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Chapter XI
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 a n integer valued function J ( n ) defined on the set of positive integers with the following property. l f 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 I < J ( 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 I G 1 . However the result is false for fields whose characteristic divides I G 1 . For example let F be an algebraically closed field of characteristic p > O and let G, = SL2(p"').Each G,, has a faithful F-representation of degree 2 but a normal abelian subgroup of G, has order at most 2 while I G,, I can be arbitrarily large. This section contains a proof of t h e following analogue of (1.1) which was first conjectured by 0. H. Kegel.
THEOREM 1.2 (Brauer and Feit [1966]). Let p be a prime. There exists a n integer valued function f ( m , n ) = f , ( 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"' be the order of a S,-group of G. Then G has a normal abelian subgroup A with I G : A 1 < f ( m , n ) . Isaacs and Passman [ 1964 have used Jordan's theorem to show that if 444
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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 J ( n ) ) .In a similar manner J. F. Humphreys [I9721 has used (1.2) to prove an analogous result in characteristic p . For a related result see J. F. Humphreys [1976]. The various proofs of Jordan's theorem (1.1) yield different values for J ( 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 , ( G )denote the F [ G ]module which affords the principal Brauer character. Let B , ( G )denote the principal p-block. If G is a finite group and P is a &-group of G, then the center Z ( P ) of P is a S,,-group of Cc;( P )and Burnside's transfer theorem implies that C, ( P ) is the direct product a p-group and a p'-group. Thus P C , ( P ) = P x H for some p'-group H. Suppose that V is an indecomposable F [ G ] module with dimF Vf 0 (modp). Then P is a vertex of V. Let V denote the F[N,(P)] module which corresponds to V in the Green correspondence. The crux of the proof of (1.2) is contained in the next result. LEMMA1.3. 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 with L in Bl(G) and L# L I ( G ) . (iii) Let P be a &,-group of G. Let N = Nc;( P ) . There exists an irreducible constituent L of V* @ V with d i mf L f 0 (mod p ) and L # L l ( G )such that i f H n isthekernelofLHthenI H : H o ( < J ( n ) ,where/(n) isdefinedby (1.1).
PROOF.Assume that G satisfies neither (i) nor (ii). Let L I = L , ( G ) , L,. . . . , L , 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 , ( G )in W is equal to dimF Inv,; W. It clearly may be assumed that W is indecomposable. If W is not in B , ( G ) the result is trivial. Suppose that W
is in B , ( 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,,, 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 = L, @ L , with 1 i, j f s. Thus Inv, W = HomF,(;,(Lt.L,). For i = j , Schur’s Lemma implies that dimF Inv, W = 1. Thus by (111.2.2) and the previous paragraph dim,L, f 0 (mod p ) for 1 < i s. For i# j , Schur’s Lemma implies that Inv, W = (0). Since (ii) is excluded it follows that no irreducible constituent of W is in B , ( G ) . Hence by (IV.4.14), L I , .. . , 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 L, for some i. Let b be the length of a composition series of Y.Thus L l ( G ) occurs with multiplicity b in LT @ Y . Since every irreducible constituent of L T @ Y is a constituent of L T @ L , and so of V * @ V @ V * @ V, it follows that b = dim, HomF,,,(L,, Y ) .Let Y,,be the socle of Y and let b,, be the length of a composition series of Yo.The previous argument applied to Yo shows that b,, = dim, HornFIC;,(L,, Yo).However HomFIGl(L,, Y)= HornFIc;,(L,,Y o ) Thus . b = b,, and so Y = Yo.As Y is indecomposable this implies that Y = L,. Consequently V* @ V is completely reducible. Let
By Schur’s Lemma a I = 1. Hence #y (111.2.2) dimF Vf 0 (mod p ) . Thus P is a vertex of V. Hence (111.5.7), (V.6.2) and (1.4) imply that
