Is the intersection of Frattini subgroup and a Sylow subgroup contained in the Frattini subgroup of the Sylow subgroup?

Question

Suppose $G$ is a finite group, $P$ is a Sylow p-subgroup of $G$. Is it always true, that $\Phi(G) \cap P$ is a subgroup of $\Phi(P)$? Here $\Phi(G)$ is the Frattini subgroup of $G$.

I managed to solve the problem for the following cases:

1. $P \cong C_{p^n}$ for some $n$. Then, because if $p\mid |G|$, then $p\mid |G/\Phi(G)|$, $\Phi(G) \cap P \cong C_{p^m}$, where $m < n$. Thus it is a subgroup of $\Phi(P)$.
2. $G$ is nilpotent. Then $G$ is the direct product of its Slow subgroups: $G = Syl_{p_1}(G) \times … \times Syl_{p_n}(G)$. Thus $\Phi(G) = \Phi(Syl_{p_1}(G))\times … \times \Phi(Syl_{p_n}(G))$. And that means $\Phi(G) \cap P = \Phi(P)$

However, I do not know, how to solve this problem in general.

Answer 1

A counterexample is a nonsplit extension $G$ of an elementary abelian group $N$ of order $8$ by $H={\rm GL}(3,2)$, with the natural induced action of $H$ on $N$. This is $\mathtt{SmallGroup}(1344,814)$ in the small groups database.

Since the extension is nonsplit, we have $N \le \Phi(G)$ (in fact they are equal). But if we choose $P \in {\rm Syl}_2(G)$, then $N \not\le \Phi(P)$.