How to find $Hol(\Bbb Z_2 \times \Bbb Z_2)$?
where $Hol(\Bbb Z_2 \times \Bbb Z_2)=(\Bbb Z_2 \times \Bbb Z_2)\rtimes_{id} Aut(\Bbb Z_2 \times \Bbb Z_2)$. Now we know that $Aut(\Bbb Z_2 \times \Bbb Z_2)\cong S_3$
$G=\operatorname{Hol}(H)=(\Bbb Z_2\times\Bbb Z_2)\rtimes_\varphi S_3$:$$\varphi:S_3\rightarrow\operatorname{Aut}(\Bbb Z_2\times\Bbb Z_2)\cong S_3$$$$\varphi(12)=\sigma_{(12)}:x\mapsto y,y\mapsto x,xy\mapsto xy$$$$\varphi(123)=\sigma_{(123)}:x\mapsto y,y\mapsto xy,xy\mapsto x$$$$\vdots$$
But how to prove that this group will be in fact $S_4$?
Best Answer
The honest way: Your group $\operatorname{Hol}(\mathbb{F}_2\times\mathbb{F}_2)=(\mathbb{F}_2\times\mathbb{F}_2)\rtimes\operatorname{Aut}(\mathbb{F}_2\times\mathbb{F}_2)$ acts on $\mathbb{F}_2^2$ by $(v,A)\colon u\mapsto Au+v$. Easy to check it is really an action, and that it has trivial kernel. Since $\operatorname{Hol}(\mathbb{F}_2\times\mathbb{F}_2)$ and $S_4$ both has 24 elements, it follows that $\operatorname{Hol}(\mathbb{F}_2\times\mathbb{F}_2)=S_4$.
The not-so-honest way: Note that $V\lhd S_4$ and $S_4/V\cong S_3$. There is also a splitting $S_3\to S_4$ so $S_4\cong V\rtimes S_3$. Check this construction agrees with $\operatorname{Hol}(\mathbb{F}_2\times\mathbb{F}_2)$. (This is "not-so-honest" because it relies on you knowing in advance $S_4\cong V\rtimes S_3$)