where each indecomposable direct summand of S has a vertex properly contained in P. By (111.7.7) i,= L I ( N )and L, is not in B1(N)for i 3 2. By (111.3.7) V p x H -- @:-,(U,@ X t ) , where X I , .. . , X , are distinct irreducible F [ H ] modules and U , , . . ., U, are F [ P ] modules which are conjugate under the action of N. Thus
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where S‘ is an F [ P x H ] module which does not contain L , ( P x H ) as a constituent. By (1.5) L , ( N ) [V * @ V . Thus L , ( P x H ) [( V * @ V ) P x Hand so L I ( P ) U : @ U, for some j . Since the modules U, are conjugate under the action of N, this implies that L , ( P ) I U ; @ U, for j = 1 , . . . , e. Thus ~ L , ( PH X ) I ( V * @ V),,x,,. Suppose that e > 1. Since a , = 1, (1.5) implies that L I ( P X H)I T I . x tfor , some indecomposable direct summand T of S. Thus L l( P ) I T p . Since a vertex of T is properly contained in P this contradicts (111.4.6). Thus e = 1. Consequently
1
VPXH =U@ X
(1.6)
where X is a n irreducible F [ H ] module a n d U is an F [ P ] module such that L , ( P ) U * @ U. Let d = dim, 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 L, for all i. Thus by (IV.4.14) L, is in BI(N) for all i and so L, is in B , ( G )by (111.7.7) and (V.6.2). T h u s s = 1. Hence by (1.4) n = 1 contrary t o assumption. Thus d > 1. Let M be an irreducible constituent of V and let D b e t h e kernel of M . By (111.2.13) P C D . Thus N / D is a p’-group. Hence by the remark following (1.1) there exists a subgroup A,,with D C A,,a N such that A o / D is abelian and I N : A , , /< J ( n ) . Let A l = A,,n H and D , = D fl H. Then A , a N,I H : A I 1 < J ( n ) and A J D , is abelian. By (1.6) M,, = U X for some positive integer u. It follows that D , is the kernel of X . If A is a n abelian group and X is an F [ A J module then X * @ X contains L , ( A )with multiplicity at least dim,X. Apply this remark with A = A l / D I .Thus ( X * @ X ) , , , contains L l ( A I )with multiplicity at least d 3 2. As X is irreducible, X* @ X contains L , ( N )with multiplicity 1. Thus there exists an irreducible constituent Z# L , ( H ) of X * @ X such that L l ( A , ) l Z 2 ,By , . Clifford’s theorem (111.2.12),A , is in the kernel of Z. Since L I ( P ) @ Zl ( V * @ V ) , , , by (1.6) it follows from (1.5) that L , ( P ) @ Z T,,,,, for some direct summand T of @ a , i , @ S. By (111.4.6) P is a vertex of T. Thus by (1.5) T = L, is irreducible. Since Z# L , ( H ) , if1. Let Hi,be the kernel of (L,)t,.Since A , belongs t o the kernel of Z and A ,
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LEMMA1.7. Let V be a n irreducible F [ G ] module with dim, V = n > 1. Suppose that G contains no normal subgroup of index p. Then there exists a n irreducible constituent L of V* @ V @ V* @ V such that
CHAFER XI
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where Go is the kernel of L . PROOF.Let L be an irreducible F [ G ]module with dimFL = d. Let G,, be the kernel of L and let (p be the trace function afforded by L. Suppose that (p has exactly e algebraic conjugates in F. Thus the subfield of F generated by all + ( x ) as x ranges over G has p e elements. By (1.19.2) G/G,, is isomorphic to a subgroup of GLd ( p ' ) . Thus 1 G : G,,ld I GLd (p')I S 1 G L ( p )1 . 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 # Go as L # L , ( G ) . By (IV.4.9) and (IV.4.18), (p has at most IPI' algebraic conjugates in F. Thus the result follows from the previous paragraph and the fact that d < n'. Suppose that (1.3)(iii) holds. Choose L accordingly. As above G # Go. Let B , ,. . . , B , be all the p-blocks of G which contain algebraic conjugates of (p. Since dimFL = d < n' the remarks above show that it suffices to prove that (p has at most 1 P I'J(n) algebraic conjugates. Thus by (IV.4.18) it suffices to show that a s J ( n ) . Let be the trace function afforded by L. By (111.7.7) there are exactly a p-blocks of N, ( P )which contain an algebraic conjugate of By (1.3)(iii) the group N, ( P ) / H , has , less than J ( n ) p-blocks. Thus by (V.4.3) a < J ( n ) as required. 0
6
6.
