subgroups containing the normalizers of Sylow subgroups normalize themselves

Let $G$ be a finite group, and $S$ a Sylow subgroup. Let $M$ be a subgroup such that $N_G(S)\subset M$. Then $M=N_G(M)$.

Proof.

By order considerations, $S$ is a Sylow subgroup of $M$. Since $M$ is normal in $N_G(M)$, by the Frattini argument, $N_G(M)=N_G(S)M=M$. ∎

Title	subgroups containing the normalizers of Sylow subgroups normalize themselves
Canonical name	SubgroupsContainingTheNormalizersOfSylowSubgroupsNormalizeThemselves
Date of creation	2013-03-22 13:16:47
Last modified on	2013-03-22 13:16:47
Owner	bwebste (988)
Last modified by	bwebste (988)
Numerical id	6
Author	bwebste (988)
Entry type	Corollary
Classification	msc 20D20