Boolean Powers of Groups

A group is ^-separating if a Boolean power of the group has a unique Boolean algebra. It is proved that a finite subdirectly irreducible group is S-separating if and only if it is non-Abelian. Suppose B is a Boolean ring and G is a group. Let B[G] denote the group ring of G with coefficient ring B. The Boolean power G [B] is defined to be the set of those elements 2e,.g,. EB[G] such that (1) 2e,= 1, (2)eieJ = 0iigi^gj. The support of the element is the set {g¡ E G\e¡ =£ 0}. G[B] is a group considered as a subgroup of the group of units of B[G]. Consult the survey article [1] for basic properties of Boolean powers. A group G is said to be 5-separating if G[B] at G[B'] implies B at B' for all Boolean rings B and B' [1]. In this note we prove: (1) No Abelian group is 5-separating; (2) A finite subdirectly irreducible group is fi-separating if and only if it is non-Abelian. On the other hand, given a group G, the group G X G is not 5-separating. It is known that no finite Abelian group is 5-separating. Neumann and Yamamuro have proved that a countable non-Abelian simple group is /?-separating [6], while Jonsson has proved that a countable centerless indecomposable group is 5-separating [3]. After this paper had been submitted, the author discovered that G. Bergman [7] had proved that no Abelian group is 5-separating. Because of its simplicity, we have retained our original proof of this fact. If B is a Boolean ring, then there is a partial order < on B defined by e < / if ef = e. Now consider B as a vector space over Z2, the two-element field. We say that B has a totally ordered basis if B has a totally ordered basis (under < ) as a vector space over Z2. It is well known that every countable Boolean ring has a totally ordered basis [5]. ~Received by the editors February 20, 1980. 1980 Mathematics Subject Classification. Primary 06E99, 20E10; Secondary 08B99.