Suppose $G$ is a finite simple group whose order is divisible by $p^2$. Show there is no subgroup with index $p$.

Question

Suppose $G$ is a finite simple group such that $p^2$ divides $|G|$ for some prime $p$. Show there is no subgroup of $G$ of index $p$.

We can write the order of $G$ as $|G|=p^2m$. I want to use the third Sylow theorem, but there is no restriction that $p$ does not divide $m$. So, I rewrote the order as $|G|=p^nm_1$ where $n\geq2$ and $p$ does not divide $m_1$. But, I don't know how to convert this into a proof showing the index of a subgroup can't be $p$.

Answer 1

If $G$ has a subgroup $H$ of index $p$, left multiplication on cosets $G/H$ gives a non-trivial homomorphism $G \to S_p$. This must be injective as $G$ is simple. But $p^2$ doesn't divide $p!$ the order of $S_p$.