Let $p$ be a prime number. I am interested in classifying all subgroups $H \subset \text{SL}_2(p)$ such that $H \cong C_3 \rtimes C_4$, where $C_3$ and $C_4$ denote the cyclic groups of order 3 and 4 respectively.
I hope that all such subgroups are conjugate to each other, but so far I haven't had any success with showing this. Frankly, I do not yet know for sure whether or not this is true, but I believe it to be true for $p = 5$ and $p = 7$, which I hope is not just a coincidence. To be even more precise, I would hope that $N_G(H) / Z(G) \cong \text{Aut}(H)$, where $G=\text{GL}_2(p)$ and $N_G(H)$ is the normaliser of $H$ in $G$ and $Z(G)$ is the center of $G$. I have already proved that the centraliser of $H$ is given precisely by the center of $G$. The proof used the fact that $H$ is not abelian and that we are working with $2 \times 2$ matrices.
The group $H$ has an element of order 6, so I figured a good start would be to show that all elements in $\text{SL}_2(p)$ that have order 6 are conjugate to each other (which again seems to be true for small primes), but I have not yet succeeded in doing this either. Assuming this for a moment, a natural choice for this element of order 6 would be the matrix $ a = \begin{bmatrix} 0 & 1 \\ -1 & 1 \end{bmatrix} $ and the second generator of the group must then satisfy $b^2 = a^3 = -1$ and $ba = a^{-1}b$. It is not hard to see that it would then follow that $ b = \begin{bmatrix} x & z-x \\ z & -x \end{bmatrix}$ where $x^2+z^2 = xz - 1$. Perhaps one could use this to prove the statement, but I'm not sure.
Does anyone have any ideas? Or would anyone know any sources in which the subgroup structure of $\text{SL}_2(p)$ or $\text{GL}_2(p)$ is treated? Thanks in advance.
Here is a quick answer - I can add more detail later if necessary.
The answer is that there are two such classes when $p \equiv \pm 1 \bmod 12$, and a single class otherwise.
The subgroups of ${\rm SL}(2,p)$ are well understood. I find it easier to consider their images in ${\rm PSL}(2,p)$, and the subgroups isomorphic to $H$ are precisely the preimages of subgroups of ${\rm PSL}(2,p)$, that are isomorphic to the dihedral group $D_6$ of order $6$.
Excluding the case $p=3$, such subgroups are contained in maximal dihedral subgroups of ${\rm PSL}(2,p)$ of order $(p-1)$ or $(p+1)$ (depending on which is divisible by $3$), and there is a single conjugacy class of each of these.
Within these subgroups, when $p-1$ or $p+1$ is divisible by $12$, then there are two classes of subgroups isomorphic to $D_6$, and one otherwise.