The maximal quotient rings of regular group rings III
Abstract
We give a new proof of the main result of [1] which does not use the classification of the finite simple groups.
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Publicacions Matem`atiques, Vol 40 (1996), 15–19. THE MAXIMAL QUOTIENT RINGS OF REGULAR GROUP RINGS III Ferran Ced´ o To the memory of Andreu Pitarch Abstract We give a new proof of the main result of [1] which does not use the classification of the finite simple groups. Introduction. In [1] the proof of the main result, [1, Theorem 2.3], is based implicitly on the classification of the finite simple groups through the use of [6, Theorem]. We give here a proof of [1, Theorem 2.3] which does not use [6, Theorem]. The proof. Let Gbe a group. Recall that ∆(G)={g∈G|[G: CG(g)] <∞} is a characteristic subgroup of G. A group Gis an FC-group (finite conjugate group) if G=∆(G). A group Gsatisfies Min (minimal condition on subgroups) if every non-empty set of subgroups of G, partially ordered by inclusion, has a minimal element. A group satisfies Min-pfor the prime pif each of its p-subgroups satisfies Min. Throughout, Kdenotes a field and Ga locally finite group with no elements of order char(K). Thus K[G] is a regular group ring (cf. [9, Theorem 3.1.5]). Lemma 1. Suppose that Gis an FC-group satisfying Min-pfor all primes pand that the type Ifpart of Qr(K[G]) is non-zero. Then Gis abelian-by-finite. Proof: Suppose that Ghas no abelian subgroup of finite index. By [9, Lemma 6.3.3] and [1, Lemma 1.3(ii)], we may assume that Gis countable. By [10, Lemma 6] and [1, Lemma 1.3(ii)], we may assume G=∞ i=1 Hi/H,
16 F. Ced´ o where ∞ i=1 Hiis a (weak) direct product of finite groups Hi.Wemay also assume that H∩Hi=1for all i. Since Hand Hiare disjoint normal subgroups of ∞ i=1 Hi, they commute and we have H≤Z∞ i=1 Hi= ∞ i=1 Z(Hi). By [1, Lemmas 1.3(ii), 1.4 and Proposition 1.1], we may assume that the Hiare not nilpotent and every proper subgroup of Hiis abelian. By [8], Hi=Piti, where Piis a normal Sylow pi-subgroup of Hi and ti∈Hiis an element of order qni i(ni≥1), where qiis a prime. Furthermore, tiis not normal in Hiand tqi i=Z(Hi). Since the type Ifpart of Qr(K[G]) is non-zero, by [3, Lemma 4.2], there exists a non-zero abelian idempotent e∈K[G]. Thus there exists an integer n, such that Supp e≤n i=1 HiH/H ≤G. Since Gsatisfies Min-pfor all primes p, it is easy to see that there are infinitely many distinct qi. Let qjbe such that j>nand Supp ehas no element of order qj. Let ˜ tj=1+tj+···+tqnj j−1 j. Note that ˜ tj/qnj j is idempotent. Since (x, tj) = 1 for all x∈Supp e,e˜ tj/qnj j∈eK[G]e is idempotent. Since eis abelian and (x, h) = 1 for all x∈Supp eand h∈Hj, we have e˜ tjh=eh˜ tj for all h∈Hj. Using the fact that qnj j−1 i=0 K[Supp e]ti jis a direct sum, we deduce that xtj∈Supp e˜ tjfor all x∈Supp e. Thus, given x∈Supp e and h∈Hj, there exist y∈Supp eand msuch that xtjh=yhtm j. Hence tjht−m jh−1=x−1y∈Supp e∩Hj=Supp e∩Z(Hj). Since Z(Hj)is aqj-group and Supp ehas no element of order qj, we see h−1tjh=tm j. But tjis not normal in Hj, which is a contradiction, so the lemma is proved. Let Jbe a right ideal of K[G]. As in [3], we define α(J, F ) = dim(J∩ K[F])/|F|for each finite subgroup Fof G, and α(J) = sup α(J, F ), where Franges over all finite subgroups of G. We denote by π(G) the set of all primes psuch that Ghas an element of order p.Ifπis a set of primes we say that Gis a π-group if π(G)⊆π.
