It is proved that every group of the form G = AB with subgroups A and B each of which has a cyclic subgroup of index at most 2 is metacyclicby-finite.
Mathematics Subject Classification (2000). Primary 20D40.Keywords. Products of groups, metacyclic group, dihedral group.
Introduction.Let the group G = AB be the product of two subgroups A and B, i.e. G = {ab | a ∈ A, b ∈ B}. It seems intuitively clear that the structure of the group G should somehow be determined by that of the factors A and B. For instance, if A and B are abelian, then G is metabelian by a well-known theorem of N. Ito [1, Theorem 2.1.1] obtained more that fifty years ago. Note that almost all results concerning the structure of the group G = AB with infinite abelian subgroups A and B are based on this theorem. One of the first is due to P. Cohn [2] who proved that in the case of infinite cyclic subgroups A and B there exists a non-trivial normal subgroup of G which is contained in A or B. This implies in particular that G has a subgroup H of finite index whose derived subgroup H and the factor group H/H are both cyclic, in other words, G is metacyclic-by-finite.