**Solution to Abstract Algebra by Dummit & Foote 3rd edition Chapter 4.2 Exercise 4.2.10**

Solution: 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 $G \cong S_3$.