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## Classify groups of order 6

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\$.