Let G be a finite group, S a
commutative ring with one and S[G] the group ring of G over S. If H is a group
with G≅H then clearly S[G]≅S[H] where the latter is an S-isomorphism. We study
here the converse question: For which groups G and rings S does S[G]≅S[H] imply
that G is isomorphic to H?
We consider first the case where S = K is a field. It is known that if G is abelian
then Q[G]≅Q[H] implies that G≅H where Q is the field of rational numbers. We
show here that this result does not extend to all groups G. In fact by a simple
counting argument we exhibit a large set of nonisomorphic p-groups with isomorphic
group algebras over all noncharacteristic p fields. Thus for groups in general the
only fields if interest are those whose characteristic divides the order of the
group.
We now let S = R be the ring of integers in some finite algebraic extension
of the rationals. We show here that the group ring R[G] determines the
set of normal subgroups of G along with many of the natural operations
defined on this set. For example, under the assumption that G is nilpotent, we
show that given normal subgroups M and N, the group ring determines the
commutator subgroup (M,N). Finally we consider several special cases. In
particular we show that if G is nilpotent of class 2 then R[G]≅R[H] implies
G≅H.
