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

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.

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

• So the number are the same and any Sylow $p$-subgroup of $SL_n(\Bbb F_p)$ is also Sylow $p$-subgroup of $GL_n(\Bbb F_p)$. So every Sylow $p$-subgroup of $GL_n(\Bbb F_p)$ lies in $SL_n(\Bbb F_p)$. Apr 11, 2022 at 7:17
• Yes; looking other way - $SL_n$ is sitting normally in $GL_n$, and one Sylow-$p$ of $GL_n$ is inside $SL_n$, hence all should be inside $SL_n$, and so their totality in $GL_n$ and $SL_n$ is same. Apr 11, 2022 at 7:20

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 $$\mathbf{GL}_n(\mathbb{F}_p)$$ .