Exercise 14.4.7

Proof. Recall the context:

Here $g,h\in \mathrm{GL}(2,{𝔽}_p)$ are such that

By (14.23),

Let $[m]\in C(H)$. There are four cases. $\text{∙}$ If $\mathit{mg}=\mathit{gm},\mathit{mh}=\mathit{hm}$, then by Lemma 14.4.3,

If $b\ne 0$, then $g=b^{-1}(c-a)I_2+b^{-1}\mathit{dh}\in \mathrm{Vect}(I_2,h)$, thus $\mathit{gh}=\mathit{hg}$. This is impossible since $\mathit{gh}=-\mathit{hg}\ne 0$ and $p>2$. Thus $b=0$, and $m=a{I}_2$, so that $[m]=[I_2]\in H$. $\text{∙}$ If $\mathit{mg}=-\mathit{gm},\mathit{mh}=\mathit{hm}$, then

thus $\mathit{mh}$ is in the centralizer of $g$ and $h$. By the first bullet, $[\mathit{mh}]=[I_2]$, and using ${[h]}^2=[I_2]$, we obtain $[m]=[h]\in H$. $\text{∙}$ If $\mathit{mg}=\mathit{gm},\mathit{mh}=-\mathit{hm}$, exchanging $g$ and $h$ in the previous case, we obtain similarly $[m]=[g]\in H$. $\text{∙}$ If $\mathit{mg}=-\mathit{gm},\mathit{mh}=-\mathit{hm}$, then, using (10),

Therefore $\mathit{mgh}$ commutes with $g$ and $h$. By the first bullet, $[\mathit{mgh}]=[I_2]$, thus $[\mathit{mg}]=[-(\mathit{mgh})h]=[-h]$, and $[m]=[-m{g}^2]=[\mathit{mg}][-g]=[-h][-g]=[h][g]=[g][h]\in H$.

We have proved $C(H)\subset H$. Since $H$ is Abelian $H\subset C(H)$. Thus $C(H)=H$. □