The following is an exercise, to prove simplicity of a group $G$, which satisfies conditions -
i) $G$ is transitive subgroup of $S_p$ (symmetric group, $p$ is prime)
ii) $|G|=p\cdot m\cdot k$,
iii) $m\equiv 1\pmod p$ and $m>1$,
iv) $k<p$ is prime.
I was not getting any direction to prove the simplicity of $G$; if $N$ is normal subgroup other than $1$, we can show that $N$ should contain a $p$-cycle (or Sylow-$p$ subgroups of $G$). If $H=Stab(1)$, then $HN=G$. But, after it, I was not able to proceed further. Any hint?
Let $P \in {\rm Syl}_p(G)$. Then, since we know that $N$ is transitive, we can choose $P$ such that $P \le N$. (In fact $N$ must contain all Sylow $p$-subgroups of $G$.)
Since $|N_{S_p}(P)| = p(p-1)$, we have $|N_G(P)| = ps$ for some divisor $s$ of $p-1$, and $|G| = pst$, where $t$ is the number of Sylow $p$-subgroups of $G$, so $t \equiv 1 \bmod p$.
So $st = mk$ and, since $m\equiv 1 \bmod p$, we have $s \equiv k \bmod p$ and, since $s,k < p$ this implies $s=k$ and $m=t$.
Now, since $N \lhd P$, it contains all Sylow $p$-subgroups of $G$, so $m$ divides $|N|$ and, if $N \ne G$, then the only possibility is $|N|=pm$, with $N_N(P) = P$.
But then $N$ has a normal $p$-complement by Burnside's Transfer Theorem, which would be characteristic in $N$ and therefore normal in $G$, contradiction (using $m>1$).