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

This article gives the statement, and possibly proof, of a particular subgroup of kind of subgroup in a group being self-centralizing. In other words, the centralizer of the subgroup in the group is contained in the subgroup
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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, $N\leq C_{G}(N)$. 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.