarily be .1 1 = {{I, i } 1 2 :oS i :oS p } and .1 = {{i, j } 1 i t- I, j � I}. Thus in particular GI is transitive on .1. Suppose that the result is false. Then G1 is not transitive on the set of ordered pairs (i, j) with 2 :oS i � j :oS p. Since G1 is transitive on .1 this implies that there are 2 orbits rJ, r2 and (i, j) E r1 if and only if U, i ) E r2 • As x E G1 and x interchanges some pairs the last possibility cannot occur. 0
PROOF. Since X i s projective and Inv G X � (0) i t follows from section 1 that
7. Characters of . degree less than
XN = V (l, p). Hence by (2.9)
(p - I )j2 (p - 3)/2 E9 V ( a 2i, p ), A � = E9 V ( a 2i + \ p ) . i =O i=O Since a (x ) = - 1 it follows that (A +)P(x), (A - )P(x) affords the Brauer character ! (p + 1 )1] 0 , ! (p - 1)1] I respectively. Since A + , A - are direct summands of X 0 X they are both projective. This implies the result. 0 A�=
LEMMA 6.7. (X, (35') = 1 . PROOF. B y (6.5) A - affords the character f3 ? B y (6.6) A - is projective. Thus f3 ? ao(Po + a , (P + 01 where (X, (1) = O. The Frobenius reciprocity theorem implies that (f3 ?, 1 G ) 0 and so ao = O. By (6.S) (iii) al = (f3 ?, X ) � O. By (6.6) (f3 ?)P(x) = C , 1] 1 for some positive integer CI and so (P P(x ) = C1]1 for some positive integer c. By (6.5) A + affords the character f3 5' + (Po. By (6.6) A + is projective. Thus f3 5' + (Po = bo(Po + bl (P + 00 where (X, (0) = (l G, (0) = O. By (6.6) ( {3 5' + (Po)P(x) Co1]o for some positive integer co. If bl � 0 this implies that (P P(x) c' 1] 0 for some positive integer c ' contrary to the previous para graph. Therefore f3 5' + (Po = bo(Po + 00 • Consequently by the Frobenius reciprocity theorem ( {3 5', X ) = (f3 5', 1 G ) = 1. 0 =
=
=
=
PROOF OF (6.4). Choose the notation so that G acts on {I, . . . , p } as a transitive permutation group and x leaves 1 fixed. Let GI be the subgroup of G consisting of all elements which leave 1 fixed. Thus x E G 1 • The permutation representation of G on the set of all unordered pairs {i, j} is transitive and the subgroup consisting of all elements which leave {1, 2}
p-1
Suppose that a Sp -group of G has order p and G is not of type L 2 (p ). Let X be a faithful irreducible character of G. By (3.3) X (l) 3 �(p - 1) for p 3 13. This already strengthens earlier results of Brauer [1942b] and Tuan [1944] in case p 3 13. Sections 7-10 are devoted to the proof of the following three results which strenghen the above mentioned inequality. The first of these includes the results of Brauer [ 1942b] and Tuan [1944]. The second includes earlier results of Brauer [1966c] and Hayden [1963] . Eor the third result we will follow the proof given by Blau [197Sb]. Actually (7.1) is a consequence of (8. 1) below which is a more general result. THEOREM 7.1 (Feit [1967bD. Let G be a finite group whose center Z has odd order. Let p ? 7 be a prime and let P be a Sp -group of G. Assume that I p l = p and P :A G. Let � be a faithful irreducible character of G. Then one of the following occurs. (i) G/Z = PSL2 (P ), �(1) = ! (P ± 1). (ii) �(1) 3 P - 2. THEOREM 7.2 (Blau [1971bD. Let G be a finite group and let Z be the center of G. Let p > 7 be a prime and let P be a Sp-group of G. Assume that I P 1 = p and P 11 G. Let � be a faithful irreducible character of G and let t denote the number of conjugate classes of elements of order p. Then one of the following occurs. (i) G/Z = PSL2 (P ), �(1) = ! (P ± 1), t = 2. (ii) �(1) 3 p - 1. (iii) �(1) 3 P + � � Vp + t p :oS t 2 - 3t + 1.
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THEOREM 7.3 (Feit [1967b], [1974]). Let G be a finite group and let Z be the center of G. Let p > 5 be a prime and let P be a Sp - group of G. Assume that I P I = p, P 11 G and G has a faithful irreducible character � with �(1) = P - 2. Then p = 2a + 1 and G = SL2 (p - 1) x Z. By using the results of Feit [1964] it can be shown that once (7. 1), (7.2) and (7.3) are proved then the same conclusions hold even when the assumption that I P I = P is dropped. In case p = 2 or 3, (7. 1), (7.2) and (7.3) are true and trivial. If p = 5 (7.2) and (7.3) are false as a covering group ;:\6 of A6 has a faithful irreducible character of degree 3. This group has 2 classes of elements of order 5 but 5 > 22 - 3.2 + 1. In case p = 7, (7. 1) and (7.2) are false as a covering group _ of A7 has a faithful irreducible character of degree 4. This group has 2 A7 classes of elements of order 7 but 7 > 22 - 3.2 + 1. As a consequence of (7.3) we can get the following result which strengthens (6.4) in a special case. THEOREM 7.4 (Ito [1960a]). Let p > 2 be a prime. Let G be a permutation group on p letters and let P be a Sp - group of G. Assume that I No (P) I = 2p. Then either P <J G or p - 1 = 2a and G = SL2 (p - 1). PROOF. If P = 3 or 5 this can easily be verified. Suppose that p > 5. The principal p -block B of G contains irreducible characters of degree 1 and p - 1. Since I No (P) : P I 2 these are all the nonexceptional characters in B. Thus by (VII.2.15) (iii) an exceptional character in B has degree p - 2. The result follows from (7.3). 0 =
Throughout the remainder of this chapter the notation introduced in section 1 will be used. Also the following notation will be used. G = N = No (P). e Co (P). e = I N : e I. t is the number of conjugate classes of elements of order p in G. Thus te = p - 1. We will be concerned with groups G which satisfy the following conditions. (i) G = G' and GIZ is simple where Z is the center of G. (ii) P > 5. 1 < e < p - 1 . There exists a cyclic subgroup E of order e in N such that N = Ee and E n e = (I). (iii) e = P x Z, Z is cyclic and ( I Z I, e ) = 1. (iv) G is not of type L2 (p). There exists a faithful irreducible character � of G with �(1) = p - e. =
CHARACfERS OF DEGREE LESS THAN P
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LEMMA 7.5. It suffices to prove (7. 1), (7.2) and (7.3) for groups G which satisfy ( ) * .
PROOF. Suppose that G is a counterexample to (7. 1), (7.2) or (7.3) of minimum order. We will show that G satisfies ( ) Since G is a counterexample, �(1) < (p - 1). Let Go be the subgroup of G generated by all elements of order p in G. Theorem B of Hall-Higman (VII. I0.2) implies that Op ' (Go) is in the center of Go. Since P 11 Go it follows that Go/Go n Z is simple and Go = Gb. If G "I Go then the minimality of 1 G 1 implies that Gol Go n Z is isomorphic to PSL2 (P ) for p > 7 or to SL2 (2a ) for 2a = p - 1 > 4. The Schur multipliers of these groups are well known. Since Go is generated by elements of order p, this yields that Go is isomorphic to one of PSL2 (p ), SL2 (P ) for p � 7 or SL2 (2a ) for 2a = p - 1 > 4. None of these groups admit an outer automorphism which stabilizes a character of degree less than p - 1. Hence G = GoZ and G is not a counterexample. Consequently G = Go. Thus G I Z is simple, G = G' and (i) is satisfied. Since G is a counterexample G is not of type L 2 (p). Thus by (3. 1) �(1) > ! (P - 1). By (3.2) e = P x Z. Hence by (VII.2.16) «1) = p - e and (iv) is satisfied. As G has a faithful irreducible character, Z is cyclic. Furthermore I Z I I �(1) = p - e. Hence ( I Z I, e ) = 1 and (iii) is satisfied. Since G = G' 1 < �( l ) < p - 1. Thus 1 < e < p - 1 by (iv). Nle is a group of automorphisms of P and so is cyclic. As NIP is abelian this implies the existence of a cyclic subgroup E of NI Z of order e. Thus (ii) is satisfied since (e, 1 Z 1 ) = 1. 0 * .
For the rem'ainder of this section assume that ( ) is satisfied. *
Since � is faithful there exists a faithful irreducible character 71 of Z such that �(z ) = 71 (z )�(1) for z E Z. There is a one to one correspondence between p-blocks of G of defect 1 and characters 71 u, 0 � u � I Z 1 - 1. Hence by the first main theorem on blocks (111.9.7), G has exactly 1 Z i p -blocks of defect 1. Let Bo, B1, denote all the p -blocks of defect 1 where the notation is chosen so that for a character 8 in Bu , 8z = 8 (1)71 u. Thus Bo is the principal p-block. Since NIP is abelian, the index of inertia of Bu is equal to e for O � u � I Z I - 1. Let A I, . . . , At denote all the irreducible characters of NI Z which do not have P in their kernel. For 0 � u � 1 Z 1 - 1 let ��ul, . . . , ��u ) denote the exceptional characters in •
•
•
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Define 8 (u ) = ± 1 by L: = I ?jU)(I) == 8(u ) (mod p ) . Thus 8 (� ) = 8o for the block Bu as defined in section 2 of Chapter VII. By (VII .2 . 17) the notation can be chosen so that for 0 � u � I z 1 - 1 S ?j u ) (y Sz ) = - 8 ( U )Ai (y )71 U ( z ) (7.6) for 1 � s � P - 1 , z E Z, 1 � i � t. Bu .
7.7. Let X be an R -free R [G] module which affords the character g. Suppose that gz = g (I)71 u . Let g = {3 + Y + P where (3 = L; =l hj?t >, y is a character in Bu which is orthogonal to every ?Ju ) and is orthogonal to every character in Bu. Let h = L; = I hj. Assume that p > 7, e � 3 and t � 3. Then the following hold. (I) Suppose that X is indecomposable and g (l) == e (mod p ) or g (l) == e + 1 (mod p). Then one of the following occurs. (i) 8 ( u ) = 1 ; h � t - 1 . (ii) 8 (u ) = - 1 ; h � 1 . (II) Suppose that X = WI EB W2 , where each � is indecomposable and g (l) == 2e (mod p ). Then one of the following occurs. (i) 8 (u ) = 1 ; h � t - 2. (ii) 8 (u ) = - 1; h � 2 .
LEMMA
PROOF. Let be a faithful irreducible character of P. For 0 � i � P - 1 and for any R -free R [ G] module Y let as ( Y) be the multiplicity of S as a constituent of the character afforded by YP• If Y affords the character 8 let as (8) = as ( Y). If 8 is an irreducible constituent of p then 8z = 8 (1)71 u and 8 is not in Bu . Hence 8 is in a block of defect O. Thus as (8) = ao (8) for O ::s; s � p - 1. Therefore as ( g ) - ak ( g ) = as ( {3 ) - ak ((3 ) + as ( y ) - ak ( y ) (7.8) for 0 � s, k � P - 1 . If 8 is an irreducible constituent of y then by (VII.2.17) as (8) = ad 8) for l � s, k � p - l and ao (8) - as (8 ) == 8 (1) (mod p ) for l ::s; s ::S; p - 1. Therefore a o ( y ) - al (-y ) == y (l) (mod p ) . Thus (7.6) implies g (I) == - 8(u )eh + ao( y ) - as ( Y ) (mod p ) for l � s � p - 1. (7.9) By (7.6) E
E
{ (Z hjAj ) - ak (Z hjA )}
as ( {3 ) - ak ( {3 ) = - 8 ( U ) as
j
for O ::s; s, k � P - 1 .
(7.10)
CHARACTERS OF D EGREE LESS THAN P
7]
-1
369
In proving the result it may be assumed that h � t - 1 otherwise there is nothing to prove. Hence by (7. 10) there exists m with 1 � m ::s; p - 1 such that ao ( {3 ) - am ((3 ) = O. Thus by (7.8) ao ( g ) - am ( g ) = ao ( Y ) - am ( y ). Hence (7.9) yields that (7. 1 1) g (l) == - 8 (u )eh + ao( g ) - am ( g ) (mod p ). (I) Since X is indecomposable it follows from (VII.I0.9) that l ao( g) - am ( g ) I � 1 . Let g (l) = e + c (mod p) where c = 0 or 1 . B y (7.1 1) e + c == - 8 (u )eh + c ' (mod p) where c ' = 0 , ± 1 . Thus {1 + 8 (u )h }e + c - c ' == 0 (mod p). If 1 {1 + 8 (u )h }e + c - c ' l � p then h � (t - 1) as I c - c ' I ::s; 2 < e and the result is proved. Thus it may be assumed that { I + 8 (u)h }e = c ' - c. Hence c ' - c == 0 (mod e ) and so c ' - c = O. Thus 8 ( u )h = - 1 and so 8 (u ) = - 1, h = 1 as required. ( II ) By (V II .I0.9) l a () ( g ) - a", (g) I ::S; 2. By (7. 1 1) 2e == - 8(u )eh + c (mod p) with c = 0, ± I , ± 2. Thus {2 + 8 ( u )h }e - c == 0 (mod p ). If 1 {2 + 8(u)h }e - c l � p then h � t - 2 as 1 c 1 � 2 and the result is proved. Thus it may be assumed that {2 + 8 (u )h}e = c. Hence c 0 (mod ) and so c = 0. Therefore 2 + 8 ( u )h = 0 and so 8 ( u ) = - 1, h = 2. 0 ==
e
7. 12. Suppose that the assumptions of (7.7) are satisfied. Assume furthermore that 8 (u ) = 1 and ?\u ) ( l) > (p - 1). Then the following hold. ( I ) If g (l) e (mod p) then g ( l ) � (t - 1) ?\U) (1) + p - 1 . If g ( l ) e + 1 (mod p) then g (l) � (t - 1)? \ u ) (l). ( II ) If g (1) == 2e ( mod p ) then g (l) � (t - 2) ?\U ) (1) + p - 1. LEMMA
==
==
PROOF. If h � r, all the statements are trivial. Suppose that h � t - 1. If g (l) == e + 1 (mod p), then h = t - 1 by (7.7) and the result is clear. Suppose that g(l) e (mod p ) . By (7.7) h = t - 1. Therefore e == g (l) == - e (t - 1) + y (I) == e + 1 + y(1) (mo d p ). Hence y (l) == - 1 (mod p) and so y (l) � P - 1 . The result is proved in this case. Suppose that g(l) == 2e (mod p). If h � t - 1 the result is clear. Suppose that h < t - 1 . Thus by (7.7) h = t - 2. Therefore 2e == g(l) == - (t - 2)e + y (l) == 1 + 2e + y (l) (mod p). Hence )' (1) - 1 (mod p ) and so )' (1) � P - 1 . This implies the result. 0 ==
==
7. 13. Let f.L be an irreducible character of N/C For 1 � i � P - 1 and 0 � u � I Z 1 - 1 let W(i, u ) denote the indecomposable R [GJ module
LEMMA
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370
[7
------such that W(i, u ) = V( T/ u /Lo< 2i + 1). If K is replaced by a suitable finite extension field then the following hold. (i) For 0 � i < He - 1) there exists an R -free R [G] module M(i, u ) such that M(i, u ) = W(i, u ) EB W(e - 1 - i, u ). (ii) There exists an R -free R [ G] module M (u ) such that M (u ) = WO (e - 1), u ) if e is odd and M(u ) = W(�e, u ) if e is even. PROOF. By (I.17. 12) there exists an R -free R [N] module Y(u ) for O � u � I z l - 1 such that Y(u ) = V( T/ U/La (e- l l/2, e ) if e i s odd and Y(u ) = V( T/ u /La e/2, e + 1) if e is even. Let M(u ) be defined by M(u ) = Y(u ). By (I1I.S.8) M(u ) has the required properties. By (1.17.12) there exists an R -free R [N] module X(i, u ) for O � u � I z l - 1, O � i � He - 1) such that X(i, u ) = V( T/ u /La i, 2e ). By (VI.2.8) there exists an exact sequence o � V (T/ U /La \ 2i + 1) � V ( T/ U /La \ 2e ) � V( T/ U p,a - i - 1 , 2(e - 1 - i ) + l) � O. As a e = 1 it follows from (1 . 1 8 .2) that if K is replaced by a suitable 'finite extension field then there exists an R -free R [N] module Y (i, u ) such that
Y(i, u ) = V( T/ u /La \ 2i + l) EB V( T/ u/La e- i -1, 2(e - 1 - i ) + 1). Let M(i, u ) be defined by M(i, u ) = Y(i, u ). The result follows from (III.S.8). 0 �
The next result is implicit in Brauer [1966c]. LEMMA 7. 14. Suppose that 0 � u, v, w � 1 z 1 - 1 with w = u + v (mod 1 z 1 ). Assume that the following conditions are satisfied. (i) o ( u ) = o (v) = o ( w ) = l. (ii) � \m ) (l) < p - 1 for m = u, v, w. Let �iU) �)v ) L� =l hijd �W ) + r, where r is orthogonal to each � �w ) . Then L� =l h;jk � t. =
PROOF. Let 'P h . " ., 'Pe be all the irreducible characters of E where 'Pj = 'P L Then lPj = 'P [E for j = 1, . . "' e are all the principal indecomposable charac ters of PE. By abuse of notation let (A'i )PE = A i . For fixed m, Uint ) } is a set of irreducible characters, any two of which agree on p i-elements. Thus for m = u, v or w (�i m ) pE = lP[( m ) - Ai, where f(m ) does not depend on i. Hence
8]
PROOF OF (7, 1)
371
(7. 15) Let PE, PEP be the character afforded by the regular representation of E, EP respectively. Then AilPs = { ( Ai )E'Ps yE ( P tE = PP . E E Similarly =
lP[(U ) lP[(v)
of
= { ( lPf( U )
E {(t + E 'Pf (v)Y = PE 'Pf(u ) 'P f ( v ) YE = ( tPE tE + ('P f(u )+f(v ) tE = tppF. + lPf (U )+f(v ) .
For each i, (Ai ) p is a sum of e distinct nonprincipal irreducible characters P. Thus
Hence A iAj = ()l + () 2 where ()l is a sum of at most e principal indecompos able characters of PE and ()2 is a linear combination of the ( Ai ) PE. Since A iAj vanishes on E - { l } it follows that () l = 0 or () l PPF.. Therefore (7. 15) becomes =
(7. 16) where 0 = 0 or 1. The set { lPs } U {Ai } is a basis for the additive group of integral linear combinations of irreducible characters of PE. Furthermore if a character of PE is expressed in terms of this basis then the coefficient of lPs is nonnegative for ,all s. By (7. 16) the coefficient of any lPs in (� :u ) �jv) )PE is at most t. Since (� kW ) ) PE = lPf( w ) - Ak the result follows. 0 8. Proof of (7. 1)
This section contains a proof of the following result which implies (7. 1). THEOREM 8 . 1 . Suppose that condition ( ) of section 7 is satisfied with p > 7 and � a faithful irreducible character. Assume that there exists u, v, w with , 0 � u, v, w � 1 z I � 1 such that w == u + v (mod 1 z I ) and the following conditions are satisfied. (i) o (u ) = o (v) = o ( w ) = l . (ii) � = � \ u ) , �\U ) (l) < p - 1 and �\v ) (l) < p - l . Then e = 2. *
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372
[8
8]
PROOF OF (7. 1 )
373
LEMMA 8.2. Statement (7. 1) follows from (8.1).
Furthermore
PROOF. Suppose that (8. 1) has been proved. In proving (7. 1) it may be assumed that ( ) is satisfied by (7 .S). Let u = Then YJ 2 u has the same order as YJ U since I z I is odd. Hence there is a field automorphism which sends YJ U to YJ 2 u. Thus the exceptional character in B 2 u is algebraically conjugate to ( and so 8 (2u ) = 8 ( u ) = 1. Now (7. 1) follows from (8. 1). 0
p 2 - e (2p - e ) = (p - e )2 = dimR XjU ) 0 Xjv) � 2: dimR W(k, w ). k= (J Suppose that e is odd. Then e� p 2 _ e (2p - e ) � k�= 2 dimR { W(k, w ) EB W(e - k - 1, w )} 0 + dimR W(H e - 1), w ) � H e - l) (t - 2) (2p - e ) + He - l)(p - 1) + (t - 1) (2p - e ) + (p - 1). Hence p 2 � (2p - e ){e + ! et - � t - e + 1 + t - l} + He + 1) (P - 1) = (2p - e )d (e + 1) + H e + l) (p - 1) = He + 1){2pt - te + p - I} = pt(e + 1). Thus p � t (e + 1) = te + t = P - 1 + t al)d so t � 1 contrary to the fact that e < p - 1 . Stlppose that e is even. Let f = dimR W O e - 1, w). Then
*
v.
For the rest of this section assume that the hypotheses of (8. 1) are satisfied. Furthermore assume that e > 2. A contradiction will be derived from this situation. Let xjm ) be an R -free R [GJ module which affords ( jm) for m = u, 1 � i � t such that Xjm ) is indecomposable. Let Aj = A � . Thus ( jU ) (y ) = ( J V ) ( y - l ). Furthermore there exist irreducible characters v, v ' of N/C such that v,
LEMMA 8.3 . (\w ) (l) = p - e. PROOF. By (2.7) e-l
V(T] wvv'a k , 2k + l) EB A, k =(J where A is a projective R [N] module. Let f-t = vv' and let W(k, w ) be defined as in (7.13). Thus (Xju ) 0 Xjv ) )N = E9
e-l
(
e- l
Xju ) 0 Xjv) = E9
W(k, w ) EB A ' k =(J for some projective R [G] module A '. Suppose that (\w ) (l) I: p - e. Thus (\w ) (l) � 2p - e. By (7. 12) and (7. 13) dimR W(k, w ) + dimR W(e - 1 - k, w ) � (t - 2)(\w ) (l) + p - 1 � (t - 2) (2p - e ) + p - 1 for 0 � k � � (e - 1). If e is even then dimR W(e /2, w ) � (t - l )(\w ) (l) � (t - 1) (2p - e ). If e is odd then dimR W(He - 1), w ) � (t - l)(\w ) (l) + p - 1 ;?; (t - 1) (2p - e ) + p - 1 .
Hence
!
p 2 - e (2p - e ) � (��-2 k =(J dimR { W(k, w ) EB W(e - 1 - k, w )} + dimR WO e, w ) + f � (! e - l) (t - 2) (2p - e ) + (� e - l) (p - 1) + (t - 1) (2p - e ) + f = (2p - e ){! et - e - t + 2 + t - I } + (� e - 1) (p - 1) + f.
pet + p = P 2 � (2p - e ) (� et + 1) + G e - 1 )et + f = pet - !e2t + �e2t - et + f + 2p - e. Therefore o � - et + p - e + f = 1 - e + f.
Hence f � e - 1 < � (p - 1). Furthermore f e - 1 (mod p) and so f > 1 as e > 2. Thus P is not in the kernel of WO e - 1 , w ). Hence by (3.1) G is of type L 2 (p) contrary to assumption. 0 ==
[8
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374
LEMMA 8.4. There is an irreducible Brauer character 'P such that (\W) = 'P as a Brauer character. PROOF. Since 8 (w ) = 1, the principal Brauer character does not occur as a constituent of (\w ) . If the result is false then some nonprincipal irreducible Brauer character has degree at most � (p - e ) < � (p - 1). Thus by (3. 1) G is of type L 2 (p) contrary to assumption. 0 LEMMA 8.5. Let 'P be defined as in (8.4). Let 8 be the unique nonexceptional irreducible character in Bw which has 'P as a constituent. Then 8 (1) ;:-; 2p - 1. PROOF. If the result is false then 8 (l) = P - 1 as 8(1) - 1 (mod p). Thus 8 = 'P + 'P I as a Brauer character and 'P l(l) = e - 1. The principal Brauer character can occur as a constituent of 8 with multiplicity at most 1. Since e > 2, 'P 1 (1) > 1 and so there exists a non principal irreducible Brauer character in Bw of degree at most e - 1 < Hp - 1). Thus by (3. 1) G is of type L 2 (p) contrary to assumption. 0 Let h = � hijk , (�U ) (lv ) = a8 + i:= l hijk ( �w ) + r; k k where r i s orthogonal t o 8 and to all ( ); ) . ==
=1
LEMMA 8.6. (i) h ::::; t. (ii) If e is odd, a + h ;:-; He - l) (t - 2) + t - 1. (iii) If e is even, a + h ;:-; � e (t - 2) + 1. PROOF. (i) This follows from (7. 14) and (8.3). Let 'P be defined as in (8.4) . For any R -free R [ G] module Y let n ( Y) be the multiplicity with which 'P occurs as a constituent of the Brauer character afforded by Y. Thus a + h = n (Xj u ) 0 Xjv) ). Let W(k, w ) be defined as in (7.13). Thus by (2.7) e-I
XjU) 0 Xjv) = EB W(k, w ) EB A k =0 for some projective R [G] module A. (ii) By (7.7) and (7.13) 2 a + h ;:-; (e f0/ {n ( W(k, w )) + n ( W(e - k - 1, w ))} k + n ( WO (e - 1), ) ) ;:-; He - 1) (t - 2) + (t - 1). =
w
8]
PROOF OF (7. 1)
375
(iii) By (7.7) and (7. 13) 2-2 a + h ;:-; e/2: {n ( W (k, w )) + n ( W (e - k - 1, ))} + n ( WO e, w )) k =O ;:-; O e - l) (t - 2) + (t - 1) = O e ) (t - 2) + 1. 0 w
LEMMA 8.7. If 1 ::::; k ::::; p - 1 then (p - l)r(y k ) = a (p - 1) - he + ep - e 2 • PROOF. By (7.6) (jU) (y )(jV) (y ) = As (y )As (y -l) for suitable s. Thus (jU ) (y )(jv ) (y ) = e + g (y ), where g (y ) is a sum of e 2 - e primitive p th roots of 1. Let Tr denote the trace from the field of p th roots of 1 to the rationals. Since rz = r(1)YJ w and r is orthogonal to all (�w ) it follows that r(y ) is rational. As 8 (y S ) = - 1 it follows from (7.6) that e (p - 1) - e 2 + e = Tr{( lU ) (y )(jV ) (y )} = - a (p - 1) + he + (p - 1)r(y ). This implies the result. , LEMMA 8.8. h + e
==
0
1 (mod t).
PROOF. Divide the equation in (8.7) by e and read modulo t.
0
LEMMA 8.9. r(l) > (p + l){a + e - 1 - (e + h - l)/t} ;:-; (p + l)a. PROOF. Since rz = r(l)YJ w and r is orthogonal to 8 and all ( �w ) it follows that where the first sum is over characters Xk in Bw with Xk (1) 1 (mod p), the second sum is over characters Xk in Bw with Xk (1) - 1 (mod p) and ro is a sum of characters in blocks of defect O. Therefore if 1 ::::; s ::::; p - 1 r (y S ) = L' Ck - L" Ck ::::; L c £ . Thus by (8.7) ==
L' Ck
i
;:-; r(y ) = a + e + (1 - e - h ).
==
CHAPTER VIn
376
[8
As the principal character of G occurs in �l u ) �jv ) with multiplicity at most 1 this implies that r (l) � 2:' CkXk (1) � { ( L ' Ck ) - 1 } (p + 1) + 1 > (p + l){a + e - 1 + t1 (1 - e - h )}. This proves the first inequality. If the second inequality is false then a > a + e - 1 + (1/t) (1 -:- e - h ). Hence by (8.6) (i) (e - 1) < t1 (e + h - 1) = t1 (e - 1) + th � t1 (e - 1) + 1 . Thus ( t - l)(e - 1 ) < t. Since e � 3 this implies that t < 2 and so t = 1 contrary to the fact that e < p - 1 . 0 LEMMA 8. 10.
P
� 13 and t � 4
or
p = 13 and t = 3.
PROOF. If P = 11 then e = 5. Thus �\U ) (I) = 6 < � (p - 1) contrary to (3. 1). Thus P � 13. Hence by (3.3) P - e � � (p - 1) and so p + 3 � 4e. Thus et = p - 1 � 4e - 4 and so t � 4 - 4/ e. Thus t � 4 if e > 4. If e � 4 and t � 3 then p = 13 and t = 3. 0 LEMMA 8. 1 1 . e is even. PROOF. Suppose that e is odd. Thus t is even. Let (8.6) (ii)
C
= ! et - ! t - e. By
a � C + t - h. By (8.9)
(p - e )2 = � l u ) (1 ) � jv) (1) = a () (1) + h (p - e ) + r (1 ) > a (2p - 1) + h (p - e ) + a (p + 1) = 3pa + h (p - e ).
Therefore (p - e )2 > 3P (c + t - h ) + h (p - e ) = 3pc + t (p - e ) + ( t - h ) (2p + e ). Hence by (8.6) (i) (p - e ) (p - e - t) > 3pc + (t - h )(2p + e ) > 3(p - e )c.
8]
PROOF OF (7. 1)
377
Therefore Thus
et - e - t + 1 = P - e - t > 3c = � et - � t - 3e. O > ! et - ! t - 2e - 1 = ! t(e - 1) - (2e + 1).
This implies that + 1) _6_ t < 2(2e e - l = 4+ e - l · Hence either t = 4 or e = 3 and t = 6. If e = 3, t = 6 then p = 19, c = 3. Thus (S. 12) implies that 160 > 171. Hence t = 4. Now c = e - 2 and (8. 12) implies that (3e + 1) (3e - 3) > 3(4e + l) (e - 2). Thus 0 > e 2 - 5e - 1 and so e � 5. Since 21 = 4.5 + 1 is not a prime e = 3. Thus by (8.8) h 2 (mod 4) and so h = 2 by (8.6) (i). Now (S. 12) implies that 60 > 39 + 58. This contradiction establishes the result. 0 ==
Define the integer b by b = ! e (t - 2) + 1 - t = ! et - e - t + 1 .
(8. 13)
B y (8.6) (iii) and (8. 1 1) a � b + t - h. LEMMA 8. 15. e > t. PROOF. Suppose that e � t. By (S.6) (i) and (8.8) h = t + 1 - e. Thus by (8.9), r(l) > (p + l) (a + e - 2) and so (p - e y > a (2p - 1) + h (p - e ) + (a + e - 2) (p + 1) = 3pa + (t + 1 - e ) (p - e ) + (e - 2) (p + 1). Therefore by (8. 14) (p - e ) (p - t - 1) > 3pb + 3P (e - 1) + (e - 2) (p + 1) > (p - e ) {3b + 3(e - 1) + (e - 2)}. Hence et - t = P - t - 1 > � et - 3e - 3t + 3 + 4e - 5 = � et + e - 3t - 2. Therefore
(8. 12)
O > ! et + e - 2t - 2 = t O e - 2) + e - 2.
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378
Hence (e /2 - 2) < 0 contrary to the fact that e ;?: 4 as e is even. LEMMA S.16. t = 3, e
=
8]
0
4 and p = 13.
