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 $1979$ )
 A: 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.

A: 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.
