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 $N \not\le \Phi(P)$.

  • 1
    $\begingroup$ This counterexample raises the question under what conditions would indeed $P \cap \Phi(G) \subseteq \Phi(P)$ for a $P \in Syl_p(G)$. $\endgroup$ – Nicky Hekster Jan 25 at 21:29

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