Sylow $p$-subgroup of $Z(G)$

Question

$G$ be a finite group, and let $p$ divides $|Z(G)|$, then show that each Sylow $p$-subgroup of $G$ contains the Sylow $p$-subgroup of $Z(G)$ .

Answer 1

Let $\;P_Z\;$ be the Sylow $\;p$-subgroup of $\;Z(G)\;$ (why does it have only one?), and let $\;S_P\;$ be the set of all Sylow $\;p-$ subgroups of $\;G\;$.

By Sylow's Theorems,

$$\exists\,Q\in S_P\;\;s.t.\;\;P_Z\le Q$$

But we also know that

$$\forall\,P\in S_p\;\;\exists g\in G\;\;s.t.\;\; P= Q^g:= g^{-1}Qg$$

and since $\;P_Z\lhd G\;$ (why?), we finally get

$$P_Z=P_Z^g\le Q^g=P$$