Sylow $p$-subgroups of $SO_3(\mathbb{F}_p)$ Let $G=SO_3(\mathbb{F}_p)$.  We have $|G|=p(p^2-1)$.  Let $n_p$ be the number of Sylow $p$-subgroups of $G$.  Is it true that $n_p=p+1$?
We have the two conditions


*

*$n_p\equiv 1\mod p$

*$n_p\mid p^2-1$.


The first condition gives us $n_p=1+kp$, and then the second gives $m(1+kp)=p^2-1$.  For fixed $p$, this equation at least has the two nonnegative integer solutions $(k,m)$ given by $(0,p^2-1)$ and $(1,p-1)$ which leads to the two possibilities $n_p=1$ and $n_p=p+1$.  Are there more solutions to this equation?
Next, I'd like to rule out the possibility that $n_p=1$.  Is $SO_3(\mathbb{F}_p)$ simple?
Any help in solving my initial question is appreciated, even if it doesn't follow my thinking.

(compiling my comments so the question will have an answer...)
The first question has an easy answer: you can easily rule out any factors $kp+1$ of $p^2-1$ with $k\gt 1$.  Suppose $m\cdot (kp+1) = p^2-1$; then since $p^2-1\equiv -1$ and $kp+1\equiv 1\pmod p$, it must be the case that $m\equiv -1$.  Since $m\gt 0$, then $m\geq p-1$.  But then $m\cdot (kp+1) \gt m\cdot (p+1)\geq (p-1)\cdot(p+1)=p^2-1$.
As for the second question, rather than breaking out the heavy machinery you're probably better off just trying to exhibit two distinct $p$-Sylow subgroups: try finding a particular subgroup and then find an element in $SO_3(\mathcal{F}_p)$ that doesn't conjugate it back to itself.  If you find the right subgroup, you should be able to just conjugate it by a permutation matrix to find another distinct one.
