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

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 $$p^2 \nmid |G|$$.

## closed as off-topic by Derek Holt, Namaste, Rebellos, jgon, user10354138Nov 20 '18 at 1:04

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Solution: 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 $$|G|| p!$$. 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|\$.