1
$\begingroup$

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):

enter image description here

$\endgroup$
5
$\begingroup$

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 $gh \in N_G(P)$.

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.