Show that $SL_2(F_3)/Z(SL_2(F_3)) \cong A_4$

Question

Show that $SL_2(F_3)/Z(SL_2(F_3)) \cong A_4$

I know that $|SL_2(F_3)/Z(SL_2(F_3))|= 12$.

Then if the quotient group has a normal subgroup of order $4$ then it is isomorphic to $A_4$.

Suppose that it has a normal subgroup of order $3$ then we need to find a contradiction.

Suppose that $V$ is the normal subgroup of order $3$ in $SL_2(F_3)/Z(SL_2(F_3))$ then $V$ is a normal subgroup of order $6$ in $SL_2(F_3)$. Also we see that $V$ must be congruent to $S_3$.

Then how do we proceed after this? Is there a way to proceed from here. There is another answer on the site, here, which doesn't seem intuitive to me.

Any help would be appreciated.

Answer 1

$PSL_2(\Bbb F_3)$ has more than one Sylow $3$-subgroup so that it must be isomorphic to $A_4$ - see the following post:

A group of order $12$ either has a normal $ 3$-Sylow subgroup or is isomorphic to $A_4$