Problem 5.5.5 ($\mathrm{Hol}(Z_2\times Z_2) \simeq S_4$)

Proof. As usual, we identify $H$ and $K$ as subgroups of $G=H{⋊}_{\phi }K$.

(a)

Let $H=Z_2\times Z_2$ and let $G=\mathrm{Hol}(H)=H⋊\mathrm{Aut}(H)$. Since $H\simeq \mathbb{Z}∕2\mathbb{Z}\times \mathbb{Z}∕2\mathbb{Z}$ and $\mathrm{Aut}(H)\simeq {\mathrm{GL}}_2({𝔽}_2)$, $$G\simeq (\mathbb{Z}∕2\mathbb{Z}\times \mathbb{Z}∕2\mathbb{Z}){⋊}_{\phi }{\mathrm{GL}}_2({𝔽}_2),$$

where

$$\phi \{\begin{array}{ccc}\hfill {\mathrm{GL}}_2({𝔽}_2)\hfill & \hfill \to \hfill & \mathrm{Aut}(\mathbb{Z}∕2\mathbb{Z}\times \mathbb{Z}∕2\mathbb{Z})\hfill \\ \hfill A=\left(\begin{array}{cc}\hfill a\hfill & \hfill b\hfill \\ \hfill c\hfill & \hfill d\hfill \end{array}\right)\hfill & \hfill \to \hfill & {\phi }_A\{\begin{array}{ccc}\hfill \mathbb{Z}∕2\mathbb{Z}\times \mathbb{Z}∕2\mathbb{Z}\hfill & \hfill \to \hfill & \mathbb{Z}∕2\mathbb{Z}\times \mathbb{Z}∕2\mathbb{Z}\hfill \\ \hfill \left(\begin{array}{c}\hfill h\hfill \\ \hfill k\hfill \end{array}\right)\hfill & \hfill \mapsto \hfill & \left(\begin{array}{cc}\hfill a\hfill & \hfill b\hfill \\ \hfill c\hfill & \hfill d\hfill \end{array}\right)\left(\begin{array}{c}\hfill h\hfill \\ \hfill k\hfill \end{array}\right)\hfill \end{array}\hfill \end{array}$$

and the law is given by

$$\left(\left(\begin{array}{c}\hfill h\hfill \\ \hfill k\hfill \end{array}\right),A\right)\left(\left(\begin{array}{c}\hfill h^{\prime }\hfill \\ \hfill k^{\prime }\hfill \end{array}\right),A^{\prime }\right)=\left(\left(\begin{array}{c}\hfill h\hfill \\ \hfill k\hfill \end{array}\right)+A\left(\begin{array}{c}\hfill h^{\prime }\hfill \\ \hfill k^{\prime }\hfill \end{array}\right),A{A}^{\prime }\right)$$

As in Exercise 4, $K={\mathrm{GL}}_2({𝔽}_2)$ is a non abelian group of order $6$, so ${\mathrm{GL}}_2({𝔽}_2)\simeq S_3$.

$${\mathrm{GL}}_2({𝔽}_2)=\left\{\left(\begin{array}{cc}\hfill 1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 1\hfill \end{array}\right),\left(\begin{array}{cc}\hfill 1\hfill & \hfill 1\hfill \\ \hfill 1\hfill & \hfill 0\hfill \end{array}\right),\left(\begin{array}{cc}\hfill 0\hfill & \hfill 1\hfill \\ \hfill 1\hfill & \hfill 1\hfill \end{array}\right),\left(\begin{array}{cc}\hfill 1\hfill & \hfill 1\hfill \\ \hfill 0\hfill & \hfill 1\hfill \end{array}\right),\left(\begin{array}{cc}\hfill 0\hfill & \hfill 1\hfill \\ \hfill 1\hfill & \hfill 0\hfill \end{array}\right),\left(\begin{array}{cc}\hfill 1\hfill & \hfill 0\hfill \\ \hfill 1\hfill & \hfill 1\hfill \end{array}\right)\right\},$$

This shows that

$$\mathrm{Hol}(H)\simeq H⋊K,K={\mathrm{GL}}_2({𝔽}_2)\simeq S_3.$$

Then $|G|=|Z_2\times Z_2|\cdot |S_3|=24.$

(b)

Let $X=\{\mathit{𝑎𝐾}\mid a\in G\}$ be the set of left cosets of $K$ in $G$. Then $|X|=|G:K|=4$. $G$ acts on X by $g\cdot (\mathit{𝑎𝐾})=(\mathit{𝑔𝑎})K$, and the associated homomorphism is $$\psi \{\begin{array}{ccc}\hfill G\hfill & \hfill \to \hfill & S_X\simeq S_4\hfill \\ \hfill g\hfill & \hfill \to \hfill & {\psi }_g\{\begin{array}{ccc}\hfill X\hfill & \hfill \to \hfill & X\hfill \\ \hfill \mathit{𝑎𝐻}\hfill & \hfill \mapsto \hfill & \mathit{𝑔𝑎𝐾}\hfill \end{array}\hfill \end{array}$$

