Abstract. Ifà : G ! G is an involution on the finite group G, then à extends to an involution on the integral group ring Z½G. In this paper, we consider whether bicyclic units u A Z½G exist with the property that the group hu; u à i generated by u and u à is free on the two generators. If this occurs, we say that ðu; u Ã Þ is a free bicyclic pair. It turns out that the existence of u depends strongly upon the structure of G and on the nature of the involution. One positive result here is that if G is a nonabelian group with all Sylow subgroups abelian, then for any involution à , Z½G contains a free bicyclic pair.
Free pairs of bicyclic unitsLet R be a commutative ring with 1 and let R½G denote the group ring of G over R.For the most part, we will be concerned with integral group rings where R ¼ Z, or with group algebras where R is a field. Furthermore, we will mainly be interested in finite groups G, although some of our results do hold more generally. If B is a finite subgroup of G, we letB B A R½G denote the sum of the elements of B in R½G. Since ð1 À bÞB B ¼B Bð1 À bÞ ¼ 0 for any b A B, we see that group ring elements of the form ð1 À bÞaB B, with a A G, have square 0. Hence 1 þ ð1 À bÞaB B is a unit in the ring R½G with inverse 1 À ð1 À bÞaB B. When B ¼ hbi is cyclic, elements of the form u ¼ 1 þ ð1 À bÞaB B are known as bicyclic units. It is easy to see that 1 þ ð1 À bÞaB B ¼ 1 if and only if b a ¼ a À1 ba A B and hence if and only if a A N G ðBÞ, the normalizer of B. In particular, if G is a Dedekind group, namely a group with all subgroups normal, then R½G has no nontrivial bicyclic units. Now suppose that à : G ! G is an involution, that is, an antiautomorphism of order 2. Then à extends to an involution of R½G. In particular, if u is a unit of R½G, then so is u à , and we are interested in the nature of the subgroup hu; u à i of the unit group that is generated by these two elements. It was shown in [6] that if