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.
Best Answer
$P=$ 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 $\frac{|\mbox{SL}_n(\mathbb{F}_p)|}{|B_0|}$.
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 $|\mbox{SL}_n(\mathbb{F}_p)|=|\mbox{GL}_n(\mathbb{F}_p)|/(p-1)$.