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GROUP ACTIONS and GROUP EXTENSIONS 1. Introduction Let G Be a Finite Group. an Extension of G by an Abelian Group M, 0 →

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GROUP ACTIONS and GROUP EXTENSIONS 1. Introduction Let G Be a Finite Group. an Extension of G by an Abelian Group M, 0 → TRANSACTIONS OF THE AMERICAN MATHEMATICAL SOCIETY Volume 352, Number 6, Pages 2689{2700 S 0002-9947(00)02485-5 Article electronically published on February 24, 2000 GROUP ACTIONS AND GROUP EXTENSIONS ERGUN¨ YALC¸IN Abstract. In this paper we study finite group extensions represented by spe- cial cohomology classes. As an application, we obtain some restrictions on finite groups which can act freely on a product of spheres or on a product of real projective spaces. In particular, we prove that if (Z=p)r acts freely on (S1)k,thenr ≤ k. 1. Introduction Let G be a finite group. An extension of G by an abelian group M, 0 ! M ! Γ ! G ! 1 is called special if its restrictions to cyclic subgroups of G do not split. Extensions of this kind arise naturally in many contexts including group theory, classification of flat manifolds [13], and group actions on special manifolds [3]. In this paper we establish inequalities for such extensions, and then look at the free group actions on the k-torus and products of real projective spaces. In both of these cases short exact homotopy sequences for corresponding coverings are special extensions. As an application, we solve special cases of some well known problems. We first consider extensions with M = Zk, a free abelian group of rank k.In this case a group extension 0 ! Zk ! Γ ! G ! 1 is special if and only if Γ is torsion free. Our main result is the following : Theorem 3.2. Let G be a finite group, and let 0 ! Zk ! Γ ! G ! 1 be an extension of G.IfΓ is torsion free, then rp(G) ≤ k for all p jjGj. r Here rp(G) denotes the p-rank of G, the largest integer r such that (Z=p) ⊆ G. As an application, we have Corollary 5.2. If G=(Z=2)r acts freely on X ' (S1)k,thenr ≤ k. The existence of this inequality for free (Z=p)r actions on (Sn)k was stated as a problem by P.E. Conner after P.A. Smith's results on finite group actions on spheres. Under the assumption of trivial action on homology, this problem was solved by G. Carlsson [7], [8]. Later W. Browder [5] by using exponents, and Benson and Carlson [4] by using machinery of varieties for modules, gave alternative proofs of this result. The homologically nontrivial case involved some module theoretical difficulties overcome by Adem and Browder in [2]. They obtained the inequality Received by the editors January 30, 1998. 1991 Mathematics Subject Classification. Primary 57S25; Secondary 20J06, 20C15. Key words and phrases. Group extensions, special classes, products of spheres, cohomology of groups. Partially supported by NATO grants of the Scientific and Technical Research Council of Turkey. c 2000 American Mathematical Society 2689 License or copyright restrictions may apply to redistribution; see https://www.ams.org/journal-terms-of-use 2690 ERGUN¨ YALC¸IN r ≤ k except in the cases p =2andn =1; 3; or 7. Corollary 5.2 solves the p =2 and n = 1 cases, leaving the other two cases open. Adem and Browder also questioned whether or not the stronger inequality ≤ G r dimFp Hn(X; Fp) holds [2]. Later Adem and Benson established this inequality for permutation modules [1]. In section 5, we give a different proof of this result for n =1. We then look at the central extensions 0 ! M ! Γ ! G ! 1whereG is a finite group, M is a free abelian group, and G acts on M trivially. We show that if such an extension is special, then G must be abelian. As a topological application, we conclude that nonabelian groups cannot act freely on the k-torus (S1)k with a trivial action on homology. Next, we consider the central extensions (1) 0 ! E ! Q ! G ! 