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

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.

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 $$N \not\le \Phi(P)$$.
• This counterexample raises the question under what conditions would indeed $P \cap \Phi(G) \subseteq \Phi(P)$ for a $P \in Syl_p(G)$. – Nicky Hekster Jan 25 at 21:29