=
Thus / Z / / 9. If S is a subset of G let SO denote its image in GO = G /Z. There exists an element x E N of order e = 4 such that XO also has order 4 since ( / Z /, 4) = 1. Suppose that Bu + v is the principal block. Thus YJ u+v = 10 and GO has an irreducible character ��u+V) of degree 9. Let (X�u+v) )NO = V(a j, 9). By (1. 1) a (x Y = a (x )9j = detv (ai,9) (x ) = 1 . Thus a j = 1No and s o InvNo (X)u+v)) i- (0). Hence
> a (2p - 1) + h (p - e ) + a (p + 1) = 3pa + h (p - e ) (S. 17) ;?: 3pb + t(p - e ) + (t - h ) (2p + e ).
Hence
(p - e ) (p - e - t) > 3pb + (t - h ) (2p + e ) > 3(p - e )b.
379
The proof of (S. l) will be completed by showing that the case in (S. 16) cannot occur. For the rest of this section it will be assumed on the contrary that e = 4, t 3 and p = 13. By (3. 1) (�U )(l), (�v) (l) and (�u+V) (l) are all at least S. Hence (�u) (l) = ��V) (l) = ��u+v) (l) = P --C. e = 9.
PROOF. Suppose that t ;?: 4. By (S.9) and (S. 14) (p - e )2 ;?: a (2P - 1) + h (p - e ) + r (1 )
Thus
PROOF OF (7. 1 )
(S. lS)
et - e - t + 1 = P - e - t > 3b = � et - 3e - 3t + 3. Therefore O > ! et - 2e - 2t + 2 = to e - 2) - 2(e - 1). By (S. 10) and (S.15) e i- 4. Therefore 12 - 4 = 4 + -(S.19) -t < 4e e -4· 4 e -Since e > t this implies that t � 6. Thus by (S. 10) t = 4, 5 or 6. By (S. 1 1) e is even. If t = 6 then (S.19) implies that e < 10 and so e = S contrary to the fact that te + 1 is a prime. If t = 4 then b = e - 3. Hence (S. lS) implies that (3e + 1) (3e - 3) > 3(4e + l)(e - 3). Thus 0 > e 2 - ge - 2 and so e < 10. Hence e = 6 or S contrary to the fact that te + 1 is a prime. If t = 5. Then b = �e - 4 and (S.lS) implies that (4e + 1)(4e - 4) > 3(5e + l)(�e - 4). Therefore 0 > 13e 2 - S7e - 16 and so e < S. Thus e = 6. Hence p = 31, b = 5. Now (S.lS) implies that 35 > (5 - h )6S. Hence h ;?: 5 and so h = 5 as h � t. By (S. 14) a ;?: 5. Hence (S.9) and (S. 17) imply that 252 ;?: 61a + 125 + 32(a + 3) ;?: (61 + 32)5 + 125 + 96. Hence 125 ;?: 93 + 25 + 19 which is not the case. By (S. 10) the result is established. 0
Thus Invoo (X�u+V)) i- (0) contrary to (S.4). Therefore B is not the u+v principal block. By (1.1) �
Thus
v (x ) = detxju)(x ) = 1,
(X� U ) )N Hence by F .7)
=
Vi
= detx\v) (x ) = 1.
V( YJ u, p - e ),
(X�u) 0 X}v))N 3 12 (S.20) = EB V( YJ u+va k , 2k + l) EB EB V(YJ " +va \ p). k =8 k =O Therefore 3 X(u) '<::Y Xcv) ffi 'CD L 2 k + l EB M, k =O where M is projective, each LZ k + 1 is indecomposable and (L2k +1 )N = V( YJ u+va k , 2k + 1) EB Tk with Tk projective. Since Bu+v is not the principal block it follows from (3.1) that Tk i- (0) for all k. Thus by (S.20) Tk is indecomposable for at least 3 values of k. By (1.1) detv(7)u+vak,2k + l) (x ) = 1 for all k. Hence I
tC\
J
�
380
[9
CHAPTER VIn
detV ( TJu+v'"k,p ) (x ) = 1 for at least 3 values of k with B � k � 12. However (1.1) implies that detV(TJu+ v",k,p ) (x ) = - a (x )k. Since - a (x t = 1 if and only if k == 2 (mod 4) it is impossible to find 3 values of k with B � k � 12 with the desired properties. 0 9. Proof of (7.2)
PROOF OF (7.2). By (7.5) it may be assumed that condition ( ) of section 7 is satisfied. If / Z / is odd the result follows from (7. 1). Thus it may be assumed that / Z / is even. Therefore e is odd as ( / Z /, e ) = 1. Hence t is even as te = p - 1 . I; is in B ) . If 8 (2) = 1 the result follows from (B. 1). Thus it may be assumed that 8 (2) = - 1. Let 'P be a nonprincipal irreducible Brauer (mod p ) with character which is constituent of 1;\2) . By (VII.10.B) 'P (1) 1 � � e. Thus � ! (p - 1). Since G is not of type L 2 (p) it follows from (3.1) that 'P (1) > p. Let L be an R [G] module which affords 'P. As 'P (1) > p there exists an integer k such that V(1] 2a \ p) I LN• Let X be an R -free R [G] module which affords I; such that X is indecomposable. Thus XN = V(1]V, p - e) for some irreducible character v of N/C. By (2.7) *
n
n
== n
a
e-)
p- I
;=0
j =2 e
(X 0 X)N = EB V(1] 2v2a i, 2i + 1) EB EB V(1] 2 v2a j, p). _
_
(9. 1)
Let IL = v2 and let W(i, 2) be identified as in (7.13). Thus (9.1) implies that e-l
X (29 X = EB W(i, 2) EB A i =O
for some projective R [G] module A. Let M(i, 2), M(2) be defined as in (7. 13). By (7.7) the character afforded by M(2) has an exceptional character in B2 as a constituent and the character afforded by M(i, 2) has the sum of two exceptional characters in B 2 as a summand for 0 � i � ! (e - 1). Thus L occurs with multiplicity at least 1 in W(He - 1), 2) and with multiplicity at least 2 in W(i, 2) EB W(e - 1 - i, 2) for 0 � i � He - 1). Consequently L occurs with multiplicity at least 2(!(e - 1)) + 1 = e in X (29 X. Since V(1] 2a \ p) is both proje�tive and injective it follows that eV(1] 2a k , p) I (X (29 X)N. Hence by (9. 1) e � t - 1 . Furthermore e � t - 2 unless a k = v2•
PROOF OF (7.3)
1 01
381
Suppose that e > t - 2. Thus e = t - 1 and a k = v2• Let X (29 X = + A EB A - where A + is the space of symmetric tensors and A is the space of skew tensors. By (2.9), W(i, 2) and W(e - 1 - i, 2) are both summands of A + in case i is odd and are both summands of A in case i is even. Since e = t - 1, V( 1] 2 a k , p) occurs exactly twice as a summand of each W(i, 2) EB W(e - 1 - i, 2) and exactly once as a summand of W(! (e - 1), 2). Therefore V (1] 2 a k, p) occurs as a summand of A N strictly more times than it occurs as a summand of A -::.. This contradicts (2.9). Thus e � t - 2. Since e is odd and t is even, e � t - 3. Therefore (9.2) p = et + 1 � ( t - 3)t + 1 = t2 - 3t + 1 . Furthermore e � (p - l)/e - 3 and so e 2 + 3e - (p - 1) � O. Hence e � - � + Vp + i Since 1;(1) = p - e the result follows from this inequality and (9.2). 0 -
-
10. Proof of (7.3)
Suppose that the assumptions of (7.3) are satisfied. Assume furthermore that condition ( ) of section 7 is satisfied. As 1;(1) = p - 2, it follows that e = 2. Thus / Z / is odd as (e, / Z / ) = 1. Hence 1] is algebraically conjugate to 1] 2. Since I; is in B l this implies that 1;�2)(1) = 1;(1) = p - 2 for i = 1 , . . . , t. In particular 8 (2) = 8 (1) = 1 . Thus 1;\2) does not have the principal Brauer character as an irreducible constituent. Since G is not of type L 2 (P) it follows from (3.1) that 1;\2) is irreducible as a Brauer character. The Brauer tree for B 2 looks as follows 1;�2) g X 'P *
�
Then X = 'P is irreducible as a Brauer character. Hence 'P (1) == 1 (mod p ). Observe that a 2 = 1 as e = 2. Let X be an R -free R [G] module which affords 1;. Let X (2) be an R -free R [G] module which affords l;i2) . Let L be an R -free R [G] module which affords X. Thus f affords 'P. By (I.17.12) there exists an R -free R [G] module Y which affords L;:� 1;�2) such that Y is indecomposable. Let X (29 X = A + EB A , where A + is the space of symmetric tensors and A is the space of skew tensors. -
-
382
CHAPTER VIII
[10
(i) X(2) = V( YJ 2a, p - 2) (ii) L = V(YJ 2, 1).
LEMMA_ 10. l .
(i) Suppose the result is false. Thus X (2) = V( YJ 2 , P - 2). By (2.9) (A -)N = V(YJ 2 a, 3) EB � (p - 5) V(YJ 2 a, p). By (2.7) H O(N, (I), HomR (X (2), A -» = H O(N, (I), X(2) * 0 A -) i' (0). Thus by (111.5. 10) HO(G, ( 1 ), HomR (X(2 , A - » i' (0) and so HomR[ G ] (X(2), A - ) i' (0). Since X (2) is irreducible this implies that X(2) is isomorphic to a submodule of A -. Consequently V( YJ 2, P 2) = X�) is isomorphic to a submodule of A N . This is however impossible as InvEP (A EP) = (0). This proves (i). Suppose_ )hat (ii) is false. Then L = V(YJ 2 a, 1) and so (111.5 . 1 3 ) and (I11.5. 10) by Thus HomR[N] (L, X (2) � (O). HomR[ G ] (L, X (2) � (0) which is not the case. 0
PROOF.
-
-
,......."
>
-
LEMMA 10. 2 . PROOF.
A
=
¥.
I
Hence ¥ = V(YJ 2 a \ 3) for k = 0 or 1. By definition HomR[ G ] (L, ¥) = (O). Hence by (111.5.10) HO (N, ( I), L � 0 YN ) = HO(N, (I), HomR (L, Y» = (0). Thus by (10. 1) (ii) k i' 0 and so Y = V(YJ 2 a, 3 ) . Hence by (2.9) ¥ I A . Since dimR ¥ = ! (p - 2) (p - 3 ) = dimR A it follows that A - = ¥. 0
=1
-
PROOF. Clearly - 1 + (n + l)p. ! (p - 1) (P - 2). LEMMA
np with n � ! (p - 3 ) . g ( l ) =
- 1 + (n
+ l)p.
X ( l ) = 1 + np for some n. Thus g ( l ) = P - 2 + 1 + np = By (2.9) and (10. 1) L I A +. Consequently 1 + np ::oS This implies that n � ! (p - 3 ) . 0
lOA. Let x be a {2, p}' element in G, x � z. Then ((x ) = O.
Let () + , () - be the character afforded by A + , A - respectively. Then it is well known that () + (x) = l r (x ) + ((x 2). Hence by (10.2) () + (x ) = Hp - 3)( ;2)(X ) + ((x 2) and so PROOF.
383
Let be the automorphism of the field of I G Ith roots of 1 over the rationals which fixes all 2" th roots of 1 and squares all m th roots of 1 for m odd. Thus ((X 2) = ( (x ) and ( \2) (X ) = ( <7 (x ). Hence C (x ) = (p - 2) Cr (x ). Choose k so that a k = 1. Thus Ck (x ) = (p - 2)2 k- 1 ( <7 k (x) = (p - 2)2L I ((x ) . Hence if ( (x ) i' 0 then CL I (x ) = (p - 2yLl and so I ((x ) 1 = p - 2 = ((1). Hence x E Z contrary to assumption. 0 a
<7
(i) I G Z 1 1 2bp (p - 2)2 for some positive integer b. z I = 1 then I G I = 2bp (p - 2) for some positive integer b. PROOF. L et be a prime with i' 2, p. Let Q be a Sq -group of G. By (lOA). I (Q I ' I b I I Q n z lw)' I Q(�Q�� I · Thus I Q : Q n I I (p - 2y. This proves (i). If I z 1 = 1 th�n ( (0 , 1 0 ) = ((1)1 I Q I by (lOA). Thus I Q I I (p - 2). This proves (ii). (ii) If I
:
LEMMA 10.5.
q
q
�
z
I2 2: i () ) ( 1 ) = - 2(t - 1) == 3 (mod p).
LEMMA 10.3. x (1) = 1 +
PROOF OF (7.3)
�
As t = ! (p - 1) it follows that -
10J
0
PROOF OF (7.3). By (7.5) it may be assumed that condition ( ) of section 7 is satisfied. Suppose first that I = 1. By (10.5) I G I = p 2b (p - 2) for some positive integer b. Since G is not of type L 2 (p) the result follows from (5.3) . Thus it may be assumed that I z 1 . Hence X ( I) i' 1 as G = G' and so n > 0. Suppose that n is even. Then by ( 10.5) 1 + np I (p - 2)2. Let d = (1 + np, (p - 2» . Thus 2n + 1 0 (mod d ). Hence 1 + np I (2n + 1 )2 . If 1 + np = (2n + 1 )2 then p = 4(n + 1) which is not the case. Thus 1 + np � ! (2n + lY. This implies that p � 2n + 2 contrary to ( 10.3 ) . Suppose that n is odd. Hence by ( 10.3) and (10.5) {(n + l)p - 1} = ((1) I (p - 2)2 . If d = (p - 2, (n + l )p - 1) then 2n + 1 0 (mod d ) and so {(n + l)p - 1} 1 (2n + 1Y. If (n + l )p - 1 = (2n + lY then p = 4n2 n+ +4n1 + 2 = 4n + _2_ n+l Hence = 1 as (n + 1) 1 2 and so p = 5 contrary to assumption. Therefore {(n + l )p - 1} ::oS H2n + 1)2 and so p � 2n + 1 contrary to ( 10.3). 0 *
Iz
I>
==
==
n
384
CHAPTER VIn
[11
1 1. Some properties o f permutation groups The material in this section, except for (1 1 .8), is independent of the previous results of this chapter and is included h ere as it will be needed in the next section. We will state several results without proof. If G is a permutation group on a set D, let ax denote the image of a under the action of x for a E D, x E G. If .1 is a subset of G let G", denote the subgroup of all elements in G which leave every element of .1 fixed. In other words GLl = {x I x E G, ax = a for all a E L1 }. If .1 = {a} let Ga = G{ a } . For a E D let a G = {ax I x E G}. A proof of the following result can be found in Wielandt [ 1 964] (13.1).
1 1. 1 (Jordan). Let G be a primitive permutation group on D. Let .1 be a subset of D with 1 � 1 .1 1 < i D 1 - 1 . If Gil is transitive on D - .1 then G is doubly transitive on D.
THEOREM
If G is a permutation group on D then Wielandt has defined the group G (2) to consist of all permutations on D which preserve all the orbits of G in D x D. Clearly G � G (2) . Suppose that G is a permutation group on D. Let .1 be a subset of D and let H be a subgroup of G such that ax E L1 for a E .1, x E H. For x E H define the permutation x il as follows ax il = ax, If a E .1, if
a E D - .1,
ax il
=
[1969] (6.5).
THEOREM 1 1 .2 (Dissection Theorem). Let G be a permutation group on D. Let .1 be a subset of D and let H be a subgroup of G such that ax E .1 for a E .1, x E H. Assume that for all a E .1, b E D - .1 , H = HaHb. Then
H il
X
H fl -Ll � H( 2) .
PROOF. We will first show that if x E H then X il E H(2 ). It suffices to prove that if (a, b ) E D x D then there exists z E H such that (a, b )z = (a, b )X Ll. There are three cases . If a, b E .1 let z = x. If a, b E D - .1 let z = 1 . Suppose that a E .1, b E D - .1 . B, y assumption x = XaXb with Xa E Ha, Xb E Hb. Let z = Xb. Then
385
(a, b )x il = (ax, b ) = ( aXaXb, b ) = (axb, b ) = (a, b )Xb. This proves that x Ll E H (2). Let x E H then x = x Llx n - il . Since X Ll E H(2) it follows that x n-Ll E H (2) . Thus HilH f2 - Ll = H Ll X H f2- Ll � H (2 ) . 0 We will next state some results without proof about rank 3 permutation groups. These are due to D.G. Higman [1964] though special cases had previously been considered by Wielandt [1956]. We will restrict our attention only to those results which are necessary for the considerations of the next section. Let G be a rank 3 permutation group on D. Assume that I D I is even. Let I G X + f) be the character afforded by the permutation representation of G on D where X, f) are irreducible characters of G. Since I D I is even, X (I) /: 8 (1) and so X, 8 are rational valued. For a E D let {a }, .1 (a ), r(a ) be the orbits of Ga where .1 (ax ) = .1 (a )x and r(ax ) = r(a )x for x E G. Let k = I .1 (a ) l , 1 = l r(a ) l. Since I G I is even, k /: and so .1 and r are self paired in the sense that
+
I
.1 (a )
=
{ax I ax - 1 E .1 (a )},
Define the integers
fL,
A by
I d (a ) n d (b ) 1
r (a ) = {ax I ax - 1 E r (a )}.
C . {I + +A + + (A -fL)(k + I) + k + I �
if
b E .1 (a),
if
b E r(a ).
(1 1 .3)
Then A, fL are independent of the particular choice of
a.
Let H il = {x il I x E H}. The following result is due to Wielandt
SOME PROPERTIES OF PERMUTATION GROUPS
1 1]
I r(a ) n r(b)
1=
- k fL - 1
1-k
1
if
b E r(a ),
if
b E .1 (a ).
a, b.
Furthermore
(11 .4)
The following conditions are satisfied. 2 Let d = (A - fL ) + 4(k - fL ) . Then d is a perfect square and
}
8 (1) 2k = + X (I)
2 Yd
2
.
(11 .5)
If furthermore s, t are the characteristic values of the incidence matrix associated to the orbit .1 then
}
s = (X - fL ) ± Yd (11.6) t 2 · Let p be a prime. Let G be a primitive permutation group on 2p letters . In case p = 5, examples of this situation are provided by A s and Ss acting on the set of 2-element subsets of {1 , . . . , 5}. All other known examples of
386
[1 1
CHAPTER VIn
such groups G are doubly transitive groups. The main object of this and the next section is to investigate such groups G which are not doubly transitive. We begin by stating without proof the following result which is the starting point of all work done on the structure of such groups.
[1956], [1964]). Let p be a prime. Let G be a primitive permutation group on n where / n / = 2p. Assume that G is not doubly transitive. Then 2p = m 2 + 1 for some integer m. Furthermore G has rank 3 on n and the following conditions are satisfied, where the notation is that introduced above. (i) X (I) = p, 8 (1) = P - 1. (ii) k = � m (m - l), I = � m (m + l). (iii) s + t = - 1. THEOREM 1 1 .7 (Wielandt
THEOREM 1 1 .8 (Ito [1962cD. Let p, G, n be as in ( 1 1 . 7). Let P be a Sp -group of G. Assume that p > 3 and / N G (P) / = 2p. Then p = 5 and G = A s . PROOF. The Brauer tree of the principal p -block of G is as follows
1 p-l
0-----0----0
.
Thus there exists an irreducible character ( of G with ((1) = P - 2. Since Op' ( G ) = (I) it follows that ( is faithful. By (7.3) P = 2a + 1 for some a and G = SL2 (p - 1). Since G has a subgroup of index 2p the known properties of PSL2 (p - 1) imply that p = 5 and G = SL2 (4) = A s . 0 LEMMA
PROOF. Hence
11.9 (Ito [ 1 967a]). Let p, G, n be as in (11 .7). Then f.L = A + 1 = � (m - 1)2. A = H m + l) (m - 3), By (11 .6) and ( 1 1 . 7) (iii) A - f.L = - 1. By (11.5) and ( 1 1 . 7) '
± Vd = ± Vd(X ( I ) -
8(1)) = 2k - (k + I) 4( k - f.L ) = d - (A - f.L ? = m 2 - 1
=
- m.
PERMUTATION GROUPS OF DEGREE 2p
1 2]
387
LEMMA 1 1 . 10 (Ito [1967a]). Let p > 3, G, 12 be as ( 1 1 . 7). Then the following hold. (i) Ga is faithful in its action on r(a ). (ii) Let H be the kernel of the action of Ga on .1 (a ). Then H is an elementary abelian 2-group and every orbit of H on r(a ) has cardinality 1 or 2 . . m
PROOF. (i) Let H be the kernel of the action of Ga on r(a ). Let {cJ, . be an orbit of H in .1 (a). Since H � GT(a ) it follows that
. . , Cn }
r* = r(a ) n r(cJ) = . . , = r(a ) n r(cn ). By (11.4), ( 1 1 . 7) and (11.9), / r /* = m + A + 1 = m + f.L. Let i � j. Then r* � r(Ci ) n r(Cj ). By (11 .4), (1 1 . 7) and (11 .9) / r(Ci ) n r(Cj ) / :!S m + f.L. Hence r(Ci ) n r(Cj ) = r* for i � j. Thus {r(Ci ) - r*} is a collection of pairwise disjoint subsets of .1 (a ) and so n O m (m + 1) - m - H m - 1)2} = n O m (m + 1) - m - f.L } :!S � m (m - l). This implies that n :!S 2m/(m + 1) < 2 and so n = 1. Thus H acts trivially on Ll (a ) and so H = (I). (ii) Let {cJ , . . . , cn } be an orbit of H on r(a ). Since H � G4 (a) it follows that .1 * = .1 (a ) n .1 (c J ) = . , . = .1 (a ) n .1 (cn ). By (11 .3) and (11.9), 1 .1 /* = f.L = A + 1 . Let i � j. Then .1 * � Ll (Ci ) n .1 (Cj ). B y (1 1 .3) and (11 .9) / .1 (Ci ) n .1 (Cj ) / :!S f.L. Hence .1 (Ci ) n Ll (Cj ) = .1 * for i � j. Thus {.1 (Ci ) - Ll *} is a collection of pairwise disj oint subsets of r(a ) and so n O m (m - 1) - � (m - 1)2} = n O m (m - 1) - f.L } :!S ! m (m + 1). As p > 5, m > 3. Thus n :!S 2m/(m - 1) < 3. Thus n = 1 or 2. Hence every orbit of H on n has size 1 or 2 and so H is an elementary abelian 2-group.
0
and so by (1 1 .7) (ii) Thus
4f.L = 4k + 1 - m 2 = m 2 - 2m + 1 = (m - If 4A = 4f.L - 4 = m 2 - 2m - 3 = (m + l) (m - 3). 0
The next result is a weak form of a theorem of Ito but is sufficient for what is needed in section 12. Ito actually proved that H = (I) for p > 5 .
12. Permutation groups of degree
2p
Let p be a prime. Let G be a primitive permutation group on 12 where / 12 / = 2p. Assume that G is not doubly transitive on n. By ( 1 1 .7)
2p = m 2 + 1 for some integer following result.
m.
The object of this section is to prove the
THEOREM
prime.
[12
CHAPTER VIII
388
12.1
(L. Scott
[1969], [1970], [1972]). If m > 3 then m is not a
We will give Scott's proof as presented in [1970], [1972] which is a simplification of an earlier proof given in [1 96 9] This proof depends on the results in section 1 1 as well as on the results in Chapter VI. In particular (12. 1) implies that if p < 313 then p = 5, 41 or 1 13. The cases p = 41 or 1 13 have been shown to be impossible in Scott [1976] by special arguments. Thus it is known that if G exists with p > 5, then
12J
PERMUTATION GROUPS OF DEGREE 2p
389
Let A be the normal closure of x in M. Then A fixes all letters in D - 8. Since Mb is transitive on .1i n 8 and A <J M it follows that .1i n 8 � b A for i = 1, 2. Clearly b E b A . Hence E � b A • Thus by Jordan's theorem (11 .1), G is doubly transitive on D contrary to assumption. 0
.
p ? 313.
Throughout the remainder of this section it is assumed that (12.1) is false and G exists for some prime q = m > 3. A contradiction will be derived from this assumption. Let G( 2) be defined as in section 1 1 . Since G is not doubly transitive on D it follows that 0(2 ) is not doubly transitive on D. Thus G may be replaced by G( 2). Hence it may be assumed that G = G(2). Let 0 b e a Sq -group o f G.
LEMMA
12.2. I 0 I = q.
PROOF. Suppose that the result is false and 1 0 1 > q. Choose a E D with o � Ga . Since the orbits of Ga on D - {a } have cardinalities � q (q - 1) and � q (q + 1) by (1 1.7), it follows that a is the only point in D fixed by O. Furthermore every orbit of 0 on D - {a } has cardinality q since � q (q + 1) < q 2 . Let b E D - {a }. Then l O : H I = q, where H = Ob. Let 8 be the set of fixed points of H on D. Choose x E 0 - H. If c E D - 8 then c H = C O as I c H I = I c O l = q. Thus 8 is a union of orbits of O. For d E 8, C Od = C H = C O. This implies that QcOd = 0 for c E D - 8 and d E 8. Thus Wielandt's Dissection Theorem (11.2) implies that O E X O n-E � G. Thus it may be assumed that x fixes all letters in
D - 8. Let b E 8 - {a }. Let {b}, .1 1 = .1 (b), .1 2 = r(b) be the orbits of Gb on D. Thus x � Gb• Suppose that .1i � D - 8 for some i. Thus Gb �
Let 10 + 8 + X be the character afforded by the permutation representa tion of G on D. By (11 .7) x (l) = p = Hq2 + 1) and 8 (1) = P - 1 = Hq2 - 1). Let C = Co (0), N = No (0). Thus C = 0 x A for some group A. Since a is the only fixed point of 0 it follows that N � Ga .
12.3 X and 8 are nonexceptional characters in some q-block B of G which is not the principal q -block of G. .
LEMMA
.
PROOF. Since N � Ga , (111.5.6) implies that 1 0 + 8 + X = (l oa ) 0 = 10 + <1>, where <1> is the character of a projective Qq [ G ] module. Since X (I), 8 (1) are not divisible by q it follows that <1> = X + 8 and X, 8 are in the same q -block B. X, 8 . are rational valued, thus they are not exceptional charac ters in B. Sinc'e X (I) � ± 1 (mod q ) as q > 3, B is not the principal q -block. 0
12.4. There exists a character I/J of A with 1/J (1) = � (q - 1) such that 8c = I/J + Po{3, Xc = (Po - 1 0 ) I/J + Po, where {3 is a character of A and Po is the character afforded by the regular representation of O.
LEMMA
PROOF. By
(VII.2. 17)
8c = ( apo ± 1 0 ) 1/J + Po {3 for some integ�r a. Since Ie is a constituent of 8c by the Frobenius reciprocity theorem and Ie is not a constituent of tf; it follows that (3 ;1 O. Furthermore I/J (l) ± 8(1) =+= � (mod q ). Thus 1/J(1) ? Hq - 1). If a ;l 0 ==
then
==
Hq2 - 1) = 8 (1) ? (q - l H (q - 1) + q = H q2 + 1). Thus a = 0 and so the + sign must occur. Hence 1/J (1) - � (mod q). As X i s i n the same block as 8 i t follows from (VII.2.17) that Xc = (a ' Po - 1)1/J + Po{3 ' since X (1) + 8 (1) 0 (mod q ). Since I e is a constituent of Xc it follows that {3 ' ;I O. Clearly a ' ;I O. Hence ==
==
390
CHAPTER VIII
Hq 2 + 1) = x(1) = (a ' q 1)I/I{1) + q/3 ' (1). As /3 ' (1) 3 1 and I/I (1) 3 Hq - 1) this is easily seen /3 ' (1) = 1 and 1/1(1) = Hq - 1). Hence /3 = 1A. 0
[ 12
-
to imply that
a' =
LEMMA 12.5. Let CPo be the principal indecomposable character of Ga corresponding to the trivial Brauer character. Then A is in the kernel of
By the Frobenius reciprocity theorem XOa contains l O as a a constituent. Let R be the integers in a suitable extension field of Qq- Let X be an R -free R [ G ] module which affords X. Thus by (111.7.7) XOa = Y EB Yo where Yo is proj ective and Y is in a p -block of Ga which corresponds to B under the Brauer correspondence. Since Inv Oa Y = (0) by (V.6.2) it follows that Invoa ( Yo) = (0). Now (12.4) implies that rankR Yo = q. Hence Yo affords CPo and CPo (l) = q. Thus A is in the kernel of CPo . 0
LEMMA 12.6. A is an elementary abelian 2-group which acts trivially on .1 (a). Furthermore all the orbits of A on r(a ) have cardinality 1 or 2.
H denote the kernel of CPo, where CPo is defined as in (12.5). H is a q '-group. By (12.5) A c.: H. Since / .1 (a ) / = ! q (q - 1), / r(a ) / = ! q (q + 1) and H <J Ga, the orbits of H on .1 (a ) have cardinality relatively prime to the cardinality of the orbits of H on r(a ). Thus if b E d (a ), then Hb is transitive on all the orbits of H in r(a ). Therefore if b E {a } U .1 (a ) and c E r(a ) then H = H�c. Thus by Wielandt's Dissec tion Theorem (11 .2), G � H(2 ) = HJ X H2 = H where HJ acts trivially on .1 (a ) and H2 acts trivially on r(a ). By (11 . 10) H2 = (1) and H = H1 is an elementary abelian 2-group such that the cardinality of every orbit on T(a ) is 1 or 2. The result follows as A c.: H. 0 PROOF.
Let
Clearly
LEMMA 12.7. The multiplicity of 1A in (10 + () + X)A is at most Hq 2 + q + 2). PROOF.
n be the multiplicity of 1A in (10 + () + X )A. n = (lA, 1A + ()A + XA ) � 1 + () (1) - 1/1 (1) + q = q + 1 + Hq 2 - 1 - q + 1) = 4 (q 2 + q + 2). 0
Let
By
(12.4),
It is now easy to show that (12.6) and (12.7) are incompatible. By (12.6) the multiplicity n of 1A in (10 + () + X)A satisfies
CHARACTERS OF DEGREE P
13]
391
n 3 1 + Hq 2 _ q ) + Hq 2 + q ) = H3q 2 _ q + 4). Hence (12.7) implies that (3q 2 _ q + 4) � 4n � 2q 2 + 2q + 4. Consequently q2 - 3q � 0 and so q � 3 contrary to completes the proof of (12. 1). 0 13.
Characters of degree
assumption. This
p
In this section groups which satisfy the following conditions will be considered. (i) (ii)
(iii)
A Sp-group P of G has order p and C o (P) = P. , Let N = No (P). Then N = EP with / E / = e, E n P = (1) and No (E) = 2e. Furthermore E is a T.1. set in G, (i.e. E n E X = ( 1 ) for x � No (E )). Every irreduCible character of G which is not in the principal p-block has degree p.
() *
The object of this section is to pr�ve the following result .
THEOREM 13.1 . Suppose that ( ) is satisfied. Then either G is of type L 2 (p) or p = 3 and G /03' ( G ) = PS L2 (5) . *
Before proving
(13.1)
we deduce two consequences.
COROLLARY 13.2 (Richen [1972]). Suppose that ( ) is satisfied and G has a unique irreducible character of degree p. Then p 3 2 and G = PSL2 (p ) . Conversely if p 3 5 then PSL2 (P) satisfies ( ) and has a unique irreducible character of degree p. *
*
An irreducible character X of G is in the principal block Bo of G if and only if Op ' ( G ) is in the kernel of X and X is in the principal block of G /Op , ( G ). Suppose that Op , ( G) � (1). Let a = / Op , ( G ) / and let bp = / G : Op , ( G ) / . There is a unique irreducible character () of G which does not have Op ' ( G ) i n its kernel and () (1) = p. Thus
PROOF.
