If abelian $P\in{\rm Syl}_p(G)$, and $H\le P$ and $H^g\le P$, show $g\in G$ is the left product of an element of $N_G(P)$ with an element of $C_G(H)$ 
If abelian $P \in{\rm Syl}_p(G)$, and $H \leq P$ and $H^g \leq P$, show $g \in G$ is the left product of an element of $N_G(P)$ with an element of $C_G(H)$

First, I set $g=nc$, where I hope to find that $n \in N_G(P)$ and $c \in C_G(H)$. I am trying to do this by contradiction, assuming either $n \notin N_G(P)$ (which implies $n \notin N_G(H) )$ or $c \notin C_G(H)$. In the first cast ($n \notin N_G(P)$, $c \in C_G(H)$), we have that $H^g = (H^n)^c \neq H^c = H$. However, $g$ is arbitrary (as far as I know) in $G$, so I can not use that $P$ is abelian to find that $H^g = H$. I know that $P$ must be the unique Sylow $p$-subgroup but I can't see how that will help me.
I am not sure where to go from here.
 A: In his (unfortunately deleted) answer, Arturo Magidin proved that $g = cn$ with $c \in C := C_G(H)$ and $n \in N:= N_G(P)$.
I have found a counterexample that shows that it is not necessarily the case that $g=nc$ for $c \in C$ and $n \in N$.
Let $G=A_7$ and $P = \langle (1,2,3), (4,5,6) \rangle \in {\rm Syl}_3(G)$, and let $H = \langle (1,2,3) \rangle \in P$.
Note that I am taking $H^g$ to mean $g^{-1}Hg$ and I compose permutations from left to right.
Then with $g = (1, 6)(2, 4)(3, 5, 7)$, we have $H^g = \langle (4,5,6) \rangle \le P$.
We have $g = cn$ with $c = (1, 3, 2)(4, 6)(5, 7) \in C$ and $n = (1, 4)(2, 5, 3, 6) \in N$, but it can be checked that $g \not\in NC$. (I checked this on the computer. Since $|NC|=|CN| = 144$, it could be tedious to check it by hand.)
A: Note. I had run into an issue when writing this, then mistakenly assumed $g$ fixed $H$ set-wise and that got me the right expression and a couple of comments below. When I fixed that error, I could not get the expression of $g$ as a product in the order in which I interpret the original question to be requiring: the product of an element in the normalizer times an element in the centralizer. Derek Holt has given a counterexample to show that it may be impossible to achieve that; since he refers to my answer, I'm adding this note and undeleting.

Following on Derek Holt’s suggestion: note that $H\leq P$, and $H\leq P^{g^{-1}}$. Since $P$ is abelian, $P\leq C_G(H)$, and since $P^{g^{-1}}$ is also abelian, $P^{g^{-=1}}\leq C_G(H)$.
Now note that $P^{g^{-1}}$ and $P$ are both Sylow $p$-subgroups of $C_G(H)$ (you should prove this). Therefore, they are conjugate in $C_G(H)$; that is, there exists $c\in C_G(H)$ such that $P^{g^{-1}c}=P$. Therefore, $g^{-1}c\in N_G(P)$.
But that gives the wrong product order: $g^{-1}c\in N_G(P)$, so $c^{-1}g=n$ for some $N\in N_G(P)$, $g = cn$.
