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

Statement
Suppose $$G$$ is a simple non-Abelian group. Suppose $$p$$ is a prime divisor of the order of $$G$$. Let $$n_p$$ be the $$p$$-Sylow number, i.e., the number of $$p$$-Sylow subgroups of $$G$$. Then, we have:

$$|G| | n_p!$$.

Related facts

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

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

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

To prove: $$|G| | n_p!$$.

Proof:


 * 1) By fact (1), we can find a $$p$$-Sylow subgroup $$P$$ of $$G$$.
 * 2) By fact (2), the number $$n_p$$ of $$p$$-Sylow subgroups of $$G$$ equals the index $$[G:N_G(P)]$$.
 * 3) $$N_G(P)$$ is a proper subgroup of $$G$$: Note that if $$N_G(P) = G$$, then $$P$$ is normal in $$G$$, forcing $$P$$ to be trivial or the whole group $$G$$. But $$P$$ is nontrivial, since $$p$$ divides the order of $$G$$. If $$P = G$$, then $$G$$ is a group of prime power order, which is not simple non-Abelian, because by fact (4), it is not centerless.
 * 4) The order of $$G$$ divides $$[G:N_G(P)]!$$: This follows from fact (3), and the previous step that established that $$N_G(P)$$ is a proper subgroup of $$G$$.
 * 5) The result now follows combining steps (2) and (4).