Hence
abp = / G / = / G : Op, (G) / + p 2 = bp + p 2. (a - l)b = p. Since p l' b this yields that b = 1
and
a = p + 1.
392
CHAPTER VIn
[ 13
Hence N = P and so E = (I). Thus 2 = I N G ( E ) I = I G I which is not the case. Therefore Op' (G) = (I). Hence by (13 . 1) N = G or there exists H <1 G with either H = PSL2 (p ) or p = 3 and H = PSL2 (5). Since G has an irreducible character of degree p, N� G. As C G (P) = P it follows that I G : H I � 2. Let () be the irreducible character of G of degree p. If G � H then () G is either irreducible of degree 2p or the sum of two distinct irreducible characters of degree p. Neither of these possibilities can occur by assumption. Thus G = H. Since PSL2 (5) has two irreducible characters of degree 3 it follows that G = PSL2 (P ). The groups PSL2 (2) and PSL2 (3) do not satisfy condi tion (ii) of ( * ) . Thus p � 5. The converse is well known. 0
Let G be a transitive permutation group of prime degree p which has a unique irreducible character of degree p. Let P be a Sp -group. Assume that NG (P) = PE with P n E = (I), I NG (E ) I = 2 1 E I and E a T.I. set in G. Then p = 5, 7 or 1 1 and G = PSL2 (P ). COROLLARY 13.3 (Ito [ 1 963bD.
PROOF. It is well known that PSL2 (P ) does not have a subgroup of index p if p > 1 1 . Thus the result follows from (13. 2 ). 0
The rest of this section contains a proof of (13.1). Suppose that the result is false. Let G be a counterexample of minimum order. The following notation will be used. P = (y ), E = ( z ) , H = NG (E), u E H - E. {I = A I , . . . , AJ is the set of all irreducible characters of E with A � = Ai . {JL l , JL �, . . . , JLI' JL � } are the remaining irreducible characters of E. Thus s + 2 t = e. Furthermore JL :: is irreducible for all k and A � is the sum of two distinct irreducible characters of H for all i. LEMMA 13 .4. G is simple. PROOF. Clearly G/Op,(G) satisfies ( * ) . Thus the minimality of G implies that Op, (G) = (l) . Let Go be the subgroup of G generated by all p elements i n G. I f Go does not satisfy ( * ) then N G ( E ) n Go = E n Go. Hence E n Go is a Hall subgroup of G. By Burnside's transfer theorem E n Go is isomorphic to a quotient group of Go . Thus E n Go = (I). Another application of Burnside's transfer theorem yields that Go has a normal p -complement and so G is p -solvable contrary to assumption. Thus G satisfies ( * ) and so G = Go is simple. 0
CHARACTERS OF DEGREE P
13J
393
LEMMA 13.5. If e is od/1 then s = 1 and JL :: = JL :: for 1 � k � t. PROOF. Since E is a T.!. set with I H 1 = 2 e it follows that E is a Hall subgroup of G. Hence (13.4) and Burnside's transfer theorem imply that H is a dihedral group. The result follows. 0
LEMMA 13.6. Let () be an irreducible character with (} (1) = p. Then «(}E, JLk ) = 0 for all k and «(}H, Ai ) = 1 for exactly one value of i with 1 � i � s. PROOF. JL i: for
(}N is a principal indecomposable character of N. Thus (}N = A � or some i or k. Since (}E is a class function (}N � JL i:. Thus (}N = 'A f. 0
LEMMA 13.7. If 1 � i � s, 1 � k � t then II (Ai - JLk ) G W = 3. PROOF. Since E is a T.I. set and ( Ai - JLd (1) = 0 it follows that
II ( Ai - JLk ) G 112 = " ( Ai - JLk )H W = 3.
0
LEMMA 13.8. If t � 0 and 1 � k � t then G (A I - JLk ) = 1 + () - Xk,
where (), Xk are irreducible characters, ,(}N = A i", (Xk )N = JL i: + JL �, where JL � is considered to be a character of N /P = E. = 1 + a - {3 for irreducible characters a, (3 of G. Since ( A 1 - JLk ) G is p -rational it follows that a (1), {3 (1) = O, ± 1 (mod p ) and 1 + a (l) = (3 (1). Since t � 0 i t follows that e � 3 and s o p > 3 . Thus
PROOF. By (13.7) (A I - JLk t
either a l (1)= O (mod p ) or {3 (I) = O (mod p ). By (13.6) and Frobenius reciprocity (111.2.5) (3 (1) � p. Hence a (1) = p and {3 (1) = P + 1 . Let () = a, Xk = {3. By (13.6) and Frobenius reciprocity (}N = A i". Direct computation yields «A l - JLk t )N = A l + A i" - (JL � + JL i:).
Henc� (XdN = 1 + (}N - « A 1 - JLk t )N is as required.
0
LEMMA 13 .9. t = O. PROOF. Suppose not. Let (), Xk be defined as in (13.8).
Assume first that e is odd. By (13.5) Xk = Xk . Thus Xk is on the real stem of the Brauer tree T of the principal block. As e is odd, T, and hence the real stem has an even number of vertices . Thus not both end points
394
CHAPTER VIII
[13
correspond to characters with degree d == 1 (mod p ). Since 1 G is an end point, Xk cannot be an end point and so Xk is reducible as a Brauer character. Let 'P be an irreducible Brauer character with 'P (1) � � Xk (1) = � (p + 1) which is a constituent of Xb Since Xk (1) == 1 (mod p) 'P is not the principal Brauer character. Hence 'P is faithful by (13.4). If P > 7 then � (p + l ) < Hp - 1) and so G is of type L 2 (P ) by (3. 1) contrary to assump tion. Since e is odd and e > 1 , p > 5. Suppose that p = 7. Then e = 3 and T , looks as follows , 1
a
8
+ 2t
e � a (1) � ! (p + 1) it follows that p � 7. Suppose that p = 7. 6 = e � a (1) � 4, which is not the case.
e I (p - 1) this implies that
Suppose finally that (A 2 - J.Lk )G = - Xk + a - {3. If a (1), {3 (1) � 0 (mod p ) then a (1), {3 ( 1 ) == ± 1 (mod p ) as (A 2 - J.Lk )G i s p -rational. Hence - 1 ± 1 ± 1 == 0 (mod p) and so p � 3 contrary to the fact that e � s + 2 t � 4. By (13.6) {3 (1) i' p. Hence a (1) = P and so (3 ( 1 ) = 1 contrary to (13 .4). 0 LEMMA 1 3 . 10.
e
=
2.
PROOF. By (13.9) t = O. Thus H is abelian. By' (13.4) and Burnside's transfer theorem H is a 2-group. Let x be the unique involution in the cyclic group E. If e i' 2 then (x ) is a characteristic subgroup of H. Hence N G (H) C N G « x ) = H. Thus H is a S 2-grOUP of G and Burnside's transfer theorem implies that G has a normal 2-complement contrary to (13 .4). D LEMMA 1 3 . 1 1 . There exist pairwise ters a, {3, y of G such that
PROOF OF (13. 1). Suppose first that one of a, {3, or y in . (13. 1 1) is an . ex�e � tlO�al char �cter in the principal block. Since (A I - A2)G is p -rational thIS ImplIes that 2 (P - 1) = (p - l )/e � 2 and so p = 5. The Brauer tree of the principal block is �o
=
Since
395
PROOF. Since E is a T.!. set it follows that II (A I - A2)G W = II (A I - A 2) H W = 4. The result follows as (A I - A2)G (1) = O. D
b
where the numbers denote the degrees of the corresponding characters. Thus either a = 3 or b = 3. The classification of all finite groups with a faithful complex representation of degree 3 implies that G = PSL2 (7) contrary to assumption. Hence e is even. Therefore s is even and so s � 2. By Frobenius reciprocity (J.L �, Xk ) = 1 and (A ?, Xk ) = 0 for all i. Hence (A2 - J.Lk )G = - Xk + a ± {3 for irreducible characters a, {3 of G . Suppose that (A2 - J.Ld G = - Xk + a + (3. Then a (l) + (3 (1) = P + 1 and so it may be assumed that a (1) � ! (p + 1). Since (l , (A 2 - J.LdG ) = 0 it G follows from (13.4) that a is faithful. Hence G is of type L 2 (p) by (3 . 1 ) if P > 7 contrary to assumption . Since 4�s
CHARACTERS OF DEGREE P
13]
distinct nonprincipal irreducible charac
1
a+1 a
Thus if a, {3, y are all in the principal block then 1 ± (a + 1) ± 2a = 0 and so a = 2 contrary to the fact that a + 1 == ± 1 (mod 5). Hence one of the degrees of a, {3, 'Y is 5. Thus the exceptional degree is 3 as 1 + 5 = 2.3 and so the classification of the finite groups with a faithful 3-dimensional complex representation yields that G = PSL2 (5) contrary to assumption . Thus none of a, {3, y is exceptional in the principal block. Since e = 2 by (13. 1 0) there are only 2 nonexceptional characters in the principal block unless p = 3 in which case there are 3 non exceptional characters. If p = 3 then one of a, (3, y has degree 3 and so G = PSL2 (5) contrary to assumption by inspection of the groups with a faithful complex representation of degree 3 . Suppose that p > 3 then a t least 2 o f the degrees a ( 1 ) , (3 (1), y (l) are p. Thus exactly two of them are p and the other degree is 1, 2p - 1 or 2 p + 1 . The case o f degree 1 i s impossibly b y (13.4). Hence the possible Brauer trees are as follows �
1 2p + 2
2p + 1
o-----cr----o
1 2p
-
1 2p - 2
If 8 is an irreducible character with 8 ( 1) = P then (8, (A I - A 2) G ) i' 0 by (13.6). Hence G has exactly 2 irreducible characters of degree p. Since (2p + 26) I I G I for 6 = 1 or - 1 this implies that 2 2 o == 1 G 1 = 2p + 1 + (2p + 6 f + � (p - 1) (2p + 26 ) ==
Hence
p
=
4
(mod p
5 and 6
=
+ 6 ).
- 1 as
p > 3.
Therefore
1 G 1 = 50 + 1 + 81 + 128 = 260. However 2 p + 2 6 8 and so (2p + 26 ) l' I G I .
lishes the result.
=
D
This contradiction estab
THE STRUCTURE OF A ( G )
1J
CHAPTER IX
Throughout this chapter R is either a field of characteristic p or the ring of integers in a finite extension K of Qp . Let R denote the residue class field. Thus R = R in case R is a field. For any R [ G ] module V let ( V) denote the isomorphism class of R [ G ] modules which contains V. Let C be a field of characteristic O. The representation algebra Ac (R [ G ] ) was defined in Chapter II, section 4. We will frequently write Ac ( R [ G ]) = A ( G ) if no special reference to C or R is required. This chapter contains results related to the structure of A ( G ). Let Sj be a nonempty set of subgroups of G. Let A( l ) ( G ) = Ac� ( R [ G ]) be defined as in (IIA.3). If H is a subgroup of G let AH ( G ) = A{H}(G). By (IIA.3) A�(G) is an ideal of A ( G ) and if ( V) ranges over the isomorphism classes of indecomposable R -free R [H]-projective R [ G ] modules for H E f;) then {( V)} is a C-basis of A,\) ( G ). It is natural to ask for conditions under which Ac (R [ G ]) is semi-simple. In section 2 it is shown that this is the case if a Sp -group of G is cyclic and R = R is a field. (A complete proof is not given as a result of Lindsey is assumed.) In case p = 2 and 8 ,f I G I then Ac (R [ G ]) is also semi-simple . The reader is referred to Conlon [ 1 965] , [ 1 966] , [ 1 969] ; Donovan and Freislich [1976] , and earlier papers of Basev [ 1 961] and Heller and Reiner [ 1961] which contain a classification of R [P] modules in case p = 2 and P is a noncyclic group of order 4. For related results see Wallis [1969] , Hernaut [ 1 969] , Muller [ 1 974a] , [1974b], Erdmann [1979a] , Donovan and Freislich [1978] . In this connection we mention some other known results which will not be treated here. If R = Zp is the ring of p -adic integers and G has a cyclic Sp -group then Ac (Zp [ G] ) is semi-simple. See Reiner [ 1 966b]. If G has noncyclic Sp-groups then Ac (Zp [ G ]) contains nonzero nilpotent elements. See 396
397
Reiner [ 1 966a] , Gudivok, Goncarova and Rudko [1971] . This has been generalized to the case that R is a finite extension of the p -adic integers by Gudivok and Rudko [ 1 973] . If R = R is a field and p � 2 it has been shown by Zemanek [ 1 971] that if a Sp -group of G is not cyclic then Ae (R [ G ]) contains nilpotent elements. Zemanek [1973] has also investigated the case that G is of order 8. See also Yamauchi [1972] , Bondarenko [ 1 975] , Ringel [ 1 975] . Various types of representations and modules have been studied. See for instance Carlson [1974] , [ 1 976a] , [ 1 976b] or Janusz [ 1 970] , [1971], [1972] or Johnson [ 1 969a], [ 1 969b] . In sections 3 and 4 yet another type of module will be discussed.
1 . The structure of A ( G)
A (G) is a commutative algebra . Furthermore dime A ( G ) is finite if and only if G has a cyclic Sp -group. This section contains some results concerning the structure of A ( G ). Let P be a p -group in G. Let !S(P) be the set of all proper subgroups of P. Define Wp ( G ) = Ap ( G )IA 5( p ) ( G ). By (III.5.9) Wp ( G ) = Wp (N G (P)).
THEOREM 1 . 1 (Conlon [ 1 967] , [ 1 968]). A (G) = EB Wp (N G (P)), where P
ranges over a complete set of representatives of the conjugate classes of p -groups in G and the sum is a ring direct sum. The proof given here is a simplification of Conlon's proof. As an immediate consequence of ( 1 . 1) one gets
COROLLARY 1 . 2 (Green [ 1 964]). A ( G ) is semi-simple if and only if
. Wp (N G (P))
is semi-simple for every p -group P in
G.
The next result was proved by Lam [ 1 968] in case R is a field . Related results may be found in Conlon \[1967] , [ 1 968] ; Wallis [ 1 968] .
THEOREM 1 .3 . (i) Let P be a p-group in G. Then AI' (N G (P)) is isomorphic to
a subalgebra of EB Ap (H), where H ranges over all subgroups of NG (P) such that P � H and HIP is cyclic. (ii) A ( G ) contains a nonzero nilpotent element if and only if there exists a p-group P in G and a subgroup H of N G (P) with HIP a cyclic p i_group such that A (H) contains a nonzero nilpotent element.
[1
CHAPTER IX
398
Por future reference the following elementary result is included here. LEMMA 1 .4. Assume that R = R is an algebraically closed field. Let P be a
p-group and let H be a p '_group. Then A (P x H) = A (P) (9 A (H). PROOF. Clear by (IIL3 .7).
1]
is
THE STRUCTURE OF A (G)
split.
Since
U (9 VI I U (9 v.
U (9 VI
0
is
R [ T]-projective
399
it
follows
that
LEMMA 1 .7. Let T <J G. (i) Let x, y E A (R [GIT]) with {3 (x ) = {3 (y ). If u E A-r (R [0]) then
ux = uy. (ii) AT (R [G]) has an identity for multiplication.
0
The rest of this section is devoted to the proofs of ( 1 . 1) and (1 .3). Some preliminary results are proved first. Let A O(R [OJ) = A �(R [OJ) ' be the Grothendieck algebra. Then A O(R [GJ) may be identified with the algebra of C-linear combinations of Brauer characters afforded by R [G] modules. If V is an R [ G] module let f3v be the Brauer character afforded by V. Define the algebra homomorph ism f3 : A (R [GJ) � A O(R [GJ) by f3 « V)) = f3v for an R [O] module V.
PROOF. (i) Repeated application of (1 .6) in case R = R is a field yields the result. (ii) Let Vo be the irreducible R [ G] module which affords the principal Brauer character 'Po. Thus ( Vo) E A (R [ G I T]). By (1 .5) there exists x E A(1)(R [G/T]) � AT (R [O]) with f3 (x ) = 'Po. Hence by (i) ( V)x = ( V) ( Vo) = ( V) for all ( V) E AT (R [OJ). 0
LEMMA 1 .5 . A O (R [GJ) = A (1) (R [ GJ) = A (1) ( R [GJ).
LEMMA 1 .8. Let T <J G. Assum,e that R is not a field. Let U be an R -free
PROOF. The second isomorphism follows from (1. 13 .7). Let {'P; } be the set of Brauer characters afforded by the irreducible R [ G] modules. Let ctJi be the Brauer character afforded by the principal indecomposable R [G] module which corresponds t o 'Pi . Given i, then b y (1. 19.3) there exist irreducible Brauer characters 'Pij such that 'Pi = 2:j 'Pij . Thus ctJi = 2:j ctJij where ctJij corresponds to 'Pij . By (lV.3.3) this implies that ('Pi, ctJk )' = oi km where m -I- 0 depends on i. This implies that {ctJi } is a basis of A O(R [GJ). Hence the restriction of f3 to A(I)(R [ O J) is an isomorphism from A (1) (R [OJ) to A O(R [GJ). 0 The next result which generalizes (IIL2.7) first appeared [ 1 967] in case R is a field.
III
Conlon
LEMMA 1 .6. Let T <J O. Let U be an R -free R [ T]-projective R [O] module.
Let V be an R -free R [ G I T] module and let VI be a pure submodule of V. Then U (9 V = ( U (9 VI) EB ( U (9 VI VI).
PROOF. Since T is in the kernel of V and VI is a pure submodule, the sequence is a split exact sequence. Tensoring with UT shows that
R [ T]-projective R [ G] module. Let VI , V2 be R -free R [ G I T] modules such that VI and V_ 2 afford the same Brauer character. Then ( U (9 VI) = ( U (9 V2)' PROOF. By (lII.2.7) () (9 VI = () (9 V2 • Let f; be the natural projection from U (9 V; onto U (9 VI = U (9 V; . Since ( U (9 V; )y = (dim V; ) U there exists an R [ T]-homomorphism h such that the following diagram is commutative
�
( U (9 V2)y
1 12
( U (9 VI)y � ( U (9 VI)y � O . Since U (9 V2 is R [ T]-projective there exists an R [ G ]-homomorphism g : U (9 V2 � U (9 VI with f l o g = f2 . Let MJ be the kernel of f l . Thus MJ + g ( U (9 Y2 ) = U (9 VI and MI = 1T( U (9 VI) � Rad( U (9 VI)' Hence g is an epimorphism. Let M be the kernel of g. Since g (U (9 V2 ) is R -free, M is a pure submodule of U (9 V. Therefore rankR M + rankR ( U (9 VI ) = rankR M + rankR g ( U (9 V2) = rankR ( U (9 V2) = rankR ( U (9 VI)' Thus rankR M 0 and so M = (0). Hence g is an isomorphism of U (9 V2 onto U (9 VI . 0 =
CHAPTER IX
400
LEMMA 1 . 9 . Let t
ux = uy. (ii) AT (R [G]) has an identity for multiplication.
[1
E AT (R [G]) then
PROOF. If R is a field the result follows from (1.7). Suppose that R is not a field. (i) This is a direct consequence of (1.8). (ii) Let Vo be the R -free R [G] module which affords the principal character. Then ( Va) E A (R [G I T]). By (1.5) there exists x E A (1)(R [ G I T]) C: AT (R [G]) with f3 (x ) = f3 « Vo)). If u E AT (R [ G]) then ux = u ( Va) = U by (i). Thus x is the desired identity for multiplication. 0
� be a complete set of representatives of the conjugate classes of p-groups in G. For D E � let �(D) = {P I P E �, p x c: D for some x E G}. If P E � let Sp (G) be the subspace of A (G) spanned by all ( V) such that P is a vertex of V. Then for each D E � there exists an ideal A b( G) of A (G) with A b( G) = WD (N G (D)) (as algebras ) such that
LEMMA 1 . 10. Let
AD ( G ) = LD) Sp (G) = L ) A �(G). PE'-/5( PE'B(D
PROOF. Induction on I D I . If 1 D 1 = 1 then S( I )( G) = A(I)( G) is an ideal of A (G). Let A(l)(G) = A( 1 )(G). Suppose that I D I > 1 . B y definition
AD (G) = LD SP (G) = SD (G) + LD) Sp (G). PE'13( ) PE'-/5 P� (D
Do E �(D), Do -I- D
then by induction LPE'B ( Do) Sp (G) = L PE 'B ( Do) A �(G) for suitable ideals A �(G) of A (G) with A �(G) = Wp (NG (P)). If DJ, D2 E �(D); D1, D2 -I- D then by definition If
and so A � c: E9 PE'-/5( D I ) n'B ( D2 ) SP (G) for P C: � ( D I ) n � ( D2) ' Furthermore AD (G) = SD (G) E A0( D)(G) and
A CS( D ) (G) = EB A �(G ) and AD (G)IA0 ( D)(G) = WD (NG (D)) . D PE'B P�D( )
1]
THE STRUcrURE OF A ( G )
401
By (1.9) Wp (N G (P)) is a ring with multiplicative identity for any P E �. Thus A0 ( D)( G) has a multiplicative identity. Therefore there exists an idempotent e E A (G) such that A8 ( D ) (G) = AD (G)e. Let A b(G) = AD (G) n A (G ) (l - e ). Then A b(G) is an ideal of A (G) and A b(G) n A0( D)(G) = (0). Since 1 - e maps onto 1 in A (G)IA0 ( D)(G) it follows that and
A b(G) = AD (G)IA0( D)(G) = WD (NG (D)) AD (G) = A b(G) E9 A8( D)(G).
0
PROOF OF (1 . 1). Let D be a Sp -group of G. Then AD (G) = A (G) and the result follows from (1. 10). 0 LEMMA 1 . 1 1 . (i) Suppose that R is not a field. Let e be the character afforded by an R -free R [ G] module. Then () = Lai1] F where each ai is rational and each 1]i is the character afforded by an R -free R [H] module for some cyclic subgroup H of G. (ii) Suppose that R = R is a field. Let 'P be the Brauer character afforded .
by an R [ G ] module. Then 'P = Lai t/J F where each ai is rational and t/Ji is the character afforded by an R [H] module for some cyclic p '-subgroup H of G.
PROOF. If R or K is a splitting field for G both results are clear as the dimension of the space of all Lai 1] F, Lai t/J F is the number of classes, p -singular classes respectively. If e T = () or 'P T = 'P for T ranging over a group M of field automorphisms then
PROOF OF (1 .3). (i) It may be assumed that G = N G (P), that is to say P
[2
CHAPTER IX
402
( V) = ( V) ( Vo) = 2: aj ( V)( YP)
=
(0).
This shows that t is a monomorphism as required. (ii) A nilpotent element in E9 A p (H ) must have all its coordinates nilpotent. Thus the result is an immediate consequence of (i), (1 . 1) and
(1 .10). D
2J
A ( G ) IN CASE A Sp -GROUP OF G IS CYCLIC AND R IS A FIELD
403
informa tion. By using these methods Rudko [1968] was able to prove the followin g result. See also Renaud [1978]. THEOREM 2.4. Let P be a cyclic p -group. Assume that p I- 2. Let C = Q and
let L be the real subfield of the field of p th roots of l over Q. Then A (P) = Q EB A where A is the direct sum of 2( I P / - 1)/(p - 1) copies of L.
We will prove (2. 1) in this section modulo the following result which we state without proof. See Lindsey [1974], Lemma 2.6.
We will not pove (2.4) here but in case / P / = P it follows from (VII.2.7) . This can be seen as follows. Let £ be a primitiv e (2p )th root of 1. By (VII.2.7) the linear map sending V(I, t) to ( £ t - £ -1 )/( £ - £ - 1 ) is a ring homomo rphislIl· The result is a simple consequ ence of this. Related results can be found in Gudivok and Rudko [1973], Butler [1974], Carlson (1975] and lakovlev [1972]. See also Lam and Reiner (1969J, Santa [1971]. Let P be a cyclic p -group. For 1 � i � I P I let V; be the indecomposable F[ P] module with dimF V; = i. Then' ( VI) . . . , ( V,p,) is a basis of A (G). Each v� is serial and V; = Rad( V;+I) for 1 �' i � / P i - I . The proof of (2.2) depends on two element ary lemmas .
LEMMA 2.2. Let P be a cyclic p -group with I p i > 1 and let I P : PI I = p. If
LEMMA 2. 5. For 1 � s, t � / P / , V, @ V; is the direct sum of min(s, t)
2.
A ( G)
in case a
Sp- group
of
G
is cyclic and R is a field
Throughout this section A (G) = A (F[ G]) where F is a field of charac teristic p. The following result is due to O'Reilly [1964], [1965]. THEOREM 2. 1 . Let G be a group with a cyclic Sp -group. Then A (G) is
semi -simple.
MI and M2 are indecomposable F[ P] modules then MI @ M2 = Mo EB E9�= 1 Vj , where Mo is F[Pd-projective, each V; is indecomposable and dimF V; I- dimF Vi for i I- j. If I P I = p then (2.2) follows from (VIII.2.7). In the general case it is a good deal more complicated. For further results see Renaud [1979] . It will first be necessary to prove
THEOREM 2.3 (Green [1962b]). Let P be a cyclic p-group. Then A (P) is
semi -simple.
The proof of (2.3) given by Green depends on getting generators and relations for A (P) and is quite complicated. We will here give a short elegant proof due to Hannula, Ralley and Reiner [1967]. O'Reilly's proof of (2. 1) was simplified by Lam [1968] . In fact this simplification was the motivation for (1.3). However the proof still depended on Green's method and used results about generators and relations for A (P). Ultimately results concerning the tensor products of F[ G ] modules are of course needed but the proof given here uses only (2.2). It should be pointed out however that Green's method yields more
nonzero indecomposable F[ P] modules.
PROOF. Let Vs @ V; be the direct sum of n nonzero indecomposable modules. Since V'; = V; it follows that
n = dimF Invp ( V-. @ V ';) = dimF HomF[p] ( Vs , V; ).
The structure of min(s, t). D
V,·
and
Vr
shows that dimF HomF[p]( Vs , V; ) =
LEMMA 2 .6. Let m be an integer and let f(X) = L: = I min(i, j)X;Xj be a real quadratic form. Then f(X) is a positive definite form. PROOF. It " is easily seen that
f(X) = (XI + . . . + Xm Y + (X2 + · · · + Xm Y + · · · + X� . D PROOF OF (2.3). It may be assumed that C is the field of real numbers. It suffices to show that if u E A (P) with u 2 = 0 then u = O. Let u = L bs ( Vs ). Let ( V, ) ( V; ) = Lj aS1j ( Vi ). By (2.5) Lj astj = min( s, t). Thus o = U2
=
bs btastj ( Vi ) 2: s ,t
404
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[2
and so �S.I bsbt as1j = 0 for all j. By summing over j this implies that �S.I min(s, t)bsbt = O . Hence by (2.6) bs = 0 for all s and so u = O. 0
3]
I p : PI I = p· (i) For 1 ::::; 1 ::::; 1 P 1 and 'P an irreducible character of H there exists an indecomposable F[N] module V('P, I ), unique up to isomorphism, such that 'P is afforded by the socle of V and dimF V('P, I) = I. Every nonzero indecomposable F[N] module is isomorphic to some V ('P, I). V('P, I ) is F[PI]-projective if and only if I 0 (mod p ) and V('P, I) = V('P, 1)* if and only if 'P * a /-I = 'P where a is defined as in Chapter VII, section 1 . Furthermore V('P, l)p is indecomposable. (ii) If M I and M2 are indecomposable F[N] modules with M� = M for i = 1 , 2 then M I @ M2 = EB�= I Vi EB Mo where Mo is F[PI].:.projective and each V; is indecomposable with V � = Vi ' (iii) Suppose that 2 / 1 CH (P) I . For each I with 1 ::::; I ::::; 1 p i there exist exactly 2 irreducible characters 'P of H with 'P * a /-I = 'P. Each such 'P has Ho in its kernel where 1 CH (P) : Ho 1 = 2. ==
PROOF. (i) This is a special case of the results of Chapter VII, section 2. (ii) By (i) (M )p is indecomposable for i = 1, 2. By (2.2) M I @ M2 = EB �= I Vi EB Mo where Mo is F[PI]-proj ective and dimF Vi � dimF Vi for i � j. As (MI @ M2)* = MI @ M2 this implies that EB �= I Vi = ��= I V � . Thus V; = V� by the unique decomposition property and the fact that dimF V; � dimF Vj for i � j. 2 (iii) As H is cyclic, 'P * = 'P - I and so 'P *a /-1 = 'P if and only if 'P = a /- I . H Since a / :CH( P )/ = 1 there are exactly two choices for 'P. Clearly Ho is in the kernel of 'P. 0 PROOF OF (2. 1). It may be assumed that C is the field of complex numbers. Let FI be the algebraic closure of F. Then A (F[ G D is a subalgebra of A (FI[ GD. Thus it may be assumed that F = FI is algebraically closed as A (FI[ GD has finite C-dimension. Furthermore it s'uffices to show that A (F[ GD has no nonzero nilpotent elements. By (1 .3) it may be assumed that G = PH with P <J G where P is a cyclic p -group and H is a cyclic p '-group. Suppose that p = 2. Then G = P x H. Thus by (1 .4) A ( G) =
405
A (P) @ A (H).
Clearly A (H) is semi-simple. By (2.3) A (P) is semi-simple. Thus A (G) is semi-simple as required. Thus it may be assumed that p � 2. Let h be the order of (x ) = H and let x - I yx = y I . Define
The next result is required for the proof of (2. 1). LEMMA 2.7. Assume that F is algebraically closed. Let N = PH with P <J G where P is a cyclic p -group with 1 p i > 1 and H is a cyclic p '_group. Let
PERMUTATION MODULES
G,
= (x, y / y / P / = X 2h
= 1,
x - ' yx = y t ).
Then G is a homomorphic image of G I and so A (G) � A (G1). Hence it suffices to show that A ( G,) is semi-simple. Therefore by changing notation it may be assumed that 2 / 1 CH (P) I . The proof is by induction on 1 P I . If 1 P 1 = 1 then A (H) = C[ H] is semi-simple. Suppose that 1 p i > 1. Let 1 P : p, 1 = p. Let I = ApI (G). It suffices to show that I and A (G)/ I are semi-simple. The map sendi ng ( V) to ( l/p ) ( VG ) for any F[PIH] module V is easily seen to induce an isomorphism from A (PI G) onto 1. Thus by induction I is semi-simple. It remains to show that A (G)jI is semi-simple. Let S be the subspace of A (G) spanned by all ( V) with V indecompos able and V* = V. By (2.7) (ii) S + 1/1 is a subalgebra of A (G)/ 1. If 1 ::::; I ::::; 1 p i then (2.7) (ii) and (2.7) (iii) imply the existence of an indecom posable F[ G ] module V with dimF V = I and ( V) E S. Thus if V is an indecomposable F[ G ] module with ( V) E S and X is an indecomposable F[ H] module then the map sending ( V) @ (X) to ( V @ X) defines an algebra homomorphism of «S + 1)/1) @ A (H) onto A (G )/1. Since A (H) is semi-simple it suffices to show that (S + 1)/1 is semi-simple. Let Z = CH (P). By (2.7) (ii) and (iii) the map sending V to VCo (P) defines a one to one linear map of S onto A (P x Z). Since V is F[PI]-proj ective if and only if VCo (P) is F[Pd-projective, it follows that (S + 1)/1 = A (P x Z )/A pl (P x Z). Thus it suffices to show that A (P x Z) is semi-simple. This follows directly from ( 1 .4) and (2.3). 0 3.
