Abelian permutable complement to core-free subgroup is-self-centralizing

Statement
Let $$G$$ be a group, $$H$$ be a core-free subgroup of $$G$$, and $$N$$ be an Abelian subgroup of $$G$$ such that $$NH = G$$. Then, $$C_G(N) = N$$, i.e., $$N$$ is self-centralizing.

(Note that, for other reasons, it is also true that if $$NH = G$$, then $$N$$ and $$H$$ intersect trivially. This can be shown by the stronger version of the statement that Abelian normal subgroup and core-free subgroup generate whole group implies they intersect trivially).

Related facts
Abelian normal subgroup and core-free subgroup generate whole group implies they intersect trivially

Proof
Given: A group $$G$$, a core-free subgroup $$H$$, an Abelian subgroup $$N$$

To prove: $$C_G(N) = N$$

Proof: Since $$N$$ is an Abelian subgroup, $$N \le C_G(N)$$. Moreover, since $$NH = G$$, we see that if $$nh \in C_G(N)$$, for $$n \in N, h \in H$$, then $$h \in C_G(N)$$. Thus, in order to show that $$C_G(N) = N$$, it suffices to show that no nontrivial element of $$H$$ centralizes every element of $$N$$.

Suppose $$h \in H$$ commutes with every element of $$N$$. Then, I claim that for any $$g \in G$$, $$h \in gHg^{-1}$$. Indeed, any $$g \in G$$, can be written as $$n'h'$$, with $$n' \in N, h' \in H$$. Thus, $$gHg^{-1} = n'Hn'^{-1}$$. But $$h = n'hn'^{-1} \in nHn^{-1} = gHg^{-1}$$.

Thus, $$h$$ is in every conjugate of $$H$$, so $$h$$ is in the normal core of $$H$$. Since $$H$$ is core-free, $$h$$ is the identity element.