Every finite non-solvable group has a minimal simple group as subquotient

Statement
Suppose $$G$$ is a finite group that is not solvable. Then, there exists a fact about::minimal simple group arising as a subquotient of $$G$$.

Facts used

 * 1) uses::Subquotient is transitive: A subquotient of a subquotient is a subquotient.

Proof
Given: A finite non-solvable group $$G$$.

To prove: There exists a minimal simple group arising as a subquotient of $$G$$.

Proof: We prove this by induction on the order of $$G$$. We assume that the result has been proved for all groups of order strictly less than the order of $$G$$. We make three cases:


 * 1) $$G$$ is itself a minimal simple group -- it is simple non-Abelian and every proper subgroup is solvable: In this case, $$G$$ is itself the desired subquotient.
 * 2) $$G$$ is not simple: Since $$G$$ is not solvable by assumption, $$G$$ must have a composition factor that is a simple non-Abelian group. Since any composition factor is a subquotient, $$G$$ has a subquotient of smaller size that is also not solvable. By induction, this subquotient has a minimal simple group as subquotient, so by fact (1), $$G$$ has a minimal simple subquotient.
 * 3) $$G$$ is simple but not minimal simple: Since $$G$$ is not solvable, $$G$$ must be a simple non-Abelian group. Since it is not minimal simple by assumption, there is some subgroup of $$G$$ that is not solvable. By the inductive assumption, this subgroup has a minimal simple subquotient, so by fact (1), $$G$$ has a minimal simple subquotient.