Permutation modules
The set of all isomorphism classes of indecomposable R [G] modules is in general very large and complicated. This section and the next contains the definition of some special types of R [ G] modules. For H a subgroup · of G let Vo(H) denote the R -free R [0] module of R -rank 1 with Vo(H) = Inv H ( Vo(H)). Thus if R is a field then Vo(H) affords the principal Brauer character and if R is the ring of integers in the p-adic number field K then VO(H) @R K affords the trivial character. An R [ G ] module is a transitive permutation module if it is isomorphic to Vo(H)G for som� subgroup H of G. A direct sum of transitive permutation modules is called a permutation module.
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LEMMA 3 . 1 . There are only finitely many isomorphism classes of transitive permutation modules. PROOF. Clear as G has only a finite number of subgroups. 0 LEMMA 3.2. Let V be a permutation module. Then the following hold. (i) V = V*. (ii) V is an algebraic module. (iii) If M is a subgroup of G then VM is a permutation module. (iv) If G is a subgroup of M then V M is a permutation module. PROOF. (i) Clear by (II.2.6). (ii) This follows from (II.5.3). (iii) Clear by the Mackey decomposition (II.2.9). (iv) Immediate from the definition . See (II.2. 1) (iv).
0
LEMMA 3.3. Let Vand W be permutation modules. Then the following hold. (i) V EB W is a permutation module. (ii) V (9 W is a permutation module. PROOF. (i) Immediate by definition. (ii) This follows from the Mackey tensor product theorem (II.2. 10). 0 LEMMA 3 .4. Suppose that G is a p-group. (i) A transitive permutation module is indecomposable. (ii) If V is a permutation module and W I V then W is a permutation
module.
PROOF. (i) This follows from (III.3. 8). (ii) This is a consequence of (i). 0 COROLLARY 3.5. Let Q be a subgroup of the p-group P and let V be an R [ Q] module. Then V is a permutation module if and only if v P is a
permutation module.
PROOF. If V is a permutation module then so is v P by definition. Suppose that vP is a permutation module. Then so is ( VP )o. By the Mackey decomposition (II.2. 10) V I ( V P ) o. Hence V is a permutation module by (3.4). 0
4J
ENDO-PERMUTATION MODULES FOR p-GROUPS
407
LEMMA 3.6. Let V be a transitive permutation module. Then the following
hold.
(i) rankR (Inv G ( V)) = 1 . (ii) V = WI EB Wz with WI indecomposable such that Inv G ( W2) = (0) and Inv G ( WI) "f= (0). PROOF. (i) This follows from (II.3.4). (ii) Immediate by (i). 0 The class of indecomposable modules defined in (3.4) are of interest and have been studied by Scott [1971], [1973] . Some related results can be found in Dress [1975] .
4. Endo-permutation modules for
p- groups
Throughout this section P is a p -group. Vo(H) is defined as in the previous section. An endo -permutation R [P ] module is an R -free R [P] module V such that V* (9 V = HomR ( V, V) is a permutation module. An endo -trivial R [P] module is an R -free R [P] module V such that V* (9 V = HomR ( V, V) = Vo(P) EB U, where U is a projective R [P] module. Since any proj ective R [P] module is a permutation module it follows that an endo-trivial R J P] module is an endo-permutation R [P] module . Clearly a permutation module for R [P] is an endo-permut ation R [P] module. These concepts were introduced by Dade [1978a] , [1978b], who also classified the endo-permut ation R [P] modules in case P is abelian. See also Carlson [1980a] . We will here only present basic elementary properties of endo-permut ation R [P] modules and compare some of these with the correspondin g properties for permutation modules. The next result · shows that endo-permutation R [P] modules arise naturally at least in case R = R is a field. THEOREM 4. 1. Let G
= HP where H = Op,(G) and P is a Sp -group of G. Let F = R be a splitting field of G and let V be an irreducible F[ H] module with G = T( V). Then V extends uniquely to an irreducible F[ G ] module W and Wp is an endo -permutation F[P ] module.
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CHAPTER IX
[4
PROOF. The existence and uniqueness of W follows from (111.3. 16). Under the action x � y -1xy, F[H] becomes a permutation module M for F[ 0]. Let e be the centrally primitive idempotent of F[H] correspond ing to e. Then V* 09 V = eF[H] I MH • Since T( V) = 0, eF[H] = W 09 W* as F[ 0] modules and W* 09 W I M. Thus W� 09 Wp I Mp and so W� 09 Wp is a permutation module for F[P] by (3.4). 0 LEMMA 4.2. Let O � V � U � W � 0 be an exact sequence of R -free R [P] modules with U projective. Then V is an endo -permutation (endo
trivial) module if and only if W is an endo -permutation (endo -trivial) module. PROOF. Immediate by (III.S. 12). 0
It follows from (4.2) that for instance every factor in a proj ective resolution of Vo(P) is an endo-permutation module . In this way it can be shown that if P is neither cyclic nor a genera'lized quaternion group then there exist infinitely many endo-permutation R [P] modules. Also there exist endo-permutation R [P] modules V which are not algebraic and such that V� V*. This is in contrast to (3.1) and (3.2) (i), (ii).
4]
ENDO-PERMUTATION MODULES FOR p -GROUPS
409
LEMMA 4.6. Let V be an endo -permutation R [P] module and let 0 be a
subgroup of P. Then Vo is an endo -permutation R [0] module.
PROOF. Clear by definition. 0 In general the direct sum of endo-permutation modules need not be an endo-permutation module in contrast to (3.3) (i). Two en do-permutation R [P] modules V, W are compatible if V EB W is an en do-permutation R [P] module . LEMMA 4.7. Let V, W be endo -permutation R [P] m'odules. The following
are equivalent. (i) V and W are compatible. (ii) V* 09 W = HomR ( V, W) is a permutation module. (iii) W* 09 V = HomR ( W, V) is a permutation module.
PROOF. Since ( V* 09 W)* = W* 09 V, (ii) and (iii) are equivalent. Since
( V EB W) 09 ( V* EB W*) = ( V 09 V*) EB ( V 09 W*) EB ( W 09 V*) EB ( W 09 W*),
Do there exist only finitely many isomorphism classes of endo -permutation R [P] modules with V = V*?
the equivalence of (i) with (ii) and (iii) now follows from (3.3) (ii). 0
This is an open question. It follows from Dade's results that the answer is affirmative if P is abelian.
LEMMA 4.8. Let V be an indecomposable endo -permutation R [P] module
LEMMA 4.3. Let V be an endo -permutation R [P] module such that V" =
PROOF. Let .0 be the set of all subgroups Q of P with Q ;i P. By (111.4.9) HO ( P, .0, V* 09 V) ;i (0). Thus there exists an indecomposable component W of V* 09 V with HO (P, .0, W) ;i (0). Since V is an endo-permutation module, W = Vo(O ) for some subgroup Q of P. By (II.3.4) HO ( P, Q, W) = (0). Hence HO(P, .0, W) = (0) if Q E .0. Thus Q g .0 and so Q = p. 0
( vn )* for some positive integer n. Then V is algebraic.
PROOF. As V 2 n = vn 09 ( vn )* is a permutation module the result follows from (3.2) (ii). D LEMMA 4.4. If V, W are endo -permutation R [P] modules then so are V*,
V 09 W and HomR ( V, W) = V* 09 W.
PROOF. Clear by definition.
D
The next result indicates that compatibility of two endo-permutati on modules is a rare phenomenon. LEMMA 4.9. Let V, W be indecomposable endo -permutation R [P] modules
LEMMA 4.5. Let V be an endo -permutation R [P] module. If W I V then W
is an endo -permutation module.
PROOF. Since W* 09 W I V* 09 V the result follows from (3.4).
with vertex P. Then Vo(P) I V* 09 V.
0
with vertex P. Then V and W are compatible if and only if V = W.
PROOF. If V = W then V and W are compatible by (4.7). Suppose that V and W are compatible. Let S) be the set of all subgroups Q of P with Q ;i P. There exist nonnegative integers ao, bo for Q � P such that
CHAPTER IX
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[4
V* ® W = apVo(P) EB EB ao ( Vo( O)r, O E.'Q
Thus
v ® V* = bp Vo(P) EB EB bo ( Vo( O )r· O E:i;;J
CHAPTER X
apV EB ffi ao ( Vo r = v ® V * ® W = bpW EB ffi bo ( Wo )G. oEe O E:i;;J Hence any indecomposable component of V ® V * ® W with vertex P is isomorphic to V. By (4.8) a p I- 0 and so V = W. 0 In general endo-permutation modules do not behave well with respect to induction. The next result is concerned with this situation .
Let 0 be a subgroup of P and let V be an endo -permutation R [ 0 ] module. The following are equivalent. (i) vP is an endo -permutation R [P] module. (ii) The endo -permutation R [ 0 x n o ] modules Voxn o and VOx no are
LEMMA 4. 10.
compatible for all x E P.
PROOF. The Mackey tensor product theorem (II.2. 10) implies that vP ® ( VP)* is a permutation module if and only if ( Vox no ® Vdn or is a permutation module for all x E P. Thus (i) is equivalent to (ii) by (3.5) and (4.7). 0
As might be expected the theory developed so far becomes somewhat simpler when applied to p -solvable groups. On the one hand certain results are true for p -solvable groups which are not true in general, on the other hand some questions which remain open in the general case can be settled for p -solvable groups. This chapter is primarily concerned with p -solvable groups though some of the results are proved in a more general context. The notation introduced at the beginning of Chapter IV will be used throughout this chapter.
1. Groups with a normal p ' -subgroup
The first three results in this section are slight refinements of results of Fong [ 1 960], [ 1 961],J1962] . The method used to prove (1 . 1) is adapted from Serre [1977] , it has its roots in the work of Schur. I am indebted to Watanabe [1979] who suggested that part of the proof of ( 1 . 1) which shows that Nx is nonempty for all x. This fills a gap in an earlier version of the result. See also Nobusato [1978] . For a related result see Tsushima [ 1 978c] .
Let H <J G. Let ? be an irreducible character of H. Assume that G = T(?), the inertia group of ? Let F be an algebraically closed field such that char F l' ' H ' . Let V be an irreducible F[ H] module which affords ? Then the following hold. (i) There exists a finite group {; and an exact sequence (1) '-;' Z � {; 4 G � (1),
LEMMA 1 . 1 .
41 1
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[1
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where Z is a cyclic group in the center of 0 and I Z I I I H 1 2 . Also 0 contains a normal subgroup H = H such that ZH = Z x H = f- I (H). The group 0 depends only on G and �, in particular it is independent of the choice of F. (ii) Let FI be the subfield of F generated by a primitive I H 1 2 root of unity. There exists an FI[ 0] module VI such that if V VI 0FJ F then f( Vff ) = V. Furthermore if W is an irreducible F [ G ] module with V a constituent of WH then W = V 0 W for some absolutely irreducible F [ 0 /H] module W. (iii) Let L1 (F) be the set of all Brauer characters afforded by irreducible F [ G ] modules W such that V is a constituent of WHo Let J (F) be the set of all Brauer characters afforded by F [ 0 /H] modules U such that Z is in the kernel of V 0 u. Then the map sending W to W defined by (ii) induces a one to one mapping from L1 (F) onto J (F). =
F' be a field with FI C F' C F. Let V' be an F' [ H] module which affords { Since char F' ,{' I H I, V' is irreducible. Let A be a representation with underlying module V'. Let SF' {det A (y ) l y E H}. Clearly S = SF' is independent of the choice of F'. For x E G let Nx be the set of all linear transformations z on V' such that z -I A ( y ) z A (x -I yx ) for all y E H and such that det z E S. Let � (1) = d. Let x E G. We will first show that there exists a linear transformation z E GLd (FI ) with det z = 1 and z l A (y )z = A (x - l yx ) for all y E H. It clearly suffices to prove this in case the order of x is q a for some prime q and some integer a � 1 . Let I H I = h . Let Fo b e the subfield o f FI generated b y a primitive h th root of 1. It may be assumed that A (y ) E GLd (Fo) for all y E H by a PROOF. Let
=
=
-
classical result of Brauer which follows directly from (IV. 1 . 1) (iii), see e.g. Feit [ 1 967c] (16.3). Since T(�) = G there exists Z o E GLd (Fo) such that z 0 1 A (y ) z o = d A (x - l yx ) for all y E H. Let a = (det zoyl and let z = azo. Then det z = 1 , I z - A (y)z = A (x -1yx ) for all y E H and z d = a dz g E GLd (Fo). Further more z q a = {31 is a scalar for some {3 E Fl. Thus {3 d = 1 and so
z q a E GLd (Fo).
If q ,{' d then z
GLd (Fo).
E GLd (Fo) C GLd (FI) as z q a and z d are both in
Suppose that q I d. Then (z, A (y ) l y E H) is a finite group whose exponent divides q ah. Thus it may be assumed by Brauer ' s theorem that z E GLd (Fo(E )), where E is a primitive q " th root of 1 for some n. As q I h, [Fo( E ) : Fo] is a power of q. Since a E Fo(E ) it follows that [ Fo( a ) : Fo] is a power of q. Assume first that either q -1 2 or q = 2 and 4 1 h. Then Fo(E ) is a cyclic
1]
413
GROUPS WITH A NORMAL p i-SUBGROUP
extension of Fo. As a d E Fo it follows that a = CEo for some c E Fo and some q " th root of 1, Eo such that Fo(Eo) = Fo ( a ) . Let q b = (d, q " ). Then [ Fo ( a ) : Fo] = [Fo(Eo) : Fo] I q C. Hence a E Fo(Eo) C FI and so z E GLd (FI). Suppose now that q = 2, 2 / h and 4 ,(' h. Then there exists Ho <1 H with I H : Ho 1 = 2. As 2 / d it follows that � = � r; for some irreducible character �0 of Ho. The previous argument applied to Ho, �o and the representation Ao which affords �o yields the existence of Zl E GL(dI2) (FI) with det Zl = 1 and Z ;-l A (y )Z l = A (x - I yx) for all y E Ho. Let u be an element of order 2 in H. Since I G : T(�o) I = 2 it follows that T(�o) <1 G and [x, u ] E Ho. Let z r be the element of GL(dI2) (FI) corresponding to X U for �� and A � . Then Zl 0 Z= E GLd (FI)
( 0 zr)
has the required properties. Therefore Nx is n onempty for all x E G. If Z l , Z 2 E Nx then Z ;-I Z 2 is a scalar by Schur ' s lemma. Thus Nx is a finite set and NI consists of scalars. Furthermore Nx is independent of the choice of F'. Define O fF') to be the group of all ordered pairs (x, z ) with x E G and z E Nx• Thus O fF') is a finite group. Furthermore O f F ) = O(FI). Define the map f : O (Fl ) � G by f« x, z )) = x. Thus there exists an exact sequence '
(1) � NI � O ( FI ) 4 G � (1) .
Since NI consists of scalars, it follows that NI is a cyclic group in the center of G and I NI I I �(l) I H I. Let Z = NI and let H = {(y, A (y )) l y E H}. Thus H <1 O (FI ) and f(H) = H. Furthermore Z n H = « 1, 1) . Thus ZH =
z x H.
For v E V', (x, z ) E O (Fl) define v (x, z ) = vz . Let V' denote the F' [ O (Fl )] module constructed this way. For FI = F' let VI = V'. For F = F' let V = V'. Then V = VI 0FJ F. If Y E H then v (y, A (y )) = vy. Hence f CVH ) = V. Let F(°l, F(p) be algebraically closed fields of characteristic 0 , p respec tively where p ,(' 1 H I . To complete the proof of (i) it suffices to show that O(F�O») = O (F�P»). Let K be the extension of Qp generated by a primitive I H 12 root of 1. Thus O (K) = O(F�O»). Let R be the ring of integers in K and let R be the residue class field. Then O ( R ) = O(F�P»). Let X be an R -free R [ O (K)] module such that XK = V'. Since p ,(' I H I, XH is irreducible. From this it follows easily that O (K) = O ( l�. ). This completes the proof of (i). (ii) Let W be an irreducible F [ G ] module such that V is a constituent of WHo Thus W is an F[ 0] module where w (x, z ) = wx for w E W. Define the vector space W = Hom F[H] ( V, W).
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Since V may be identified with V it follows that if
v (x, z ) = vz.
v
E V then
By Clifford's Theorem (III.2. 12), WH = n V for some integer n. Thus the linear map g : V ® W ---,). W defined by g (v ® h ) = vh is an isomorphism. For h E W, (x, z ) E G define h (x, z ) by v {h (x, z )} = {(vz -I)h }x for v E V. Then for v E V, h E W
{g(v ® h )} (x, z ) = (vh ) (x, z ) = (vh )x and so
g {( v ® h ) (x, z )} = g { v (x, z ) 0 h (x, z )} = g {vz 0 h (x, z )} = (vz ) {h (x, z )} = (vh )x = {g(v 0 h )} (x, z ). Since V and W are F[ G ] modules, this implies that W is an F[ G J module and g is an F[ G ]-isomorphism. If (y, A (y )) E R then for v E V, h E W, v {h (y, A (y ))} = {(vy - l )h }y = vh. Thus W has R in its kernel and so W is an F[ G IIl] module. W is irreducible as W = V 0 W is irreducible. (iii) For any module X let {3x denote the Brauer character afforded by X. Then {3 w = {3v{3w by (ii). Suppose that W and W' both have H in their kernel and {3v{3w = {3v{3w' . If {3w -I {3w ' there exists x E G such that (3 v (xy ) = 0 for all y E H. Therefore A (x )A (y ) has trace 0 for all y E H. Since {A (y ) l y E H} spans the full matrix ring this implies that A (x ) = 0 which is impossible. Thus {3w is uniquely determined by {3w. Hence (3w ---,). {3w defines a one to one mapping from Ll (F) to .1 (F). It only remains to show that if U is an irreducible F[ G I R ] module such that Z is in the kernel of V ® U then V 0 U is irreducible. Suppose that WI, . . . , Wm are the composition factors of V 0 U. Then V is a constituent of ( � )H for all i. By (ii) � = V 0 Wi, where each "HI; is irreducible. This implies that {3v®u = {3v 2:�� 1 {3w ; and so {3u = 2:�� 1 {3w, . Hence m = 1 by (lV. 3.4) and so V 0 U is irreducible. 0 The group G I R defined in ( 1 . 1) is called the representation group of the
character {
THEOREM 1 .2 (Fong [1961]). Let p be a prime. Let H <J G with p ,f I H I·
Let B (G) be a block of G and let B (H) be a block of H which is covered by B ( G). Let ? be an irreducible character of H in B (H) and let G, fI, Z be defined as in ( 1 . 1) corresponding to the group T(?). Let G (?) be the
1]
415
GROUPS WITH A NORMAL p '-SUBGROUP
representation group of { Then there exists a block B of G (?) such that the following conditions are satisfied. (i) There exists a block B ( T(?)) of T(?) such that if (} is an irreducible character or an irreducible Brauer character in B (?) then (} = (J 5' for some irreducible character or irreducible Brauer character respectively in B (T (?)). (ii) Let TJ 1] denote the one to one mapping defined in ( 1 . 1) for char F = O. Thus (J ---,). 0 defines a one to one mapping from the set of all irreducible characters in B (G) onto the set of irreducible characters in B. This mapping preserves heights and sends the set of all p -rational irreducible characters in B (G) onto the set of all p -rational irreducible characters in B. (iii) Let TJ ---,). 1] denote the one to one mapping defined in ( 1 . 1) for char F = p. Thus (J ---,). (Jo defines a one to one mapping from the set of all irreducible Brauer cha�acters in B (G) onto the set of all irreducible Brauer characters in B which preserves heights. (iv) With respect to the one to one mappings defined in (ii) and (iii), B (G) and B have the same decomposition matrix and the same Cartan matrix. (v) B and B (G) have isomorphic defect groups. ---,).
PROOF. Since p .{ I H I, ? is the unique irreducible character in B (H) and ? = !/J is irreducible as a Brauer character. Furthermore T(?) T(B (H)). =
A field automorphism which preserves all p 'th roots of unity preserves
blocks. Thus by (V.2.5) it suffices to prove the result in case G = T(?) since p -conjugate characters of T(?) induce p -conjugate characters in G. Let Lio = .1 (F), .1 0 = .1 (F) be defined as in ( 1 . 1) where char F = O. Let .1p = Ll (F), .1p = .1 (E) be defined as in ( 1 . 1) where char F = p. Let V afford to, tp in case of char F = 0, p respectively . Thus to (x ) = tp (x) for p '-elements x in G. Furthermore to is p -rational by ( 1 . 1) (ii). Hence (J = to o if (J E .10, ( 1 . 3) (J = tp o if (J E .1p. If (J E .10 then (J is p -rational if and only if 0 is p -rational as to is p -rational and the map (J ---,). 0 is one to one. If (} is an irreducible character or an irreducible Brauer character in B (G) then by (V .2.5) (iv) (J E .10, .1p respectively. Let
S = { o I (J is in B ( G), (J E .10 U Llp }. If X E .10, 'P E Llp then by ( 1 . 3 ), d (X, 'P) = d (X, iP) where d denotes the appropriate decomposition number. This and (lV.4.2) imply that S consists of all the irreducible characters and irreducible Brauer characters in a set
[1
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416
of blocks B1, , Bm of G (t). Since the decomposition matrix of B ( G ) is indecomposable by (1. 1 7.9) it follows that m = 1. Let B = BI. Let p Q be the order of a Sp-group of G. Then pQ is the order of a Sp -group of G(t). Let d be the defect of B (G). Since t (l ) Ji O (mod p ), (1.3) and (IV.4.5) imply that d is the defect of B. Thus (1.3) implies that if () E ..10 U Lip and () is in B ( G) then () and e have the same height. Let Y be an irreducible R [ G ] module in B ( G) which affords a Brauer character of height O. Thus a defect group D of B (G) is a vertex of Y. As Y affords a Brauer character of height 0, a defect group 15 of B is a vertex of Y. As Y = V Q9 Y it follows that Y is R [ 15 ]-projective. Hence D is conjugate to a subgroup of 15 in G. As I D I = 1 15 I = p d this implies that • • •
D
=
15
.
0
Let G, H = Op, (G), t be as in (1 .2). Thus a Sp-group of G (t) has order at most that of a Sp -group of G. Furthermore if G is p -solvable and H = Op, (T(t» then either p 1" I G(t) I or Op'p (G(t» = z x P for some p -group P rf (I) since Z = Op' (G(t» is in the center of G (t). These two facts will make it possible to prove some results about p -solvable groups by induction on the order of a Sp -group.
Let p be a prime and let G be a p-solvable group. Let H = Op, (G) and let t be an irreducible character of H such that H = Op, ( T(t» . Let p Q be the order of a Sp-group of the representation group G(t) of r Then every block of G (t) has defect a.
LEMMA 1 .4 (Fang [1962]).
PROOF. If a = 0 the result is trivial. Suppose that a rf O. Let P = Op (G(t» . Thus P -I (I). As Z = Op' ( G(t» is in the center of G (t) it follows that C O U ; ) (P) = P X Z. By (IV.4. 17) G (t) has ex�ctly I Z I blocks of defect a. By (V.3.6) and (V.3.10) G(t) has exactly I Z I blocks. 0
1J
GROUPS WITH A NORMAL p i-SUBGROUP
417
B.
Thus (ii) follows from (i). It only remains to show that if Op, (G) = (1) then B is the unique block of G. Since Op, (G) = (I) it follows that Co (Op (G» � Op (G). The result now follows from (V.3.1 1). 0 kernel of
As an immediate Corollary of (1 .5) one gets the following result which like (1 .5), is false for groups in general.
Let p be a prime and let G be a p -solvable group. Let V1 and V2 be R [ G] mol1ules in the principal block of G. Then every direct summand of V1 Q9 V2 is in the principal block of G.
COROLLARY 1.6.
Let p be a prime. The following statements are equivalent. (i) Every irreducible character of G is irreducible as a Brauer character in characteristic p. (ii) G = Op, (G)P, where P is an abelian p -group.
THEOREM 1.7 (Richyfl [1972]).
PROOF. Let H = Op, (G). (i) :::} (ii). By (1. 17.9) each block contains a unique irreducible Brauer character. Thus G has a normal p -complement by (IV.4. 12) and so G HP f? r P a Sp -group of G. Since the hypotheses are satisfied by P = G IH It follows that every irreducible character of P has degree 1 and so P is abelian. (ii) :::} (i). Let () be an irreducible character of G and let t be an irreducible co� stituent of ()H . Let e be defined as in (1.1). By (1 .2) it suffices to prove that () is irreducible as a Brauer character. Since G(t)/Z = P for a central p '-group Z in G (t) it follows that G (t) = Z x P is abelian and so every irreducible character of G (t) is irreducible as a Brauer character. 0 =
See Osima [1942] for results related to (1 .7). The following result is somewhat simpler and appears to have been known for some time though it first appeared explicitly in Fang and Gaschiitz [ 1 961] as did (1 .6), both for solvable groups. THEOREM 1 .5. Let p be a prime and let G be a p-solvable group. (i) The principal block is the only block of G if and only if Op' (G) = (I). (ii) An R [G] module V is in the principal block of G if and only
Op, (G) is in the kernel of V. PROOF. Let
B
-
be the principal block of G. By (IV.4.12)
if
Op, (G) is in the
THEOREM 1.8 (Hamernik and Michler [1972]).
Let p be a prime and let G be a p -solvable group. Let F be an algebraically closed field of characteristic p. Let W be an irreducible F[ GJ module. Let
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Suppose that Op (G) = D � (1). By (111.4. 1 3) W is an F[ G ID ] module with vertex PI D. The result follows by induction. Suppose that Op (G) = (I). Thus H = Op' ( G) � (1). Let V be an irreduc ible F[H] module with V I WH By induction and (V.2.5) it may be assumed that T( V) = G. By ( 1 . 1) (ii) W = V 0 W for an irreducible F[ {; IH] module W and dimp V = dimF V -1= 0 (mod p ). Since Op ({; IH ) � (I) , induction and the previous paragraph applied to {; Ifl implies that o
p C = (dimF W)p = (dimF W)p = p a �\ where I Po l = p bo and Po is a vertex of W. Let X be a source of W. By Frobenius reciprocity (III.2.5) W I V 0 xG = ( VPo 0 xf. Hence W is F[Po]-projective and so b =S boo Thus c = a - bo =S a - b. 0 For a related result see Cliff [1979] . It follows from (1.8) that question (II) of Chapter IV, section 5 has an affirmative answer for p -solvable groups. However the next section contains stronger results in this direction. For groups in general (1 .8) is false . For instance if G = II , the smallest Janko group and p = 2 then I G 12 = 8 but G has a nonprojective irreducible Brauer character of degree 56. See Fong [1974] . As another example let G = SL2 (4) and p = 2. Let W be the natural 2-dimensional representation of G in characteristic 2 and let P be a Srgroup of G. Since Wp is faithful and P is abelian, Wp is not induced from a representation of a proper subgroup. Thus P is a vertex of W by (III.3.8). The statement of (1.8) asserts that if G is p -solvable then the vertex of an absolutely irreducible F[ G] module is as small as possible. If R is the ring of integers in a p -adic number field K then the analogous statement for R [ G ] modules W with WK absolutely irreducible is not true. In answer to a question raised in an earlier version of this material Cline [ 1971], [1973] suggested the following counterexamples. Let p be a prime. Let P be an extra special p -group of order p3 which admits a cyclic group E of automorphisms with I E I = p + 1 such that [E, Z(P)] = (1) and E acts transitively on the set of all subgroups of index p in P. Thus P is of exponent p if p � 2 and P is the quaternion group if p = 2. Let G = PE. Then G has a faithful irreducible character X with X (l) = p. Let V be an R -free R [ G ] module which affords X. We will show that P is the vertex of V. Without loss of generality it may be assumed that R contains a primitive I G Ith root of 1. V is absolutely indecomposable as X is irreducible. If the
2]
BRAUER CHARACTERS OF p -SOl.VABLE GROUPS
419
result is false then a vertex D of N has index p in P. Since V I ( VD ) G it follows from (III.3.8) that Vp = w P for an R [D ] module W of rank 1 . Then D is i n the kernel o f Vp = WP• This i s however impossible since P is the smallest normal subgroup of G which contains D. 2. Brauer characters of p -solvable groups
Let 'P be an irreducible Brauer character of a p -solvable group G. Swan [1960] observed that it is a very simple consequence of Fong's work that there exists an irreducible character X of G such that X = 'P as a Brauer character. This result is now known as the Fong-Swan Theorem. The following refinement is due to Isaacs [1974] , though the proof is different from his and is essentially the same as Swan's proof of the Fong-Swan Theorem. THEOREM 2 . 1 . Let G be a p -solvable group. Let 'P be an irreducible Brauer character of G. There exists a p-rational irreducible character X of G such that X = 'P as a Brauer character.
PROOF. Induction on I G : Op' ( G) I . If G = Op' (G) the result is clear. Let H = Op' ( G). Since p � I H I, 'PH is an ordinary character of H. Let ? be an irreducible constituent of 'PH. By (V.2.5) and (1 .2) it suffices to prove the result for the group G (?). Thus by induction it may be assumed that I G : H I = I G (?) : Op , ( G (?)) I . Thus Op, ( G (?)) = Op' (Z), where Z is the center of G (?). Hence Op ( G (?)) � (I). Therefore it may be assumed that Op (G) � (1). By (III.2. 13) Op (G) is in the kernel of 'P and so 'P is an irreducible Brauer character of GlOp ( G). Thus the result follows by induction. 0 As an immediate Corollary of (2. 1) one gets a strengthening of (IV.6.11) for p -solvable groups. COROLLARY 2.2. Let G be a p-solvable group. Let B be a block of G. Then
the number of irreducible p-rational characters in B is greater than or equal to the number of irreducible Brauer characters in B. The character X in (2. 1) need not be unique. For instance let p = 2 and let G be an elementary abelian 2-group. However the next result shows that this is a peculiarity of the prime 2.
