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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 $p^2 \nmid |G|$.

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closed as off-topic by Derek Holt, Namaste, Rebellos, jgon, user10354138 Nov 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 $|G|$.

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