Exercise 14.4.14

Proof. (a) First $\left\{\left(\begin{array}{cc}\hfill \mu \hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill \nu \hfill \end{array}\right)\mid \mu ,\nu \in {𝔽}_p^{\text{∗}}\right\}\subset C(g)$, since for all $\mu ,\nu \in {𝔽}_p^{\text{∗}}$,

Conversely, if $h=\left(\begin{array}{cc}\hfill \mu \hfill & \hfill \xi \hfill \\ \hfill \eta \hfill & \hfill \nu \hfill \end{array}\right)\in C(g)$, since $g\notin {𝔽}_p^{\text{∗}}{I}_2$, Lemma 14.4.3 shows that $h=a{I}_2+\mathit{bg}$ for some $a,b\in {𝔽}_p$, thus

so that $\xi =\eta =0$, and $h=\left(\begin{array}{cc}\hfill \mu \hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill \nu \hfill \end{array}\right)$.

We have proved

(b) If $m=\left(\begin{array}{cc}\hfill a\hfill & \hfill b\hfill \\ \hfill c\hfill & \hfill d\hfill \end{array}\right)\in N(C(g))$, then the text p. 452 shows that

thus $\left(\begin{array}{cc}\hfill \mathit{a\alpha }\hfill & \hfill \mathit{b\beta }\hfill \\ \hfill \mathit{c\alpha }\hfill & \hfill \mathit{d\beta }\hfill \end{array}\right)=\left(\begin{array}{cc}\hfill \mathit{\mu a}\hfill & \hfill \mathit{\mu b}\hfill \\ \hfill \mathit{\nu c}\hfill & \hfill \mathit{\nu d}\hfill \end{array}\right),$ that is $\left(\begin{array}{cc}\hfill a(\alpha -\mu )\hfill & \hfill b(\beta -\mu )\hfill \\ \hfill c(\alpha -\nu )\hfill & \hfill d(\beta -\nu )\hfill \end{array}\right)=0$.

If $\alpha =\mu$, then $\alpha \ne \nu$ (since $\mu \ne \nu$), and $\beta \ne \mu$ (since $\alpha \ne \beta$), then $b=c=0$.

Similarly, if $\beta =\mu$, then $\beta \ne \nu$ and $\alpha \ne \mu$, thus $a=d=0$.

It remains the case where $\alpha \ne \mu$ and $\beta \ne \mu$. Then $a=b=0$, but then $m\notin \mathrm{GL}(2,{𝔽}_p)$: this is in contradiction with $m\in N(C(g))\subset \mathrm{GL}(2,{𝔽}_p)$.

To conclude,

(c) Let $m\in {(M_2)}_0$. Then

In the first case, for all $c=\left(\begin{array}{cc}\hfill \mu \hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill \nu \hfill \end{array}\right)\in C(g)$, since $m,c$ are diagonal, $\mathit{mc}=\mathit{cm}$, and $\mathit{mc}{m}^{-1}=c\in C(g)$, and $m\in N(C(g))$.

In the other case, $m=\left(\begin{array}{cc}\hfill 0\hfill & \hfill b\hfill \\ \hfill c\hfill & \hfill 0\hfill \end{array}\right)$, and $m^{-1}=\left(\begin{array}{cc}\hfill 0\hfill & \hfill c^{-1}\hfill \\ \hfill b^{-1}\hfill & \hfill 0\hfill \end{array}\right)$, then

Thus, for all $c\in C(g)$, $\mathit{mc}{m}^{-1}\in C(g)$, therefore $m\in N(C(g))$.

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