the Frattini subgroup of the Fitting subgroup of a group whose Frattini subgroup is trivial

Question

$G$ is a finite group. If $\Phi(G)=1$, do we have $\Phi(F(G))=1$?

In general, I think, for a normal subgroup $N$ of $G$, we have $\Phi(N)\le \Phi(G)$. But I was stuck.

Let $M \le G$ be some maximal subgroup. We want to prove that $\Phi(N)$ is contained in every maximal subgroup of $G$. If $\Phi(N)\le M$, then we are done. If $\Phi(N)$ is not contained in $M$ then since $M$ is maximal in $G$, we have $M\Phi(N) = G$. Hence by the Dedekind modular law we have $N = N \cap M\Phi(N) = (N\cap M)\Phi(N)$. But what to do next? Thank you!

Answer 1

No, you are not "done" if $\Phi(N) \leq M$.

From the top: There are two possibilities. Either $\Phi(N)$ is contained in every maximal subgroup of $G$, in which case $\Phi(N)$ is also contained in the intersection of all those maximals which is by definition the Frattini subgroup of $G$ (and now you are done)

or

there exists a maximal subgroup $M$ such that $\Phi(N)$ is not contained in $M$. But $\Phi(N)$ is normal in $G$ (because it is characteristic in $N$ which is normal in $G$) so $M\Phi(N)$ is a group which properly contains $M$ and thus $G=M\Phi(N)$. By Dedekind's lemma we now have $N = (N \cap M)\Phi(N)$ and by the standard property which says that the Frattini subgroup is the set of non-generators this yields $N=N\cap M$ and thus $N \leq M$. This is a contradiction, however, since $\Phi(N) \leq N \leq M$, but we are assuming that $\Phi(N) \nleq M$.

Answer 2

I’ve understood and accepted @the_fox’s answer. I want to write something for myself.

I was stuck exactly because I didn’t know the idea of non-generator and the fact that the Frattini subgroup of a group is the set of all non-generators of this group. I searched on this site, but nearly all of the proofs used Zorn’s Lemma. I want to add a proof in the finite case without using the lemma.

Define $X$ to be the set of all non-generators of $G$. We want to prove $X=\Phi(G)$.

Since $1$ is a non-generator of $G$, $X$ is not empty.

For any $x\in X$ and any maximal subgroup $M$ of $G$, by the definition of non-generator, $\langle M,x\rangle=M$; otherwise, if $M<\langle M, x\rangle$, then $\langle M,x\rangle=G$ (since $M$ is maximal), contradicting the definition of non-generator. Hence $x\in M$ and it implies $X\subseteq \Phi(G)$.

For any $g\in \Phi(G)$ and any maximal subgroup $M$ of $G$, since $g\in M$, we have $\langle M,g\rangle=\langle M\rangle=M\neq G$. For any subgroup $H$ of $G$, $H$ is contained in at least one of the maximal subgroups of $G$. Without generality, $H\le M$. Then $\langle H,g\rangle\le\langle M,g\rangle\neq G$. Hence $g$ is a non-generator and it implies $\Phi(G)\subseteq X$.

Thus $X=\Phi(G)$.