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

Statement
Suppose $$G$$ is a group, $$N$$ is an Abelian normal subgroup and $$H$$ is a core-free subgroup such that $$NH = G$$. Then, $$N \cap H$$ is trivial.

Core-free subgroup
A subgroup of a group is termed core-free if its normal core in the whole group is trivial, or equivalently, if it does not contain any nontrivial normal subgroup.

Applications

 * Abelian minimal normal subgroup and core-free maximal subgroup are permutable complements

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

Proof
Given: A group $$G$$, an Abelian normal subgroup $$N$$, and a core-free subgroup $$H$$ such that $$NH = G$$

To prove: If $$x \in N \cap H$$ then $$x$$ is the identity element

Proof: We first show that $$x$$ is in every conjugate of $$H$$. For this, pick $$g \in G$$ and consider $$gHg^{-1}$$.

Since $$G = NH$$, we can find $$n \in N, h \in H$$ such that $$g = nh$$. Then, $$gHg^{-1} = nhHh^{-1}n^{-1} = nHn^{-1}$$. Further, since $$N$$ is Abelian, $$x = nxn^{-1} \in nHn^{-1}$$, so $$x \in gHg^{-1}$$.

Thus, the element $$x$$ is in every conjugate of $$H$$, so $$x$$ is in the normal core of $$H$$, but by assumption, the normal core of $$H$$ is trivial. Hence $$x = e$$.