Let $G$ be some finite group and $p$ some prime divisor of the order of $G$. Suppose $O_p(G) = 1$, i.e. the largest normal subgroup with order $p$ is trivial, and $G / N$ is a $p$-group for some normal subgroup $N$. Then $\Phi(N) = \Phi(G)$, i.e. the Frattini subgroups of $N$ and $G$ are equal.
This is an exercise from The Theory of Finite Groups by Kurzweil and Stellmacher.
By an application of the Frattini argument we find that every Sylow subgroup of $\Phi(G)$ is normal in $G$, hence as $O_p(G) = 1$ we know that $p$ does not divide the order of $\Phi(G)$. Also as $G/N$ is a $p$-group we can deduce $\Phi(G) \le N$. Now maybe it is true that if the Frattini subgroup of the whole group is contained in some normal subgroup $N$, then $\Phi(G) \le \Phi(N)$ might hold. I tried to show that $\Phi(G) \le M$ for every maximal subgroup $M$ of $N$, for if not we have some $x \in \Phi(G)$ such that $\langle x, M \rangle = N$. But that is all I can deduce, I do not see how to get a contradiction from that…
The other inclusion is a previous exercise, which I solved. For if $N \unlhd G$, then $\Phi(N) \unlhd G$. Let $M \le G$ be some maximal subgroup, if $\Phi(N)$ is not contained in $M$ then $M\Phi(N) = G$, hence $N = N \cap M\Phi(N) = (N\cap M)\Phi(N)$ by Dedekinds modular law, which implies $N = N\cap M$, or $N \le M$, which gives $\Phi(N) \le M$ contrary to our assumption. Hence $\Phi(N) \le M$ for every maximal subgroup $M$ of $G$.
So any suggestions how to show the implication $\Phi(G) \le \Phi(N)$?
Best Answer
Since $\Phi(N) \le \Phi(G)$, we have $\Phi(N/\Phi(N))=1$ and $\Phi(G/\Phi(N)) = \Phi(G)/\Phi(N)$.
Assume for a contradiction that $\Phi(G) \ne \Phi(N)$, and so $\Phi(G/\Phi(N)) \ne 1$. Let $M/\Phi(N)$ be a minimal normal subgroup of $\Phi(G)/\Phi(N)$. Then $M/\Phi(N)$ is an elementary abelian $q$-subgroup of $G$ for some prime $q$ and, since we know that $p$ does not divide $|\Phi(G)|$, we have $q \ne p$. Since $\Phi(G) \le N$, we have $M \le N$.
Since $\Phi(N/\Phi(N)) = 1$, $M/\Phi(N)$ has a complement in $N/\Phi(N)$ by Exercise 2 on page 76 of the book. Now by Gaschütz's Theorem (3.3.2 in the book), since $|M/\Phi(N)|$ and $|G:N|$ are coprime, $M/\Phi(N)$ has a complement in $G/\Phi(N)$. But that contradicts $M/\Phi(N) \le \Phi(G/\Phi(N))$.