Let be the subgroup generated by elements of the form i.e. commutator subgroup. By Burnside's lemma , the center, is non-trivial. Therefore, . But then the group has order and therefore it is abelian. Thus, we must have that . We are told that is not abelian therefore, and . Suppose that and choose . Then , the centralizer, contains properly, and is a subgroup of . Thus, Lagrange's theorem forces . But that would mean the index of in is . This means that the conjugacy class of is just and so - a contradiction! Thus, we see that . The only remaining possibility is that . However, is a non-trivial subgroup of which forces .