I need show PSL(2,13)=SL(2,13)/{id, -id} contains a subgroup isomorphic to A_4. I wanted to work with the elements of order 3 in PSL(2,13) (since A_4 is generated by the 3-cycles), but I don't know how to explicitly relate the elements of order 3 in PSL(2,13) with the alternating group. Does anyone know a better way to approach this problem?
Thank you in advance.
It is useful to know that the order of an element in ${\rm SL}(2,q)$ is determined by its trace. We want to generate a subgroup of ${\rm SL}(2,q)$ of order $48$ isomorphic to ${\rm SL}(2,3)$; then its image in ${\rm PSL}(2,q)$ will be isomorphic to $A_4$.
To do that, it would be enough to find two elements $x,y$ of order $3$ (trace $-1$) in ${\rm SL}(2,q)$ with product of order $4$ (trace $0$), because then we would have $(xy)^2 = -I$, and the images of $x,y$ in ${\rm PSL}(2,q)$ would satisfy the relations of the presentation $\langle x,y \mid x^3=y^3=(xy)^2=1 \rangle$ of $A_4$ that was mentioned in the comment by ancientmathematician.
This is possible for all odd prime powers $q$, but in the case when $q \equiv 1 \bmod 4$, there is a solution $$x = \left(\begin{array}{rr}-1&-1\\1&0\end{array}\right),\ \ y = \left(\begin{array}{rr}0&i\\i&-1\end{array}\right),\ \ xy = \left(\begin{array}{cc}-i&1-i\\0&i\end{array}\right), $$ where $i^2=-1$. So, in the case $q=13$, we can take $i=5$.