## 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.