# 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 |