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Suppose $G$ is a finite simple group in which every proper subgroup is abelian. If $M$ and $N$ are distinct maximal subgroups of $G$ show that $M \cap N = 1$.

My plan for this problem is to use abelianess of proper subgroups of $G$ to produce a map out of $G$ with kernel $M \cap N$. I am not sure if I am on the right track.

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    are there such simple groups with all proper subgroups abelian?2012-09-15
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    Apparently only $\Bbb Z/p\Bbb Z$ for prime $p$; see [this blog post](http://crazyproject.wordpress.com/2010/06/07/a-nonabelian-group-whose-every-proper-subgroup-is-abelian-is-not-simple/).2012-09-15
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    In fact it is true that if a finite group has even **one** maximal subgroup that is abelian, then $G$ is solvable.2012-09-15

2 Answers 2

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HINT: Let $H$ be the normalizer of $M\cap N$ in $G$. Since $M$ and $N$ are Abelian, $M\subseteq H$ and $N\subseteq H$. What does this tell you about $H$?

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    ah, this means $H = G$. Thanks a lot.2012-09-15
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    @HongshanLi: You’ve got it.2012-09-15
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In 1903 Miller and Moreno proved that a non-abelian group of which all proper subgroups are abelian, must be solvable. Hence the only group satisfying your conditions is a cyclic group of prime order. See the article here, where much more is proved.

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    More is true: If every proper subgroup of a finite group is nilpotent, then the group is solvable. This is the [Schmidt-Iwasawa theorem](http://groupprops.subwiki.org/wiki/Schmidt-Iwasawa_theorem).2012-09-15
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    Yes, actually this question belongs a sequence of questions that ultimately lead to Miller and Moreno's result.2012-09-15