Let $p$ be an odd prime such that $2$ is not a square mod $p$.
I want to determine all conjugacy classes of elements of order $p$ in $PSL_2(F_p)$.
Since each such element has an eigenvalue $1$ and determinant $1$ it's upper triangular form is
$$E_s:= \left(\begin{matrix} 1 & s \\ 0 & 1 \end{matrix}\right)$$
The author claims that there are only two conjugacy classes of elements of order $p$: namely with $s$ square and not square mod $p$.
The question is why?
My attempts: Let $C:= \left(\begin{matrix} a & b \\ c & d \end{matrix}\right) \in PSL_2(F_p)$. Then it induces the conjugation of $E_s$:
$$ \left(\begin{matrix} a & b \\ c & d \end{matrix}\right) \left(\begin{matrix} 1 & s \\ 0 & 1 \end{matrix}\right) \cdot \frac{1}{ad-bc} \cdot \left(\begin{matrix} a & b \\ 0 & a^{-1} \end{matrix}\right)^{-1} = $$ $$\frac{1}{ad-bc} \cdot \left(\begin{matrix} (ad+sac +bc) & sa^2 \\ (2cd+sc^2) & (-bc+sac+ad) \end{matrix}\right) $$
The goal woulb be that the latter matrix is conjugated to a matrix of the sheape
$$\left(\begin{matrix} 1 & k \\ 0 & 1 \end{matrix}\right)$$
or
$$\left(\begin{matrix} 1 & k^2 \\ 0 & 1 \end{matrix}\right)$$
for $k$ not square.
Is there an elegant way to deduce it?
Source: Szamuely's "Galois Groups and Fundamental Groups" in the excerpt below (page 133):
Assuming that 2 is not a square mod $p$ is irrelevant and a distraction - this is true for all odd primes $p$ - in fact it's true for all powers of odd primes. (For powers of $2$ there is a single class.)
Since the Sylow $p$-subgroups $P$ are abelian, the conjugacy classes are the same as in $N_G(P)$, and you can take $N_G(P)$ to be the image of the group of upper triangular matrices of determinant 1. Then it's a straightforward calculation.
The fact that $P$ abelian implies conjugacy in $P$ is controlled by $N_G(P)$ is a standard exercise in the application of Sylow's theorem. Let $x,y \in P$, $x^g=y$ with $g \in G$. Then $P^g,P \in {\rm Syl}_p(C_G(y))$, so $\exists h \in C_G(y)$ with $P^{gh} = P$, and then $x^{gh} = y$ with $gh \in N_G(P)$.