Exercise 4.5.40 (Number of Sylow $p$-subgroups of $\mathrm{GL}_2(\mathbb{F}_p$)

Proof. Put $G={\mathrm{GL}}_2({𝔽}_p)$. By Exercise 39,

and the order of a $p$-Sylow subgroup of ${\mathrm{GL}}_2({𝔽}_p)$ is $p$.

The two obvious Sylow $p$-Sylows are, using Exercise 39,

There are many solutions to the conditions of Sylow’s Theorem $n_p\equiv 1(\mathit{𝑚𝑜𝑑}p)$ and $n_p\mid {(p-1)}^2(p+1)$, among them $1,{(p-1)}^2,p+1,{(p-1)}^2(p+1)$, so we try another method, by computing $N_G(H)$.

Let $A=\left(\begin{array}{cc}\hfill a\hfill & \hfill b\hfill \\ \hfill c\hfill & \hfill d\hfill \end{array}\right)\in {\mathrm{GL}}_2({𝔽}_p)$, so that $\mathit{𝑎𝑑}\ne 0$, and write $M_t=\left(\begin{array}{cc}\hfill 1\hfill & \hfill t\hfill \\ \hfill 0\hfill & \hfill 1\hfill \end{array}\right)\in H$. If $A\in N_G(H)$, then $A{M}_1{A}^{-1}=M_s$ for some $s\in {𝔽}_p$, thus $A{M}_1=M_{s}A$. This gives

therefore

so $c+d=d$, and

Conversely, suppose that $c=0$, so that $A=\left(\begin{array}{cc}\hfill a\hfill & \hfill b\hfill \\ \hfill 0\hfill & \hfill d\hfill \end{array}\right)$, where $\mathit{𝑎𝑑}\ne 0$. Then, for all $t{\in }_p$

so $A=\left(\begin{array}{cc}\hfill a\hfill & \hfill b\hfill \\ \hfill 0\hfill & \hfill d\hfill \end{array}\right)\in N_G(H)$. This shows that

Since $|{𝔽}_p^{\times }|=p-1$ and $|{𝔽}_p|=p$, this gives

Since $n_p=|G:N_G(H)|$ by Sylow’s Theorem, we obtain

The number of Sylow $p$-subgroups of ${\mathrm{GL}}_2({𝔽}_p)$ is $p+1$. □