Exercise 4.2.10 (Groups of order 6)

Proof. Let $G$ be a non-abelian group of order $6$.

By Cauchy Theorem, there is an element $a\in G$ of order $2$ and an element $b$ of order $3$, so $a\ne 1,b\ne 1$. Put $H=\{1,a\}=⟨a⟩$ and $K=\{1,b,b^2\}$ (so $|H|=2$ and $|K|=3$). By Lagrange’s Theorem, $|H\cap K|$ divides $2$ and divides $3$, thus $|H\cap K|=1$, so $H\cap K=\{1\}$. This implies that the six elements $1,b,b^2,a,\mathit{𝑎𝑏},a{b}^2$ are distinct, so

If $\mathit{𝑎𝑏}=\mathit{𝑏𝑎}$, then $(a^i{b}^j)(a^k{b}^l)=(a^k{b}^l)(a^i{b}^j)$ for all integers $i,j,k,l$, thus $G$ is abelian, in contradiction with the hypothesis. So $\mathit{𝑎𝑏}\ne \mathit{𝑏𝑎}$. Therefore $\mathit{𝑏𝑎}{b}^{-1}\ne a$. Moreover $\mathit{𝑏𝑎}{b}^{-1}\ne 1$, otherwise $a=1$. Thus $\mathit{𝑏𝑎}{b}^{-1}\notin H$, so $H$ is not normal in $G$.

There is a nonnormal subgroup $H$ of $G$ of order $2$.

Let $G$ acts on the set $A$ of left cosets of $H$. Then $|A|=|G:H|=3$, and let ${\pi }_H:G\to S_A\simeq S_3$ be the corresponding representation. By Theorem 3, the kernel of ${\pi }_H$ is a subgroup of $H$ normal in $G$ (the core of $H$ in $G$). The only subgroups of $H$ are $\{1\}$ and $H$, but $H$ is not normal in $G$, hence

Therefore the representation is faithful, and so

Every non-abelian group of order $6$ is isomorphic to $S_3$.

(If $G$ is abelian, $G=\mathit{𝐻𝐾}\simeq H\times K\simeq Z_2\times Z_3\simeq Z_6$, but this is another story.) □