Every finite group whose every proper subgroup is abelian is solvable

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

We proceed by induction on the width of G; recall that the width of a finite group G is the number of prime divisors of |G| including multiplicity.

For the base case, if G has width 1 then G has prime order; thus G is simple and abelian, hence solvable.

For the inductive step, suppose every group of width at most k and all of whose proper subgroups are abelian is solvable. Let G be a group of width k+1, all of whose proper subgroups are abelian. If G is abelian, then G is solvable. If G is not abelian, then by this previous exercise G is not simple. Let N be a nontrivial proper normal subgroup of G. Note that N and G/N have width at most k.

Every proper subgroup of N is a proper subgroup of G, hence is abelian. By the induction hypothesis, N is solvable.

By the Lattice Isomorphism Theorem, every subgroup of G/N has the form H/N for some subgroup H with N \leq H < G. Since H is abelian, H/N is abelian. Again by the induction hypothesis, G/N is solvable.

Since N and G/N are solvable, G is solvable.

By induction, the conclusion holds for finite groups of arbitrary width.

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