Finite non-abelian and every proper subgroup is abelian implies metabelian

Statement
Suppose $$G$$ is a non-Abelian finite group such that every proper subgroup of $$G$$ is Abelian. Then, $$G$$ is a fact about::metabelian group: it has solvable length two. In fact, $$G$$ has an Abelian maximal normal subgroup that has prime index.

Related facts

 * Finite non-abelian and every proper subgroup is abelian implies not simple
 * Slender nilpotent and every proper subgroup is abelian implies Frattini-in-center

Facts used

 * 1) uses::Finite non-abelian and every proper subgroup is abelian implies not simple
 * 2) uses::Fourth isomorphism theorem
 * 3) uses::Abelianness is quotient-closed
 * 4) uses::Simple abelian implies cyclic of prime order

Proof
Given: A finite non-Abelian group $$G$$ in which every proper subgroup is Abelian.

To prove: $$G$$ is a metabelian group: it has solvable length two.

Proof: We prove this in two steps:


 * 1) Pick a maximal normal subgroup: Let $$N$$ be a maximal normal subgroup of $$G$$. Then, $$N$$ is an Abelian group by assumption, and $$G/N$$ is simple.
 * 2) The quotient group has the property that every proper subgroup is Abelian: By the fourth isomorphism theorem (fact (2)), every proper subgroup of $$G/N$$ is the image under the quotient map of a proper subgroup of $$G$$ containing $$N$$. By assumption, every proper subgroup of $$G$$ is Abelian, so by fact (3), we obtain that its image is Abelian. Thus, every proper subgroup of $$G/N$$ is Abelian.
 * 3) The quotient group is a simple Abelian group: By step (1), $$G/N$$ is simple, and by step (2), every proper subgroup of $$G/N$$ is Abelian. If $$G/N$$ were non-Abelian, we would obtain a contradiction to fact (1). This forces $$G/N$$ to be Abelian. Note that by fact (4), we also obtain that $$G/N$$ is cyclic of prime order.

Thus, we have found an Abelian normal subgroup $$N$$ such that $$G/N$$ is also Abelian, yielding that $$G$$ is metabelian, i.e., has solvable length two.