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[2
THEOREM 2.3 (Isaacs [1 974]). Let p � 2 and let G be a p-solvable group. Let 'P be an irreducible Brauer character of G. Then there exists a unique irreducible p-rational character X of G such that X = 'P as a Brauer character. The following preliminary lemma is needed for the proof of (2.3). LEMMA 2.4. Suppose that p � 2 . Let P be a p-group with P <J G. Let X be a p-rational character such that X = 'P as a Brauer character where 'P is an irreducible Brauer character. Then P is in the kernel of x. PROOF. Let R be the ring of integers in a finite unramified extension K of t� e p -adic number such that R is a splitting field for R [ G ] . Let V be an R [ G ] module which affords 'P and let U be the corresponding indecom posable projective R [ G ] module . Let Qj be the character afforded by U. Thus ( Qj, X) 1. As X is p -rational, K (X ) = K. Hence by a basic property of the Schur index (see e.g. Feit [1 967c] ( 1 1 .4» UK = W EB W' where W affords X and no composition factor of W' affords X. Let X = u n W, where as usual U is identified with U 0R R � UK = U 0R K. Thus X affords X and X = V. Let x E P. By (III.2. 13) V(x - 1) = (0). Hence X(x - 1) � pX. Conse quently either x = 1 or x is of the form 1 + p k a for some linear transforma tion a with coefficients in R and a t= 0 (mod p ), and some integer k > 2. Since p� 2 x p " == 1 + p k+n a (mod p n + k + l ) . =
Since x p " = 1 for some n � 0 this implies that a assumption. Thus x = 1 . 0
==
0 (mod p ) contrary to
PROOF OF (2.3). Induction on I G : Op' (G) I. If G = Op' (G) the result is clear. The existence of X follows from (2. 1). Let H = Op' ( G). Let � be an irreducible constituent of 'PH. By (V.2.5) and (1.2) it suffices to prove the result for the group G (�). By induction it may be assumed that I G : H I = I G (�) : Op,(G(�» I . Thus Op, (G (�» = Op' (Z), where Z is the center of G (�). Thus Op (G(�» � (I). Therefore it may be assumed that Op (G) � (I). Let XI, X2 be p -rational characters such that X l = X2 = 'P as Brauer characters. By (2.4) Op (G) is in the kernel of X l and X2 . Hence X l and X2 are characters of GlOp ( G ). Thus by induction X l = X2. 0 THEOREM 2.5 (Isaacs [1974]). Suppose that p � 2. Let G be a p-solvable group. The following statements are equivalent.
PRINCIPAL INDECOMPOSABLE CHARACTERS OF P -SOLVABLE GROUPS
3]
421
(i) Every irreducible p-rational character of G is irreducible as a Brauer
character. (ii) For each p -singular element x in G with x Xpxp, where Xp is the p-part of x there exists a power y of x such that I (xp ) I = I (y ) I but y is not conjugate to xp in CG (xp) =
PROOF. It follows from (IV.6. 10), or more directly from Brauer's com binatorial lemma quote9 in (IV.6. 10), that the number of irreducible p -rational characters of G is equal to the number of conjugate classes that contain an element z whose p -part x is conjugate to every power of x that has the same order a� x in C G (zp') where Zp' is the p '-part of z. Thus condition (ii) holds if and only if the number of p -rational irreducible characters of G is equal to the number of irreducible Brauer characters of G. By (2. 1) this is true if and only if condition (i) holds. 0 COROLLARY 2.6. Let G be a p -solvable group and let 'P be an irreducible Brauer character of G. Let be an automorphism of the field Q( 'P ). Then 'P T is an irreducible Brauer character. T
PROOF. Clear by (2. 1). 0 (2.6) is false for groups G in general. For instance SL2 (1l) has a unique irreducible Brauer character 'P for p = 11 of degree 2. However YS E Q('P ). Further results related to (2.3) and (2.5) can be found in Isaacs [1978] . Cliff [1977] has given an alternative treatment of the original Fong-Swan theorem and has obtained information about indecomposable modules of p-solvable groups. Gagola [1975] has generalized the Fong-Swan theorem and shown that if the vertex of an absolutely irreducible F[ G] module is contained in a normal p -solvable subgroup, then this module can be lifted to one in characteristic O.
3.
Principal indecomposable characters of p -solvable groups
The first three results in this section are due to Fong [1962].
P <J G with I p i = p n. Let 'P be an irreducible Brauer character of G /P. Let Qj, Qj o be the principal indecomposable character of G, G/P respectively, which corresponds to 'P. Then $ (1) =
LEMMA 3 . 1 . Suppose that
p n Qj O( l ) .
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[3
PROOF. See (lV.4.26). 0 THEOREM 3.2. Let G be a p -solvable group. Let 'P be an irreducible Brauer character of G and let
3]
PRINCIPAL INDECOMPOSABLE CHARACTERS OF p -SOLVABLE GROUPS
423
PROOF. The proof is by double induction on 1 G : Op' ( G) 1 and 1 G I . If G = Op' (G) the result is trivial. Let 'P be the irreducible Brauer character corresponding to
No result like (3.2) is true for groups in general. For instance if G = As and p = 3, there exists a principal indecomposable character of G of degree 9. COROLLARY 3.3. Let G be a p-solvable group and let F be a field of characteristic p. Then Ann(J(F[ G))) F[ G]c, where c is the sum of all p-elements in G. =
PROOF. Let F be a finite extension field of F which is a splitting field of F[ G ] . Let VI, V2 , be a complete set of representatives of all the irreducible F[ G] modules. Let I be the annihilator of EB Vj. Then 1 = J(F[ G )). By (VI.4.S) and (3.2) Ann( I ) = F[ G]c. Since c E F[ G] and J(F[ G)) = J(F[ G)) @FF the result follows. 0 •
•
•
It is well known that a p -solvable group G contains a p -complement M. That is to say there exists a p '-subgroup M of G such that G = PM and P n M = (1) where P is a Sp -group of G. Furthermore any p '-subgroup of G is conjugate in G to a subgroup of M. THEOREM 3.4. Let G be a p -solvable group and let M be a p -complement in G. Let
COROLLARY 3.S. Let G be a p-solvable group and let M be a p -complement in G. Let 'P be an irreducible Brauer character of G with p � 'P (1). Then 'PM
is irreducible.
PROOF. Let
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424
[4
3.6. Let G be a p -solvable group with an abelian Sp -group P and an abelian p -complement. Then all the decomposition numbers are 0 or 1.
COROLLARY
PROOF. It suffices to show that if l/J is a principal indecomposable character of G then ( l/J, (J) � 1 for every irreducible character (J of G. Since a p-complement is abelian, l/J(1) = I P I by (3.4). Hence l/Jp is the character afforded by the regular representation of P. Since P is abelian, l/Jp is multiplicity free. Thus also l/J is multiplicity free. 0
For results about general metabelian groups see Basmaji [1972].
4. Blocks of
p -solvable
groups
LEMMA 4.1 . Let G be a p -solvable group and let B be a block of G with a cyclic defect group. Let K be a splitting field for G and all its subgroups. Then the Brauer tree of G is a star with the exceptional vertex (if any ) at the center. Every indecomposable R [G] module in B is serial.
PROOF. By (2.1) every edge has a vertex which is an end point and corresponds to a p-rational character; Thus this vertex is not exceptional by (2.3). This proves the first statement. The second statement follows from (VII.2.21). 0
Section 9 of Chapter VII has examples which show that (4.1) is false for groups G in general even if a Sp-group is cyclic. The remaining results in this section are all due to Fong [1960], [1961], [1962]. In special cases some of these were proved by Berman [1960]. See also Michler [1973b].
4.2. Let G be a p -solvable group. Let B be a block of G and let D be a defect group of G. Let Z(D) denote the center of D and let I D : Z(D ) I = P k . Then the height of any irreducible character or irreducible Brauer character in B is at most k.
THEOREM
By (2. 1) it suffices to prove the result for an ordinary irreducible character; We will first prove this for blocks B whose defect group D is a Sp -group of G. This will be done by induction on I G I . If I G I = 1 the result is clear; Suppose that I G I > 1 . PROOF.
BLOCKS OF P -SOLVABLE GROUPS
Let () be an irreducible character in B. Let N be a maximal normal subgroup of G. Then Clifford's theorem (111.2.12) implies that (IN = e ��= l Xi, where {xd is a set of irreducible characters of N which are conjugate under the action of G. In particular if (Jo is an irreducible character of height 0 in B then (Jo(l) }i 0 (mod p) and so by (IV.4.10) the blocks of N which are covered by B have a Sp-group of N as a defect group of N. Suppose that p .r I G : N I . Then p .r em and so the. same power of p divides (J (1) and Xi (l). Since D c;;;;, N the result follows by induction. Suppose that p I I G N I. Thus p = I G : N I as G is p -solvable. By (111.2.14) e = 1 . Thus m = 1 or p. If m = p then (J = X l . Hence by (V.1 .2) and (V. 1.6) Z(D) c;;;;, D n N. Thus if DJ is the center of D n N then I D n N_: DI I p � I D : Z(D) I . Hence the result follows by induction. Sup pose that m = 1 . Then by induction the height of (IN is most I D n N : Z(D) n N I � I D : Z(D ) I. This implies the result in case JJ is a Sp -group of G. The proof for general G is by induction on I G : H I, where H = Op' (G). Let , be an irreducible character of H which is in a block of H that is covered by B. By (V.2.S) and (1.2) it suffices to prove the result for the group G ({). By induction it may be assumed that I G : H 1 = I G (,) : Op, (G(,») I· By (1.4) every block of G (,) has a Sp-group as a defect group. The result follows from the first part of the proof. 0 4]
\"
:
425
0
r +
Suppose that F is a field of characteristic p and G is p-solvable. Let B be a block of G and let V be an irreducible F[ G] module in B with vertex P. If V is absolutely irreducible then (4.2) implies that I Z(D) I � I P I · Hamernik and Michler [1976] Theorem 3.2 have strengthened this to prove that Z(D) c;;;;, o P (even if V is irreducible but not necessarily absolutely irreducible). 4.3. Let G be a p-solvable group. Let B be a block of G with an abelian defect group. Then every irreducible character in B has height O.
COROLLARY
PROOF.
Immediate by (4.2). 0
4.4. Let G be a p -solvable group. The following are equivalent. (i) G has a normal abelian Sp-group. (ii) The degree of every irreducible character of G is relatively prime to p.
THEOREM
PROOF.
Let P be a Sp-group of G.
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426
[4
(i) =? (ii). Immediate by (III.6.9) and (4.3). (ii) =? (i). The proof is by induction on 1 G I. If 1 G 1 = 1 the result is trivial . Suppose that 1 G 1 > 1 . Let N be a maximal normal subgroup of G. Thus by Clifford's theorem (III.2. 12), the degree of every irreducible character of N is relatively prime to p. Suppose that p 1" 1 G : N I. Then by induction P <J N and P is abelian. Thus P <J G. Hence it may be assumed that 1 G : N 1 = p. Let Po = P I I N. Suppose that Po � (I) . Thus by induction Po <J N and so Po <J G. By induction GIPo has a normal Sp -group. Hence P <J G. If P is not abelian there exists an irreducible character X of P with p I X (l). Let 0 be an irreducible character of G such that X is a constituent of O. By Clifford's theorem (III.2. 12) X ( l) 1 0 (1) and so p 1 0 (1) contrary to assumption. Thus it may be assumed that Po = (1). Hence I P 1 = p. Let 0 be an irreducible character of G. Since p 1" 0 (1), ON = � is irreducible . Thus if P = (x ) then C = � for every irreducible character of N. Hence Brauer's combinatorial lemma (see e.g. Feit [1967c] (12. 1» implies that the map sending y to x - ' yx preserves all the conjugate classes of N. Thus if y E N the number of elements of N conjugate to y is not divisible by p. Hence H = eN (P) meets every conjugate class of N and so N = U y -'Hy. Thus N = H and so G = PH = p x H. 0
5]
PRINCIPAL SERIES MODULES FOR p -SOLVABLE GROUPS
5. Principal series modules for
p -solvable
427
groups
Let G be a p -solvable group. Let G = Go :J G1 :J Gn = (1) be a principal series of G. Then for each i, Gi IGi+] is either a p i_group or an elementary abelian p -group. If Gi 1Gi+1 is an elementary abelian p -group then GJGi +1 is an irreducible F[ G ] module, where F is the field of p elements. Such a module is called a principal series module of G with respect to p or more simply a principal series module of G. A principal series module of G is irreducible but not necessarily absolutely irreducible. The next two results were originally proved for solvable groups. The first of these has been gen �ralized by Cossey and Gaschiitz [1974] . •
•
•
THEOREM 5 . 1 (Fong and Gaschiitz [1961]). A principal series module of the
p -solvable group G is in the principal block.
PROOF. By definition Op ( G ) is in the kernel of a principal series module of G. Thus the result follows from ( 1 .5). 0 ,
COROLLARY 4.5. Let
Let G be a p -solvable group and let M be a p -complement in G. For a group N with M � N � G let Vo(N) denote the F[ N] module which affords the principal irreducible Brauer character, where F is the field of p elements. Thus Vo(N) G is isomorphic to a submodule of Vo(Mf. Since Vo(M) G is the principal indecomposable module corresponding to Vo( G) it follows that Vo(M) G has Vo( G) as its socle. Thus Vo(N) G has Vo( G) as its socle and so Vo(N) G is an indecomposable module in the principal p -block of G. Consequently the following result is a generalization of (5. 1).
PROOF. By (IV.4. 12) Op'( G) is in the kernel of B. Thus it may be assumed that Op ( G ) = (I) . By (1 .5) B is the unique block of G. The result follows from (4.3) and (4.4). 0
THEOREM 5 .2 (Green and Hill [1969]). Let G be a p -solvable group and let
G be a p -solvable group. Let B be the pricipal block of G. Then a Sp -group is abelian if and only if every character in B has height O.
,
G be a p-solvable group and let B be a block of G of defect d. If Cij is a Cartan invariant of B then Cij ::<S; p d.
THEOREM 4.6. Let
PROOF. Let 'P be an irreducible Brauer character of G in B. Let H = Op' ( G) and let � be an irreducible constituent of 'PH. By (V.2.5) and (1.2) it may be assumed that G = G (�). Thus by (1 .4) a Sp -group of G has order p d. Let 4>, be the principal indecomposable character of G corres ponding to 1 G. By (3.2) 4>1(1) = p d. The result follows from (IV.4. 15) (ii).
0
G. Let N = N G (M). Let Vo(N) be defined as above. Then every principal series module of G is a constituent of Vo(N) G.
M be a p -complement in
PROOF. Let HIS be a principal factor of G which is a p -group. Let GO = GIS and let A ° denote the image in GO of any subset A of G. Then Vo(NO) G O = Vo(NS) G . Since N° � N G o (MO) it follows that Vo(N G o (MO » G O � Vo(� ) G O
=
Vo(NS )G � Vo(N) G .
Thus it suffices to prove the result for the group GO. Hence by changing notation it may be assumed that G = GO and S = (I). Define W = {[ y ] l y E H} with [y]x = [ x 1 yx ] for x E G, y E H. Thus W is an F[ G ] module isomorphic to the principal series module H. -
CHAPTER X
428
Suppose that N n H"I (I). For x E N fl H, [x, M] � M n H (1). Since N n H"I (l) this implies that InvM ( WM) "I (O). Since N/M is a p group which acts on WM it follows that InvN ( WN) "I (0) and so Homp[H] ( Vo(N), WN) "I (0). By Frobenius reciprocity (III.2.5), this yields that Homp[ G ] ( Vo(N)G, W) "I (0). Since W is irreducible we get that W is a constituent of Vo(N) G in this case. Thus it may be assumed that N n H = (l). Let X I, . . . , Xs be a cross section of NH in G. Then {x;y J 1 :s; i :s; s, y E H} is a cross section of N in G as N n H = (I). Thus {NYXi 1 1 :s; i :s; s, y E H} is an F -basis of Vo(N)G. The subspace of Vo(N)G with F-basis {N(y - l)xi 1 1 :s; i :s; s, y E H, y "I I} is an F[ G] module V. It is the kernel of the natural map of Vo(N)G onto Vo(NH)G. Define an F-linear map f : V � W by f (N( y - l )xi ) = [Y]Xi = [x � l yxd . It is easily verified that if X E G, y E H then f(N(y - l)x) [X - l yX] . Furthermore if z E G then f (N( y - l)xz ) = [ Z - l x -1 yxZ ] = [x -1 yx]z = f (N( y - l)x)z. Thus f(vz ) = f(v ) z for all v E V, E G and so f is an F[G] homomorphism. Since f"l 0 and W is irreducible it follows that W is a constituent of V, and so of Vo(N) G . D
[6
=
=
z
COROLLARY 5.3. Let G be a p-solvable group. Let CPI be the principal indecomposable character of G corresponding to the principal irreducible character of G. Let qJ be an irreducible Brauer character which is a . constituent of the Brauer character afforded by principal series module. Then qJ is a constituent of CPl.
PROOF.
Clear by (5.2).
a
D
THE PROBLEMS OF CHAPTER IV, SECTION 5 FOR P -SOLVABLE GROUPS
6]
LEMMA 6.1 . Let G be a p -solvable group. Suppose that the character table of G is given and it is known which conjugate classes of G consist of p '-elements. Then the table of irreducible Brauer characters is uniquely determined. PROOF. For any class function 0 on G let 0' denote the restriction to the set of p '-elements of G. Let I be the number of p'-classes of G. In view of (2.1) it suffices to prove the following result. Let {Xi 1 1 :s; i :s; I}, {�j 1 1 :s; j :s; l } be sets of irreducible characters of G such that for every irreducible character 0 of G, 0 ' is a linear combination of the x i with nonnegative integral coefficients and 0 is a linear combination of the �i with nonnegative integral coefficients. Then {X; } = {�j}. The proof of this result is quite simple. Let X : = �j aij�j and let �j = �i bijX :· Then xi �j,k aij bkjX �. The set {X : 1 1 :s; i :s; I} is a set of linearly independent functions on the set of p '-elements in G. Hence I
=
2: aijbkj
(6.2) ()ik. j Suppose that aij "l O. There exists k with bkj "l O. Hence (6.2) implies that k � i and aij = bij = 1 . Thus � j = X : . As i may be chosen arbitrarily this implies the result. D =
Insofar as problems (V) and (VI) are concerned the following results can be proved.
LEMMA 6.3. Let P be a Sp -group of G. Assume that POp, (F) <J G. Then the following are equivalent. (i) There exists a conjugate class C of defect 0 in Op'( G). (ii) G has a block of defect o. (iii) POp, (G) has a block of defect o.
(i) =? (ii). This follows from (IV.5. 1). (ii) =? (iii). Let 0 be an irreducible character in a block of defect 0 of G. Let H Op, (G). By a result of Reynolds (see e.g. Curtis and Reiner [1962], p. 364) OPH = e �� l Tli' with em I I G : PH I and {Tli } a set of conjugate irreducible characters of PH. Thus 0 (1) = emTl l(l) and p ,( em. Hence I P I I Tl l (l). (iii) =? (i). Suppose that the result is false. Let 0 be an irreducible character in a block of defect 0 of G = POp, ( G). Let x be a p '-element in G. Thus x E Op' (G): Since I G CG ( x ) I 0 (x )/0 (1) is an algebraic integer PROOF.
6. The problems of Chapter IV, section 5 for
p -solvable
groups
This section contains a survey of results related to the problems of Chapter IV, section 5 for p -solvable groups. By (2. 1) problems (I) and (II) have an affirmative answer if G is p -solvable. The next result shows that problems (III) and (IV) almost have an affirmative answer for p -solvable groups.
429
=
=
:
430
CHAPTER X
[6
and / P / � / G : Co (x ) / it follows that O (x) == 0 (mod p ). Thus 0 is irreduc ible as a Brauer character and O (x ) == 0 (mod p ) for all p '-elements x of G. This contradicts (IV.3 . 1 1). D In the general case that G is a p -solvable group, neither of conditions (ii) or (iii) of (6.3) implies the other. The following examples are due to W. Willems. Let P be a Srgroup of GL2 (3) and let V be the underlying elementary abelian group of order 9. Let G be the split extension of V by GL2 (3). It is straightforward to verify that G has an irreducible character of degree 16 but PV has no 2-block of defect O. Thus G satisfies (6.3) (ii) but not (6.3) (iii). The group SL2 (3) is isomorphic to a subgroup of SL2 (5). Let V be the underlying elementary abelian group of order 5 2 . It is easily seen that SL2 (3) acts as a group of fixed point free automorphisms on V. Hence the semidirect product H = VSL2 (3) is a Frobenius group. Let P be a S2-grouP of SL2 (3). Then VP has three 2-blocks of defect 0 and H has exactly one 2-block of defect O. Let G = H 1. Z2 be the wreath product, where / Z2 / = 2. Then it can be seen that G has no 2-block of defect 0 but if Q is a Srgroup of G then VQ = 02, (G)Q has six 2-blocks of defect O. (The existence of at least one 2-block of defect 0 follows from (6.3).) The next result is a technical preliminary. LEMMA 6.4. Let p, q be distinct odd primes and let F be the field of q elements. Let P be a p -group and let V be a faithful irreducible F [ P ] module. Then there exists v E V with vx I- v for all x E P - {I}. PROOF. Induction on / P / . Suppose that Q
6]
THE PROBLEMS OF CHAPTER IV, SECTION 5 FOR P -SOL VABLE GROUPS
be assumed that P is abelian. Since V is irreducible it follows that cyclic and if v E V, v I- 0 then vx I- v for all x E P - (I). D
43 1
P is
THEOREM 6.5 (Ito [ 1951 a] , [ 195 1b]). Suppose that 1 G / is odd and Op, (G) and G /0p' ( G) are nilpotent for some prime. Then the following are
equivalent. (i) G has a block of defect o. (ii) If P is a Sp-group. of G then P n p x (iii ) Op (G) = (I).
= (1)
for some x E G.
PROOF. (i) � (ii). This follows from (1II.8. 14). (ii) � (iii) This is obvious. (iii) � (i) Let G be a counterexample of minimum order. Thus G has no block of defect O. Let H = Op, (G). Let P be a Sp -group of G. Suppose that G /H is not a p -group. By induction PH has a block of defect- 0 and so by (6.3) G has a block of defect ° contrary to assumption. Thus G = PH. Let Ho be a minimal normal subgroup of G. Since G /Ho has no block of defect 0 , induction implies the existence of an element x E P, x I- 1 such that [x, H ] C Ho. As H is nilpotent this shows that Ho is not in the Frattini subgroup L of H. Since Ho n L
.
.
m.
It is easy to see that (6.5) is false if p = 2. For instance let P be a S 2-grouP of GL2 (3). Let H be an elementary abelian group of order 9 and let G be the split extension of H by P where P acts on H as a subgroup of GL2 (3). Then P is transitive on the set H - {I}. This is easily seen to imply that 2 1 1 T(/:) 1 for every irreducible character ( of H. Thus by (6.3) G has no block of defect O. In case p = 2 there are some conditions on G which imply that (6.5) is true. See e.g. Ito [195 1a] , [1951b] . Another result similar to (6.5) will be proved after the next lemma.
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432
[6
If G is a permutation group on a set S let G at" 'am denote the stabilizer of all the elements a I , , am in S. .
•
.
LEMMA 6.6. Let G be a permutation group on a set S. Let F be a field of characteristic p. For A � G let A = 2:x EA X E F[ G ] . Then p I I Ga,b I for all a, b E S if and only if ( 2:aEs F[ G ] Ga )2 = (0). PROOF. If a, b E S and { xd is a cross section of Gab in Gb then GaGb = GaGab (2: x; ) = I Gab I Ga ( 2: x; ).
-
I
I Gax GaxI and so GaxGb = 0 by (6. 7) . Thus ( a2:ES F[ G ] Ga) 2 (0).
Suppose that p
xEGx
(6.7)
=
Gab I for all
a, b E S. Then
GaGb
=
0 by (6.7). For any
=
Suppose conversely that ( 2:a Es F[ G ] Ga )2 = (0). Then I Gab I Ga ( 2: x; ) = 0 by (6.7) for all a, b E S. Since Gax; n GaXj is empty for i I: j this implies that
p l l Gab l .
0
THEOREM 6.8. Suppose that G is p-solvable and p-radical. Then the following are equivalent. (i) G has a p-block of defect O. (ii) If P is a Sp-group of G then P n p x = (1) for some x E G. PROOF. Let F be a field of characteristic p which is a splitting field of F[ G ] . Let c be the sum of all the p -elements in G. By (VI.4.5) and (3.3) Ann(J(F[ GJ) = F[ G ] c. By (VI.5 .6) c 2 is the sum of all the centrally primitive idempotents of defect 0 in F[ G ] . As G is p -radical (VI.6.3) implies that Ann(J(F[ GJ) = 2:x E o F[ G ] ( 2:y EPX y ). Thus G has no blocks of defect 0 if and only if 2:xEo F [ G ] ( 2:y EPX y ) = (0). Let { x; } be a cross section of P in G. Let S = {PXi } and apply (6.6) to G acting on S by right multiplication. The stabilizer of PXi is x � I PXi. Hence by (6.6) { 2:xEo F[ G ] ( 2:Y EPX y )}2 = (0) if and only if X � I PXi n xj l pxj l: (I) for all i, j. 0 The hypotheses of (6.5) and (6.8) are somewhat stringent. However the following example due to Ito [ 1 95 1 b ] shows that it may be difficult to weaken these hypotheses substantially. Let p, q be primes such that q == 1 (mod p) and (qP - l)J(q - 1) = pro Then qP == 1 (mod r) and q ]i 1 (mod r). Thus r == 1 (mod p ) and there exists a Frobenius group M with a cyclic Frobenius kernel C of order r where y X = y q for y E C and some fixed x E M - C. Let X be a faithful
7]
IRREDUCIBLE MODULES OF p -SOLVABLE GROUPS
433
irreducible character of M. Then X (1) = p. Let F be the field of q elements. There exists a finite extension field FI of F and an irreducible FI [M] module VI which affords X. It is easily seen that the trace function afforded by VI has its values in F on group elements. Thus by (1. 1 9.3) there exists a faithful irreducible F[ M] module V with dimF V = p. Let P be a cyclic group of order p which acts faithfully as scalars on V. Then V becomes an F[ M x P ] module. Let G be the split extension of V by M x P. It is straightforward to verify that if 8 is a faithful irreducible character of G and � is an irreducible constituent of 8v then p I I T (�) I . This implies that p 2 ,( 8 (1). If 8 is an in-educible character of G which is not faithful then clearly p 2 ,( 8 (1). Hence G has no block of defect 0 but Op ( G ) = (I). Incidentally the groups described in the previous paragraph are exam ples of groups with a normal p -complement which are not p-radical. Hence in particular there exists groups G with H <J G such that H and G JH are p -radical while G is not. Saksanov [1971] has observed that SL2 (3) is not 3-radical. This provides another example. For other results related to problems (V) and (VI) see Tsushima [1974] . Problems (VII) and (VIII) are answered in the affirJJ?ative by (4.6). A good deal is known about problem (IX) for p -solvable groups. One direction is settled by (4.3). The other direction is settled for the principal block by (4.5). Fong [1963] has also shown that if F is p -solvable and p is the largest prime dividing I G I then a block in which every irreducible character has height 0 has an abelian defect group. The fact that G is assumed to be p -solvable doesn't seem to help at all with problem (X). Very little is known about this for p -solvable groups that is not known in general. See Gow [ 1980] for some estimates. For problems (XI) and (XII) see the remarks in Chapter IV, section 5. 7. Irreducible modules of
p -solvable
groups
The following result is due to Berger [1976] , [1979] . THEOREM 7. 1 . Let F be an algebraically closed field and let G be a solvable group. Then every irreducible F[ G] module is algebraic. If char F = P and G is p -solvable then the conclusion of (7. 1) is presumably also true. We will "almost" prove this in the following sense. A finite simple group G is well behaved if for every prime p ,( I G I, the Sp -group of the automorphism group of G is cyclic.
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434
The main object of this section is
to
[7
prove the following result.
THEOREM 7.2. Let F be an algebraically closed field with char F = p. Let G
be a p-solvable group such that every simple group which is a factor group of a subgroup of G is well behaved. Then every irreducible F[ G ] module is algebraic.
Every known finite simple group is well behaved so that once the finite simple groups are classified it will presumably be possible to deduce the conclusion of (7.2) for all p -solvable groups. The proof of (7.2) given here follows that in Feit [1980]. It is somewhat simpler than Berger's original proof. Some preliminary results are proved first. Of these (7.8) is of independent interest and is related to, though different from, some results of Dade [197 2] . It is perhaps curious that although (7.2) refers only to p -solvable groups, the proof uses properties of groups that are not p -solvable. LEMMA 7.3. Let q be the power of a prime with q = 3 (mod 4) and let K = Fq. Let V be a vector space of dimension 2n over K and let f be a nondegenerate alternating bilinear form on V. Suppose that J E Sp(f) Sp2n (q ) with J2 = - 1. Then there exists a, b E K and an integer m > 1 such that if =
y = aI + bJ then y 2m v, w E v.
=
-
I and
r =
-
f, where
r (v,
w ) = f(vy, wy ) for
PROOF. Since q = 3 (mod 4) it follows that the K -algebra generated by J is isomorphic to Fq2. Hence there exist c, d E K such that if x = cI + dJ then x q + 1 = 1. Since Jq = J- I = J this implies that -
dJ) (cI + dJ q ) = (cI + dJ) (cI - dJ) = c l I - d2J2 = (c2 + d2)1. Hence c 2 + d2 = 1 Since J E Sp(f) f(v, wJ) = f(vJ, wJ2) = f(vJ, w ) - 1 = (cI +
-
.
-
and so
r (v, w ) = f(cv, cw ) + f(cv, dwJ) + f(dvJ, cw ) + f(dvJ, dwJ)
(c 2 + d l)f(v, w ) + cd {f(v, wJ) + f(vJ, w )} = (c 2 + d2)f(v, w ) = f(v, w ). Thus r = - f. Let y be the 2-part of x. Then y lm = I for some m > 1 as q + 1 0 (mod 4) and r = f. 0 =
-
-
=
-
7]
IRREDUCIBLE MODULES OF p-SOLVABLE GROUPS
435
LEMMA 7.4. Let p be a prime, let q be the power of an odd prime distinct from p and let K = Fq. Let V be a vector space of dimension 2n over K and let f be a nondegenerate alternating bilinear form on V. Let P be a Sp-group of Sp(f). Assume that P acts irreducibly on V. Let c E K, c � O. Then one of the
following holds. (i) There exists x E GL( V) such that r = cf, and x commutes with every element of P. Either some power of x is equal to cd2 I for some d E KX with cd2 � 1 or c = 1 and x = 1. (ii) P = 2, c � a 2 for a E K and P acts absolutely irreducibly on V. -
PROOF. If c = a 2 with a E K then x = aI satisfies (i). Thus it may be assumed that c � a2 for any a E K. If x satisfies (i) then for a E K\ ax satisfies (i) if c is replaced by ca 2. Thus it suffices to prove the result for any fixed nonsquare c in K X . Hence it may be assumed that C 21 = - 1 for some integer t � O. In particular c = - 1 if q = 3 (mod 4). Let G = GL( V). Suppose that (ii) does not hold. By Schur's Lemma C o (P) F;k as finite division rings are fields. If p � 2 then k is even as dim V = 2n is even. If p = 2 then P does not act absolutely irreducibly by assumption and so k is even since 4> (2 m ) is a power of 2. Hence in any case Co (P) contains a unique cyclic subgroup A of order q 2 1 which contains all nonzero scalars. Choose J E A with J2 = c1. If r = cf then x = J satisfies (i). Suppose that r � cf. Then g = cf r � O. Clearly g is a P-invariant alternating bilinear form. Since P acts irreducibly on V it follows that g is nondegen erate. By definition gl = - cg. There exists z E GL( V) with r = g. As Sylow groups are conjugate, z may be chosen so that pz = P. Let Jo = JZ. Then 10 E Co (P) and so Jo E A. Thus by changing notation it may be assumed that J E A, J2 = cI and r = cf. If - 1 = a 2 for some a E K then x = aJ satisfies (i). Suppose that - 1 � a 2 for all a E K. Hence q = 3 (mod 4) and c = - 1. Thus J 2 = - I and r = f. Hence j E Sp(f). Let x = y be defined as in (7.3). Then x satisfies (i). 0 =
-
-
-
LEMMA 7.5. Let q be the power of an odd prime and let K = Fq. Let V be a vector space over K and let f be a nondegenerate alternating bilinear form on V. Let V = VI EB V2 with V2 = vt and dim VI = dim V2 = 2n. Let f be the restriction of f to \1;. Let Pi be a S 2-groUP of Sp(fi ) Sp2 n (q). Assume that Pi acts absolutely irreducibly on \1;. Then P I may be identified with P2 . Let P = {(y, y ) y E P I = P2} Then there exists x E GL( V) such that r = - f, x =
l
.
