Prove that every nonabelian group of order 6 has a nonnormal subgroup of order 2. Use this to classify nonabelian groups of order 6.
Let be a nonabelian group of order 6.
We claim that if is any element of order 2 and is any element of order 3, then and do not commute. Proof of claim: Suppose otherwise that . Then , and we have . Thus is cyclic, hence abelian, a contradiction.
Now by Cauchy’s Theorem, there exist such that and , and . Thus , so that is not normal.
Now acts on the left cosets of by left multiplication. Let be the permutation representation induced by this action. Certainly , so that . Now is a nonnormal subgroup of prime order and is normal, so that . Thus is injective. Because , is an isomorphism. Thus .