# proof of Frattini argument

Let $g\in G$ be any element. Since $H$ is normal, $gS{g}^{-1}\subset H$. Since $S$ is a Sylow subgroup of $H$, $gS{g}^{-1}=hS{h}^{-1}$ for some $h\in H$, by Sylow’s theorems. Thus $n={h}^{-1}g$ normalizes $S$, and so $g=hn$ for $h\in H$ and $n\in {N}_{G}(S)$.

Title | proof of Frattini argument |
---|---|

Canonical name | ProofOfFrattiniArgument |

Date of creation | 2013-03-22 13:16:05 |

Last modified on | 2013-03-22 13:16:05 |

Owner | bwebste (988) |

Last modified by | bwebste (988) |

Numerical id | 4 |

Author | bwebste (988) |

Entry type | Proof |

Classification | msc 20D20 |