On self-injective group rings

Abstract

Conditions are given under which the self-injectivity of the group ring AG implies the finiteness of G. It has been known for some time that if A is a self-injective ring (associative with 1) and G is a finite group, then the group ring AG is selfinjective; and conversely that if AG is self-injective then A is self-injective and G is locally finite [4]. Whether or not G must actually be finite, has been studied by Gentile [2] who obtained an affirmative answer in case A is a commutative ring which is torsion-free as a Z-module. His result is included in the following theorem (see Corollary). We denote the Jacobson radical of a ring A by Rad A, and use o(H) to denote the order of a group H. THEOREM. If AG is a seif-injective group ring and o(H) is a unit in A/Rad A for allfinite subgroups H of G, then (A is seif-injective and) G is finite. PROOF. Since AG is self-injective, AG/Rad(AG) is self-injective and (Von Neumann) regular [5] and similarly since AG self-injective=-A selfinjective, it follows that A/Rad A is regular. Since G is locally finite, Rad A = Rad(AG) nA [1] so (Rad A)G c Rad(AG) and therefore AG/Rad (AG) AG/(Rad A)G (A/Rad A)G Rad(AG)/(Rad A)G Rad((A/Rad A)G)' Since o(H) is a unit in the regular ring A/Rad A for all finite subgroups H of G, it follows that (A/Rad A)G is regular [1], and therefore has zero radical. Thus AG/Rad(AG)-(A/Rad A)G and it suffices to consider a regular self-injective group ring AG. It is easily shown that when AG is self-injective, so is AH for all subgroups H of G (see e.g. [2]), so without loss of generality we assume G is countable. Then the fundamental ideal A of AG is countably generated, and since AG is regular, a result of Kaplansky [3] shows that A is Received by the editors April 26, 1971. AMS 1970 subject classifications. Primary 16A26, 16A52; Secondary 16A30.