# 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 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.

## 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 $[G:H]!$.