Maximal quotient rings 17 Lemma 2. Suppose that Gsatisfies Min-pfor all primes pand that the type Ifpart of Qr(K[G]) is non-zero. Then Gis abelian-by-finite. Proof: By [9, Lemma 6.3.3] and [1, Lemma 1.3(ii)], we may assume that Gis countable. By [1, Lemma 1.3(ii)] and Lemma 1, ∆(G)is abelian-by-finite. By [9, Lemma 12.1.2], ∆(G) has a characteristic abelian subgroup Aof finite index. Let σ=π(∆(G)/A). Since the type Ifpart of Qr(K[G]) is non-zero, by [3, Lemma 4.2], there exists a non-zero abelian idempotent e∈K[G]. Let H=Supp e and τ=π(H). Hence π=σ∪τis a finite set of primes. Now Ais the direct product A=Apof its p-primary parts. Let Aπ=p/∈πAp. Then Aπis characteristic in G. Consider ¯ G=G/Aπ. Let δ∈K[G]. We denote by ¯ δthe image of δin K[¯ G]. By [4, Lemma 7.6], ¯eis an abelian idempotent, and clearly it is non-zero. Let pbe a prime such that α(¯eK[¯ G]) >p −1and p/∈π. We shall see that p/∈π(¯ G). Suppose that p∈π(¯ G). Then there exists g∈G\∆(G) with o(g)=pn and o(¯g)=p. Let ˜g=1+g+···+gpn−1. Since K[G] is regular, there exists β∈K[G] such that ˜ge =˜geβ˜ge. By squaring it, we see that eβ˜ge is an idempotent in eK[G]e.By[2, Lemma 2.1], Supp eβ˜ge ⊆∆(G)H. Let H1=Supp eβ˜ge ∪Supp e. Thus p/∈π(¯ H1). Using the fact that we have a direct sum p−1 i=0 ¯giK[¯ H1], we deduce that pn−1¯e=pn−1eβ˜ge. Since p=0inK,¯e=eβ˜ge.Thus¯eK[¯ G]∼ =˜geβK[¯ G]. By [3, Lemma 1.2(iv)], α(¯eK[¯ G]) ≤α(˜gK[¯ G]). But an easy calculation shows that α(˜gK[¯ G]) = p−1. This contradicts the choice of p.Thusp/∈π(¯ G). Hence π(¯ G) is finite. Since ¯eis a non-zero abelian idempotent, by [3, Lemma 4.2], the type Ipart of Qr(K[¯ G]) is non-zero. By [1, Proposition 1.2], [ ¯ G:∆(¯ G)] <∞ and ∆( ¯ G)is finite. By [7, Theorem 3.13], ∆( ¯ G) satisfies Min-pfor all primes p.By[1, Lemma 1.4], [∆( ¯ G):Z(∆( ¯ G))] <∞.Thus¯ G is abelian-by-finite. By [1, Lemma 1.3(ii)], we may assume that ¯ Gis abelian. Let π1=π∪π(¯ G) and let Aπ 1=p/∈π1Ap. Then π(G/Aπ 1)⊆π1.By [9, Lemma 12.4.12], there exists a π1-subgroup Qof Gwith G=Aπ 1Q. By [1, Lemma 1.3(ii), Proposition 1.2 and Lemma 1.4], Qis abelianby-finite. By [1, Lemma 1.3(ii)], we may assume that Qis abelian. Since π(Q) is finite and satisfies Min-pfor all primes p,Qhas a minimal
18 F. Ced´ o subgroup of finite index. Thus by [1, Lemma 1.3(ii)], we may assume that Qcontains no proper subgroup of finite index. We shall see that Gis abelian. Let qbe a prime such that q/∈π1. By [1, Lemma 1.3(ii), Proposition 1.2 and Lemma 1.4], AqQhas an abelian normal subgroup Bof finite index. Now [Q:Q∩B]<∞, thus Q≤B.NowBis the direct product B=Bpof its p-primary parts. Let Bπ1=p∈π1Bp. Since Qis a π1-group, Q≤Bπ1.Thus AqQ=AqBπ1. Since Bπ1is a normal subgroup of AqBπ1,AqQis abelian. Hence G=Aπ 1Qis abelian. Theorem 3 ([1, Theorem 2.3]).The type Ifpart of Qr(K[G]) is non-zero iff [G:∆(G)] <∞and ∆(G)is finite. Furthermore, in this case the type Ifpart of Qr(K[G]) is isomorphic to Qr(K[G/M]), where M=∩Land the intersection is over all subgroups Lof Gof finite index. Proof: The proof of the “if” part and the second part is as in [1]. Suppose that the type Ifpart of Qr(K[G]) is non-zero. Suppose that [G:∆(G)] = ∞or |∆(G)|=∞.By[1, Lemma 1.3(i)], there exists a non-zero central idempotent u∈K[G] such that uK[G] has bounded index of nilpotence. Thus, by [9, Theorem 5.3.15], uK[G] does not satisfy any polynomial identity. By [1, Lemma 2.2], there exists an irreducible uK[G]-module Vwith representation ρ:K[G]→End Vsuch that ρ(G) has no abelian subgroup of finite index. Since uK[G] has bounded index of nilpotence, by [4, Corollary 7.10], V is finite dimensional over its commuting ring. By [5, Lemma 2.4], ρ(G) satisfies Min-pfor all primes p. Let ϕ:K[G]→K[ρ(G)] the natural projection. It is easy to see that ϕ(u) is a non-zero central idempotent of K[ρ(G)]. By [4, Proposition 7.7], ϕ(u)K[ρ(G)] has bounded index of nilpotence. By [4, Corollary 7.4 and Theorems 7.20 and 10.24], the type Ifpart of Qr(K[ρ(G)]) is non-zero. By Lemma 2, ρ(G) is abelian-byfinite, a contradiction, thus [G:∆(G)] <∞and |∆(G)|<∞. Acknowledgements. This work was partially supported by the DGICYT through grant PB92-0586. References 1. F. Ced´ o, On the maximal quotient ring of regular group rings, J. Algebra 115 (1988), 164–174. 2. F. Ced´ o, The maximal quotient ring of regular group rings. II, Proc. Amer. Math. Soc. 104 (1988), 357–362.
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