Classify groups of order 6

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 G be a nonabelian group of order 6.

We claim that if x is any element of order 2 and y is any element of order 3, then x and y do not commute. Proof of claim: Suppose otherwise that xy = yx. Then |xy| = 6, and we have G = \langle xy \rangle. Thus G is cyclic, hence abelian, a contradiction.

Now by Cauchy’s Theorem, there exist x,y \in G such that |x| = 2 and |y| = 3, and xy \neq yx. Thus y\langle x \rangle = \{ y, yx \} \neq \{ y, xy \} = \langle x \rangle y, so that \langle x \rangle is not normal.

Now G acts on the left cosets of \langle x \rangle by left multiplication. Let \varphi : G \rightarrow S_3 be the permutation representation induced by this action. Certainly \langle x \rangle = \mathsf{stab}(\langle x \rangle), so that \mathsf{ker}\ \varphi \leq \langle x \rangle. Now \langle x \rangle is a nonnormal subgroup of prime order and \mathsf{ker}\ \varphi is normal, so that \mathsf{ker}\ \varphi = 1. Thus \varphi is injective. Because |G| = |S_3| = 6, \varphi is an isomorphism. Thus G \cong S_3.

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