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• W2091238957 abstract "We give a result on Cartan invariants of the group algebra kG of a finite group G over an algebraically closed field k, which implies that if the Loewy length (socle length) of the projective indecomposable kG-module corresponding to the trivial kG-module is four, then k has characteristic 2. The proof is independent of the classification of finite simple groups. 0. INTRODUCTION AND NOTATION Let kG be the group algebra of a finite group G over a field k of characteristic p > 0. By a kG-module we always mean a right kG-module. In this paper we discuss Cartan invariants of kG, especially those of the projective cover P = P(kG) of the trivial kG-module kG. Let j be the Loewy length of P, that is, j is the least positive integer t such that pJt = 0 where J is the Jacobson radical of kG. It is well-known that the structure of G is completely determined provided j 0. Assume that S is a simple kG-module which is self-dual, that is, the dual module S* = Homk(S, k) of S is isomorphic to S itself as kG-modules. If p is odd, then there is a simple kG-module T such that T is self-dual and the Cartan invariant c(S, T) with respect to S and T is odd. Corollary. Let k be an arbitrary field and G a finite group. If the Loewy length of the projective indecomposable kG-module corresponding to the trivial kG-module is four, then k has characterstic 2. Received by the editors August 17, 1994 and, in revised form, February 7, 1995. 1991 Mathematics Subject Classification. Primary 20C05, 20C20. Supported in part by the Alexander von Humboldt Foundation, the Mathematical Prizes Fund, the University of Oxford and the Sasakawa Foundation. ( 1996 American Mathematical Society 2319 This content downloaded from 157.55.39.215 on Tue, 30 Aug 2016 04:58:29 UTC All use subject to http://about.jstor.org/terms 2320 SHIGEO KOSHITANI Note that the theorem and the corollary do not depend on the classification of finite simple groups. Remark. It should be remarked that if G is the symmetric group on 4 letters and if k has characteristic 2, then the Loewy length of the projective indecomposable kGmodule corresponding to the trivial kG-module is four. The situation for p-solvable groups is treated completely in our previous paper [2]. Throughout this paper we use the following notation and terminology. By a kG-module we mean a finitely generated right kG-module. We write kG for the trivial kG-module, and P(kG) for its projective cover. We denote by J(kG) the Jacobson radical of kG. Let M be a kG-module. Then j(M) denotes the Loewy length of M, namely, j(M) is the least positive integer j such that M.J(kG)i = 0, and Soc(M) denotes the socle of M (see [3, I ?8]). We write M* for the dual module of M, that is, M* = Homk(M, k) and this is also a right kG-module (see [3, I ?6]). Then, M is called self-dual if M M* as kG-modules. For kG-modules M and N, we denote diMk [HOmkG (M, N)] by [M, N]G. For simple kG-modules S and T, c(S, T) denotes the Cartan invariant with respect to S and T (usually, a notation CS,T is used instead). Other notation and terminology follow the books of Feit [1] and Landrock [3]. 1. CALCULATION OF DETERMINANTS In this section we give several lemmas on elementary computation of determinants for matrices over a field of characteristic 2. Throughout this section all entries of matrices are elements in a field of characteristic 2. This assumption is essential here. Lemma 1.1. For non-negative integers m and n, let A be a square matrix of size m + 2n + 2 of the form 1 ... m 1 1* ... n n* u v 1 / aii a ... an an 0 0 m am, am, *** amn amn 0 0 A ?0 bo1 ... bom Col do, Con don uo VO 0* bol ... bOm do, Co1 don COn Uo Uo n bnl ... bnm Cn 1 dn1 Cnn dnn un Vn n bnl ... bnm dn1 Cni ... dnn Cnn Un Vn where S is an (m x m) -matrix. Then, det A = 0 . Proof. We prove the lemma by induction on n. If n = 0, then det A = 0 since two rows indexed by 0 and 0* are the same. Assume n ? 1. Expand A with respect to the last column indexed by v. Then, by symmetry, it is enough to show that the This content downloaded from 157.55.39.215 on Tue, 30 Aug 2016 04:58:29 UTC All use subject to http://about.jstor.org/terms CARTAN INVARIANTS 2321 determinant of the following square matrix B of size m + 2n + 1 is zero: 1 ... .. m 1 1* ... n n* u 1 z all all aln aln 0 : L} ~ : : : : : : m amni aml amn amn 0 0 + 0* 0 ... 0 eol eol ... eon eOn 0 1 bil ... bim Cii di, Cln di, Ul 1* bil ... bim di, Cii ... dln Cln Ul n bnl ... bnm Cnl dnl Cnn dnn Un n* bnl ... bnm dnl Cnl ... dnn Cnn Unr where eoi = coi + doi for i = 1,..., n. Now, expand B with respect to the final column indexed by u. So, again by symmetry, it suffices to prove det C = 0 for the following square matrix C of size m + 2n: 1 ... m 1 1* ... n n* 1 z all all aln aln S m aml ami amn amn 0 + 0* 0 ... 