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

==Statement==

Suppose <math>G</math> is a finite [[simple non-Abelian group]] and <math>p</math> is a prime dividing the order of <math>G</math>. Let <math>n_p</math> denote the <math>p</math>-[[Sylow number]] of <math>G</math>: the number of <math>p</math>-Sylow subgroups of <math>G</math>. Then, the [[order of a group|order]] of <math>G</math> divides <math>n_p!/2</math>.

==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==

# [[uses::Sylow subgroups exist]]
# [[uses::Prime power order implies not centerless]]
# [[uses::Sylow number equals index of Sylow normalizer]]
# [[uses::Order of simple non-Abelian group divides half the factorial of index of proper subgroup]]

==Proof==

'''Given''': A finite simple non-Abelian group <math>G</math>, a prime <math>p</math> dividing the order of <math>G</math>. <math>n_p</math> is the number of <math>p</math>-Sylow subgroups.

'''To prove''': The order of <math>G</math> divides <math>n_p!/2</math>.

'''Proof''':

# Let <math>P</math> be a <math>p</math>-Sylow subgroup of <math>G</math> (such a <math>P</math> exists by fact (1)).
# <math>P</math> is not normal in <math>G</math>: Since <math>p</math> divides the order of <math>G</math>, <math>P</math> is nontrivial. Thus, <math>P</math> is a nontrivial normal subgroup of <math>G</math>. Also, <math>P</math> cannot equal <math>G</math> because then <math>G</math> 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.
# The subgroup <math>N_G(P)</math> is a proper subgroup of <math>G</math> with index <math>n_p</math>: The index being <math>n_p</math> follows from fact (3). The ''proper'' part follows from the previous step, which concluded that <math>P</math> is not normal in <math>G</math>.
# The order of <math>G</math> divides <math>n_p!/2</math>: This follows from the previous step and fact (4).