# All Sylow $p$-subgroups of $GL_2(\mathbb F_p)$?

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.

• First, what is the order of the group? Nov 17, 2016 at 19:50
• $p(p+1)(p-1)^2$, is this correct?
– user377174
Nov 17, 2016 at 19:52
• Yes, that is correct. So what could the number of Sylow subgroups be? Nov 17, 2016 at 19:53
• @DietrichBurde Yes, that's extremely helpful. Thank you.
– user377174
Nov 17, 2016 at 20:05

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 $$n_p=(G:N_G(P))=\frac{p(p+1)(p-1)^2}{p(p-1)^2}=p+1.$$