Classification of order 6 groups (D&F 4.2.10)

Prove that every non-abelian group of order 6 has a nonnormal subgroup of order 2. Use this to classify groups of order 6.

Let \(G\) be a non-abelian order 6 group. By Cauchy's Theorem, there are 2 subgroups \(H = \{1, h\}\) and \(K = \{1, k, k^2\}\). Note that \(h \neq k,k^2\) since \(h\) has order 2 but \(k\) and \(k^2\) do not. Also, \(K \trianglelefteq G\) as it has index 2. In particular, \(hkh^{-1} \in K\). However, \(k \neq 1\) so \(hk \neq h\) and thus \(hkh^{-1} \neq 1\). Furthermore \(hkh^{-1} \neq k\). To see why, observe that there are 2 cosets \(1K\) and \(hK\) in \(G\). This means that every element in \(G\) can be written in terms of \(h\) and \(k\), so \(G = \left<h,k\right>\). But \(hkh^{-1} = k\) implies that \(h\) and \(k\) commute, which means that all elements of \(G\) commute, contradicting the fact that \(G\) is non-abelian. So the only remaining possibility is \(hkh^{-1} = k^2\), or \(khk = h\). But \(khk^{-1} = khk^2 = hk \not\in H\), and so we conclude that \(H\) is nonnormal.

It is easily verified that if \(hkh^{-1} = k\), then \(G = \left<hk\right>\) and so \(G \cong Z_6\). On the other hand suppose \(hkh^{-1} = k^2\). Then \(\ker \pi_H \neq H\) as \(H\) is not normal in \(G\). So \(\ker \pi_H = 1\), i.e. \(\pi_H\) is injective. Since \(\pi_H(G) \le S_3\) (because there are 3 cosets of \(H\)) and \(|\pi_H(G)| = 6 = |S_3|\), it follows that \(\pi_H\) is an isomorphism. Thus \(G \cong S_3\).