Exercise 14.4.15

Proof. (a) Here $g=\left(\begin{array}{cc}\hfill \alpha \hfill & \hfill \beta \hfill \\ \hfill 0\hfill & \hfill \alpha \hfill \end{array}\right)\in \mathrm{GL}(2,{𝔽}_p)$, where $\beta \ne 0$ since $g\notin {𝔽}_p^{\text{∗}}{I}_2$.

Let $m=\left(\begin{array}{cc}\hfill \mu \hfill & \hfill \nu \hfill \\ \hfill \xi \hfill & \hfill \eta \hfill \end{array}\right)\in \mathrm{GL}(2,{𝔽}_p)$. Since

we obtain, using $\beta \ne 0$,

Therefore

(b) Let $h=\left(\begin{array}{cc}\hfill a\hfill & \hfill b\hfill \\ \hfill c\hfill & \hfill d\hfill \end{array}\right)\in N(C(g))$. Since $g\in C(g)$, we obtain $\mathit{hg}{h}^{-1}\in C(g)$, thus there are some $\mu ,\nu \in {𝔽}_p,\mu \ne 0,$ such that

This gives

that is

If $c\ne 0$, then $\alpha =\mu$. Thus $\mathit{c\beta }+\mathit{d\alpha }=\mathit{d\mu }=\mathit{d\alpha }$, so $\mathit{c\beta }=0$, which is impossible since $\beta \ne 0$. Therefore $c=0$, and every $h\in N(C(g))$ is of the form

Conversely, if $h=\left(\begin{array}{cc}\hfill a\hfill & \hfill b\hfill \\ \hfill 0\hfill & \hfill d\hfill \end{array}\right),\mathit{ad}\ne 0$, and $n=\left(\begin{array}{cc}\hfill \mu \hfill & \hfill \nu \hfill \\ \hfill 0\hfill & \hfill \mu \hfill \end{array}\right)$ is any element in $C(g)$, then

We have proved

which is the same as (14.31):

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