# 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 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.

## 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).