Past year paper question:
Let $P$ be a Sylow p-subgroup of a finite group $G$.
Define $Z_G(P)= \{ g \in G \mid pg=gp \quad \forall p \in P \}$
Suppose $x,y \in Z_G(P)$ such that $gxg^{-1}=y$ for some $g \in G$. Show that there exists $n\in N_G(P)$ such that $nxn^{-1}=y$.
I'm guessing I need to to use the fact that $gxg^{-1}=y$, and maybe find some $z$ such that $zgxg^{-1}z^{-1}=y$ and $(zg) \in N_G(P)$. Then $zg$ would be the $n$ that I need to find.
$zgxg^{-1}z^{-1}=y$ would mean that $zyz^{-1}=y$ so $z$ must commute with $y$. Also, $(zg) \in N_G(P)$ would mean that $zgPg^{-1}z^{-1}=P$. But how does one proceed from here?
Some other things:
I get that $yp=py \implies (gxg^{-1})p = (gxg^{-1})p$
I get some easy facts that $P \triangleleft N_G(P), Z_G(P) \leq N_G(P)$, and hence $PZ_G(P)$ is a subgroup of $N_G(P)$. And maybe I need to define some clever group action but I think of one that leads somewhere.
I'll write $C_G(P)$ for the centraliser of $P$ in $G$, which is what you have defined as $Z_G(P)$. We are given $x, y \in C_G(P)$ which are conjugate in $G$ and we want to show that they are in fact conjugate in $N_G(P)$.
So let $y=x^g$ for some $g \in G$ and note that since $x \in C_G(P)$ we have $y=x^g \in C_G(P)^g = C_G(P^g)$. Then $P^g \leq C_G(y)$, thus $P^g$ is a Sylow $p$-subgroup of $C_G(y)$. We also have that $y \in C_G(P)$ implies $P \leq C_G(y)$ so $P$ is also a Sylow $p$-subgroup of $C_G(y)$.
We deduce that $P$ and $P^g$ are conjugate in $C_G(y)$ by Sylow's theorem, so $(P^g)^c = P$ for some $c \in C_G(y)$. Since $(P^g)^c = P^{gc}$ we see that $gc$ normalises $P$, thus $gc \in N_G(P)$ and also $x^{gc}=y^c=y$, so $x, y$ are conjugate in $N_G(P)$, which is what we wanted to prove.