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, $G=\mathrm{\Phi }(G)N_G(S)=⟨\mathrm{\Phi }(G)\cup N_G(S)⟩$. But the Frattini subgroup is finite and formed of non-generators, so it follows that $G=⟨N_G(S)⟩=N_G(S)$. 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
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