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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? –  Marshal Kurosh Sep 15 '12 at 6:34
    
Apparently only $\Bbb Z/p\Bbb Z$ for prime $p$; see this blog post. –  Brian M. Scott Sep 15 '12 at 6:49
    
In fact it is true that if a finite group has even one maximal subgroup that is abelian, then $G$ is solvable. –  user641 Sep 15 '12 at 6:56

2 Answers 2

up vote 6 down vote accepted

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. –  Hongshan Li Sep 15 '12 at 23:17
    
@HongshanLi: You’ve got it. –  Brian M. Scott Sep 15 '12 at 23:23

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. –  Urban Sep 15 '12 at 8:37
    
Yes, actually this question belongs a sequence of questions that ultimately lead to Miller and Moreno's result. –  Hongshan Li Sep 15 '12 at 23:15

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