Number of Sylow $p$-subgroup of $\mbox{SL}_n(\Bbb F_p)$.

Question

What is the number of Sylow $p$-subgroup of $\mbox{SL}_n(\Bbb F_p)$ where $\Bbb F_p$ is finite field of order $p$?

This problem is known for $\mbox{GL}_n(\Bbb F_p)$. By checking the order, strictly upper triangular matrix $P$ is a Sylow $p$-subgroup of $\mbox{GL}_n(\Bbb F_p)$. I know the normalizer of $P$ in $\mbox{GL}_n(\Bbb F_p)$ is a group of upper triangular matrices $T$. Since the order of $T$ is $(p-1)^n p^{1+2+\cdots n-1}$, the number of Sylow $p$-subgroup of $\mbox{ GL}_n(\Bbb F_p)$ is $$n_p = {(p^n-1)(p^n-p)\cdots(p^n-p^{n-1})\over (p-1)^np^{1+2+\cdots+(n-1)}}.$$ Since $P$ is also Sylow $p$-subgroup of $\mbox{SL}_n(\Bbb F_p)$, all I need to do is compute the number of $\det =1$ elements in $N:=N_{\mbox{GL}_n(\Bbb F_p)}(P)$. I tried to use the fact that for any given $A\in N$, I can correct one value in the diagonal to make $\det A =1$. But I don't know how to get further.

Answer 1

$P=$ the set of upper triangular matrices over $\mathbb{F}_p$ with diagonal entries $1$.

$P$ is Sylow $p$-subgroup of $\mbox{GL}_n(\mathbb{F}_p)$ and of $\mbox{SL}_n(\mathbb{F}_p)$.

$B=$ set of upper triangular invertible matrices over $\mathbb{F}_p$

($B$ is normalizer of $P$ in $\mbox{GL}_n(\mathbb{F}_p)$; $|B|=|P|(p-1)^{n}$.)

$B_0:=$ subset of $B$ of matrices with determinant $1$.

($B_0$ is normalizer of $P$ in $\mbox{SL}_n(\mathbb{F}_p)$; $|B_0|=|P|(p-1)^{n-1}$.)

The number of Sylow-$p$ subgroups in $\mbox{SL}_n(\mathbb{F}_p)$ is $\frac{|\mbox{SL}_n(\mathbb{F}_p)|}{|B_0|}$.

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Consider map $\det:B\rightarrow \mathbb{F}^*_p$, $A\mapsto \det(A)$. This is surjective homomorphism with kernel $B_0$, hence $|B_0|=|B|/(p-1)$. Similar way, you can see $|\mbox{SL}_n(\mathbb{F}_p)|=|\mbox{GL}_n(\mathbb{F}_p)|/(p-1)$.

Answer 2

The Sylow $p$-subgroups of $\mathbf{GL}_n(\mathbb{F}_p)$ are the same as the ones in $\mathbf{SL}_n(\mathbb{F}_p)$ . Moreover, for any matrix $A$ in $\mathbf{GL}_n(\mathbb{F}_p)$ of order $p^k$ ( $A^{p^k}=I$ , the identity matrix) for some integer $k$ , $\det(A)=\bar1$ .

Consider the determinant homomorphism $\det:\mathbf{GL}_n(\mathbb{F}_p)\to\mathbb{F}_p^{\times}$ , write $\det(A)=\bar a$ where $A\in\mathbf{GL}_n(\mathbb{F}_p)$ is of order $p^k$ for some integer $k$ .

Since $\bar a\in\mathbb{F}_p^{\times}$ , $\bar1=\bar a^{\varphi(p)}=\bar a^{p-1}\ \ \Rightarrow\ \ \bar a=\bar a^p$ . Then by induction we see also that $$ \bar{a}^{p^k}=(((\bar{a}\overbrace{\ ^p)^p)^{\cdots})^p}^{k\ \text{terms}\ \text{of}\ p}=\bar a\ . %\bar a^{p^k}=((\bar a\overbrace{\left.\left.\left.\ \phantom{\left(\bar a\right)}^p\right)^p\right)^{\cdots}\right)^p}^{\phantom{\overset{}{}}k\ \text{terms}\ \text{of}\ p} %\bar a^{p^k}=\left(\left(\left(\bar a^p\right)^p\right)^{\cdots}\right)^p $$ i.e., $$ \bar a=\bar a^{p^k}=\det(A)^{p^k}\overset{\text{homo}}{=}\det\left(A^{p^k}\right)=\det(I)=\bar1\ , $$ which implies that $A$ is also in $\mathbf{SL}_n(\mathbb{F}_p)$ . Hence $P\leqslant\ker\det=\mathbf{SL}_n(\mathbb{F}_p)$ holds for any $p$-subgroups $P$ in $\mathbf{GL}_n(\mathbb{F}_p)$ .