I want to show that $PSL(2,13)$ has no subgroup of prime index,where $PSL(2,13) = \frac{SL(2,13)}{\brace-I,I}$.
We have the below fact.
《If $G$ be a simple group and $H$ be a subgroup of $G$ such that $|G:H|=n$, ($n$ is greater than 1)then $G$ is embeded in $A_n$.》
Now I know that $PSL(2,13)$ is simple. Also $|PSL(2,13)|= 2^2×3×7×13.$ If $H$ be a subgroup of $PSL(2,13)$ with prime index $p$ then $p$ can be 2,3,7 or 13.
If $p=2$ then $H$ is normal in $PSL(2,13)$ and it is impossible as $PSL(2,13)$ is a simple group.
If $p=3$ then according to that fact $PSL(2,13)$ should be embeded in $A_3$ and it is impossible because of the order of the groups.
Also if $p=7$ we have the same result as 3.
If $p=13$ then $PSL(2,13)$ should be embeded in $A_{13}$.
I don't know how to show that it doesn't happen.
Suppose $G = {\rm PSL}(2,13)$ embeds in $A_{13}$. Then clearly $G$ acts transitively on the $13$ points.
Now a Sylow $7$-subgroup $P$ of $G \le A_{13}$ would be generated by a single $7$-cycle, and so would fix $6$ points.
By a standard argument using Sylow's Theorem, if a finite group (in general) $G$ acts transitively on a set $X$ and $P$ is a Sylow $p$-subgroup of a point stabilizer $G_\alpha$ for $\alpha \in X$, then $N_G(P)$ acts transitively on the fixed point set of $P$.
So in this case, $N_G(P)$ acts transtiviely on the six fixed points of $P$ and hence $|N_G(P)|$ is divisible by $3$. But the normalizer of $P$ in $G$ has order $14$, contradiction.