I am trying to determine the conjugacy classes of $SL(2,3)$ (there are 7) and it is so long and boring, that I am starting to question all of my life choices leading up to this moment.
I've read about $PSL(2,3)\cong SL(2,3)/Z(SL(2,3))\cong SL(2,3)/\{1,-1\}$ , meaning that $A_4$ is isomorphic to a quotient subgroup of $SL(2,3)$. So, my guess is that it suffices to determine the conjugacy classes of $A_4$, which is significantly easier, but then I still need to find and apply the isomorphism to these elements, and I'm unsure if the calculation is still faster that way.
Do you have any tips?
Thanks in advance!
Let $A= \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}$, $B= \begin{bmatrix} 1 & -1 \\ 0 & 1 \end{bmatrix}$, $C= \begin{bmatrix} -1 & 1 \\ 0 & -1 \end{bmatrix}$, $D= \begin{bmatrix} -1 & -1 \\ 0 & -1 \end{bmatrix}$, $E= \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix}$ and $I$ be the identity matrix. It is straight forward, that $Z(G)$ is $\{I, -I\}$.
The first observation, we make is that $A$ and $B$ are order $3$ elements, $C$ and $D$ are order $6$ and $E$ is order $4$. Now, if two elements have different order, then they cannot be in the same conjugacy class, hence it is enough to check, that $X^{-1}AX \neq B$ and $X^{-1}CX \neq D$ for $X\in SL(2,3)$. If we suppose equality, we get in both cases that $a^2=-1$ in the first row, second column of the matrices, which is contradictory in $\mathbb{F}_3$.
$C_G(A)=C_G(B)=C_G(C)=C_G(D)=\{\begin{bmatrix} a & b \\ 0 & a \end{bmatrix}\in G: a,b\in \mathbb{F}_3, a^2=1 \}$ is similarly easy to check.
With the use of the class equation the size of the conjugacy class containing $E$ is less than, or equal to 4.
$S=\{I, -I, \begin{bmatrix} 0 & -1 \\1 & 0 \end{bmatrix}, \begin{bmatrix} 0 & 1 \\-1 & 0 \end{bmatrix}\}$ is a set containing $4$ such elements.
I'm not sure this is a 100% correct, but if so, it is much easier, than I first thought.