CHAPTER X
436
[7
commutes with every element of P and some power of x is equal to - d 2 I for some d E K with - d 2 � 1 . PROOF. I f - 1 = a 2 for some a E K, the result follows from (7.4). Suppose that - 1 � a 2 for any a E K. Hence q == 3 (mod 4) and d 2 � 1 for all d E K. We may identify VI with V2 and fl with f2 . Define the linear transformation J on V = VI EB V2 by J : (v!, V 2) � ( - V2, V I ) . It is easily seen that J 2 = I and r = f. Furthermore J commutes with every element of P. If x = Y is defined as in (7.3 ) then x has the desired properties. 0 -
-
LEMMA 7.6. Let p be an odd prime. Let K = F2, let V be a vector space over K and let f be a nondegenerate quadratic form on V. Let P be a p-group with p � O(f). Assume that P acts irreducibly on V. If x E Z(P) {I} then vx � v -
for all v E V, v � O. PROOF. Clear. 0
7]
PROOF. (i) This is well known. (ii) The existence of a unique irreducible Brauer character 'PA afforded by an F[ 0 ] module such that A is a constituent of ('PA )Z is well known, as are all the other properties of 0 in the statement. Then G = T( 'PA ). The existence and uniqueness of XA now follow from (III.3.16). (iii) Straightforward verification. 0 = PO, V, Z, Z(c ) be defined as in (7.7). Let A, J.L be linear characters of Z with A � 1, J.L � 1 . Then one of the following occurs. (i) Let Af.L be the character of PZ defined by AJ.L (xz ) = AJ.L (z ) for x E P, Z E Z. Then
LEMMA 7.8. Let P, 0, G �
A
XA @ X�
=
and let F be an algebraically closed field of characteristic p. (i) Let f(x, y ) = [x, y ] . Then f defines nondegenerate alternating bilinear form from V to Z = F;. If q = 2 then f(x ) = x 2 defines a nondegenerate quadratic form on V. Let A ( 0) denote the group of all outer automorphisms of 0 and let Ao( 0) denote the subgroup consisting of all automorphisms which fix all the elements of z. Then Ao( 0) <J A (0) and A (0)/Ao( 0) = Aut(Z) is cyclic of order q - 1 . If q � 2 then Ao( 0) = Sp(f) = Sp2n (q ). If q = 2 then Ao( 0) = O(f) = 02 n (2) is an orthogonal group. In any case a subgroup H of 0 is abelian if and only if the image of H in V is isotropic. (ii) Let P be a p-group with P � Ao(O) and let G = OP be the semidirect product. If A is a linear character of Z with A � 1, then (up to isomorphism ) there exists a unique irreducible F[ G] module XA such that A is a constituent of the character afforded by (XA ) Furthermore (XA )o is irreducible and every irreducible F[ 0] module which does not have Z in its kernel is isomorphic to some (XA ) If H is a maximal abelian subgroup of 0 then H = Ho x z, / Ho l = q n and ,\" 0 = (XA )o where '\"(hz ) = A ( ) for h E Ho. (ii) For i = 1, 2 let Oi be extra special with Zi = Z( Oi ). Then ZI = Z2 = Z. Let c E KX and let A I , A 2 be linear characters of Z with A l = A �c. Let 0 = 0 1 X 02 and let Z(c ) = {(z, ) I z E Z} � Z(O). Then XA1 @ XA2 is an irreducible F[ 0 ] module with kernel Z (c ). Furthermore 0/ Z (c) is extra special. a
z.
0.
Z
Z
c
�
A (Af.L ) G.
(ii) P = 2. Let Oi = 0 for i = 1, 2. Let 00 = (0 1 X 02)!Z( - 1), let Po = {(x, x ) I x E P}, let Go = PoOo . Let XA20 be the J:reducible F[ Go] module such that (X A 20 ) = (XA ) @ (XA ) 2 • Let A be the character of PoZ(Oo)/Z( - 1) such that A « )) = A ( ) for x E P, E Z. Then 00
LEMMA 7.7. Let q be a prime and let 0 be an extra -special q -group with / 0 / = q 2n + l . Let Z = Z(O) and let V = O/Z. Let p be a prime distinct from q
437
IRREDUCIBLE MODULES OF p -SOLVABLE GROUPS
01
4
XZ I ' XZ 2
0
2
4
Z I Z2
Zi
PROOF. The proof is by induction on / 0 / . Without loss of generality it may be assumed that P is a Sp -group of Ao( 0). Suppose that V contains a nonisotropic proper P-invariant subspace. Let W be minimal among such spaces . Then W = W n W-L EB Wo for some P-invariant space Wo � (0) with Wo n W� = (0). The minimality of W implies that W = Woo Hence if W = WI then V = WI EB W2 , where each Wi is P-invariant, WI � (0) and wt- = W2 . Since P is a Sp-group of Ao(O) i t follows that P = P I X P2 , where Pi acts trivially o n "'f for i � j. There exist extra special groups with R /Zi = Wi for i = 1 , 2 where Zi = Z(R ), such that ( P I H1 x P2H2)/Zo = PO for some subgroup Zo of ZI x Z2. Thus XA = XA1 @ XA2 and X� = X�1 @ X� , where XAi , X�, are irreducible F[ PiR ] modules which do not have Zi in their kernels. Observe that if p = 2 and (i) is satisfied for a group then also (ii) is satisfied. Thus by induction it may pe assumed that either (i) or (ii) is satisfied for the groups PiR for) = 1, 2. Hence either XAi @ X�, = (�yiHi for i == 1 or 2 or XA�O @ XA10 = (A 4 ) G,o for i = 1 or 2. Since PHI Z2 n PH2 Z1 = PZ1 Z2 and POHl OZ(H20) n POH20Z(HlO) = POZ (HIO x H20) it follows from the tensor product theorem that (A;t YI H] @ (fit y2 H2 = (,Q )G and (fl)G[o @ (f!)G20 = (A4)G and the re sult is proved in either case. Hence it may be assumed that V has no proper nonisotropic P-invariant subspace.
438
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Suppose that (O)� V I � V with VI a P -invariant subspace. Then VI � v t . Thus vt = VI E9 Vo with Vo P-invariant. A s Va n v t = (0) and Vo is isotropic this yields that Vo = (0). Hence VI is a maximal isotropic subspace of V. Furthermore V = VI E9 V2 with V2 a maximal isotropic subspace which is P-invariant. Let M be the inverse image in 0 of \1;. Then M is abelian . Furthermore Mi = Mo x Z, where P normalizes Mia. Let A, Ii respectively be the character of PM! , PM2 respectively with PMlO, PM20 in its kernel such that A (xz ) = A (z ) for x E PMlO, Z E Z and Ii (xz ) = /L (z ) for x E PM20, Z E Z. Then A O = XA and Ii ° = Xw As PM, n PM2 = PZ, the Mackey tensor product theorem (II.2. 10) implies that XA
0 X�
= A°
0 Ii ° = (A;;, ) 0.
-c.
l
=
Y Q(O)P(O)Z(H)
-(' H ) Q(O)P(O)Z(H) _ - ( /\/L - ( /\/L -(' P(O) Z(H» _
Since
0 (0) P(O)Z(H)/Z (c ) = Go the result is proved.
IRREDUCIBLE MODULES OF p -SOLVABLE GROUPS
439
Let p � q be primes. Let 0 be an extra special q -group and let P be a p-group contained in Ao( 0). Let F be an algebraically closed field of characteristic p and let V be an irreducible F[ PO] module which does not have Z( 0) in its kernel. Then V is an algebraic module.
LEMMA 7.9.
PROOF. By (II.5.3) and (7. 8) either V 0 V or V 0 V 0 V 0 V is alge braic. Hence by definition
V is algebraic.
0
In effect the content of (7.8) is the assertion that if V is an irreducible F[ PO] module then the en do-permutation module Vp has the property that Vp = V� if P is odd and V� = ( V�)* if P = 2. This would be enough to
prove (7.9) which is the essential result needed for the proof of (7.2).
Thus (i) holds. Therefore it suffices to prove the result in case P acts irreducibly on V. Let A = f.-t Two cases will be considered. Case (I). There exists an automorphism (J of G such that x 0" = x for all x E P, 0 0" = 0, V 0" ;I v for v E V, v ;l 0 and z 0" = Z C for z E Z. Case (II). The conditions of Case (I) are not satisfied. Suppose that Case (II) holds. By (7.4) and (7.6) p = 2, c ;l a 2 for a E K and P acts absolutely irreducibly on V. Let c = - 1 . Let Po, 00 , Go be defined as in statement (ii). By (7.5) there exists an automorphism (J of Go such that XU = x for all x E Pa, O g = 00 , v O" ;I v for v E Vo, v ;l 0, where Vo = Oo/Z(Oo) and z O" = Z - I for z E Z. In Case (I) change the notation and let P = Po, 0 = 00, G = Go, V = Vo, XA = XA O, X� = X�o so that both cases can be handled simultane ously. 0 Let H = P(O)( 00 x 00), where prO) = {(x, x ) I x E Po}. Define 0 ( ) = {(y, y ) l y E Oo} and 0 (0") = {(y, y O" ) y E Oo}. Then prO) normalizes both 0 (0) and 0 (0"). Furthermore H/Z(c ) is extra special, Z(c ) � 0 0" and O ( O" )/Z(c) is abelian. Since H � PoOo x PoOo, there exists an irreducible F[H] module Y = (XA O 0 X�O) H with kernel Z (c ) . --Let A/L be the linear character of P(O)O (u)Z(H) with p(O) o (a) in its kernel H � dH � (z l , Z2) = A (Zl)/L (z2) = /L (z lCz2).0 Then ( � )QoxQo = Y QoxQo and so A/L = Y by (7.7) (ii). By definition 0 ( ) n Q ( O" ) � Z(H.) Thus the Mackey decomposition (II.2.9) implies that (XA O 0 X� o)Q(O)P(O)Z(H)
7]
0
Q(O)P (O)Z(H) •
PROOF OF (7.2). Let W be an irreducible F[ G] module and let 'P be the Brauer character afforded by W. It may be assumed that 'P is faithful and so Op (G) = (1). The proof is by induction on 'P (l) = dimF W. If 'P is induced by a character 'Po of a proper subgroup then 'Po is algebraic by induction and so 'P is algebraic by (II.5.3). Thus it may be assumed that 'P is not induced by a character of any subgroup. Let H be a minimal normal noncentral subgroup of G and let � be an irreducible constituent of 'PH. As G is p -solvable, H is a p '-group. Therefore G = T(�) is the inertia group of { Since H is noncentral � (1) > 1. Thus W = V 0 \tV by ( 1 . 1) �here ((1) = diriJp V. If dimF V < 'P ( l) then also dimF W < 'P (1) and so V and W are algebraic by induction. Hence W is algebraic by (II.5.2). Thus 'P (1) = �(1) and so VH is irreducible. Hence if P is a Sp -group of G, then VHP is irreducible . Furthermore V I ( VHP ) O and so by (II.5 .3) it suffices to show that VHP is algebraic. Thus by changing notation it may be assumed that G = HP. Suppose that P � Go <J G. It suffices to prove the result for Go by (II.5.3) since V I ( Voo f . We will now consider two cases depending on whether H is solvable or not. Suppose first that H is solvable. The minimality of H implies that H is an extra special q -group for some prime q ;l p. Thus W is algebraic by (7.9) . Suppose finally that H is not solvable. Let Z = Z( G) . The minimality of H implies that Z � H' and H/Z = MI x . . . X Mk, where M = M for all i and some simple group M. Then {M } is the set of all conjugate subgroups of M and P acts as a transitive permutation group on the set {M }. There exists a group Ho with Z(Ho) � Hb and Ho/Z(Ho) = M such 'that H is a homomorphic image of H, x . . . X Hk with R = Ho for all i. Thus G is a
440
.
[8
CHAPTER X
homomorphic image of G = P(HI X . , . X Hk ), and P permutes the set {H; } transitively. Therefore 'PH = I; II� I 1;;, where I;i is an irreducible character of H whose kernel contains all � for j � i. Hence 1 G : T(l;l ) 1 = k . By (111.3. 16) there exists an irreducible Brauer character 'P I of T(I;I ) with ( 'Pl) H = 1;1 . Hence 'P Y = 'P . Since 'P is primitive this implies that T(l;l) = G. Hence k = 1 and HI Z is simple. As HIZ is well behaved, a Sp -group of G is cyclic and so there are only a finite number of indecomposable F[ H] modules (up to isomorphism). Hence by (11.5. 1), W is algebraic. 0 =
=
ko( x ) X (x )ko(x ) == X (x ) k (x ) ko(x ) == == ko(x ) k (x ) k (x ) X ( l) k (x ) x (l)
Thus X H i s in the principal block o f R [ H ] .
Let X be an irreducible character in B with
Throughout this section the following notation will be used. G is a finite group. If x E G then k (x ) = 1 G : cG (x ) l . K i s a finite extension of Qp which i s a splitting field of G and all o f its subgroups. R is the ring of integers in K and 7r is a prime in R. We will be concerned with the following notation and assumptions.
=
HYPOTHESIS 8. 1 . (i) P is a Sp -group of G, p e G <J G and G G CG (P). (ii) B, B is the principal block of R [ G ] , R [ G ] respectively. (iii) A, A is the ideal of R [ G] , R [ G ] corresponding to B, B respectively. The object of this section is to prove the following result. THEOREM 8.2. Suppose that (8. 1) is satisfied. Then the algebras A and A are
=
Alperin [1976d] showed that A A in case GI G is solvable. In full generality (8.2) is due to Dade [1977] . His proof depends on his deep work on Clifford theory. See e.g. Dade [1971a] . For an alternative proof see Schmid [1980]. We will here adapt Alperin' s approach to the more general situation. We will first prove a series of lemmas.
Throughout the remainder of this section it will be assumed that (8. 1) is satisfied. LEMMA 8.3. Let H be a subgroup of G with G � H. Let X be an irreducible
character in B such that XH is irreducible. Then XH is in the principal block of
R [H].
PROOF. For x E H let ko(x ) = I H : CH (x ) l . A Sp -group of CG (x ) is con tained in G and hence in C H (x ). Thus k (x ) and ko(x ) are divisible by the same power of p and so ko(x )1 k (x ) � 0 (mod p). Since X is in B it follows from (IV 04.2) that (mod 7T).
0
LEMMA 804. Suppose that X is an irreducible character in B with T(X ) = G.
8. Isomorphic blocks
isomorphic.
441
ISOMORPHIC BLOCKS
8]
X
� XC . Then
Xc = X.
PROOF. Apply ( 1. 1) with char F = 0 and H = G. T!ms X = 8Yf where 8 is a character of G with 8c = X if G is identified with G <J G. Furthermore Yf is an irreducible character of G I G and (G I G )I Z GIG for Z in the center of G I G. Let y E G I G and let y be the image of y in GI G. By (8. 1 ) there exists x E CG (P) with y = .i where .i is the image of x in GI G. Thus X (x ) = 8 (Yo) Yf (y ) for some Yo E G. Since X E B, =
X (x ) k (x ) ==
k
)�O
(x x(l) Hence Yf (y ) � O. As Yf is irreducible and y is an arbitrary element of GIG this implies that Yf ( l ) = 1 . Hence x (l) = 8 (1) X (l). 0
(mod 7r).
=
LEMMA 8.5. Let S, S be the set of all irreducible characters in B, B respec tively. For X E S let r(x) = Xc. Then r is a one to one map from S onto S. PROOF. Induction on 1 G : G I . If 1 G : G 1 = 1 the result is trivial. By induction it may be assumed that GI G is simple. We will first show that r maps S onto S. Suppose that 1 G : G 1 = q is a prime . Let X E S and let X be an G irreducible constituent of Xc . As q is a prime either Xc = X or X = X . Since X E B it follows from (IVA.2) that if x E CG (P) then x
:��S
(
x)
== k (X ) � O
(mod 7r).
G Thus X (x ) � O and so X� X as G = G CG (P). Hence Xc = X. By (8.3) Xc = X E S. If conversely X E S then by (V.2.3) X is a constituent of Xc for some X E S. Thus X = Xc and r maps S onto S in this case.
CHAPTER X
442
[8
Suppose that GIG is a noncyclic simple group. Let X E S. Let q be a prime and let Q be a Sq -group of G. By induction QG is the inertia group of X in QG and so Q C T(X). Since q was chosen arbitrarily this implies that G = T(X) · By (V.2.3) X C Xo for some X E S. Hence X = Xo by (8.4). If conversely X E S then by (V.2.3) there exists X E S with X C Xo. By (8.4) Xo = X· Thus r maps S anto S in any case. It remains to show that r is a one to one map. Let X E S with X = Xo. Let p be the character afforded by the regular representation of G I G. Then X O = xp· Hence if 0 E S with 00 = X = Xo then 0 C X o and so 0 = XA for some irreducible character A of GIG. Since 0 (1) = X ( 1) it follows that A (1) = 1. Thus if x E Co (P) then
k (X )X (X ) A (X ) k (x )O (x) k (x ) k (x ) X (x ) (mod ) X (l) X ( l) x (l) Hence A (x ) 1 (mod ) As G = G Co (P) this implies that A = 1 0 and so X = O. 0 ==
==
==
7T
==
7T
.
.
LEMMA 8.6. Let So, So be the set of all irreducible Brauer characters in B, B respectively. Let D, b be the decomposition matrix of B, B respectively and let C, C be the Cartan matrix of B, B respectively. For 'P E So let ro('P ) = 'Po. The
following hold. (i) ro is a one to one map of So onto So. (ii) Let r be defined as in (8.5). Let r (Xu ) = Xu for Xu E S and let ro( 'Pi ) = �i for 'Pi E So. Then D = b and C = C.
443
ISOMORPHIC BLOCKS
8]
Thus if 'P E So then (IV.3. 12) implies that 'P<x,O) is in the principal block of (x, G). Suppose that 'P h 'P E So with ('P I )O = 'Po. Since ('P l)(X,O), 'P(x,O) are in the principal block of (x, G) it follows by induction that ( 'P lh, o ) = 'P (x,O). Hence in particular 'Pl (X) = 'P (x ) for an arbitrary p '-element x E G. Hence 'P I = 'P and so ro is one to one in all cases. 0 LEMMA 8.7. Let V be a projective R [G] module in B. Let E = EndR[o] ( V) and let E = EndR[o] ( Vo ). Let f : E � E be defined by restriction. Then f is an algebra isomorphism of E onto E. PROOF. Clearly f is an algebra monomorphism. It remains to show that f is an epimorphism. Suppose that h E E and a E R with a - I h an R [G] endomorphism of V. Then a - 1h EndR ( V) n EndK[o]( VK ) = E. Thus f(E ) i s a pure submodule o f the R module E. Hence i t suffices t o show that rankR E = rankR E. Let U;, � be indecomposable projective R [GJ modules in B. By (8.5) rankR HomR[o] ( U;, �) = rankR HomR[o] « U; )o, (�)o ).
Since V is the direct sum of indecomposable projective R [G] modules in B it follows that rankR E = rankR E. 0 PROOF OF (8.2). Since A ' is an algebra with unity element it is anti isomorphic with the endomorphism ring of the F[ G] module A. Similarly A is anti-isomorphic with the endomorphism ring of the F[ GJ module A. By (8.7) it suffices to show that A o A as F[ GJ modules. Let {'Pi } be the set of all irreducible Brauer characters in B. Then by (8.6) {�i } is the set of all irreducible Brauer characters in B where �i = ('Pi )0. Let Vi be a proj ective indecomposable R [GJ module corresponding to 'Pi . Since C = C by (8.6) ( U; )o is the proj ective indecomposable R [GJ module corresponding to �i. Thus =
PROOF. In view of (8.5) it is clear that (ii) will follow as soon as (i) is proved. We will prove (i) by induction on 1 G : G I . If 1 G : G 1 = 1 the result is trivial. By induction it may be assumed that GIG is simple. If 'P E So then 'Po is an irreducible Brauer character by (8.5) and (IV.4.33). By (IV.4. 10) 'Po E So. If conversely � E So then there exists 'P E So with � C 'Po by (lV.4. 10). Thus � = 'Po by the previous sentence. Hence ro maps So onto So. It remains to show that ro is one to one. Suppose that 1 G : G 1 = q is a prime. Let 'P, 'P I E So with 'Po = ('PI)O . By (111.2. 14) there exists a linear character A of GIG with 'P I = 'PA. By (IV.3. 12) 'P (x ) = Lxu ES auXu (x) for all p '-elements x in G. Hence 'PA ( X ) = Lxu ES auXuA (x ) for all p '-elements x in G. Since Xo = XAo for all X E S it follows from (8.5) that XuA � B unless A = 1 0. Since 'PA E So this implies that A = 1 0 and 'PA = 'P. Thus ro is one to one in this case. Suppose that GIG is a noncylic simple group. Let x E G. If X E S then X<X,O) is irreducible and so X< X, O) is in the principal block of (x, G) by (8.3).
Ao
=
EB 'Pi (1) ( U; )o = EB �i ( 1) ( Vi )o
=
A.
0
1]
CHAPTER XI
1.
An analogue of Jordan's theorem
One of the oldest results in group theory is the following theorem. THEOREM 1 . 1 (Jordan). There exists an integer valued function l(n ) defined
on the set of positive integers with the following property. If the finite group G has a faithful representation of degree n over the complex numbers then G has a normal abelian subgroup A with 1 G : A / < l(n ).
Several proofs of this theorem are known. See for instance Curtis and Reiner [1962] (36. 13) for a proof and references to other proofs . It is an immediate consequence of Jordan's theorem that the same conclusion holds if the field of complex numbers is replaced by any field whose characteristic does not divide 1 G / . However the result is false for fields whose characteristic divides 1 G I . For example let F be an algebrai cally closed field of characteristic p > 0 and let Gm = SL2(p m ). Each Gm has a faithful F-representation of degree 2 but a normal abelian subgroup of Gm has order at most 2 while / Gm I can be arbitrarily large. This section contains a proof of the following analogue of (1.1) which was first conjectured by O. H. Kegel. THEOREM 1.2 (Brauer and Feit [1966]). Let p be a prime. There exists an
integer valued function f(m, n) = It) (m, n) such that the following is satisfied. Let F be a field of characteristic p and let G be a finite group which has a faithful F-representation of degree n. Let p m be the order of a Sp -group of G. Then G has a normal abelian subgroup A with 1 G : A I < f(m, n ). Isaacs and Passman [1964] have used Jordan's theorem to show that if 444
AN ANALOGUE OF JORDAN'S THEOREM
445
the degree of every irreducible complex representation of the group G is bounded by some integer n then the conclusion of (1.1) holds (with a function different from l(n » . In a similar manner J. F. Humphreys [1972] has used (1 .2) to prove an analogous result in characteristic p. For a related result se�, J. F. Humphreys [1976] . The various proofs of Jordan's theorem (1 .1) yield different values for ' I (n ). None of these values seem to be anywhere near the best possible value. Since ( 1 . 1) is needed for the proof of (1.2) and several fairly crude estimates are also used in the proof of (1 .2) the value of f(m, n ) which can be derived from the given proof of (1 .2) is probably nowhere near the best possible result. Thus (1.1) and (1 .2) are both qualitative results which do not yield any useful bounds . The proof of (1 .2) will be given in a series of lemmas. Without loss of generality it may be assumed that the field F in (1 .2) is algebraically closed . For any group G let L J (G) denote the F[G] module which affords the principal Brauer character. Let B J (G) denote the principal p -block. If G is a finite group and P is a Sp -group of G, then the center Z(P) of P is a Sp -group of Cc (P) and Burnside ' s transfer theorem implies that Co (P) is the direct product a p -group and a p '_group. Thus PCc (P) = P x H for some p '-group H. Suppose that V is an indecomposable F[ G] module with dimF V¢ 0 (mod p). Th en P is a vertex of V. Let V denote the F[No (P)] module which corresponds to V in the Green correspondence. The crux of the proof of (1 .2) is contained in the next result. LEMMA 1 .3. Suppose that V is an irreducible F[ G] module with n = dimF V > 1 . Then at least one of the following holds. (i) G has a normal subgroup of index p. (ii) There exists an irreducible constituent L of V* ® V ® V* ® V �ith
L in BJ(G) and L � L J(G). (iii) Let P be a Sp -group of G. Let N = N o (P). There exists an irreducible constituent L of V* ® V with dimF L ¢ 0 (mod p) and L � L (G) such that if Ho is the kernel of LH then 1 H : Ho 1 < l(n ), where l(n) is defined by ( 1 . 1). J
PROOF. Assume that G satisfies neither (i) nor (ii). Let LJ = L J (G), L2, , Ls denote the distinct irreducible constituents of V* ® V. Let W be an F[ G] module, all of whose irreducible constituents are constituents of V* ® V ® V* ® V. We will show that the multiplicity of L J(G) in W is equal to dimF lnvo W. It clearly may be assumed that W is indecomposable. If W is not in B J (G) the result is trivial. Suppose that W •
•
•
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CHAPTER XI
[1
1]
IS
III B I (G). Since (ii) is excluded, every composItIon factor of W is isomorphic to L I(G). If x is a p '-element in G then W(x ) is completely reducible . Thus x is in the kernel of W. Since G has no normal subgroup of index p it follows that G is in the kernel of W. Hence dimF W = 1 as W is indecomposable and the result is proved. This fact will be applied to several modules . Let W = Li @ Lj with 1 ::::; i, j ::::; s. Thus InvF W = HomFI G ] (Li , Lj ). For i = j, Schur's Lemma implies that dimF Inv G W = 1 . Thus by (III.2.2) and the previous paragraph dimF Li � 0 (mod p ) for 1 ::::; i ::::; s. For i -I- j, Schur's Lemma implies that InvF W = (0). Since (ii) is excluded it follows that no irreducible constituent of W is in BI(G). Hence by (IV.4. 14), Lr, . . . , L, lie in s distinct p-blocks. Let Y be an indecomposable direct summand of V* @ V. Since all the irreducible constituents of Y lie in one p -block it follows from the previous paragraph that all the irreducible constituents are isomorphic to Li for some i. Let b be the length of a composition series of Y. Thus L I ( G) occurs with mUltiplicity b in L ; @ Y. Since every irreducible constituent of L ; @ Y is a constituent of L ; @ Li and so of V* @ V @ V* @ V, it follows that b = dimF HomF[ G ] (L , Y). Let Yo be the socle of Y and let bo be the length of a composition series of Yo. The previous argument applied to Yo shows that bo = dimF HomF[ G ] (L , Yo). However HomF[ G ] (L , Y) = HomF[ G ] (L , Yo). Thus b bo and so Y = Yo. As Y is indecomposable this implies that Y = Li . Consequently V* @ V is completely reducible. Let =
s
V* @ V = EB aiLi . i=1
(1.4)
B y Schur's Lemma a I = 1 . Hence'& Y (III.2.2) dimF V � 0 (mod p ). Thus P is a vertex of V. Hence (III .5 .7), (V.6.2 ) and (1 .4) imply that s
V* @ V = EB a;lJi E9 S, i=1
(1 .5)
where each indecomposable direct summand of S has a vertex properly contained in P. By (III.7.7) L I = L I (N) and L is not in B I (N) for i � 2. By (III.3.7) VPXH = EB� = I ( Vi @ Xi ) , where XI , . . . , Xe are distinct ir reducible F[H] modules and VI , . . . , Ve are F [ P] modules which are conjugate under the action of N. Thus
e ( V* @ V)pxH = EB ( V i @ Uj ) @ (Xi @ Xj ) i.j= 1
447
AN ANALOGUE OF JORDAN'S THEOREM
where S is an F[ P x H] module which does not contain L I (P x H) as a constituent. By (1 .5) L I (N) I V* @ V. Thus L I (P x H) I ( V* @ V)PXH and so L I (P) I Vj @ Uj for some j. Since the modules Uj are conjugate under the action of N, this implies that L I (P) I Vj @ Uj for j = 1, . . . , e. Thus I
eL I (P x H) I ( V* @ V)PXH. Suppose that e > 1 . Since al
= 1 , (1 .5) implies that L I (P x H) I TpXH for some indecomposable direct summand T of S. Thus L I (P) I Tp. Since a vertex of T is properly contained in P this contradicts ( III .4 .6) . Thus e = 1 . Consequently
(1 .6) where X is an irreducible F[ H] module and V is an F[ P] module such that L I (P) I V * @ V. Let d dimF X. If d = 1 then X* @ X = L I (H) and H is in the kernel of V* @ V. Hence by ( 1 .5 ) H is in the kernel of Li for all i. Thus by (JV.4.14) Li is in BI(N) for all i and so Li is in B I (G) by (III.7.7) and (V. 6.2) . Thus s = 1. Hence by (1.4) n = 1 contrary to assumption . Thus d > 1 . Let M b e a n irreducible constituent o f V and let D b e the kernel o f M. By (III.2. 13) P � D. Thus N/D is a p '-group. Hence by the remark following ( 1 . 1) there exists a subgroup A o with D � Ao <J N such that A o/D is abelian and I N : Ao l < J(n ). Let A I = Ao n H and DI = D n H. Then A I <J N, I H : A I 1 < J(n ) and A I /D I is abelian. By (1 .6) MH = uX for some positive integer u. It follows that D I is the kernel of X. If A is an abelian group and X is an F[ A ] module then X* @ X contains L I (A ) with multiplicity at least dimF X, Apply this remark with A A JD I . Thus (X* @ X)A, contains L I (A I) with multiplicity at least d � 2. As X is irreducible, X* @ X contains L I (N) with multiplicity 1 . Thus there exists an irreducible constituent Z � L I (H) o f X* @ X such that L I (A I ) / ZA, . By Clifford 's theorem (III.2. 12), A I is in the kernel of Z. Since L I (P) @ Z I ( V* @ V)PXH by (1 .6) it follows from ( 1 .5 ) that LI(P) @ Z I TpXH for some direct summand T of EB aiL E9 S. By (III.4.6) P is a vertex of T. Thus by (1 .5) T = Lj is irreducible. Since Z;;6 L I (H), j -I- 1. Let Ho be the kernel of (Lj )H. Since A I belongs to the kernel of Z and A I <J N it follows from Clifford's theorem (III .2. 12) that A I � Ho. Thus I H : Ho i ::::; I H : A l l < J(n ). Consequently (iii) holds for L Lj • D =
I
=
=
LEMMA 1 .7. Let V be an irreducible F[ G] module with dimF V = n > 1 .
