Order of simple non-abelian group divides half the factorial of every Sylow number

Statement
Suppose $$G$$ is a finite simple non-Abelian group and $$p$$ is a prime dividing the order of $$G$$. Let $$n_p$$ denote the $$p$$-Sylow number of $$G$$: the number of $$p$$-Sylow subgroups of $$G$$. Then, the order of $$G$$ divides $$n_p!/2$$.

Related facts

 * Order of simple non-Abelian group divides factorial of every Sylow number
 * Order of simple non-Abelian group divides 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

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

Facts used

 * 1) uses::Sylow subgroups exist
 * 2) uses::Prime power order implies not centerless
 * 3) uses::Sylow number equals index of Sylow normalizer
 * 4) uses::Order of simple non-Abelian group divides half the factorial of index of proper subgroup

Proof
Given: A finite simple non-Abelian group $$G$$, a prime $$p$$ dividing the order of $$G$$. $$n_p$$ is the number of $$p$$-Sylow subgroups.

To prove: The order of $$G$$ divides $$n_p!/2$$.

Proof:


 * 1) Let $$P$$ be a $$p$$-Sylow subgroup of $$G$$ (such a $$P$$ exists by fact (1)).
 * 2) $$P$$ is not normal in $$G$$: Since $$p$$ divides the order of $$G$$, $$P$$ is nontrivial. Thus, $$P$$ is a nontrivial normal subgroup of $$G$$. Also, $$P$$ cannot equal $$G$$ because then $$G$$ would be a group of prime power order, hence have a nontrivial center by fact (2), contradicting the assumption that it is simple non-Abelian.
 * 3) The subgroup $$N_G(P)$$ is a proper subgroup of $$G$$ with index $$n_p$$: The index being $$n_p$$ follows from fact (3). The proper part follows from the previous step, which concluded that $$P$$ is not normal in $$G$$.
 * 4) The order of $$G$$ divides $$n_p!/2$$: This follows from the previous step and fact (4).