TITLE: All Sylow $p$-subgroups of $GL_2(\mathbb F_p)$?
QUESTION [8 upvotes]: How would I use Sylow's 2nd theorem to write down all the Sylow $p$-subgroups of $GL_2(\mathbb F_p)$?
How many Sylow $p$-subgroups of $GL_2(\mathbb F_p)$ are there?
Struggling with these questions, any step-by-step solutions offered would really help my understanding.

REPLY [25 votes]: The number of $p$-Sylow subgroups in $G=GL_2(\mathbb{F}_p)$ is $p+1$.
The order of $G$ is $(p^2-p)(p^2-1)=p(p+1)(p-1)^2$. Therefore a Sylow $p$-subgroup has size $p$.
The matrix  $\left(\begin{smallmatrix} 1 & 1 \\ 0 & 1 \end{smallmatrix}\right)$ has order $p$, hence
it generates a Sylow $p$-subgroup $P$, which consists of all upper unitriangular matrices. Since all
Sylow $p$-subgroups are conjugate, any matrix of order $p$ in $G$ is conjugate to some power of 
$\left(\begin{smallmatrix} 1 & 1 \\ 0 & 1 \end{smallmatrix}\right)$. 
By Sylow III, the number of Sylow $p$-subgroups is given by $(G:N_G(P))$. Let us compute $N_G(P)$.
For a matrix $\left(\begin{smallmatrix} a & b \\ c & d \end{smallmatrix}\right)$ to lie in $N_G(P)$
means it conjugates $\left(\begin{smallmatrix} 1 & 1 \\ 0 & 1 \end{smallmatrix}\right)$ to some power
$\left(\begin{smallmatrix} 1 & a \\ 0 & 1 \end{smallmatrix}\right)$. Since
$$
\begin{pmatrix} a & b \\ c & d \end{pmatrix}\begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}
\begin{pmatrix} a & b \\ c & d \end{pmatrix}^{-1}=\frac{1}{ad-bc}\begin{pmatrix} ad-bc-ac 
& a^2 \\ -c^2 & ad-bc+ac \end{pmatrix},
$$
we see that $\left(\begin{smallmatrix} a & b \\ c & d \end{smallmatrix}\right)\in N_G(P)$ precisely when
$c=0$. Therefore $N_G(P)=\{ \left(\begin{smallmatrix} a & b \\ 0 & d \end{smallmatrix}\right) \}$ in $G$,
which has size $p(p-1)^2$. It follows that
$$
n_p=(G:N_G(P))=\frac{p(p+1)(p-1)^2}{p(p-1)^2}=p+1.
$$