p-element centralizing a Sylow p-subgroup

Question

Let $G$ be a finite group, $P$ a Sylow $p$-subgroup for a prime $p$ and $g$ a $p$-element with $gxg^{-1} = x$ for all $x \in P$. Then $g \in Z(P)$.

Is this true? How can i prove that $g \in P$ ?

Answer 1

In general the following holds:

Proposition Let $G$ be a finite group and $p$ a prime dividing the order of $G$. If $P$ is a $p$-subgroup of $G$ and $S \in Syl_p(G)$, then $C_P(S)=P \cap Z(S)$.

Proof It is clear that any element of $P \cap Z(S)$ centralizes $S$. So assume $x \in C_P(S)$. Then $x$ centralizes $S$, whence $S\langle x \rangle$ is a subgroup of $G$. Since $x$ is a $p$-element, $|S\langle x \rangle|=\frac{|S| \cdot |\langle x \rangle|}{|S \cap \langle x \rangle|}$, which is a power of $p$. Since $S$ is Sylow, and $S \subseteq S\langle x \rangle$, it follows that in fact $S=S\langle x \rangle$. And this implies $x \in S$. But $x$ centralized $S$, meaning $x \in Z(S)$.

Note: one can prove the following useful similar statement for normalizers.

Proposition Let $G$ be a finite group and $p$ a prime dividing the order of $G$. If $P$ is a $p$-subgroup of $G$ and $S \in Syl_p(G)$, then $N_P(S)=P \cap S.$