Suppose that G contains no normal subgroup of index p. Then there exists an irreducible constituent L of V* @ V @ V* @ V such that
448
CHAPTER XI
[1
1 < I G : Go I < I GLIPI 2 n4J(n)(p) I ,
where Go is the kernel of L. PROOF. Let L be an irreducible F[ G] module with dimF L = d. Let Go be the kernel of L and let iP be the trace function afforded by L. Suppose that iP has exactly e algebraic conjugates in F. Thus the subfield of F generated by all iP (x ) as x ranges over G has p C elements. By (1. 1 9.2) GIGo is isomorphic to a subgroup of GL d (p e ). Thus I G : Go I � I GL d (p e ) I � I GLde (p ) / . By assumption (1 .3) (i) does not hold. Thus by (1 .3) either ( 1 .3) (ii) or (1 .3) (iii) must hold . Suppose that (1 .3) (ii) holds. Choose L accordingly. G rf Go as L ;l: L , ( G ) . By (IV.4.9) and (IV.4. 1 8), iP has at most I p l 2 algebraic conjugates in F. Thus the result follows from the previous paragraph and the fact that d < n4. Suppose that (1 .3) (iii) holds . Choose L accordingly. A s above G rf Go. Let B" . . . , Ba be all the p -blocks of G which contain algebraic conjugates of iP. Since dimF L = d < n 4 the remarks above show that it suffices to prove that iP has at most I p 12J(n ) algebraic conjugates. Thus by (IV.4. 1 8) it suffices to show that a � J(n ). Let $ be the trace function afforded by i. By (III. 7.7) there are exactly a p -blocks of N G (P) which contain an algebraic conjugate of $. By (1 .3) (iii) the group N G (P)IHo has less than J(n ) p -blocks. Thus by (V.4.3) a < J(n ) as req uired. 0 Let G be a finite group and let V be an F[ G] module. Let 7( G, V) = (m, n, a ) where a Sp -group of G has order p m , dimF V = n and a is the multiplicity with which L , (G) occurs as a constituent of V* 0 V 0 V * 0 V. Define the partial ordering -< as follows : (m " n" a,) -< (m, n, a ) if one of the following is satisfied. (i) m, < m, n, � n. (ii) m , � m, n, < n. (iii) m , = n, n, = n, a, > a. If H is a subgroup of G, then clearly either 7 (H, VH ) = 7 (G, V) or 7(H, VH ) -< 7(G, V). Observe that if 7(G, V) = (m, n, a ) then a � n 4. Thus there are only finitely many triples 7( G" V, ) with 7( G" VI) -< 7( G, V).
LEMMA 1 . 8. Let P be a Sp -group of G and let / p I = p m . Let X be a faithful F[ G] module and let dimFX = n. Let g(m, n ) = / GLIPI2J( n ) n 4(p ) / , where
1]
AN ANALOGUE OF JORDAN'S THEOREM
449
J(n ) is defined by ( 1 . 1). Assume that G is not abelian. Then there exists Go <J G with I G : Go / < g(m, n ) such that 7(Go, XGo) -< 7(G, X). PROOF. It may be assumed that G has no normal subgroup of index p otherwise the result is trivial. Suppose first that every composition factor of X has F -dimension 1 . Let X, be a completely reducible F[ G J module with the same composition factors as X. Then P is the kernel of X, and G I P is abelian. Thus G is solvable and so G contains an abelian subgroup A with / G : A I = / p I . Let Go = nxEG x-'Ax. Then Go <J G and / G : Go l � I p i ! � g (m, n ). Since G is not abelian / P I � 1 . Thus 7( Go, XGo) -< 7( G, X). Suppose next that some irreducible constituent V of X has F -dimension at least 2. Let L and Go be defined as in (1 .7). If 7( G, X) = (m, n, a ) and 7(Go, XGo) = (m " n l , a l ), then clearly m l � m and n , = n. Since Go is the kernel of L and L ;I: L I ( G ), al > a. Thus 7(Go, XGo) -< 7(G, X). 0 LEMMA - 1 .9. There exists a function h (m, n, a ) such that if X is a faithful F[G] module, then I G : A I � h (7(G, X» for some normal abelian sub
group A of G.
PROOF. Let h (m, n, a ) = 1 if (m, n, a ) � 7( G, X) for any pair (G, X). If the result is false it is possible to choose a counterexample (G, X) so that h (m " n " a ) is defined for all (m l , n l , a l ) -< 7(G, X). Let g(m, n ) = I GLIPI2 n4J(n )(p ) I . D efine
and let
h (7( G, X» = g(m, n )ho( 7( G, X» 8( m.n) . If G is abelian let G = A. The result is clear in this case. Suppose that G is nonabelian. Let Go be defined by (1 .8). Thus Go contains a normal abelian subgroup Ao with I Go : Ao l � h (7(Go, XGo» ' Let A = nXEG x -IAox. Then I G A 1 < g(m, n )h (7(Go, XGo» 8( m.n). This implies that I G A 1 < h (7.( G, X» . 0 :
:
(1 .2) is now a direct consequence of (1 .9) if f(m, n ) is defined by
f (m, n ) = max h (m, n, a ). O::::::; a � n 4
1J
CHAPTER XII
I f a block B has a noncyclic defect group D then the situation i s vastly more complicated than in the case described in Chapter VII when D is cyclic. Such a block in particular contains an infinite number of indeco � posable modules, see for instance Hamernik [ 1 974a] , [ 1 975a] . �xcept I� very special cases when p = 2 and D contains a cyclic subgroup of mdex 2 It is hopeless to attempt to describe all the indecomposable modules in B. See Bondarenko [1975], Ringel [ 1 974], [ 1 975] . It is perhaps less obvious that there appears to be no method for constructing the irreducible Brauer characters in B. The purpose of this chapter is to study questions concerning the ordinary irreducible characters in a block B and to present some applications of this study. In this connection the concept of a basic set introduced in Chap ter . IV, section 3 plays an important role. After a general introductory sectIon most of the chapter deals with the case that p = 2 and D is a special type of 2-group. The material in this chapter originated with work of Brauer [1952] . The notation introduced at the beginning of Chapter IV is used throughout this chapter.
1. Types of blocks
The results in this section are due to Brauer [1961a], [ 1964a], [ 1 969b], [ 1 971a], [1971c], [1971 d] . See also Reynolds [1965] . For related results see Brauer [ 1 964c] . Let B be a block and let {Xu } be the set of irreducible characters in B. Let 'PB = {'Pi } be a basic set for B. Thus there exist rational integers dur such that Xu (x ) = 22i dui'Pi (x ) for all p i-elements x in G. As in Chapter IV, 450
45 1
TYPES OF BLOCKS
section 3 {dui } is called the set of decomposition numbers with respect to 'PB and {eij } is called the set of Cartan invariants with respect to 'PB where Cij = 22u duiduj . Let y be a p -element in G. Let {'P r} be the union of basic sets for all the blocks of CG (y ). Then for each Xu in B there exist algebraic integers d �i in the field of p n th roots of unity over Q for suitable n such that Xu (yx ) = 22i d �i'P r(x ) for all p I-elements x in CG (y ). The algebraic integers d �i are the higher decomposition numbers with respect to {'P r}. The set of columns d �i as 'P r ranges over a given basic set is a subsection with respect to {'P r}.
LEMMA 1 . 1 . Let B be a block of G. Let y, z be p-elements in G. Let {'P r}, {'P �} be the unions of basic sets for all blocks of CG (y ), CG (z ) respective/yo (i) If Xu is in B then d �i = 0 unless 'P r belongs to a basic set of a block B of CG (y ) with B G = B. (ii) If y is not conjugate to z in G then for all i, j L (d �i) * d �j = L (d �i)*d �j = o. U
� �B
(iii) If (c �) is the Cartan matrix of
i, j
CG (y) with respect to {'P r} then for all
Lu (d�i)*d �j = c �. PROOF. Immediate by (IV.6. 1) and (IV.6.2).
D
Given a basic set for the block B then the Cartan matrix (Cij ) of the block with respect to that basic set is the matrix of a positive definite integral quadratic form Q. If the basic set is replaced by another basic set then Q is replaced by a quadratic form which is equivalent to Q over Z. Thus to each block there is associated an equivalence class of integral quadratic forms.
LEMMA 1 .2. Suppose that p and d are given. There are only finitely many classes of integral quadratic forms which are associated to p-blocks of defect d of groups G. PROOF. Let Q be a quadratic form associated to a p -block of defect d. By (IV.4. 1 S) the dimension of Q is at most p 2d. By (IV.4. 16) every elementary divisor of the Cart an matrix is a power of p which is at most p d and so the discriminant of Q is bounded by a function of p and d. This implies the result by the reduction theory of quadratic forms. D
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452
[1
COROLLARY 1.3. There exists a bound f(P, d ) depending only on p and d
such that for each p-block of defect d of any group, a basic set can be chosen such that the Cartan invariants are at most equal to f(P, d ). PROOF. Immediate by (1 .2).
D
LEMMA 1 .4. Let B = Bo be the principal p -block of G and let % be the principal Brauer character of G. There exists a basic set for B containing 0/0 such that the Cartan invariants for this basic set lie below a bound fo(p, n ) depending only on p and n where a Sp -group of G has order p n. PROOF. Choose a basic set {'P; } for B as in (1 .3). Let % = 'Ldoi'Pi . Then d �i ::s; Cii ::s; f(p, n ). Furthermore the ideal of Z generated by all dOi is Z itself. Thus there exists a matrix of determinant ± 1 with entries in Z and first row equal to (do; ). The entries in this matrix are bounded by a function of the dO i and so by a function of p and n. D LEMMA 1 .5. Let B be a block of defect d, let y be a p -element in G and let B
be a block of C C (y) with B c = B. (i) Let {'P r} be a basic set for B and let m be the maximum of the Cartan invariants associated to {'P r}. Then the corresponding higher decomposition numbers d �i for Xu in B all belong to a finite set M(p, d, m ) depending only on p, d, and m. (ii) There exists a basic set for B such that the corresponding higher decomposition numbers d �i for Xu in B all belong to a finite set M(p, d) depending only on p and d.
1]
TYPES O F BLOCKS
453
over all the irreducible characters of B and the column index i is such that 'P ( Y, B ) = {'P r}.
Let T"b(B ) be the matrix of ordinary decomposition numbers of B with respect to the basic set 'P (y, B ). A block B of defect d is of a given type for a subsection (y, B ) if the pair of matrices Tl;( B ) , TY (B, B ) formed with respect to some basic set 'P (y, B ) of B is given. A block B is of a given type for an element y E Y if B is of a given type for all subsections (y, B ) corresponding to B. A block B is of a given type if it is of a given type for all elements y E Y. Let BO be a block of the group GO. Suppose that B and B O both have defect d. Define D O , yO, y O E yO , T"b°( B O) , T y o (B O , B O) for the group GO. Then B and B O are of the same type if the following conditions hold . (i) 1 Y I = 1 yO I and for a suitable ordering Y = { Yi }, yO = {y ?} with 1 B l (Cc ( Yi ), B ) 1 = 1 B l (Cco(y D, B O) I · (ii) After a suitable rearrangement BI(Cc ( Yi ), B ) = { Bij } and BI(Cco (y?), B O) = {B �} such that there exist basic sets for Bij and B ;� for all i, j with the property that T Yi (B, Bij ) = TIl (B O, B �) and T"bi(Bij ) = Ttl (B �) .
THEOREM 1 .6. For given p and d there exist only finitely many types of p -blocks of defect d (in the category of all finite groups). PROOF. If Y is defined a s above then from ( 1 .3 ) and ( 1 .5). D
1 Y 1 ::s; p d. Thus the result follows
If y is a p -element in G let BI(Cc (y ), B ) be the set of all blocks B of Cc (y ) with B c = B. Let D be a defect group of B. If y is not conjugate to an element of D then BI(Cc (y), B ) is empty by (IV.6.6) (iv). Let Y be a set of elements of D which consists of a complete set of
Suppose that the type of a block is given. It is natural to ask what additional information is required to compute the values of the irreducible characters in the block. Before considering this question a preliminary result will be proved. The following situation will be studied. Let D be a fixed subgroup of G with 1 D 1 = p d. Let y E D. Let K1 , , Kn denote the classes of p '-elements in Cc (y ). Assume that the following information is given. (i) I Ki 1 is known for i = 1 , . . . , n. (ii) It is known which conjugate class of G contains Ki • Denote this class by KF. (iii) It is known for which i = 1 , . . . , n the class KF has defect less than d and for which i the class KF has D as a defect group.
representatives of the conjugate classes of G which contain an element of D. For y E Y and B E BI(Cc (y ), B ) let 'P (y, B ) be a basic set for B. Define the matrix TY (B, B ) = (d �i) where the row index ranges over u as Xu ranges
LEMMA 1 .7. Let D be a subgroup of G with 1 D 1 = pd and let y E D. Suppose that (i), (ii), (iii) above are given. Suppose also that for each block B
PROOF. (i) If
(T
is an automorphism of the field generated by all d �i then (T. The result follows
« d �it)* = « d �;)*t· Thus by ( 1 . 1 ) 1 (d�it 1 2 ::s; m for all as each d � i is an algebraic integer. (ii) Clear by (i) and ( 1 .3).
D
• • •
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of Co (y) of defect at most d there is a n irreducible character X in 13 such that X ( ) is known for E Ki , l � i � n. For any block b with defect group D and for E G let Xi
Xi
X
CUB
l x (x) ( ) _ / G : CoX (x) ( ) X
-
l
for some irreducible character X in B. Then it is possible to compute (x) for any p '-element in G. In particular it is possible to determine the number of blocks of G with defect group D.
2]
SOME PROPERTIES OF THE PRINCIPAL BLOCK
455
there exists a left inverse of T"bi(13ij ) which can be found explicitly. In other words the given basic set for 13ij can be expressed in terms of the irreducible characters in Bij .' Since TYi (B, 13ij ) is known this makes it possible to evaluate the irreducible characters Xu in B on the p -sections which contain elements of Yo. 0
CUB
X
PROOF. The last statement follows from the first by (IVA.3) and (IVA.8). It
remains to prove the first statement. Let X be a p '-element in G and let C be the conjugate class of G with X E C. Since y E D it follows from (IV.6.6) (iv) that if B is any block of G with defect group D then there exists a block 13 of Co (y ) with B 0 = B . By (III.9.6) 13 has defect at most d. If CU B is defined analogously to CUB it follows that CU B (x ) = Li CU B ( Xi ) , where Xi ranges over a complete set of representatives of the conjugate classes of Co (y ) which are contained in C n Co (y ). In particular since X is a p '-element, each such conjugate class is some Ki for 1 � i � n. If C i- K� for all i, the sum is empty and CUB (x ) = O. If C = K? for some i, then it is known by assumption which of the classes Ki lie in C and CUB ( Xi ) can be computed as X ( Xi ) is known by assumption. 0
THEOREM 1 .8. Let D be a subgroup of G of order p d. Let Y be a subset of D which is a complete set of representatives of conjugate classes of G which meet D and let Yo be a nonempty subset of Y. Suppose that for every y E Y, (i), (ii) (iii) are given and that for y E Yo the values on p '-elements in C o (y ) of the irreducible characters in blocks of defect at most d of Co (y ) are known. Let Yo be a fixed element in Yo, let 13 be a block of Co (Yo) for which 13 0 = B has defect group D and suppose that the type of B with respect to every element in Yo is known. Then it is possible to find the values of the irreducible characters in B on the p -sections which contain elements of Yo. y = yo. Thus it is possible to find
(x ) for p '-elements X in G. For each Yi E Yo it is possible to determine the set {13ij } = BI(Co (Yi ), B ). For each 13ij the values of the irreducible characters in Bij are known on p '-elements in Co (Yi ). By hypothesis the matrix T"hfBij ) of decomposition numbers with respect to a suitable basic set for Bij is known. Since all the elementary divisors of this matrix are 1 by (IV.3. 1 1)
PROOF. Apply
(1 .7) for
CUB
2. Some properties of the principal block
The results in this section will be useful for applications below. See Brauer [ 1 964b] . Let {Xu } be the set of all irreducible characters in the principal block Bo of G. For an element X E G define
A (G, x ) = L
Xu i n B o
Let
A (G) = A (G, 1).
(2. 1)
Thus
A (G) = L
Xu in B o
Clearly
I Xu (x ) 1 2.
2 Xu (1) .
(2.2)
A (G, x ) � 1 Co (x ) 1 ·
LEMMA 2.3. For x E G, A (G, x ) is a positive rational integer. Let i denote the image of x in G = G /Op (G ). Then A (G, x ) = A (G, i) � I CG (i ) l . In particular A (G) = A (G) � 1 G Op'( G) 1 and the equality sign holds if and only if G is of deficiency class 0 for p. ,
:
PROOF. By (IVA.9) A (G, x ) is a rational integer. Since Bo contains the principal character, A (G, x » O. By (IVA. 12) A (G, x ) = A (G, i). The in equalities are clear and equality holds if and only if Bo is the unique block of G. 0 LEMMA 204. Let y be a p-element in G and let x be a p '-element in Co (y ). Then A (G, yx ) = A (Co (y), x ). In particular A (G, y) = A (Co (y )). PROOF. By the second main theorem on blocks (IV.6. 1) and by (V.6.2) ( ) (f> r(x ) (f> J(x -I ), A (G, yx ) = L L i,j d �i d �j * Xu in B o
where (f> ; and (f> ] range over the irreducible Brauer characters in the principal block of Co (y ). By (IV.6.2) this implies that
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[3
3]
INVOLUTIONS AND BLOCKS
457
G which contains yx. By assumption the coefficient of C in C1 C2 is O. Thus a well known formula implies that
A ( G, yX ) = � c � 'P i(X )'P HX - I ) = A ( C o (Y ) , X ) , j,j
where 'P r , 'P J range over the irreducible Brauer characters in the principal block of Co (y ) . 0 LEMMA 2.5. If H <J G then A (H) � A (G) � I G : H I A (H).
PROOF. By (IVA. 10) every irreducible character in the principal block of H is a constituent of the restriction to H of some irreducible character in Bo. This implies the first inequality. If Bo, B " . . . are all the blocks which cover the principal block of H then I G : H I A (H) = 2:X (1)2 as X ranges over the irreducible characters in all Bj • 0 3. Involutions and blocks
Involutions (i.e. elements of order 2) play a special role in group theory. Brauer [ 1 957] was the first to realize their importance for the classification of simple groups. In that same paper he showed how the study of involutions can be connected with block theory. Most of the material in the rest of this chapter is based on these ideas. The results in this section are from Brauer [1964b], [1966a] . Related results can be found in Brauer [ 1 961b], [ 1 962a], [ 1966b], [ 1 974c] . THEOREM 3 . 1 . Let G be a group of even order and let p be a prime which
divides 1 G I . Let y be a p-element in G and let tl and t2 be involutions in G. Assume that y is never the p-part of 2 ,22 where 2j is conjugate to ti for i = 1, 2. Let B be a p-block of G and let {Xu } be all the irreducible characters in B. Then for all i
(3.2)
where the d �j are the higher decomposition numbers belonging to basic sets for the blocks of Co (y ). Furthermore (3.3)
PROOF. Let C denote the conjugate class of G which contains ti for i = 1, 2. Let x be a p '-element in Co (y ) and let C be the conjugate class of
where Xv ranges over all the irreducible characters of G and x is any p I-element in Co (y ). Let {'P n be the union of basic sets for all the p -blocks of Co (y). Thus Xv (yx ) = :2:i d �j 'Pi (x ) for every p I-element in Co (y ). Hence the linear independence of the set {'P n implies that
� d �i Xv (tl )Xv (t2) = 0 . v
Xv (1)
If 'P r is in the block fJ then d �i = 0 unless Xv is in fJ 0 by the second main theorem on blocks (IV.6. 1). This proves (3.2). Then (3.3) follows by multiplying (3.2) by 'P r(x ) and summing over all 'P r in basic sets for blocks o iJ with iJ = B. 0 If y E G we will say that x inverts y if x -I yx = y -l .
COROLLARY 3A. Let y be a p-element in G. If t l and t2 are involutions in G such that no conjugate of t, inverts y then the conclusion of (3. 1) holds. PROOF. Suppose that y is the p -factor of 2 1 22 where 2i is conjugate to ti for i = 1 , 2. Then 2 1 22 = yx for some p '-element x E Co (y ). Thus 1 -I )- 1 -I Z I (y x ) 2 1 = 2 1 2 1 22Z 1 = 22Z 1 = ( 2 1 22)- = ( yx .
As y is a power of yx it follows that 2 I inverts y contrary to assumption . The result follows from (3. 1). 0 It may happen that for fixed involutions tl and t2 there exist several nonconjugate elements y such that the conclusions of (3. 1) hold. Thus new equations can be obtained as linear combinations of the given ones. If it is possible to choose such linear combinations so that the Xu (y ) are replaced by rational integers au where 2:a;, is small, then it may be pos�ible to deduce properties of G. This idea will be used frequently in this chapter. The remaining results in this section are of importance for such applica tions. Let P be a p -group contained in G. Let {Xu } be the set of all irreducible characters contained in the principal block Bo. If () is a complex valued class function on P define
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�
(3.5) au (0) = « Xu )p, O)p = I I y�p Xu (y )O (y - l ). Let a(O) denote the column (au (0)). If a = (au ) and b = (bu ) are two columns define (a , b) = Lu aub �, where * denotes complex conjugation. If 0 is a generalized character of P then au (0) is a rational integer by definition. Let A be defined as in (2.2). THEOREM 3.6. Let P be a p-subgroup of G and let N be a subgroup with P � N � Ne (P). Let U be a nonempty subset of P with N � Ne ( U). Assume
that the following conditions are satisfied. (i) If y 1 and Y2 are elements of U which are conugate in G then they are conjugate in N. (ii) w = A (Ce (y ))/ I CN (y ) I is independent of y in U. Let 0 and 11 be generalized characters of P such that 0 vanishes outside U. Then
I p I 2 S = L (X (Y ), X (z ))O(y - l )11 (Z ), y,z E P
where (X (y ), X (z )) is the inner product of the columns X (y ) = (Xu (y )) and X (� ) = (Xu (z )), and {Xu } is the set of all irreducible characters in Bo. By the second main theorem on blocks (IV.6. 1)
Lu Xu (y )Xu (z - 1 ) = L d �i(d �j)* cp ;(I) cp ] (1) = 0 if y is not conjugate to z in G. If y is conjugate to z then by (2.4) �0
(X(Y ), X(z )) = (X(Y ), X (y )) = A (G, y ) = A (Ce (y )).
Since 0 vanishes outside of U it follows that
I p I 2 S = L A (Ce (y ))O (y - l ) L 11 ( Z ), yEU
SOME COMPUTATIONS WITH COLUMNS
459
Since y can range over all of P this implies the result. 0 THEOREM 3.7. Let P be a p-subgroup of G. Let U be a subset of P. Assume that there exists an involution t in G such that no element in U is inverted by a G-conjugate of t. Let 0 be a generalized character of P which vanishes outside U. Let {Xu } be the set of all irreducible characters in Bo. Then the
following hold. au (O)Xu (t t = 0 . Lu Xu (1)
L au (O)Xu (1) = O. L au (O)Xu (t) = O.
(3. 8) (3.9) (3. 10)
If furthermore (lp, 0) I- 0 then either there exist at least two positive and two negative au (0) or G has a proper normal subgroup which contains (02,(G), t).
PROOF. Let (a( 0 ), a( 11 )) = s. By definition
(x (y ), X(z )) =
4]
PROOF. It follows from ( 3 . 4) that (3.3) holds for t = tl = t2 and all y E U. If (3.3) is multiplied by I p l - 1 0 (y - l ) and added over all y in U then the definition of au (0) shows that (3 .8) holds . By assumption 1 � U and no conjugate of t is in U. Thus by the second main theorem on blocks (IV.6. 1) the columns x (1) and X (t) for the blocks Bo are orthogonal to the column X (y ) for y E U. This proves (3.9) and (3. 10). Let Q be the diagonal quadratic form (au ( O )IXu (1)). If at most one of au (0) is positive or negative then a maximal isotropic subspace has dimension at most 1 . By (3.8), (3. 9) and (3. 1 0), (Xu (1)) and (Xu (t)) are vectors which lie in an isotropic subspace and so are proportional. If XO = Ie then Xo(l) = Xo(t) = 1 . It follows that Xu (1) = Xu (t) for all Xu in Bo with au (O) I- O. Choose Xu l- Xo with au ( O ) I- O. Then (02,(G), t) is in the kernel of Xu but G is not. Thus the kernel of Xu is the desired normal subgroup. 0
z
where z ranges over the G -conjugates of y in P. Since N � Ne ( U) it follows from (i) that z ranges over the N-conjugates of y in P. In other words z ranges over the distinct elements of the form y with x E N. Hence L X E N 11 (y X ) = I CN (y ) I L z 11 (z ). Now (ii) implies that x
4. Some computations with columns
This section contains results which indicate how computations with the columns a( 0 ) introduced in section 3 can be used to yield information
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[4
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about irreducible characters of G. These methods are closely related to those introduced in Chapter V, section 7. Only elementary results will be discussed in this section. These are primarily from Brauer [1966a] . For (4. 9) (ii) see Feit [1974]. In the sequel the case p = 2 will be studied more closely and several applications will be presented. In case p � 2 several authors have used the "method of columns" and properties of isometries proved in Chapter V, section 7 to derive results about the structure of groups. See for instance G. Higman [1973], [1974] ; Smith and Tyrer [1973a] , [ 1973b] ; Smith [ 1 974] , (1976b] , [ 1976c], [1 977] . As an example we state here one such result without proof. THEOREM 4. 1 (Smith and Tyrer [1973b]). Let p � 2. Suppose that G has an abelian Sp -group P and 1 NG (P) : C G (p) 1 = 2. If G = G ' then P is cyclic. Consider the following hypothesis. HYPOTHESIS 4.2. (i) P is an abelian p -subgroup of G and N is a subgroup with P � N � N G (P). U is a nonempty subset of P. (ii) If y E U then C G (y) has a normal p -c;omplement. (iii) If y E U and y is conjugate in G to x E P then x E U and y is conjugate to x in N. (iv) If y E U there exists a Sp -group Py of C G (y ) such that CN (y) = Py CN (P) and P = Py n CN (P). The similarity of (4.2) and (V.7.1) is evident. LEMMA 4.3. Suppose that (4.2) is satisfied. Let () and characters of P such that () vanishes outside U. Then
77
be generalized
where x ranges over a cross section of CN (P) in N. PROOF. The assumptions of (3.6) are satisfied. If y E U then A (C G (y» = 1 Py 1 by (lV.4. 12), (2.4) and (4.2) (ii). If w is defined as in (3.6) then w - I = 1 CN (P) : p i by (4.2) (iv). The result now follows from (3.6). D The following hypothesis is also relevant. HYPOTHESIS 4.4. (i) P is an abelian Sp -group of G with 1 P 1 � 1 .
4]
461
SOME COMPUTATIONS WITH COLUMNS
(ii) N G (P)/C G (P) is cyclic of order m with 1 < m < 1 P 1 - 1 . (iii) If y E P - { I } then C G (y ) n N G (P) = C G (P). LEMMA 4.5. If (4.4) is satisfied then so is (4.2) with U = P - {I} and N = N G (P). PROOF. Suppose that (4.4) is satisfied. Then (4.2) (i) and (4.2) (iv) are immediate and (4.2) (iii) is a standard property of abelian Sylow groups. Burnside's transfer theorem shows that (4.4) (iii) implies (4.2) (ii). D If (4.4) is satisfied the following notation will be used. r = ( l p l - 1 )/m . ljio = h, ljil , . . . is the set of all irreducible characters of P. Thus each ljij for j > 0 has exactly m conjugates under the action of N = N G (P). The notation will be chosen so that {ljij / 1 ::;:;; j ::;:;; r} is a complete set of representatives of the orbits of N. By definition a ( ljij) = a( ljij ) for all j and x E N. THEOREM 4.6; Suppose that (4.4) is satisfied. There exists an integer s ::;:;; m such that the principal p-block Bo of G contains exactly r + s irreducible characters (1 , . . . , ( XI = 1 G, • • • , Xs . If Y is the p -part of z in G and
y E P - {I} then
r ,
(j (z ) = d + 8 L lji j ( y ) 1 � j ::;:;; r, x
Xu (z ) = au
1 � u ::;:;; s,
where x ranges over a cross section of C G (P) in N ; 8 = and au � O. Moreover
± 1;
d, a
(d - 8 f + (r - 1) d 2 + ! a ;, = m + 1 . u =1 If furthermore t is a p i-element then (j (t) = ( I (t) for all j and (rd - 8 )(j (t) + ! auXu (t) = O. 1
I,
•
•
•
, as E Z (4.7)
(4.8)
PROOF. Let bj = a(ljio - ljij ) = a (ljio) - a(ljij ). By (4.3) and (4.5) (bi , bj ) = m + 8ij • Let Ci = bi - bl . Then (Ci , cj ) = 1 + 8ij • The result now follows from a standard argument in character theory. See e.g. Feit [ 1 967c]. 0
CHAPTER XII
462
[5
Observe that in case P is cyclic (4.6) is a very special case of the results in Chapter VII.
COROLLARY 4.9.
Suppose that (4.4) is satisfied. (i) If m = 2 then the notation may be chosen so that in (4.6) d = 0, s = 2, a2 = 0 and �j ( 1) = X2(l) + D. (ii) If m = 3 then the notation may be chosen so that in (4.6) d = 0, s = 3 and au = ± 1 for all u. If furthermore G contains an involution w hich is in no proper normal subgroup then exactly two of 0, a2, a3 are equal to - 1 and �j (1)X2(1)X3(1) is the square of a rational integer.
PROOF. 1 d 1 � 1 by (4.7) as m � 3 . Hence either d = 0 or d = ° by (4.7). Suppose that d = D. If s = 1 then (4.8) implies that (r - 1 )0�j (1) = - 1 and so �j (1) = 1 which is not the case. Thus s ;?: 2. As m < 1 P 1 - 1 it follows that r ;?: 2. Hence (4.7) implies that 2 � r � m + 2 - s � m. Thus r = 2 if m = 2. If m = 3 then 2 � r � 3 and so 1 P 1 = 5 or 7. Thus 1 p i = 7 as m I I 'p 1 - 1 . Therefore r = 2 in any case. Hence i f j ' i- j then (4.6) implies that Thus the notation may be chosen so that d = ° in all cases. (i) By (4.7) s = 2 and a2 = ± 1 . Thus (4.8) implies that - O�j (1) + 1 + a2X2 ( 1 ) = 0. Hence �j (1) = oa2X2(1) +0 and so oa2 > 0. (ii) By (4.7) s = 3 and a u = ± 1 for all u . Suppose that G contains an involution which is in no proper normal subgroup. Let () = {(t/lo - t/llt } P. By (3 .7) exactly two of 0, a2, a3 are equal to - 1. Let V be a 4-dimensional vector space over Q with diagonal quadratic form
5]
GROUPS WITH AN ABELIAN 52-GROUP OF TYPE (2"', 2m )
Another example comes from are 6, 10, and 1 5 .
463
A7 for p = 7. The corresponding degrees
5. Groups with a n abelian Sr group o f type (21n , 2 m )
This section contains a proof of the following result. See Brauer [ 1 964b] Section VI. 5 . 1 . Let G be a group w hose S2-groUP P is abelian of type (2m , 2 m ) ith m ;?: 2. If 02, ( G ) = ( 1 ) then P <J G, Co (P) = P and 1 G : P / is 1 or 3. w
THEOREM
PROOF. The proof is by induction on / G / . Let G be a counterexample of minimum order. Then 02, ( G ) = (1). Let C Co (P), N = No (P). Let t l , t2, t3 be the involutions in P. If N = C then Burnside's transfer theorem implies the result. Since N/C acts faithfully on P by conjugation it follows that / N : C / = 3 and N/C permutes { tl , t2, t3} cyclically. If y E P - {1} then Co (y ) has a normal 2-complement by induction. Thus (IV.4. 12) implies that =
A (Co (y » = A (P) = 1 p i
=
22 m
for y E P - {1}.