By Theorem 3 of Section 4.2,

$$\ker <!--\; FUNCTION\; APPLICATION\; -->(\psi )=\underset{x\in G}{\bigcap }\mathit{𝑥𝐾}{x}^{-1}.$$

Since every element $x\in G$ is of the form $x=\mathit{h𝑘}$, where $h\in H,k\in K$, and $\mathit{𝑥𝐾}{x}^{-1}=\mathit{h𝑘𝐾}{k}^{-1}{h}^{-1}=\mathit{h𝐾}{h}^{-1}$, we obtain

$$\ker <!--\; FUNCTION\; APPLICATION\; -->(\psi )=\underset{h\in H}{\bigcap }\mathit{h𝐾}{h}^{-1}.$$

Let $y=\left(\left(\begin{array}{c}\hfill u\hfill \\ \hfill v\hfill \end{array}\right),B\right)$, where $u,v\in {𝔽}_2$ and $B\in {\mathrm{GL}}_2({𝔽}_2)$, be any element of $\ker <!--\; FUNCTION\; APPLICATION\; -->(\psi )$. We must prove $u=v=0$ and $B=I$.

For all $h=\left(\begin{array}{c}\hfill r\hfill \\ \hfill s\hfill \end{array}\right)\in H$, $y\in \mathit{h𝐾}{h}^{-1}$, thus there is some $A\in {\mathrm{GL}}_2({𝔽}_2)$ (depending of $h$) such that

$$\begin{array}{llll}\hfill \left(\left(\begin{array}{c}\hfill u\hfill \\ \hfill v\hfill \end{array}\right),B\right)& =\left(\left(\begin{array}{c}\hfill r\hfill \\ \hfill s\hfill \end{array}\right),I\right)\left(\left(\begin{array}{c}\hfill 0\hfill \\ \hfill 0\hfill \end{array}\right),A\right){\left(\left(\begin{array}{c}\hfill r\hfill \\ \hfill s\hfill \end{array}\right),I\right)}^{-1}& \hfill & \\ \hfill & =\left(\left(\begin{array}{c}\hfill r\hfill \\ \hfill s\hfill \end{array}\right),A\right)\left(\left(\begin{array}{c}\hfill -r\hfill \\ \hfill -s\hfill \end{array}\right),I\right)& \hfill & \\ \hfill & =\left(\left(\begin{array}{c}\hfill r\hfill \\ \hfill s\hfill \end{array}\right)+A\left(\begin{array}{c}\hfill -r\hfill \\ \hfill -s\hfill \end{array}\right),I\right)& \hfill & \\ \hfill & =\left((I-A)\left(\begin{array}{c}\hfill r\hfill \\ \hfill s\hfill \end{array}\right),A\right).& \hfill & \end{array}$$

For every $\left(\begin{array}{c}\hfill r\hfill \\ \hfill s\hfill \end{array}\right)\in H$, $A=B$ , and $\left(\begin{array}{c}\hfill u\hfill \\ \hfill v\hfill \end{array}\right)=(I-B)\left(\begin{array}{c}\hfill r\hfill \\ \hfill s\hfill \end{array}\right)$, so

$$\left(\begin{array}{c}\hfill 0\hfill \\ \hfill 0\hfill \end{array}\right)=(I-B)\left(\begin{array}{c}\hfill 0\hfill \\ \hfill 0\hfill \end{array}\right)=(I-B)\left(\begin{array}{c}\hfill 1\hfill \\ \hfill 0\hfill \end{array}\right)=(I-B)\left(\begin{array}{c}\hfill 0\hfill \\ \hfill 1\hfill \end{array}\right),$$

thus $I-B=0$, an so $B=I$. Finally $\left(\begin{array}{c}\hfill u\hfill \\ \hfill v\hfill \end{array}\right)=(I-B)\left(\begin{array}{c}\hfill r\hfill \\ \hfill s\hfill \end{array}\right)=\left(\begin{array}{c}\hfill 0\hfill \\ \hfill 0\hfill \end{array}\right)$. This proves

$$\ker <!--\; FUNCTION\; APPLICATION\; -->(\psi )=\{1\},$$

and so the representation given by $\psi$ is faithful. Therefore $\psi$ is injective, and $|G|=|S_X|=|S_4|=24$, so $\psi$ is an isomorphism:

$$\mathrm{Hol}(Z_2\times Z_2)\simeq (\mathbb{Z}∕2\mathbb{Z}\times \mathbb{Z}∕2\mathbb{Z}){⋊}_{\phi }{\mathrm{GL}}_2({𝔽}_2)\simeq S_4.$$