Sylow subgroups of $\mathrm{SL}_2(\mathbb{Z}_p)$ and $\mathrm{SL}_2(\mathbb{F}_p)$

Question

Let $p,l$ be odd primes and $p\neq l$. I am interested in determining the $l$-Sylow subgroups of the groups $\mathrm{SL}_2(\mathbb{Z}_p)$ and $\mathrm{SL}_2(\mathbb{F}_p)$.

I have read that for $\mathrm{SL}_2(\mathbb{F}_p)$, these are all cyclic, but I have not found a reference.

Since $\# \mathrm{SL}_2(\mathbb{F}_p) = p(p-1)(p+1)$, there probably will be two case depending on wether $l\mid (p-1)$ or $l\mid (p+1)$. Furthermore any $l$-Sylow subgroup of $\mathrm{SL}_2(\mathbb{Z}_p)$ maps to a $l$-Sylow of $\mathrm{SL}_2(\mathbb{F}_p)$ under the reduction map, so both questions are related.

Answer 1

The argument for $SL_2(\mathbb{F}_p)$ is straightforward. As you say, every odd prime $\ell \neq p$ dividing the order divides either $p - 1$ or $p + 1$. $SL_2(\mathbb{F}_p)$ has a cyclic subgroup of order $p - 1$ generated by $\begin{bmatrix} x & 0 \\ 0 & x^{-1} \end{bmatrix}$ where $x \in \mathbb{F}_p^{\times}$ is a primitive root $\bmod p$, so it has cyclic subgroups of order $\ell^k$ for every prime power $\ell^k \mid p - 1$. So the Sylow $\ell$-subgroups are all cyclic in this case. Similarly $SL_2(\mathbb{F}_p)$ has a cyclic subgroup of order $p + 1$, namely the kernel of the norm map $N : \mathbb{F}_{p^2}^{\times} \to \mathbb{F}_p^{\times}$, and the same argument applied to this subgroup shows that if $\ell \mid p + 1$ then the Sylow $\ell$-subgroups are all cyclic in this case too.

For $SL_2(\mathbb{Z}_p)$ I'm not sure what the precise definition of a Sylow $\ell$-subgroup should be, but we can argue as follows. The kernel of the reduction map $SL_2(\mathbb{Z}_p) \to SL_2(\mathbb{F}_p)$ consists of matrices congruent to $I \bmod p$ (with determinant $1$); for $p$ an odd prime no such matrix can have finite order (see here; that argument doesn't directly apply to $\mathbb{Z}_p$ as written but a slight adaptation of it does), so it follows that every finite subgroup of $SL_2(\mathbb{Z}_p)$ injects into $SL_2(\mathbb{F}_p)$. So every finite $\ell$-subgroup is cyclic, with bounded order.

Answer 2

In the other answer, Qiaochu showed that every finite $\ell$-subgroup of $\mathrm{SL}_2(\Bbb Z_p)$ is cyclic. This is not sufficient to determine the (pro-)$\ell$ subgroups without further argument (as the example $\Bbb Z_\ell \subset \widehat{\Bbb Z}$ shows, in general the $\ell$-Sylow subgroup of a profinite group can be torsion-free).

However, in this particular case, it's possible to show that the $\ell$-Sylow subgroups of $\mathrm{SL}_2(\Bbb Z_p)$ are finite (and hence cyclic by the answer).

The idea is to use the profinite version of the order (and index) of profinite groups. For a profinite group $G$ with a closed subgroup $H$, the index $(G:H)$ is a supernatural number. Details on this can be found in Ribes-Zalesskii or Neukirch-Schmidt-Wingberg. For example, every infinite pro-$p$ group has order $p^{\infty}$. Consider the kernel $K$ of the projection $\mathrm{SL}_2(\Bbb Z_p) \to \mathrm{SL}_2(\Bbb F_p)$. I claim that $K$ is a pro-$p$ group. Let $$K_n:=\ker(\mathrm{SL}_2(\Bbb Z/p^n\Bbb Z) \to \mathrm{SL}_2(\Bbb F_p))$$

From the fact that taking inverse limits in the category of profinite groups is exact, we obtain that $K=\varprojlim K_n$. Hence it suffices to show that $K_n$ is a finite $p$-group for each $n$. This follows from a general computation of the order of $\mathrm{SL}_2(\Bbb Z/n\Bbb Z)$, see this answer which implies that the order of $K_n$ is a power of $p$.

Thus we obtain $\#K=p^\infty$. Hence the order of $\mathrm{SL}_2(\Bbb Z_p)$ is $$(\mathrm{SL}_2(\Bbb Z_p):1)=(\mathrm{SL}_2(\Bbb Z_p):K)(K:1)=\#\mathrm{SL}_2(\Bbb F_p)p^\infty$$

Because $\#\mathrm{SL}_2(\Bbb F_p)$ is a finite (super)natural number, we get that the supernatural order of $\mathrm{SL}_2(\Bbb Z_p)$ can only be divisible by a finite power of a prime $\ell \neq p$ and hence all $\ell$-Sylow subgroups are finite (and hence cyclic by the other answer).