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. Simple non-Abelian group is isomorphic to subgroup of symmetric group on left coset space of proper subgroup
2. 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 $[G:H]!$.