[Hungerford, p99 Q7]
Show that
If G is a nonabelian group of order (p prime), then the center of G is the subgroup generated by all elements of the form .
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 .