If $H$ is a subgroup with prime index $p$ of a finite simple group $G$, then $p$ is the maximal prime $p$ dividing $|G|$.

Question

Let $G$ be a finite simple group. Let $H$ be a subgroup of $G$ whose index is a prime $p$. Prove that $p$ is the maximal prime dividing the order of $G$ and that $p^2 \nmid |G|$.

Answer 1

Consider the transitive action of $G$ on the left cosets of $H$. This induces a homomorphism $f: G \rightarrow S_{p}$. But $G$ is simple, so $\ker(f)$ is trivial or all of $G$. Since $f$ is not the zero map $G$ injects into $S_p$, so $|G|\mid p!$. That says that any prime decomposition of $|G|$ is less than or equals to $p$. Also $p^2 \nmid p!$ so $p^2 \nmid |G|$.