Frattini subgroup is nilpotent in finite

Verbal statement
The Frattini subgroup of any finite group is nilpotent.

Symbolic statement
Let $$G$$ be a finite group and $$\Phi(G)$$ denote the intersection of all maximal subgroups of $$G$$ (the so-called Frattini subgroup of $$G$$).

Frattini subgroup
The Frattini subgroup of a (here, finite) group is the intersection of all its maximal subgroups.

Finite nilpotent group
A finite group is nilpotent if every Sylow subgroup of it is normal.

Generalizations
The result generalizes in two important respects. First, we can prove considerably stronger results about Frattini subgroups for finite groups, and more generally, for groups in which every proper subgroup is contained in a maximal subgroup. Second, we can generalize to proving the results about any Frattini-embedded normal subgroup of an arbitrary group.

For instance:


 * Frattini-embedded normal-realizable implies inner-in-automorphism-Frattini
 * Frattini-embedded normal-realizable implies ACIC

Proof
Given: A finite group $$G$$, and $$\Phi(G)$$ is the Frattini subgroup

To prove: For any Sylow subgroup $$P$$ of $$\Phi(G)$$, $$P$$ is normal in $$\Phi(G)$$.

Proof: In fact, we shall show that $$P$$ is normal in $$G$$.

Here's the idea. By applying Frattini's argument and the fact that $$\Phi(G) \triangleleft G$$, we have $$\Phi(G)N_G(P) = G$$. Now if $$N_G(P) \ne G$$, it is contained in a maximal subgroup $$M$$ of $$G$$. Since $$\Phi(G)$$ is contained in every maximal subgroup, $$\Phi(G)N_G(P) \le M$$, leading to a contradiction.

Textbook references

 * , Exercise 25, Page 199 (Section 6.2)