Control of fusion of p-elements in a subgroup containing a Sylow p-subgroup I need help on the following problem - my first attempt came to a dead end and I'm not sure where else to go. First, a definition:
Definition (control of fusion): Suppose $P \le H \le G$. Then $H$ controls fusion in $P$ with respect to $G$ if whenever two elements of $P$ are conjugate in $G$ then they are already conjugate in $H$.
Another way to phrase this: whenever $x,g \in G$ satisfy $x,x^g \in P$ then $g=ch$ for some $c \in C_G(x)$ and $h \in H$.
Problem: Let $H \le G$. Suppose that $H$ contains a Sylow $p$-subgroup $P$ of $G$ and that $H$ controls fusion in $P$ with respect to $G$. Then any two $p$-elements of $H$ that are conjugate in $G$ are already conjugate in $H$.
Idea for proof: Since $H$ controls fusion in $P$ with respect to $G$, this is already true for $p$-elements in $P$. So, consider $s \in H \setminus P$ where $s$ is a $p$-element and let $g \in G$ be an element such that $s,s^g \in H$. Since all Sylow $p$-subgroups of $G$ are conjugate, there exists some $k \in G$ such that $s^k \in P$ and some $\ell$ such that $(s^g)^\ell \in P$. Now I want to send both $s$ and $s^g$ into $P$ where we know $H$ controls fusion, use the second version of the definition above to rewrite that element as a product of an element of $C_G(s)$ and $H$, then conjugate again to pull them back into $H \setminus P$ and show that this factorization still holds there.
The problem is that there isn't necessarily one element that conjugates $s$ and $s^g$ into $P$. Can this outline of an idea be made to work, and if not what other path can be taken toward a proof?
 A: Note that since $P\leq H$, then $P$ is a Sylow $p$-subgroup of $H$, not merely of $G$.
Let $x\in H$ be a $p$-element, and suppose that there exists $g\in G$ such that $x^g\in H$. Now,  since $x^g$ is a $p$-element of $H$, and $P$ is a Sylow $p$-subgroup of $H$, we know that there is an $H$-conjugate of $P$ that contains $x^g$; that is, there exists $h\in H$ such that $x^g\in hPh^{-1}$. Therefore, $x^{gh}\in P$. 
On the other hand, since $x$ is a $p$-element, we also know that there is an $H$-conjugate of $x$ that is in $P$; that is, there exists $k\in H$ such that $x^k\in P$.
Now, $x^{gh}$ and $x^k$ are both in $P$, and they are clearly conjugate to each other in $G$. Since $H$ controls fusion in $P$, there exists $y\in H$ such that $x^{gh}=x^{ky}$.
But this gives $x^g = x^{kyh^{-1}}$, with $kyh^{-1}\in H$. That is, $x^g$ is conjugate to $x$ in $H$, as desired.

To fix your argument, all you need to do is observe that $k$ and $\ell$ can actually be chosen in $H$, since $P$ is a Sylow $p$-subgroup of $H$. Now you know that $s^k$ and $(s^g)^{\ell}$ lie in $P$, and they are conjugate to each other in $G$, hence conjugate to each other in $H$ by the control of fusion. 
I’m not sure why you want to “pull back to $H\setminus P$”. All you need is to make sure that $s$ and $s^g$ are conjugate in $H$, and if some $H$-conjugate of $s$ is $H$-conjugate to an $H$-conjugate of $s^g$, then that will do it. 
