# Frattini subgroup of a finite group is nilpotent, the

The Frattini subgroup^{} of a finite group^{} is nilpotent (http://planetmath.org/NilpotentGroup).

###### Proof.

Let $\mathrm{\Phi}(G)$ denote the Frattini subgroup of a finite group $G$. Let $S$ be a Sylow subgroup of $\mathrm{\Phi}(G)$. Then by the Frattini argument, $G=\mathrm{\Phi}(G){N}_{G}(S)=\u27e8\mathrm{\Phi}(G)\cup {N}_{G}(S)\u27e9$. But the Frattini subgroup is finite and formed of non-generators, so it follows that $G=\u27e8{N}_{G}(S)\u27e9={N}_{G}(S)$. Thus $S$ is normal in $G$, and therefore normal in $\mathrm{\Phi}(G)$. The result now follows, as any finite group whose Sylow subgroups are all normal is nilpotent (http://planetmath.org/ClassificationOfFiniteNilpotentGroups). ∎

In fact, the same proof shows that for any group $G$, if $\mathrm{\Phi}(G)$ is finite then $\mathrm{\Phi}(G)$ is nilpotent.

Title | Frattini subgroup of a finite group is nilpotent, the |
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Canonical name | FrattiniSubgroupOfAFiniteGroupIsNilpotentThe |

Date of creation | 2013-03-22 13:16:44 |

Last modified on | 2013-03-22 13:16:44 |

Owner | yark (2760) |

Last modified by | yark (2760) |

Numerical id | 16 |

Author | yark (2760) |

Entry type | Theorem |

Classification | msc 20D25 |