Finite and any two maximal subgroups intersect trivially implies not simple non-abelian

Statement
Suppose $$G$$ is a finite group with the property that any two distinct maximal subgroups of $$G$$ intersect trivially. Then, $$G$$ is not a simple non-abelian group.

Direct applications

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

Indirect applications

 * Classification of cyclicity-forcing numbers
 * Classification of abelianness-forcing numbers
 * Schmidt-Iwasawa theorem

Facts used

 * 1) The trivial subgroup is maximal if and only if the group is a group of prime order, which is a simple abelian group.
 * 2) uses::Lagrange's theorem
 * 3) uses::Group acts as automorphisms by conjugation: Thus, conjugates of a maximal subgroup are maximal and all have the same order.
 * 4) uses::Size of conjugacy class of subgroups equals index of normalizer
 * 5) In a finite non-cyclic group, every element is contained in a maximal subgroup. This basically follows from the fact that uses::cyclic iff not a union of proper subgroups.

Proof
We prove the statement by contradiction.

Given: A finite simple non-abelian group $$G$$ of order $$n$$ such that any two distinct maximal subgroups of $$G$$ intersect trivially.

To prove: A contradiction.

Proof: