If is a finite group in which every proper subgroup is abelian, show that is solvable.
We proceed by induction on the width of ; recall that the width of a finite group is the number of prime divisors of including multiplicity.
For the base case, if has width 1 then has prime order; thus is simple and abelian, hence solvable.
For the inductive step, suppose every group of width at most and all of whose proper subgroups are abelian is solvable. Let be a group of width , all of whose proper subgroups are abelian. If is abelian, then is solvable. If is not abelian, then by this previous exercise is not simple. Let be a nontrivial proper normal subgroup of . Note that and have width at most .
Every proper subgroup of is a proper subgroup of , hence is abelian. By the induction hypothesis, is solvable.
By the Lattice Isomorphism Theorem, every subgroup of has the form for some subgroup with . Since is abelian, is abelian. Again by the induction hypothesis, is solvable.
Since and are solvable, is solvable.
By induction, the conclusion holds for finite groups of arbitrary width.