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Follow this link a nonabelian group whose every proper subgroup is abelian is not simple. So if $G$ is a nonabelian finite simple group, does $G$ contain a maximal subgroup which is not abelian?

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    $\begingroup$ Yes, that follows directly. $\endgroup$ – Tobias Kildetoft Dec 17 '13 at 10:19
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By the lemma in your link, if $G$ is simple and nonabelian, then $G$ has a nonabelian proper subgroup $H$. Every proper subgroup is contained in some maximal subgroup, so $H$ is contained in a maximal subgroup $M$. All subgroups of abelian groups are abelian, so $M$ must be nonabelian.

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