It depends what is. If then , to see a proof go here. If then must be a normal subgroup. This means where . It remains for you to eliminate which of these are not possible. We begin by noting that because is abelian which would imply that a contradiction. Now for to be abelian it would imply it is a cyclic group of order six. But it is impossible for to have order since is a cycle. Thus, is not abelian which implies must be the commutator subgroup. Try doing it for .