Let G be a finite group and let V be an F [ G ] module. Let T ( G ,V) = ( m , n, a ) where a S,-group of G has order p"', dim, V = n and a is the multiplicity with which L , ( G ) occurs as a constituent of V * @ V @ V * @ v. Define the partial ordering < as follows: ( m , , n I , a < ( m , n, a ) if one of the following is satisfied. (i) m , < m , n l s n . (ii) m , s m, n l < n. (iii) m , = n, n , = n, a , > a . If H is a subgroup of G, then clearly either T ( H ,V H = ) T ( G ,V) or T ( H ,V,,) < T ( G ,V). Observe that if T ( G ,V) = ( m , n, a ) then a S n'. Thus there are only finitely many triples 7(GI,V,) with 7(GI,V,) < T ( G ,v). LEMMA1.8. Let P be a S,-group of G and let 1 P 1 = p"'. Let X be a faithful F [ G ] module and let dim, X = n. Let g ( m , n ) = IGLlp12,(,,,~(p)l, where
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AN ANAI.OCiUE OF JOKDAN'S
THEORthl
449
J ( n ) is defined by (1.1). Assume that G is not abelian. Then there exists Goa G with I G : G o (< g ( m , n ) such that T(G,,,Xc;,)< T ( G XI. , 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 I be a completely reducible F [ G ] module with the same composition factors as X . Then P is the kernel of X I and G / P is abelian. Thus G is solvable and so G contains an abelian subgroup A with 1 G : A 1 = 1 P I . Let GI,= 'Ax.Then G o a G a n d J G : G u ( s l P I ! S g ( m , n ) . S i n c e G is not abelian I P I # 1. Thus 7 ( G o ,X , , )< T ( G X , ). Suppose next that some irreducible constituent V of .Yhas F-dimension , > = ( m ,n, a ) and at least 2. Let L and GI,be defined as in (1.7). If T ( G X 7 ( G 0 ,X&) = ( m , ,n , , a , ) , then clearly m , s m and n , = n. Since G,,is the kernel of L and L 7 t . L I ( G ) ,a l > a . Thus T ( G , , X , , ) < T ( G , X ) .
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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 1 < h ( T ( G ,X ) ) for some normal abelian subgroup A of G. PROOF.Let h ( m , n, a ) = 1 if (m,n, a ) # T ( 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 l ,a ) is defined for all ( m i , n , , a , )< T ( G X , ) . Let g(m, n ) = 1 G L I , ~ I ~ , , ~.,Define (,,,(~)~
and let
h ( T ( G ,X ) ) = g ( m ,n)h,(T(G, X))Rc'rrn). 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 , Let A = abelian subgroup A(, with 1 Go: A,] G h ( T ( G OXC,")). 'A,x. Then 1 G : A I < g(m, n ) h ( T ( G o X(;c,))8'"'"'. . This implies that I G : A 1 < h ( T ( G ,X ) ) . 0
nXGc;x
(1.2) is now a direct consequence of ( I .Y) if f ( m , n ) is defined by
One of the oldest results in group theory is the following theorem.
THEOREM 1.1 (Jordan). There exists a n integer valued function J ( n ) defined on the set of positive integers with the following property. l f 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 I < J ( 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 I G 1 . However the result is false for fields whose characteristic divides I G 1 . For example let F be an algebraically closed field of characteristic p > O and let G, = SL2(p"').Each G,, has a faithful F-representation of degree 2 but a normal abelian subgroup of G, has order at most 2 while I G,, I can be arbitrarily large. This section contains a proof of t h e following analogue of (1.1) which was first conjectured by 0. H. Kegel.
THEOREM 1.2 (Brauer and Feit [1966]). Let p be a prime. There exists a n integer valued function f ( m , n ) = f , ( 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"' be the order of a S,-group of G. Then G has a normal abelian subgroup A with I G : A 1 < f ( m , n ) . Isaacs and Passman [ 1964 have used Jordan's theorem to show that if 444
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AN ANALOWE OF JORDAN'S T H ~ O R E M
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 J ( n ) ) .In a similar manner J. F. Humphreys [I9721 has used (1.2) to prove an analogous result in characteristic p . For a related result see J. F. Humphreys [1976]. The various proofs of Jordan's theorem (1.1) yield different values for J ( 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 , ( G )denote the F [ G ]module which affords the principal Brauer character. Let B , ( G )denote the principal p-block. If G is a finite group and P is a &-group of G, then the center Z ( P ) of P is a S,,-group of Cc;( P )and Burnside's transfer theorem implies that C, ( P ) is the direct product a p-group and a p'-group. Thus P C , ( P ) = P x H for some p'-group H. Suppose that V is an indecomposable F [ G ] module with dimF Vf 0 (modp). Then P is a vertex of V. Let V denote the F[N,(P)] module which corresponds to V in the Green correspondence. The crux of the proof of (1.2) is contained in the next result. LEMMA1.3. 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 with L in Bl(G) and L# L I ( G ) . (iii) Let P be a &,-group of G. Let N = Nc;( P ) . There exists an irreducible constituent L of V* @ V with d i mf L f 0 (mod p ) and L # L l ( G )such that i f H n isthekernelofLHthenI H : H o ( < J ( n ) ,where/(n) isdefinedby (1.1).
