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$\DeclareMathOperator\Syl{Syl}$Let $G$ be a finite group, $\Phi(G)$ is the Frattini subgroup of $G$. And $G/\Phi(G)$ is a simple group. Let $P\in \Syl_{p}(\Phi(G))$, where $p\in \pi(\Phi(G))$ and $\pi(\Phi(G))$ is the set of prime divisors of $|\Phi(G)|$.

Suppose that every subgroup $H$ of $P$ of order $p$ is normal in $G$.

Let $P_1\in \Syl_{p}(G)$. Then for any subgroup $H$ of $P$ with $|H|=p$, $H\leq Z(P_1)$. So $P_1\leq C_G(H)\unlhd G$.

Under the above conditions, can we use the simplicity of $G/\Phi(G)$ to conclude that $H\leq Z(G)$?

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  • $\begingroup$ (Typing tip: follow each punctuation mark (comma, period etc) with a space.) $\endgroup$
    – YCor
    Commented May 9, 2022 at 13:06
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    $\begingroup$ Any normal subgroup of order $p$ in a finite $p$-group $P$ lies in the centre of $P$. This is hardly a research problem! $\endgroup$
    – Derek Holt
    Commented May 9, 2022 at 13:12
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    $\begingroup$ I have added some paragraph breaks to make it easier to read. You have not said what $H$ is in the third paragraph. Presumably it is a subgroup of $P$ with $|H|=p$? $\endgroup$
    – Derek Holt
    Commented May 9, 2022 at 13:15
  • $\begingroup$ Yes.In the third paragraph, $H$ is a subgroup of $P$ with |$H$|=$p$. $\endgroup$
    – Bob
    Commented May 9, 2022 at 13:21

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Yes we can.

If $P_1=P$ then, by the Schur-Zassenhaus Theorem,$P$ has a complement in $G$, contradicting $P \le \Phi(G)$.

Otherwise $P_1\Phi(G)/\Phi(G)$ is a nontrivial Sylow $p$-subgroup of the simple group $G/\Phi(G)$, and its conjugates generate $G/\Phi(G)$, so $C_G(H)\Phi(G) = G$, and hence $C_G(H)=G$.

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