# A finite group with the property that all of its proper subgroups are abelian

Let $G$ be a finite group with the property that all of its proper subgroups are abelian. Let $N$ be a normal subgroup of $G$. Prove that either $N$ is contained in the center of $G$ or else $G$ has a normal abelian subgroup of prime index.

I think $G$ is solvable. http://crazyproject.wordpress.com/2010/06/08/every-finite-group-whose-every-proper-subgroup-is-abelian-is-solvable/. I hope that idea maybe usefull.

Help me some hints.

Thanks a lot.

P/s: This is a question comes from a qualifying exam in Algebra ( Wisconsin August $1979$ )

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There is a paper on this subject which describes completely these groups. You can see "ams.org/journals/tran/1903-004-04/S0002-9947-1903-1500650-9/…; –  Babak Miraftab Jan 3 '14 at 8:14
I can't see this problem in that paper. Can you notice me @Babgen ? –  chuyenvien94 Jan 3 '14 at 8:41
No, I didn't say that your problem is there. For more detail about these groups you can see this paper. –  Babak Miraftab Jan 3 '14 at 13:35
Ehm ... of course :) –  chuyenvien94 Jan 3 '14 at 13:39
You can follow the link math.library.wisc.edu/reserve/qualifying-exams.html –  chuyenvien94 Jan 3 '14 at 13:43

Or you could argue as follows. Suppose that $N \not \subseteq Z(G)$. Let $M$ be a maximal normal subgroup of $G$ containing $N.$ Then $M = C_{G}(N)$ as $M$ is Abelian and $M$ is proper, but $N \not \subseteq Z(G).$ Then certainly $M = C_{G}(M).$ Hence $M$ is in fact a maximal subgroup of $G,$ for if $M$ is contained in another proper subgroup $H$ of $G$ normal or not) then $H$ s Abelian, so $H \subseteq C_{G}(M) = M$ and $H = M.$ Hence $G/M$ is a simple group (as $M$ is maximal normal) with no proper non-identity subgroup (as $M$ is not contained in any proper subgroup of $G$). But $G/M$ contains a cyclic subgroup of prime order (you don't even need Cauchy's theorem to see this), so $G/M$ is in fact cyclic of prime order.

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+1 vote: Thank you very much –  chuyenvien94 Jan 3 '14 at 11:03
Can you explain why $G/M$ contains a cyclic subgroup of prime order? –  chuyenvien94 Jan 3 '14 at 11:06
Well, let $X$ be any finite group with more that one element. Choose $x$ to be any non-identity element of $X.$ Then $\langle x \rangle$ is cyclic of order $n >1.$ Let $p$ be a prime divisor of $n.$ Then $x^{\frac{n}{p}}$ has prime order $p.$ –  Geoff Robinson Jan 3 '14 at 12:31
I agree with you. –  chuyenvien94 Jan 3 '14 at 13:40
Use the isomorphism theorems. $M$ is not contained in any proper subgroup of $G.$ But $G/M$ has a subgroup of order $p.$ This means that $G$ has a subgroup $X$ which contains $M$ and has order $p|M|.$ But then $X$ must be all of $G$ and $G/M$ is cyclic of order $p.$ –  Geoff Robinson Jan 3 '14 at 14:02

Here's how I would argue:

• Show that the centralizer of $N$ in $G$ is also normal and contains $N$. So either $N$ is contained in the center, or we replace it with its centralizer, which is also a proper normal abelian subgroup of $G$.
• Repeating the previous step as many times as needed ends with either the conclusion that $N$ is in the center or that $N$ is its own centralizer, and also a proper normal abelian subgroup of $G$.
• In the latter case consider the homomorphism $f:G/N\to Aut(N)$ gotten by conjugation action. Show that it has to be injective.
• If $G/N$ has a proper cyclic subgroup $K/N$ show that $K/N\le \ker f$.
• Conclude that $G/N$ must be cyclic of prime order.
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+1 vote: Thank you –  chuyenvien94 Jan 3 '14 at 8:42
Yeah: As Geoff's answer explains, we don't need to do many iterations in step one. As $N$ grows, its centralizer can only shrink, so one round will do. –  Jyrki Lahtonen Jan 3 '14 at 18:26