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If $G$ is a finite group in which every proper subgroup is abelian, show that $G$ is solvable.

I'm working on this proof and already proved that a subgroup in which every proper subgroup is abelian will not be simple and therefore contains a normal subgroup.

Because this group doesn't have to be of maximal order I can't use the answer here If a proper subgroup of maximal order is solvable and normal, then the group is solvable. Generalization.

So I was thinking it should be possible to proof that the commutator subgroup is a proper subgroup and create the required derived series. $1\trianglelefteq$ $\mid G,G\mid$ $\trianglelefteq G$ which would prove solvability.

I found another proof here http://www.ams.org/journals/tran/1903-004-04/S0002-9947-1903-1500650-9/S0002-9947-1903-1500650-9.pdf which covers concepts which were not yet part of the book I'm working through. Specifically I'm not familiar with transitive substitution groups or primitive groups.

The question is from Exercise 56 in Chapter 4.5 in the book Abstract Algebra of Dummit and Foote.

tretro
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    Hint: Do induction on the order of the group by taking the quotient by the normal subgroup you found. – Tobias Kildetoft Jan 28 '18 at 09:45
  • Very nice hint thanks! Let $N$ be the normal subgroup. Then $G\mathbin{/}N$ has the same properties as $G$. I continue until I'm left with a maiximal normal subgroup. Which happens eventually because $G$ is finite. Then I use https://math.stackexchange.com/questions/2622855/if-a-proper-subgroup-of-maximal-order-is-solvable-and-normal-then-the-group-is. – tretro Jan 28 '18 at 17:44
  • You really don't need to continue. You have an abelian normal subgroup with a quotient that is solvable by induction. That is enough. – Tobias Kildetoft Jan 28 '18 at 17:59

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