$\DeclareMathOperator\Syl{Syl}$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 $H\leq Z(G)$?



The center of Sylow subgroups
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 $C\_G(H)=G$.