0 eol eol ... eOn eOn C 1 + 1* 0 ... 0 ell ell ... eln eln 2 b2l ... b2m C21 d21 C2n d2n 2* b2l ... b2m d2l C21 ... d2n C2n n bnl ... bnm Cnl dnl Cnn dnn n* bnl ... bnm dnl Cnl .. dnn Cnn where eli = cli + dii for i = 1, ..., n. Then, by carrying two rows indexed by 0 + 0* and 1 + 1* to the end, and then by taking its transpose, we have det C = 0 by induction. O Lemma 1.2. For non-negative integers m and n, let A be a square matrix of size m + 2n + 1 of the form u 1 ... m 1 1* ... n n* V 0 0 ... 0 Vi Vi ... Vn Vn,V 1 0 all all aln aln : : S : : : :. m 0 ami ami amn amn A = 1 ul bil ... bim Cii di1 Cln dln 1* Ul bil ... bim di, Cii *.. dln Cln n Un bnl ... bnm Cnl dnl Cnn dnn n* Un bnl ... bnm dnl Cnl ... dnn Cnn where S is an (mxm)-matrix. Then, det A = 0. This content downloaded from 157.55.39.215 on Tue, 30 Aug 2016 04:58:29 UTC All use subject to http://about.jstor.org/terms 2322 SHIGEO KOSHITANI Proof. First of all, expand A with respect to the first row indexed by v, and sum each pair of determinants of size m + 2n that have the same coefficient vi for = 1,...,n. Namely, we can write detA = Ein4videtBi, where each Bi is a square matrix of size m + 2n and each Bi has the same form as in Lemma 1.1 by a suitable exchanging of columns. Therefore, det A = 0 by Lemma 1.1. O 2. LEMMAS In this section we state several lemmas which will be used in the proofs of our results. Throughout this section we assume that k is an algebraically closed field of characteristic p > 0, and we fix a finite group G such that p divides the order of G. Lemma 2.1 (Webb [9, Theorem E). Let P = P(kG) and assume j(P) ?3 3. If p is odd, then P.J(kG)/Soc(P) is an indecomposable kG-module. Lemma 2.2 ([6, Lemma 1.2]). If M is an indecomposable kG-module with j(M) = 2, then M.J(kG) = Soc(M). Lemma 2.3 ([3, II Corollary 6.9]). For kG-modules M and N, [M,N]G = [N*,M*]G. Lemma 2.4 ([3, I Lemma 8.4 (i)]). For a kG-module M, [M/(M.J(kG))]* Soc(M*) as kG-modules. Lemma 2.5 (Landrock). For simple kG-modules S and T, c(S, T) = c(T, S) = c(S*, T*) = c(T*, S*). Proof. We get the assertion from [1, I Lemma 14.9 and Theorem 16.7] and Landrock's result [4, Theorem A] (cf. [3, I Theorem 9.9]). O 3. PROOFS In this section we give proofs of the theorem and the corollary in the introduction. Proof of Theorem. Let B be a block ideal of kG containing S. Since S is self-dual, Y* is a simple kG-module in B again if Y is a simple kG-module in B (see [3, I Proposition 10.8]). Thus, let So = S,Sj,...,Sm,Ti,Tj*, ..., Tn Tn* all be nonisomorphic simple kG-modules in B; and Si Si* for all i and Tj Tj * for all j. We denote by C the Cartan matrix for B, and let C be its image induced by the canonical epimorphism Z -> 2/2Z. Now, suppose that c(So, Si) is even for all i = 0, ..., m. Then Lemma 2.5 implies that C has the same form as in Lemma 1.2, so that det C = 0 from Lemma 1.2. This means det C is even, which contradicts Brauer's result [1, IV Theorem 3.9]). OI Proof of Corollary. First of all, we may assume that k is algebraically closed (see [5, Proposition 12.11]). Let J = J(kG), P = P(kG) and M = PJ/Soc(P). Assume p is odd. By Lemmas 2.1 and 2.2, M is an indecomposable kG-module with MJ = Soc(M). Then, the Theorem implies that there is a simple kG-module T such that T is self-dual and c(kG, T) is odd. On the other hand, since T and M are both self-dual, and since MJ = Soc(M), it follows from Lemmas 2.3 and 2.4 that [M/MJ,T]G = [T,(M/MJ)*]G = [T, Soc(M)]G = [T, MJ]G, This content downloaded from 157.55.39.215 on Tue, 30 Aug 2016 04:58:29 UTC All use subject to http://about.jstor.org/terms CARTAN INVARIANTS 2323 which says that the multiplicities of T in M/MJ and MJ as direct summands are the same, so that c(kG, T) is even, a contradiction. O ACKNOWLEDGEMENTS The author is so grateful to the referee for her or his nice and kind advice, which has improved the main result in the first version of the paper. A part of this work was done while the author was staying at the Institute for Experimental Mathematics, Essen University (July October 1992), and at the Mathematical Institute, University of Oxford (September 1992). He would like to thank the Humboldt Foundation, the Mathematical Prizes Fund, the University of Oxford and the Sasakawa Foundation for their financial support, and also he would like to express his great thanks to Professor G.O. Michler and the people around him, and Professor K. Erdmann for their hospitality." @default.
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