Finite non-abelian and every proper subgroup is abelian implies not simple

Statement
Suppose $$G$$ is a non-abelian finite group such that every proper subgroup of $$G$$ is an Abelian group. Then, $$G$$ is not a simple group.

Further facts about every proper subgroup being abelian

 * Finite non-abelian and every proper subgroup is abelian implies metabelian
 * Slender nilpotent and every proper subgroup is abelian implies Frattini-in-center
 * Classification of finite non-abelian groups in which every proper subgroup is abelian
 * Schmidt-Iwasawa theorem

Analogues

 * The analogous statement for infinite groups is not true. In fact, Tarski has constructed infinite groups in which every proper nontrivial subgroup is cyclic of prime order.

Applications

 * Classification of cyclicity-forcing numbers: This is a classification of all natural numbers $$n$$ such that every group of order $$n$$ is cyclic.

Facts similar to ideas used in the proof techniques

 * Union of all conjugates is proper
 * Product of conjugates is proper

Facts used

 * 1) uses::Center is normal
 * 2) uses::Subgroup of index two is normal
 * 3) uses::Finite and any two maximal subgroups intersect trivially implies not simple non-abelian

Proof
Given: A finite non-abelian group $$G$$ such that every proper subgroup of $$G$$ is abelian.

To prove: $$G$$ is not simple.

Proof: We assume that $$G$$ is simple, and derive a contradiction. Let $$n$$ be the number of elements of $$G$$.