NE implies weakly normal

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This article gives the statement and possibly, proof, of an implication relation between two subgroup properties. That is, it states that every subgroup satisfying the first subgroup property (i.e., NE-subgroup) must also satisfy the second subgroup property (i.e., weakly normal subgroup)
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Statement

Any NE-subgroup of a group is a weakly normal subgroup.

Definitions used

NE-subgroup

Further information: NE-subgroup

A subgroup $H$ of a group $G$ is termed a NE-subgroup of $G$ if the intersection in $G$ of the normalizer $N_{G}(H)$ and the normal closure $H^{G}$ is $H$ itself.

Weakly normal subgroup

Further information: Weakly normal subgroup

A subgroup $H$ of a group $G$ is termed a weakly normal subgroup of $G$ if any conjugate of $H$ that is contained in $N_{G}(H)$ is actually contained in $H$.

Proof

Given: A group $G$, a subgroup $H$ such that $H=H^{G}\cap N_{G}(H)$. A conjugate $H^{g}$ of $H$ such that $H^{g}\leq N_{G}(H)$.

To prove: $H^{g}\leq H$.

Proof: By definition of normal closure, $H^{g}\leq H^{G}$. Thus, we get $H^{g}\leq H^{G}\cap N_{G}(H)=H$, so $H^{G}\leq H$.