# 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 SubgroupsContainingTheNormalizersOfSylowSubgroupsNormalizeThemselves 2013-03-22 13:16:47 2013-03-22 13:16:47 bwebste (988) bwebste (988) 6 bwebste (988) Corollary msc 20D20