Exercise 8.1.1

Proof.

(a)

Write $a=(12)(34),b=(13)(24),c=(14)(23)$. Note that $e,a,b,c$ are even permutations, so $K=\{e,a,b,c\}\subset A_4$. They satisfy the relations

$$a^2=b^2=c^2=e,\mathit{ab}=\mathit{ba}=c,\mathit{ac}=\mathit{ca}=b,\mathit{bc}=\mathit{cb}=a.$$

This gives the same Cayley table of the Klein’s four-group $\mathbb{Z}∕2\mathbb{Z}\times \mathbb{Z}∕2\mathbb{Z}$, where we write

$$e^{\prime }=(0,0),a^{\prime }=(1,0),b^{\prime }=(0,1),c^{\prime }=(1,1).$$

The mapping ( $e\mapsto e^{\prime },a\mapsto a^{\prime },b\mapsto b^{\prime },c\mapsto c^{\prime }$) is an isomorphism.

If $\sigma \in S_4$, $\mathit{\sigma a}{\sigma }^{-1}=\sigma [(12)(34))]{\sigma }^{-1}=(\sigma (1)\sigma (2))(\sigma (3)\sigma (4))\in K$, and the same is true for $b$ and $c$, so $K$ is normal in $S_4$ (a fortiori in $A_4$).

Conclusion: $K=\{e,(12)(34),(13)(24),(14)(23)\}$ is a normal subgroup of $S_4$ included in $A_4$, and $K$ is isomorphic to $\mathbb{Z}∕2\mathbb{Z}\times \mathbb{Z}∕2\mathbb{Z}$.

(b)

So we obtain a chain $$S_4\supset A_4\supset K\supset ⟨a⟩\supset \{e\},$$

$\text{∙}$ $A_4$ has index 2 in $S_4$, so $A_4$ is a normal subgroup of $S_4$, and $S_4∕A_4\simeq \mathbb{Z}∕2\mathbb{Z}$.

$\text{∙}$ By part (a), we know that $K$ is normal in $A_4$.

As $|A_4∕K|=3$, $A_4∕K\simeq \mathbb{Z}∕3\mathbb{Z}$.

$\text{∙}$ $K$ being Abelian, $⟨a⟩$ is normal in $K$, and $K∕⟨a⟩\simeq \mathbb{Z}∕2\mathbb{Z}$.

$\text{∙}$ $⟨a⟩∕\{e\}\simeq \mathbb{Z}∕2\mathbb{Z}$.

Conclusion: $S_4$ is solvable (and also $A_4$).