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, $N_G(M)=N_G(S)M=M$.∎