(5 .2)
Let t/lo = 1 , t/l 1 , . . . be all the irreducible characters of P. Let r 2 1 ( 2 m - 1) ;?: 5. Then the notation may be chosen so that t/lo, . . . , t/lr is a complete set of representatives of the orbits of N on {t/li }. Let U = P - {1} in (3.6) then w = 3 / 1 N : P I . Hence (3.6) implies that
(a(t/li - t/lo), a(t/lj »
= Oij
(a(t/li - t/lo), a(t/lo»
= -3
=
for 1 � i, j � r, for 1 � i � r.
Define bi = a( t/li - t/lo) = a( t/I; ) - a( t/lo) for . 1 � i � r. Then
( bi , bj ) By (3 . 7) V has a 2-dimensional isotropic subspace and so V is the direct sum of two hyperbolic planes. Thus the discriminant of the form, which is equal to � 1 (1 )X2(I)X3(1 ) is a square in Q. 0 The situation described in (4. 1 0 ) occurs infinitely often. For instance, let G 1 ( q ) = PSL3(q ) and G - I(q ) = PSU3(q ). Then for E = ± 1 , G" (q ) contains a subgroup P x H which satisfies (4.4) of order (q 2 + Eq + 1 )/k, where 2 2 k = (q + Eq + 1 , 3). The degrees of the given characters are q\ q + Eq and 2 (q - 1) (q
-
E
).
Let Ci
=
=
3 + Oij
1 � i, j � r.
(5.3)
bi - br for 1 � i � r - 1. Then ( c i , Cj )
=
1 + Oij
for 1 � i, j � r - 1 .
A standard argument (see e.g. Feit [ 1 967c], §23) implies that the nonzero coefficients of the r - 1 columns C i appear in r rows and if the rows are suitably arranged, the matrix in the first (r - 1) of the rows is EI with E = ± 1 while all coefficients in the rth row are - E. By (5 .3) ( br, ci ) = - 1. Thus b r has the same coefficient b in each of the first r - 1 rows and the coefficient b + E in the r th row. There occur further
[5
CHAPTER XII
464
rows in which all the coefficients of (" . . . , Cr - l vanish. If the coefficients of br in these rows are 00, 0 " . . . then (5.3 ) implies that (5 .4) For the row corresponding to XO = 1 we have ao(t/Jo) = 1, aO(t/Ji ) = ° for i > 1. Hence the coefficient of br in this row is - 1, while the coefficient of each Cj vanishes. This shows that one OJ has the value - 1. Thus b = ° in (5.4) as r - 1 � 4. Thus if the rows are arranged suitably the nonzero coefficients of the columns b l , b2 , , br appear as follows 00, 00, . . . , 00 •
•
•
£ 0 ,...,0 0" £
°
°
£
°
(5.5)
where 00 = - 1 , XO = 1 and Xi denotes the irreducible character corre sponding to the (i + 1 )st row. In particular ai (t/Jj ) - ai (t/Jo) = OJ for i = 0, 1 , 2 and j = 1, . . . , r. Hence ai (t/J) - aj (t/Jo) = Oi for every nonprincipal irreducible character t/J of P and i = 0, 1, 2. It follows from (3.5), the definition of a-j (t/J), that if y E P then for ° � i � 2
Xi ( y ) = L ai (t/J )t/J( y ) . l'
Thus if
y E P - { I}, ° � i � 2 then Xi ( Y ) = L { ai (t/J) - ai (t/JO)}t/J( y ) = Oi L t/J( y ) = - Oi . �
���
(5 .6)
« Xi )p, 1) is an integer this implies that (5.7) Xi (1) - Oi (mod 22 m ) , i = 1 , 2. Choose t/J, as a nonprincip al character of P with t/J � = t/Jo . Then all elements of order less than 2 m belong to the kernel of t/J I and so
Since
==
I � I �p Xi (y ) (t/JI ( Y ) - t/Jo( Y )) = � I �S Xi ( y ) (t/J I ( y ) - t/JO ( Y )) ' I
ai (t/JI - t/lo) =
5]
BLOCKS WITH SPECIAL DEFECT GROUPS
465
where S is the set of all elements in P of order 2 m . As P is abelian no element in S is conjugate to its inverse in G and so in particular no element in S is inverted by a conjugate of tl • Hence (3 . 7) may be applied with U = S and () = t/JI - t/Jo . By (5.5) and (5.6) with y = t = t,
£X3(tt 02 01 XI(I) + X2 ( 1 ) + X3(1) - 0, - 1 + o I XI ( l ) + 02 X2 ( 1 ) + £X3(1) = 0, - 3 + £X3(t) = 0.
-1+
_
(5.8) (5. 9) (5. 10)
Suppose that XI ( l ) � 1 and X2 (1) � 1 . By (5.7) XI ( I ) � 15 and X2 (1) � 15. Hence (5 .8) and (5 . 10) imply that 9£ /X3(1) � 13/15. Thus £ = 1 and X3(1) � 10. By (5.9) X3(1) == 3 (mod 16) and so XlI ) = 3 contrary to (5.8). Hence Xi (1) = 1 for i = 1 o r 2. By (5.7) Oi = - 1 and s o P is in the kernel H of Xi by (5 .6). Thus H is a proper normal subgroup of G and 02,(H) � 02, (G) = (1 > and so by induction P <j H. Hence P <j G. The remaining statements are immediate as O2,( G) = (1). 0 6. Blocks with especial defect groups
Given a block B with a defect group D it is of interest to obtain information concerning the irreducible characters and irreducible Brauer characters in B, the decomposition matrix, the Cartan matrix and perhaps also the structure of certain modules and their sources. In case D is cyclic this situation has been discussed in Chapter VII. If D is noncyclic then virtually nothing is known for p odd, but much work has been done in case D is a special type of 2-group. In this sort of work the principal block is usually much easier to handle than a general block. The results of section 1 show, that given D, there are only a finite number of possibilities for various numbers attached to B. Unfortunately the number of possibilities can be very large even if D is not too complicated. Brauer [ 1 964b] was the first to make such investigations. He also used these to get information about the structure of the group G as in the previous section. The remainder of this chapter contains some further results of this sort . We will here only mention some of the literature on this subject. If B is the principal block and D is dihedral (including the 4-group as a special case) see Brauer [1964b], [ 1966a] ; Landrock [1976] ; Erdmann [ 1977b] . For the general case of D dihedral see Brauer [1971a] , [ 1 974b] ; Erdmann and Michler [ 1 977] ; Donovan [ 1 979] . For quasi-dihedral,
466
[6
CHAPTER XII
quaternion and similar 2-groups s_e e Brauer [1966a] ; Olsson [1975], [1 97 7] ; Erdmann [ 1 979b] ; Kiilshammer [ 1 980] . Olsson [1977] also has some results for p "1 2. By using the classification of simple groups with an abelian S2-grouP, questions about the principal block in case D is abelian are reduced to the study of groups of Ree type. For the study of these groups see Fong [1974] , Landrock and Michler [ 1 980a] , [1980b] . For the case of I D 1 = 8 and B arbitrary see Landrock [ 1 981 ] . In this section we will only prove some very elementary results which will be needed for the next section. See Brauer [ 1 964b] .
LEMMA 6.1. Let G have a S2-grouP P which is abelian of type (2, 2). Then either G has three or one conjugate class of involutions. In the former case G has a normal 2-complement and the irreducible characters in the principal 2-block Bo of G are the four characters of degree 1 of G /02, ( G ) = P. In the latter case the principal 2-block B o contains exactly four irreducible charac ters Xo = 1, Xl , X2, XCI. For u = 1, 2, 3 there exists Cu = ± 1 such that Xu (y ) = Cu for every 2-singular element y in G. Furthermore Xu (1) == .su (mod 4) for u = 1, 2, 3.
PROOF. If No (P) Co (P) then G has a normal 2-complement by Burn
1=
side's transfer theorem and the result follows. Suppose that No (P) "I Co (P). Then I No (P) : Co (P) 3 and G has only one conjugate class of involutions. Let t be an involution in G. Then Co (t) has a normal 2"'complement and so the principal Brauer character is the unique irreduc ible Brauer character in the principal block of Co (t). By (V.6.2) Xu ( y ) d �o for all u. By (IV.6.2) L;=o I d �o l 2 = c� = 4. Since Xu does not vanish on all 2-singular elements d �o "I 0 for all u. As d �o is rational for all u and d� = 1, it follows that Bo contains exactly 4 irreducible characters. Furthermore Cu = d �o = ± 1 for all u. As LyEP Xu (y ) == 0 (mod 4), (6.2) holds. Equation (6.3) follows from (IV.6.3) (ii). 0
=
=
COROLLARY 6.4. Suppose that G has exactly one class of involutions in (6.1). Then the notation can be chosen so that CI = 1, C3 - 1 . In that case 1 , XI , C2X2 is a basic set for Bo and the decomposition matrix and Carta'n matrix with respect to this basis are as follows.
GROUPS WITH A QUATERNION 52-GROUP
D =
(� n 0 1 0
c
1
=
1
c D 2 1
467
= (1 + &, ).
PROOF. Let x = 1 in (6.3) then the notation can be chosen so that C I = 1 and C3 = - 1 . By (6.3) 1 , Xl , C2X 2 is a basic set and D has the required form. Thus C = D 'D as required. 0 It is much more difficult to compute the decomposition matrix and Cartan matrix with respect to the basic set consisting of the irreducible Brauer characters. Landrock [1976] has done this and has shown that in case G has exactly one class of involutions in (6. 1) there are two possibilities as follows .
(� = (}
D =
(6.2) (6.3)
1 + c , X ,(x ) + C2X2 (X ) + C 3X 3(X ) = 0 for any 2'-element x in G.
=
7]
D
0 1 1 0 0 1
1
0
�), ,�)
c
=
1
c D· 2 1
(6.5)
2 C= 2 2 2 1
(6.6)
=
C D
Both of these cases occur infinitely often. A direct computation shows that if q is a prime power and G P SL2 (q ) then (6.5) occurs in case q == 3 (mod 8) and (6.6) occurs in case q == 5 (mod 8). If q == 3 c (mod 8) with C 1 the_n
=±
Xo(1) = 1, X I( 1 ) = q, X2 (1) = X 3 (1) = ! (q - .s ) .
If 'Po, 'P I , 'P 2 are the irreducible Brauer characters in Bo then 'Po(1) = 1 and
'P I (1) 'P2(1) = ! (q - 1). =
In either case Bo is the unique 2-block of G of defect 2 and so contains all the irreducible characters of odd degree. 7. Groups with a quaternion Sz- group
=
THEOREM 7. 1. Suppose that 02, ( G ) (1) and a Sz-group T of G is a (generalized) quaternion group. Then the center of G has order 2.
468
CHAPTER XII
[7
Brauer and Suzuki [1 959] first proved (7.1). Proofs of (7. 1) can also be found in Suzuki [1959] and Brauer [1964b]. All of these proofs depend on the theory of modular representations. It has been known for some time that if I T I > 8 then one can give an elementary proof using only the theory of ordinary characters. See e.g. Feit [ 1967c] Section 30. More recently Glauberman [ 1 974] has given a proof in case I T 1 = 8 which uses only ordinary characters. The proof given here is that in Brauer [ 1 964b] . It uses results proved earlier in this chapter. See also Dade [ 197 1 b] for the results in this section and the next. PROOF OF (7. 1). The proof is by induction on I G I . Let 1 T 1 = 2 n + ' with
T = (y, z I y 2n = 1, y 2n - 1 = Z 2 , z - ' yz = y - ' ) .
Let t = y 2n - 1 = Z 2 be the unique involution in T. Suppose that it has been shown that t is contained in a proper normal subgroup H of G. A Sz-group To of H contains only one involution and so is either cyclic or a quaternion group. Furthermore 0 2, (H) � 02 ,(G) = (I). If To is cyclic then H has a normal 2-complement and so To
Since a ( e ) is a column with integral coefficients this shows that there are exactly 3 nonzero coefficients. No 2-element of G of order at least 4 is inverted by any involution since an involution is in the center of any
7]
GROUPS WITH A QUATERNION 52-GROUP
469
2-group which contains it. Hence (3.7) implies that (t) is contained in a proper normal subgroup of G. Thus it may be assumed that n = 2 and T is the ordinary quaternion group. The conjugate classes of T are represented by 1, t, y, z, yz. If no two distinct elements in the above list are conjugate then G has a normal 2-comple ment by a theorem of Frobeniu s and the result is trivial. Suppose that two of these elements are conjugat e. Say x - ' yx = z. Thus x - I NG « y »)x = NG «z »). Since T is a Sz-group of NG « y ») and of NG «z ») it may be assume d that x E NG (T) . Hence x induces an automo rphism of odd order of T and so x - I zx = (yz tl. Thus G has only one conjuga te class of elemen ts of order 4 as each of y, z, yz is conjuga te to its inverse in T . Furthermore any two elemen ts of order 4 are conjuga te in CG (t). The group CG (y) has a cyclic Sz-grou p of order 4 and so has a normal �-complement. Thus the principal Brauer character 'P h is the only irreduc Ible Brauer charact er in the principal block of CG (y ) by (IV.4. 12). Let X (y ), bb be the column s (Xu (y )) , (d �o) respect ively. By (IV.6. 1) and (V.6.2) X (y ) = b6 .
(7.2)
By the second m�in theorem on blocks (IV.6. 1) (bb, bb) = 4,
(bb, X ( l )) = 0,
(7.3)
where x (1) is the column (Xu (1)). Let H = CG Y)I(t). Then f = (9, i) is a Srgroup of H with 9 2 = i2 = 1 . Furthermore H contain s only one class o f involut ions. Choose a basic set ljJo = 1, 1jJ 1 , 1jJ2 for the princip al block of H as in (6.4). Thus the Cartan invarian ts are 1 + Oij . Then ljJo, 1jJ" 1jJ2 is also a basic set for the princip al block of CG ( t ) but the Cart an invarian ts are twice those for H. Let {d �J be the set of higher decomp osition number s for this basic set and let b; be the column (d �i). Thus in particul ar Xu ( t ) =
2
i2: =O
d � iljJi (1).
(7.4)
Furthermore (b; , bJ) = c :j = 2(1 +
Oij ),
(X (1), b D = (bb, b:) = 0.
(7.5) (7.6)
For i = 0, 1 , 2 ljJi (1) is odd. If t is an irreduci ble character of T then t ( y ) == t ( t ) (mod 2). Thus Xu (y ) == Xu (t) (mod 2) for all u . By (7.2) and (7.4)
CHAPTER XII
470
[7
8]
THE Z*-THEOREM
471
as Xu ranges over all the irreducible characters of G. Thus by (IV.6.3)
bt; + bb + b� + b� == 0 (mod 2).
L
Define
Xu i n B o
Xu (s)Xu (t)2 = 0 Xu (1)
for all 2-singular elements. Let VI == (vu ). Then VI is a linear combination of the columns (Xu (s)) with s E T - {1}. Thus
u (t f . L VuX Xu (1) = O
Xu in B
Then U I , U2, U3 are columns with integral entries. By (7.3) (7.5) and (7.6)
(X (1), Uj ) = 0, (b6', Uj ) = 2, (Ui ' Uj ) = 1 + 2 oii . Thus in particular each Ui has 3 nonzero coefficients and these are ± 1 . A coefficient 1 occurs in the first row corresponding to XO = 1 . Choose the notation so that the other two nonzero coefficients O J , 0 2 of VI occur in the
second and third row. It may also be assumed that the coefficients of bt; in the first three rows are 1, OJ, 0 as (bb, VI) = 2. If Vj for j = 2, 3 has a nonzero coefficient in the second or third row it follows from (V I , Vi ) = 1 that V I - Ui has only one nonzero coefficient contrary to (X (1), UI - vj ) = O. Hence the coefficients of V2 and V3 in the second and third row vanish. Thus (7.2) and (7.4) imply that
[ bfl - b� - b�h =
( �I ) 202
Therefore
2 o 2X2(t t XI(l) + X2 (1) = 0 .
0 XI(t) 1+ I
(7.9)
If (7.7) and (7.S) are substituted into (7.9) we see that
ol{Xl(l) - XI(t)}2 = 0 XI(1) {l + o I XI (l)} and so XI (l) = X I (t). Thus t is in the kernel H of X l . Since Xl -I- 1 it follows
that
H -I-
G.
0
8. The Z*-theorem
If G is a group let Z( G) denote the center of G and let Z*(G) denote the inverse image of Z( G /0 2,( G)) in G. If O 2, ( G) = (1) then of course Z*(G) = Z( G).
,
THEOREM S. 1 . Let t be an involution in G and let T be a Srgroup of G which where [ h means that we only look at the first three rows. Therefore Hence 1 + o I X I (t) - o2X 2 (t) = O.
(7.7)
Since (X (1), U I ) = 0 it follows that
1 + o I X I (l) + 02X2(1) = O.
(7.S)
No 2-singular element in G is the product of two involutions as t is the unique involution in T. Hence if s is a 2-singular element
contains t. The following are equivalent. (i) X - I txt has odd order for all x E G. (ii) No element in T - {t} is conjugate to t in G. (iii) t E Z*( G).
This result, which is known as the Z*-theorem, is due to Glauberman [1966a] . It is of vital importance for many results connected with the classification of finite simple groups. Glauberman [ 1 966b] has also used it to derive some results about automorphism groups of simple groups. A generalization is due to Goldschmidt [1971] . If T is a quaternion group then (S. l) (ii) is obviously satisfied. Thus (7. 1) is a direct consequence of (S. l). However (7. 1) is needed for the proof of (S l ) We will follow ' Glauberman's proof quite closely. We begin with a well 'known result about involutions.
..
472
CHAPTER XII
[8
By definition a dihedral group H of order 2 is defined for n > 1 by H = (u, x 1 u 2 = X n = 1, uxu = X - I ). Thus the noncyclic group of order 4 is the dihedral group of order 4 . LEMMA 8.2. Let u -I v be involutions in G. ' Let H = (u, v ). Then the following hold. (i) I H (uv ) I = 2 and H is a dihedral group. (ii) A Sr group of H either has order 2 or is a dihedral group. (iii) If 4 ,r I H I then u is conjugate to v in H. (iv) If 4 1 I H I then I H : H' I = 4 and u is not conjugate to v in H. (v) If 4 1 I H I then either I H I = 4 or I Z(H) I = 2 and u, v are both not in Z(H). :
PROOF. (i) This follows as u - I (uv )u = VU = (UV f l .
(ii) Let x = uv and let x have order mn where m is odd and n is a power of 2. Then (u, x m ) is a Srgroup of H. As ux m u = x -m it follows that either x m = 1 or (u, x m ) is a dihedral group. (iii) As 4 ,r I H I , (u ) and (v ) are Srgroups of H and so are conjugate in H. Since u, v is the unique element of ( u ), (v ) respectively of order 2 u is conjugate to v. (iv) (UV )2 = uvu - I V - I E H'. As 4 1 I H I it follows that I H : « uv )2 ) 1 = 4. Since « UV )2) <J H this implies that « UV )2 ) = H'. Hence I H : H' 1 = 4. Fur thermore uH' -I vH' and so v -I Y - I uy for any y E H. (v) Let x = uv. Then x has order 2m for some integer m > O. Since U - I X k U = x - k the only power of x in the center of H is the involution x l» . The result follows easily. 0 The next two results handle the simple implications in (8.1). LEMMA 8.3. Conditions (i) and (ii) of (8. 1) are equivalent. PROOF. (i) => (ii). Suppose that x - I tx E T then X - I txt E T and x - I txt has odd order. Thus x - I txt = 1 and so x - I tx = t - I = t. (ii) => (i). Let x E G and let s = x - I tx. Suppose that st = x - I txt has even order. Thus 4 1 I (s, t) I . By (8.2) (v) there exists an involution z -I s, t with z in the center of (s, t). Let D be a Srgroup of (s, t) which contains t. Then z E D and D is a dihedral group by (8.2) (ii). There exists y E (s, t) with y - I sy E D. By (8. 1) (iv) y - I sy -l t. Choose x E G with x - I Dx c T. Then x - I tx E T and x - I (y - I sy )x E T. Hence by assumption X� I tx = t = X - I (y - I sy )x and so t = Y - I sy contrary to what has been shown above. 0
8]
THE Z*-THEOREM
473
LEMMA 8.4. Condition (iii) of (8. 1) implies conditions (i) and (ii). PROOF. Clearly (8. 1) (iii) implies (8.1) (i). The result follows from (8.3). 0
In view of (8.3) and (8.4) the proof of (8. 1) will be complete once it is shown that (8. 1) (i) implies (8.1) (iii). This,will be done in a series of lemmas.
Throughout the rest of this section G is a minimal counterexample to the assertion that (8. 1 ) (i) implies (8. 1 ) (iii). Thus there exists an involution t E G with tE Z*(G) such that x - I txt has odd order for every x in G. Observe that every subgroup of G that contains t satisfies (8. 1) (i) and so does every factor group G /H where t E H. LEMMA 8.5. If H <J G then 02,(H) = (1). PROOF. The minimality of G implies that O2,( G) = (1). Hence Oz,(H) � O2,( G) = (1). 0
LEMMA 8.6. t is in the center of any 2-group which contains t. PROOF. Let D be a 2-group with t E D. If x E D then x - I txt E D and
X - I txt has odd order. Hence X - I txt = 1.
0
LEMMA 8.7. T contains an involution distinct from t. PROOF. If the result is false then T is either cyclic or a quaternion group. If
T is cyclic then G has a normal 2-complement. Thus G = T by (8.5) and the result is trivial. If T is a quaternion group then t E Z*( G) by (7. 1). 0 LEMMA 8.8. Let X be an irreducible character in the principal 2-block of G. Let s be an involution in G which is not conjugate to t. Then there exists a conjugate So of s with X (ts) = x (tso) and So E Co (t). PROOF. By (8.2) (iii) st has even order. Let z be the involution which is a power of st. Let To be a Srgroup of (s, t) with t E To. By (8.2) (ii) To is a dihedral group. By (8.6) z, t are both in the center of To. Thus I To 1 = 4 by (8.2) (v). Let So be a conjugate of s in (s, t) which lies in To . Then So -I z, t, So E Co (t) and z = tso.
474
CHAPTER XII
[8
Since st = yz = zy where y has odd order it follows from (V.6.3) applied to Co (x ) that
X (ts) = X (zy ) = X (z ) = X (tso). 0 LEMMA 8.9. Let s be an involution in G which is not conjugate to t. Let s ', t'
be conjugates of s, t respectively in G. Let X be an irreducible character in the principal 2-block of G. Then x (ts) = x (t's '). PROOF. If t' = x - I tx then t's ' = x - 1 (txs 'X - 1 )x. Thus it may be assumed that t' = t as X is a class function on G. By (8. 8) there exist conjugates so, sb of s in Co (t) such that
x(ts) = x (tSo) , X (ts ') = x (tsb). Let x - 'sbx = so. Then t, x - 1 tx are both in
(8. 10)
Co (so). By assumption x - 1 txt has odd order and so by (8.2) (iii) there exists y E (t, X - I tx ) � Co (so) with y -l ty = x - 1 tx. Thus yx - I E Co (t). Therefore
(yx - I r l(tso) (yx - I ) = t(yx - 1 r 1 so(yx - I ) = txsoX - I = tsf) . LEMMA 8. 1 1 . Let s be an involution in T with s "l t. Let X be an irreducible
character in the principal 2-block of G with X "I I. If X (s ) "I 0 then X (t) = - X (I). PROOF. By (8.3) s is not conjugate to t in G. Let Co, CI , . . . be all the conjugate classes of G where Co = {I}, t E C1, S E C2 • Let C = �X ECi X in the complex group algebra of G. Let C\ (;2 = �ai�j . Thus (8. 12) Let Xj E G for all j. If w is the central character corresponding to X then w ( G ) = / G / X (xj )IX(I). Thus
X (I)
I Cj I x(xj ) X (I) = Lj a . X (I) . ]
For x E G define a (x ) =·X (x )IX (I). By (8. 9) X (Xj ) = X (ts) whenever aj "l O. Hence
/ C 1 / a (t) / C2 / a (s ) = Lj aj / G / a (ts ) = a (ts ) Lj aj / G / ·
THE Z*-THEOREM
475
Thus (8. 12) implies that
a (t) a (s) = a (ts). Since t E Z(T) by (8.6), ts is an involution in T with ts "l t. The argument above applied to ts yields that a (t)a (ts ) = a (s ). Hence a (t)2 a (s ) = a (t)a (ts ) = a (s). Now suppose that X (s ) "l O and x "l 1 . Then a (s ) "l O and s o a (t? = 1. Hence a (t) = ± 1 . Thus X (t ) = ± X (I). Suppose that x (t) = X (I). Then t is i n the kernel H o f X. A s X "I I, H"I G. Thus the minimality o f G implies that t E Z*(H). By (8.5) Z*(H) = Z(H) and so t E Z(H). Let x E G. Then x - 1 tx is an involution in H. Therefore (x - 1 txt? = 1 . Thus x -1 txt = 1 as X - I txt has odd order. Consequently t E Z(G). 0 PROOF OF ( 8 . 1 ). By (8.7) T contains an involution s "l t. By (8.3) s is not conjugate to t in G. By (IV.6.3)
0 = L Xu (s)Xu (t) = L Xu (s )Xu (I) u u
Hence x (tso) = x (tsb) and the result follows from ( 8 . 10). 0
I C1 I x (t) I C2 I x( s )
8]
where Xu ranges over all the irreducible characters in the principal 2-block . of G. Thus (8. 1 1) implies that if Xo = 1 then
2 = Xo(s ) (Xo(t) + 1) = L Xu (s ) (Xu (t) + Xu (1» = L Xu (s )Xu (t) + L Xu (s )Xu (1) = O. u u This contradiction completes the proof. 0
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SUBJECT INDEX
S UBJECT INDEX
completely primary, 33 component, 2 composition series, 3 equivalent composition series, 3 constituent, 3 irreducible constituent, 3 cover, 1 69, 249 cross section, 79
Algebra, 3 finitely generated, 3 free, 3 Frobenius, 49 R-algebra, 3 serial, 58 sym metric, 49 uniserial, 58 Alperin-McKay conjectures, 171 annihilator, 17 Artin-Wedderburn Theorem , 26, 27 ascending chain condition, 5
decomposition matrix, 67 decomposition numbers, 67 for a basic set, 148 defect, 1 24, 1 26, 1 27 defect group, 124, 126, 1 27 deficiency class, 246 descending chain condition, 5 dual basis, 47
BO is defined, 1 36 basic set, 1 48 basis, 1 block, 23 block pair, 207 extend, 207 properly extend, 207 weakly extend, 207 Brauer character, 142 Brauer corresponden ce, 136 Brauer graph, 300 Brauer homomorphism, 1 29 Brauer mapping, 129 Brauer tree, 301, 305
elementary subgroup, 1 4 1 exceptional character, 277, 279 extension, 69 finite extension, 69 unramified extension, 69 First Main Theorem on blocks, 1 37 Fitting's Lemma, 34 Fong-Swan Theorem, 419 Frobenius reciprocity, 99
canonical character, 205 Cartan invariants, 55 for a basic set, 148 Cartan matrix, 55 central character, 54 character, 141 Clifford's Theorem, 101 coherent, 224
generalized character, 141 germ, 209 Green algebra, 92 Green correspondence, 1 1 3 Grothendieck algebra, 92 group algebra, 4 500
Hall-Higman Theorem B, 309 H-conjugate, 1 23 height, 1 5 1 Hensel's Lemma, 40 higher decomposition number, 1 72 with respect to a basic set, 45 1 Higman's Theorem , 89 idempotent, 1 centrally primitive, 1 primitive, 1 inertia group of a block, 195 of a characte r, 1 95 of a module, 86 inertial index, 235, 272 intertwining number, 53 invariant, 82 inverse endomorphism ring, 4 Jacobson radical, 1 8 Jordan-Holder Theorem, 3 kernel, 86 kernel of a block, 154 Krull-Schmidt Theorem , 37 linked idempotents, 45 local ring, 33 lower defect group, 241 associated to a section, 243 Mackey decomposition, 85 Mackey tensor product theorem, 85 major subsection, 230 Maschke's Theorem , 9 1 McKay conjecture, see Alperin-McKay metric complete, 3 1 complete o n modules, 31 defined on a module, 29 equivalent metrics, 29 module absolutely indecomposable, 72 absolutely irreducible, 7 1 A-faithful, 1 7 algebraic, 93 Artinian, 6 B-projective, 1 1 com patible endo-permutation, 409 complete, 29 completely reducible, 15
decomposable, 2 dual, 46 endo-permutation, 407 endo-trivial, 407 faithful, 86 finitely generated, 1 free, 2 .s)-projective, 93 indecomposable, 2 induced, 80 injective, 50 irreducible, 2 irreducibly generated, 95 left, 1 left Artinian, 6 left Noetherian, 6 left regular, 2 Noetherian, 6 of quadratic type, 188 of symplectic type, 1 88 periodic, 96 permutation, 405 principal indecomposable, 42 principal series, 427 projective, 8 reducible, 2 regular, 2 relatively injective, 50 relatively projective, 1 1 serial, 58 torsion free, 64, 65 transitive permutation, 405 two-sided, 5 uniserial, 58 multiplicity of a lower defect group, 243 Nakayama's Lemma, 31 Nakayama relations, 99 nil ideal, 19 nilpotent ideal, 19 nonsingular element in Hom R (A, R ), 49 normal series, 3 factors of, 3 proper refinement of, 3 refinement of, 3 without repetition, 3 orthogonal idempotents, 1 p-conjugate characters, 1 77 P-defective, 1 24 p-radical group, 266
501
502
SUBJECT INDEX
p-rational character, 1 78 p-section, 1 72 1T-height, 227 1T-section, 2 1 6 primitive ideal, 1 7 principal block, 1 54 Brauer character, 1 54 character, 154 projective resolution, 95 pure submodule, 64, 65 quasi-regular, 18 radical of a module, 1 6 o f a ring, 1 8 ramification index, 69 of a module, 101 real stem, 307 regular block, 1 99 representation, 74 equivalent representations, 74 representation algebra, 92 representation group of a character, 4 1 4 Reynold ' s ideal, 258 Schanuel's Lemma, 9 Schreier's Theorem, 3 Schur index, 1 85
Schur's Lem ma, 23 Second Main Theorem on blocks, 172 section, 1 72 semi-simple, 1 9 socle, 1 6 source, 1 13 splitting field of an algebra, 53 of a module, 53 subsection, 230 with respect to a basic set, 451 symmetric element in Hom R (A, R), 49 Third Main Theorem on blocks, 207 trace, 87 relative trace, 87 trace function, 74 type of block, 453 for an element, 453 for a subsection, 453 same, 453 Type L2(p), 347 underlying module, 74 unique decomposition property, 37 vertex, 1 12 weakly regular block, 1 99 width, 252
