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

Statement
Let $$G$$ be a simple non-Abelian group and $$H$$ be a proper subgroup of finite index in $$G$$. Then, $$G$$ is finite and the order of $$G$$ divides $$[G:H]!$$: the factorial of the index $$[G:H]$$.

Related facts

 * Poincare's theorem
 * Order of simple non-Abelian group divides factorial of every Sylow number
 * Order of simple non-Abelian group divides half the factorial of index of proper subgroup
 * 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

Stronger facts

 * Order of simple non-Abelian group divides half the 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 symmetric group on left coset space of proper subgroup
 * 2) uses::Lagrange's theorem

Proof
Given: A simple non-Abelian group $$G$$, a proper subgroup $$H$$ of finite index.

To prove: $$G$$ is finite and the order of $$G$$ divides $$[G:H]!$$.

Proof:


 * 1) By fact (1), $$G$$ is isomorphic to a subgroup of $$\operatorname{Sym}(G/H)$$.
 * 2) By fact (2), the order of $$G$$ divides the order of $$\operatorname{Sym}(G/H)$$, which is $$[G:H]!$$.