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Need just hints

Let $G$ is a finite non-abelian group such that all its proper subgroups are abelian. Then there are at most two different prime numbers dividing $|G|$.

I found some ideas about such this group here,, but honestly don’t know much about Dedekind groups. Thanks.

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The Quaternions are non-abelain, have order $2^3=8$ and every proper subgroup is abelian... – user1729 Sep 13 '12 at 9:21
(In the link you give, you should perhaps note that the Quaternions are the only non-abelian group such that all its proper subgroups are both normal and abelian. I have no idea if this is relevant, but it is the intersection of the two questions...) – user1729 Sep 13 '12 at 9:25
@user1729 - give me more then $2$ prime numbers dividing $8$ – Belgi Sep 13 '12 at 9:36
@user1729: The statement says "at most two prime numbers dividing $\left| G \right|$" $-$ how many prime numbers divide $p^3$ for $p$ prime? – Clive Newstead Sep 13 '12 at 9:37
@NancyR - because he meant it to be a counter example while it is consistent with the OP question – Belgi Sep 13 '12 at 10:04
up vote 8 down vote accepted

Hard part: Show that $G$ cannot be simple. One way to do is to apply a certain fact about maximal subgroups, which is good to know. If in a finite non-abelian group intersections of distinct maximal subgroups are trivial, then the group cannot be simple. Thus it is enough for you to show that if $G$ is simple, then intersections of distinct maximal subgroups of $G$ are trivial.

After that, show that $G$ must be solvable.

If $G$ is solvable, it has a normal subgroup $N$ of prime index, say $[G:N] = p$. Since $N$ is abelian, it has all Sylow subgroups normal. Thus $G$ has all Sylow subgroups normal with the possible exception of $p$-Sylow subgroups.

If $G$ has more than two prime factors, you can prove that the $p$-Sylow subgroup is normal as well. But this it not possible, because then $G$ would be abelian.

I'm leaving a lot of details out, but I believe this approach should work. This theorem is similar to a different one:

If $G$ is a finite non-nilpotent group with all proper subgroups nilpotent, then $|G| = p^a q^b$ for distinct primes $p$ and $q$.

A proof can be found in Derek Robinson's group theory book, and I'm basically using the same idea here.

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@Babak, you should also study a classic paper (1903) of Miller and Moreno. George Miller was a very prolific group theorist, however he did not dispose of our modern notation. So do not be scared off by the style.

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Thanks so much Nicky. Thanks for the link. Honestly, I didn't want to bring up such this question here. I found out that this question is so difficult to be here and for me also. I thought about that a lot everywhere but couldn't find any clue. I didn't to get time of my friends here. :) – Babak S. Sep 15 '12 at 11:01

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