Is there an easy way to determine the conjugacy classes of $SL(2,3)$? 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!
 A: 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. 