PROOF.Assume that G satisfies neither (i) nor (ii). Let L I = L , ( G ) , L,. . . . , L , 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 , ( G )in W is equal to dimF Inv,; W. It clearly may be assumed that W is indecomposable. If W is not in B , ( G ) the result is trivial. Suppose that W
is in B , ( 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,,, 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 = L, @ L , with 1 i, j f s. Thus Inv, W = HomF,(;,(Lt.L,). For i = j , Schur’s Lemma implies that dimF Inv, W = 1. Thus by (111.2.2) and the previous paragraph dim,L, f 0 (mod p ) for 1 < i s. For i# j , Schur’s Lemma implies that Inv, W = (0). Since (ii) is excluded it follows that no irreducible constituent of W is in B , ( G ) . Hence by (IV.4.14), L I , .. . , 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 L, for some i. Let b be the length of a composition series of Y.Thus L l ( G ) occurs with multiplicity b in LT @ Y . Since every irreducible constituent of L T @ Y is a constituent of L T @ L , and so of V * @ V @ V * @ V, it follows that b = dim, HomF,,,(L,, Y ) .Let Y,,be the socle of Y and let b,, be the length of a composition series of Yo.The previous argument applied to Yo shows that b,, = dim, HornFIC;,(L,, Yo).However HomFIGl(L,, Y)= HornFIc;,(L,,Y o ) Thus . b = b,, and so Y = Yo.As Y is indecomposable this implies that Y = L,. Consequently V* @ V is completely reducible. Let
By Schur’s Lemma a I = 1. Hence #y (111.2.2) dimF Vf 0 (mod p ) . Thus P is a vertex of V. Hence (111.5.7), (V.6.2) and (1.4) imply that
where each indecomposable direct summand of S has a vertex properly contained in P. By (111.7.7) i,= L I ( N )and L, is not in B1(N)for i 3 2. By (111.3.7) V p x H -- @:-,(U,@ X t ) , where X I , .. . , X , are distinct irreducible F [ H ] modules and U , , . . ., U, are F [ P ] modules which are conjugate under the action of N. Thus
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447
where S‘ is an F [ P x H ] module which does not contain L , ( P x H ) as a constituent. By (1.5) L , ( N ) [V * @ V . Thus L , ( P x H ) [( V * @ V ) P x Hand so L I ( P ) U : @ U, for some j . Since the modules U, are conjugate under the action of N, this implies that L , ( P ) I U ; @ U, for j = 1 , . . . , e. Thus ~ L , ( PH X ) I ( V * @ V),,x,,. Suppose that e > 1. Since a , = 1, (1.5) implies that L I ( P X H)I T I . x tfor , some indecomposable direct summand T of S. Thus L l( P ) I T p . Since a vertex of T is properly contained in P this contradicts (111.4.6). Thus e = 1. Consequently
1
VPXH =U@ X
(1.6)
where X is a n irreducible F [ H ] module a n d U is an F [ P ] module such that L , ( P ) U * @ U. Let d = dim, 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 L, for all i. Thus by (IV.4.14) L, is in BI(N) for all i and so L, is in B , ( G )by (111.7.7) and (V.6.2). T h u s s = 1. Hence by (1.4) n = 1 contrary t o assumption. Thus d > 1. Let M be an irreducible constituent of V and let D b e t h e kernel of M . By (111.2.13) P C D . Thus N / D is a p’-group. Hence by the remark following (1.1) there exists a subgroup A,,with D C A,,a N such that A o / D is abelian and I N : A , , /< J ( n ) . Let A l = A,,n H and D , = D fl H. Then A , a N,I H : A I 1 < J ( n ) and A J D , is abelian. By (1.6) M,, = U X for some positive integer u. It follows that D , is the kernel of X . If A is a n abelian group and X is an F [ A J module then X * @ X contains L , ( A )with multiplicity at least dim,X. Apply this remark with A = A l / D I .Thus ( X * @ X ) , , , contains L l ( A I )with multiplicity at least d 3 2. As X is irreducible, X* @ X contains L , ( N )with multiplicity 1. Thus there exists an irreducible constituent Z# L , ( H ) of X * @ X such that L l ( A , ) l Z 2 ,By , . Clifford’s theorem (111.2.12),A , is in the kernel of Z. Since L I ( P ) @ Zl ( V * @ V ) , , , by (1.6) it follows from (1.5) that L , ( P ) @ Z T,,,,, for some direct summand T of @ a , i , @ S. By (111.4.6) P is a vertex of T. Thus by (1.5) T = L, is irreducible. Since Z# L , ( H ) , if1. Let Hi,be the kernel of (L,)t,.Since A , belongs t o the kernel of Z and A ,