1 where both E and G are elementary abelian 2-groups of rank k and r respectively. Observe that these extensions are special if and only if r2(Q)=k.Inthiscaseevery element of order 2 in Q is central, which is known as the 2C condition for Q.For special central extensions as (1), Cusick showed that the inequality r ≤ 2k must hold [9]. We first prove a refinement of this result which gives a stronger inequality r ≤ k +rkQ0 where Q0 is the commutator subgroup of Q. Then in Section 7 we obtain 0 rk Q =dimFp (im d3) 2 3 where d3 : H (E;Z) ! H (G; Z) is a differential on the Hochschild-Serre spectral sequence H∗(G; H∗(E;Z)) ) H∗(Q; Z) for (1). Applying these to the group ex- tensions associated to the group actions on products of real projective spaces, we obtain the following: r Theorem 8.3. If GQ=(Z=2) acts freely and mod 2 homologically trivially on a k ' 2mi+1 ≤ ··· finite complex X i=1 RP where mi > 0 for all i,thenr µ(m1)+ + µ(mk); where µ(mi)=1for mi even and µ(mi)=2for mi odd. This solves a special case of a conjecture on elementary abelian 2-group actions on products of projective spaces, stated by L.W. Cusick [9]. Now, we fix some notation. Throughout this paper Z(G) will denote the center of the group G,andG0 will be the commutator subgroup of G. Also,foraprime number p, Z=p will be the integers mod p, while Fp will denote the field of p elements. The results in this paper form part of a doctoral dissertation, written at the University of Wisconsin. I am indebted to my thesis advisor Alejandro Adem for his guidance and constant support. I also would like to thank Nobuaki Yagita for pointing out a mistake in an earlier version of this paper. 2. Rational representations of (Z=2)r Let G be an elementary abelian 2-group of rank r.Ifπ : G !! Z=2 is a surjection, then the composition G !! Z=2 ! Q(−1)∗ determines an irreducible QG-module which will be denoted by M(π). In fact, all nontrivial irreducible representations License or copyright restrictions may apply to redistribution; see https://www.ams.org/journal-terms-of-use GROUP ACTIONS AND GROUP EXTENSIONS 2691 arise in this way (see [2]) and therefore, any QG-module M can be expressed uniquely as ! Mt ∼ m M = Q ⊕ M(πi) : i=1 For a ZG-module L,letQL denote the corresponding rational representation Q ⊗Z L. Lemma 2.1. Let L be a Z free ZG-module such that QL is a simple QG-module. Then, for every α 2 H2(G; L) there is an index 2 subgroup H ⊆ G such that G resH α =0. Proof. We first show that when QL is an irreducible module, H2(G; L)hasexponent ∼ ∼ ∼ 2. This is clear when QL = Q because then, L = Z and H2(G; Z) = (Z=2)r.When QL is nontrivial, by the above classification, it is isomorphic to M(π)forsome π : G !! Z=2. Let K =kerπ and let C be a cyclic subgroup of G such that ∼ G = K × C. The exact sequence C ! G ! K gives a Lyndon-Hochschild-Serre spectral sequence H∗(K; H∗(C; L)) ) H∗(G; L) where even dimensional horizontal lines vanish because LC =0.SinceL is one dimensional and C acts on L by ∼ ∼ ∼ multiplication by (−1), H1(C; L) = L=2L = Z=2, which gives that H2(G; L) = 1 H (K; F2) has exponent 2. Now, consider the following long exact sequence × ! H1(G; L=2L) −→ H2(G; L) −→2 H2(G; L) −→ H2(G; L=2L) ! × which comes from the coefficient sequence 0 ! L −→2 L −→ L=2L ! 0. Since multiplication by 2 is zero on H2(G; L), there is a class α0 2 H1(G; L=2L)which ∼ 1 ∼ maps to α.But,byL=2L = F2, and hence H (G; L=2L) = Hom(G; F2). Let ⊆ 0 G 0 H G be the kernel of the map that corresponds to α .ThenresH α =0,and G hence resH α =0. By using an induction on simple summands of QM we get Lemma 2.2. Let M be a Z free ZG-module with rk M<r. Then, for any α 2 H2(G; M) there exists a nontrivial subgroup H ⊆ G such that rk H =(r − rk M) G and resH α =0. Proof. (By induction on rk M.) When QM is irreducible, this lemma follows from Lemma 2.1. Assume QM is ∼ not irreducible, then QM = S ⊕ U for some nonzero QG-modules S and U.Letq be the projection map from QM to U.SetN = q(M)andL =kerq \ M.Then we have a short exact sequence of ZG lattices q 0 ! L −→i M −→ N ! 0 ∼ ∼ where QL = S and QN = U. Consider the corresponding long exact sequence for cohomology ∗ q∗ ! H2(G; L) −→i H2(G; M) −→ H2(G; N) ! : Let α be any element in H2(G; M). Then q∗(α) 2 H2(G; N) and by induction there − G ∗ is a subgroup K in G such that rk K = r rk N and resK q (α) is zero. From the License or copyright restrictions may apply to redistribution; see https://www.ams.org/journal-terms-of-use 2692 ERGUN¨ YALC¸IN commutative diagram q∗ H2(G; M) /H2(G; N) resG resG K K ∗ q∗ H2(K; L) i /H2(K; M) /H2(K; N) ∗ G 2 0 we get q (resK α) = 0. Hence there exists an element in H (K; L), say α ,which G ⊆ maps to resK α.
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