# Conjugate elements of G that are in the center of a Sylow p-subgroup

I've had a lot of trouble with this question, as well as my advisor.

Let x and y be two elements of $Z(P)$, where $P$ is a Sylow $p$-subgroup of $G$. If $x$ and $y$ are conjugate in $G$, prove that $x$ is conjugate to $y$ in $N_G(P)$, the normalizer of $P$ in $G$.

-

Hint: Use Sylow's theorem in $C_G(y)$, the centralizer of $y$ in $G$.
If $y = x^g$ and $x,y$ are in $Z(P)$, then $C_G(y)$ contains both $P$ (because $y \in Z(P)$) and $P^g$ (since $y=x^g \in Z(P^g)$). By Sylow's theorem in $C_G(y)$, $P^g = P^z$ for some $z \in C_G(y)$, so $gz^{-1} \in N_G(P)$ and $x^{gz^{-1}} = y^{z^{-1}} = y$.
Above, $\,z\in C_G(y)\,$, right? –  DonAntonio Aug 13 '12 at 15:48