# Conjugacy Classes in $PSL_2(\mathbb{F}_p)$

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): ## 1 Answer

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 $$gh \in N_G(P)$$.

• 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? – KarlPeter Apr 13 at 22:42
• I have added a proof of that claim. – Derek Holt Apr 14 at 0:28