Order of simple non-abelian group divides half the factorial of index of proper subgroup

Statement
Suppose $$G$$ is a simple non-Abelian group and $$H$$ is a proper subgroup of finite index $$n$$. Then, $$G$$ is finite and the order of $$G$$ divides $$n!/2$$.

Related facts

 * Simple non-Abelian group is isomorphic to subgroup of symmetric group on left coset space of proper subgroup
 * Simple non-Abelian group is isomorphic to subgroup of alternating group on left coset space of proper subgroup of finite index
 * Order of simple non-Abelian group divides half the factorial of every Sylow number

Weaker facts

 * Order of simple non-Abelian group divides factorial of index of proper subgroup

Related survey articles
Small-index subgroup technique: The use of this and other results to show that groups satisfying certain conditions (e.g., conditions on the order) cannot be simple.

Facts used

 * 1) uses::Simple non-Abelian group is isomorphic to subgroup of alternating group on left coset space of proper subgroup of finite index
 * 2) uses::Lagrange's theorem

Proof
The proof follows directly from facts (1) and (2).