Let be an abelian group and let . Prove that is abelian. Give an example of a nonabelian group containing a proper normal subgroup such that is abelian.
Lemma 1: Let be a group. If such that , then . Proof: Use cancellation.
Lemma 2: Let be a group. If , then . Proof: Since has an identity element, say , we know that , , and . If , we have , a contradiction. Thus . We can easily see that .
If is abelian, every subgroup of is normal; in particular, is normal, so is a group. Now let . Then . Hence is abelian.
Consider . We saw in a previous exercise that the normalizer of in is all of , so that is normal. Since has 8 elements and has 4 elements, has 2 elements. Thus , which is abelian.