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:

$\left|G\right||n_p!$.

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. Sylow subgroups exist
2. Sylow number equals index of Sylow normalizer
3. Order of simple non-Abelian group divides factorial of index of proper subgroup
4. 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: $\left|G\right||n_p!$.

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).