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Let $G$ be a finite group and let $H\leq G$ and $g \in G$. Suppose that $H, H^g \leq P$ such that $H \unlhd N_G(P$) and $H^g \unlhd N_G(P)$, where $P$ is the Sylow $p$ subgroup of $G$. Then $H$ and $H^g$ are conjugate in $N_G(P)$.

Suppose that $H^x = H^g$ for some $g\in G$. I need to show that $x \in N_G(P)$. Now $(N_G(H))^x = N_G(H^x) = N_G(H^g)$. Since $H \unlhd P$, we have that $P \leq N_G(H)$, and so $P^x \leq (N_G(H))^x = N_G(H^g)$. Also $H^g \unlhd P$, so $P \leq N_G(H^g)$. By Sylow's Theorem, there exists a $y \in N_G(H^g)$ such that $yPy^{-1} = xPx^{-1}$. Hence $y^{-1}x \in N_G(P)$. I can't seem to show that $x\in N_G(P)$ as desired. Is this correct way to prove it?

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    $\begingroup$ Why don't you say "I am trying to prove that $H$ and $H^g$ are conjugate in $N_G(P)$"? If you just state it, it is not clear whether this is a result that you already know, or whether this is what you are asking about. As Robert Chamberlain says in his answer, your statement is any case confusing because it is equivalent to saying that $H = H^g$. $\endgroup$ – Derek Holt May 14 '17 at 9:28
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So, the confusion is that you don't need to show that $x\in N_G(P)$ whenever $H^x=H^g$. You need to show $H^x=H^g$ for some $x\in N_G(P)$.

The way I would prove this is as follows:

We have $H^g\trianglelefteq N_G(P)$ so for $p\in P$, since $P\le N_G(P)$, $H^{gp}=H^g$. Therefore $gpg^{-1}\in N_G(H)$, so $P^{g^{-1}}\in N_G(H)$. Also $H\trianglelefteq N_G(P)$, so $P\le N_G(P)\le N_G(H)$. Thus by Sylow's Theorem there is some $x\in N_G(H)$ with $P^x=P^{g^{-1}}$. This gives $P^{xg}=P$ so $xg\in N_G(P)\le N_G(H)$, hence $H=H^{xg}=H^g$.

$H$ and $H^g$ are equal so definitely conjugate.

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