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

$$\left(\begin{matrix} 1 & k \\ 0 & 1 \end{matrix}\right)$$


$$\left(\begin{matrix} 1 & k^2 \\ 0 & 1 \end{matrix}\right)$$

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):

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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 $gh \in N_G(P)$.

  • $\begingroup$ let me recapitulate your answer: since all Sylow (p)-groups are conjugate to each other all conjugacy classes can be found in an "excellent" (fixed from now) Sylow p-group $P$, namely the of $E_s$'s. Then that $N_G(P)$ coinsides with triangulars is also just is a simple observation since we have choosen $P$ beeing of such "simple shape", right? What I don't understand is why does the fact that $P$ is abelian already imply that the conjugate classen in $G$ and $N_G(P)$ of $p$ orders coinside? $\endgroup$ – KarlPeter Apr 13 at 22:42
  • $\begingroup$ I have added a proof of that claim. $\endgroup$ – Derek Holt Apr 14 at 0:28

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