Finding an element of order $p+1$ in $\text{SL}_2(\mathbb{F}_p)$ and Sylow subgroups of $\text{SL}_2(\mathbb{F}_p)$ Let $p$ be a prime number. I am trying to show that $\text{SL}_2(\mathbb{F_p})$ has an order $p+1$ element and use this to show that for every odd prime $q$, the $q$-Sylow subgroup of $\text{SL}_2(\mathbb{F}_p)$ is cyclic. I have read Sylow subgroups of $\text{SL}_2(q)$. but this is not quite what I'm looking for. I have outlined my attempt below and I would appreciate any hints in the right direction.
I have shown that $|\text{SL}_2(\mathbb{F}_p)|=p(p+1)(p-1)$. In the case where $p=2$, it follows from Cauchy's theorem that there exists an element of order $p+1$ in $\text{SL}_2(\mathbb{F_p})$. So suppose $p$ is an odd prime. Using the Sylow Theorems, I have only gotten as far as showing that there are either $1$ or $p+1$ $p$-Sylow subgroups. I'm not sure how to use these to extract an element of order $p+1$.
Here is my attempt for showing for every odd prime $q$, the $q$-Sylow subgroup of $\text{SL}_2(\mathbb{F}_p)$ is cyclic. We must have $q \leq p$ or the $q$-Sylow would be trivial. If $q = p$, then the result follows. If $q < p$, then either $q \mid p-1$ or $q \mid p+1$. If $q \mid p-1$, then $p-1 = q^m \cdot n$ for some $m, n$. We may then choose a generator $a \in \mathbb{F}_p^\times$ and conclude that
$\begin{pmatrix} a^n & 0 \\ 0 & a^n \end{pmatrix}$ generates a $q$-Sylow. If $q \mid p+1$, then if we knew there was an element $b \in \text{SL}_2(\mathbb{F}_p)$ of order $p+1$ then we could use the same logic as in the previous case to construct a generator for a $q$-Sylow. Does this reasoning work? Also, is it possible for there to be more than one $q$-Sylow because the initial phrasing of the question states "the $q$-Sylow".
 A: Yes, your reasoning almost works. You need to fix your matrix $\begin{bmatrix} a^n & 0 \\ 0 & a^n \end{bmatrix}$ slightly. Is that matrix really in SL or just in GL? What is its determinant?
If $p > 3$, then $\operatorname{SL}_2(\mathbb{F}_p)$ has more than one Sylow $q$-subgroup for every prime $q$ dividing its order, so they probably should have said "a Sylow" instead of "the Sylow".
To find elements of order $p+1$, find an irreducible quadratic polynomial, say $x^2-bx-c$. Then $X=\begin{bmatrix} 0 & 1 \\ c & b \end{bmatrix}$ satisfies that polynomial. The matrices $\mathbb{F}_{p^2}=\{ rI_2+sX : r,s \in \mathbb{F}_p\}$ form a field of order $p^2$. Just like $\mathbb{F}_p^\times$, $\mathbb{F}_{p^2}^\times$ is a cyclic group of order $p^2-1$.
Choose an element $\hat x$ of order $p+1$ in $\mathbb{F}_{p^2}^\times$, and let $x^2-\hat bx-\hat c$ be its minimum polynomial. Defining $\hat X = \begin{bmatrix} 0 & 1 \\ \hat c & \hat b \end{bmatrix}$ you get a matrix of order $p+1$. This is definitely in $\operatorname{GL}_2(\mathbb{F}_p)$, but it is even in $\operatorname{SL}_2(\mathbb{F}_p)$.
If you take a generator of $\mathbb{F}_{p^2}^\times$, then it has some determinant $d \in \mathbb{F}_p^\times$. To get the element of order $p+1$, raise it to the $(p^2-1)/(p+1)=p-1$st power. This raises the determinant $d$ to the $p-1$st power, to get $d^{p-1}=1$ as the new determinant.
So GL has a cyclic subgroup of order $p^2-1$, but the corresponding SL has a cyclic subgroup of order $p+1$.
Your matrix is part of a GL subgroup of order $(p-1) \times (p-1)$, and the corresponding subgroup of SL has order $(p-1)$.
