A group,G, contains a subgroup,Z, that is isomorphic to the integers. Z is maximal in the sense that if for any g such that some power g^m is in Z then g is already in Z. Is it true that any cohomology class ,h, in H^1(G;Z/2) that pulls back to the generator of H^1(Z;Z/2) via the inclusion homomorphism has trivial self-cup products? i.e. is h cup h = 0 in H^2(G;Z/2)?