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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 conclude that $H\leq Z(G)$?

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

$\DeclareMathOperator\Syl{Syl}$Let $G$ be a finite group, $\Phi(G)$ is the Frattini subgroup of $G$. 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 $P_1\in \Syl_{p}(G)$. We can use $|H|=p$ and $H$ is normal in $G$ to prove that $H\leq P_1$. Then can we prove $H\leq Z(P_1)$?

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

$\DeclareMathOperator\Syl{Syl}$Let $G$ be a finite group, $\Phi(G)$ is the Frattini subgroup of $G$. 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 every subgroup $H$ of $P$ of order $p$ is normal in $G$. Let $P_1\in \Syl_{p}(G)$, we can use $|H|=p$ and $H$ is normal in $G$ to prove that $H\leq P_1$. Then can we prove $H\leq Z(P_1)$?

$\DeclareMathOperator\Syl{Syl}$Let $G$ be a finite group, $\Phi(G)$ is the Frattini subgroup of $G$. 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 $P_1\in \Syl_{p}(G)$. We can use $|H|=p$ and $H$ is normal in $G$ to prove that $H\leq P_1$. Then can we prove $H\leq Z(P_1)$?