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LEMMA1.7. Let V be a n irreducible F [ G ] module with dim, V = n > 1. Suppose that G contains no normal subgroup of index p. Then there exists a n irreducible constituent L of V* @ V @ V* @ V such that
CHAFER XI
14X
where Go is the kernel of L . PROOF.Let L be an irreducible F [ G ]module with dimFL = d. Let G,, be the kernel of L and let (p be the trace function afforded by L. Suppose that (p has exactly e algebraic conjugates in F. Thus the subfield of F generated by all + ( x ) as x ranges over G has p e elements. By (1.19.2) G/G,, is isomorphic to a subgroup of GLd ( p ' ) . Thus 1 G : G,,ld I GLd (p')I S 1 G L ( p )1 . 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 # Go as L # L , ( G ) . By (IV.4.9) and (IV.4.18), (p has at most IPI' algebraic conjugates in F. Thus the result follows from the previous paragraph and the fact that d < n'. Suppose that (1.3)(iii) holds. Choose L accordingly. As above G # Go. Let B , ,. . . , B , be all the p-blocks of G which contain algebraic conjugates of (p. Since dimFL = d < n' the remarks above show that it suffices to prove that (p has at most 1 P I'J(n) algebraic conjugates. Thus by (IV.4.18) it suffices to show that a s J ( n ) . Let be the trace function afforded by L. By (111.7.7) there are exactly a p-blocks of N, ( P )which contain an algebraic conjugate of By (1.3)(iii) the group N, ( P ) / H , has , less than J ( n ) p-blocks. Thus by (V.4.3) a < J ( n ) as required. 0
6
6.
Let G be a finite group and let V be an F [ G ] module. Let T ( G ,V) = ( m , n, a ) where a S,-group of G has order p"', dim, V = n and a is the multiplicity with which L , ( G ) occurs as a constituent of V * @ V @ V * @ v. Define the partial ordering < as follows: ( m , , n I , a < ( m , n, a ) if one of the following is satisfied. (i) m , < m , n l s n . (ii) m , s m, n l < n. (iii) m , = n, n , = n, a , > a . If H is a subgroup of G, then clearly either T ( H ,V H = ) T ( G ,V) or T ( H ,V,,) < T ( G ,V). Observe that if T ( G ,V) = ( m , n, a ) then a S n'. Thus there are only finitely many triples 7(GI,V,) with 7(GI,V,) < T ( G ,v). LEMMA1.8. Let P be a S,-group of G and let 1 P 1 = p"'. Let X be a faithful F [ G ] module and let dim, X = n. Let g ( m , n ) = IGLlp12,(,,,~(p)l, where
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AN ANAI.OCiUE OF JOKDAN'S
THEORthl
449
J ( n ) is defined by (1.1). Assume that G is not abelian. Then there exists Goa G with I G : G o (< g ( m , n ) such that T(G,,,Xc;,)< T ( G XI. , 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 I be a completely reducible F [ G ] module with the same composition factors as X . Then P is the kernel of X I and G / P is abelian. Thus G is solvable and so G contains an abelian subgroup A with 1 G : A 1 = 1 P I . Let GI,= 'Ax.Then G o a G a n d J G : G u ( s l P I ! S g ( m , n ) . S i n c e G is not abelian I P I # 1. Thus 7 ( G o ,X , , )< T ( G X , ). Suppose next that some irreducible constituent V of .Yhas F-dimension , > = ( m ,n, a ) and at least 2. Let L and GI,be defined as in (1.7). If T ( G X 7 ( G 0 ,X&) = ( m , ,n , , a , ) , then clearly m , s m and n , = n. Since G,,is the kernel of L and L 7 t . L I ( G ) ,a l > a . Thus T ( G , , X , , ) < T ( G , X ) .
nXtc;x
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 1 < h ( T ( G ,X ) ) for some normal abelian subgroup A of G. PROOF.Let h ( m , n, a ) = 1 if (m,n, a ) # T ( 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 l ,a ) is defined for all ( m i , n , , a , )< T ( G X , ) . Let g(m, n ) = 1 G L I , ~ I ~ , , ~.,Define (,,,(~)~
and let
h ( T ( G ,X ) ) = g ( m ,n)h,(T(G, X))Rc'rrn). 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 , Let A = abelian subgroup A(, with 1 Go: A,] G h ( T ( G OXC,")). 'A,x. Then 1 G : A I < g(m, n ) h ( T ( G o X(;c,))8'"'"'. . This implies that I G : A 1 < h ( T ( G ,X ) ) . 0
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(1.2) is now a direct consequence of ( I .Y) if f ( m , n ) is defined by