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 $H^g \le N_G(H)$.

To prove: $H^g \le H$.

Proof: By definition of normal closure, $H^g \le H^G$. Thus, we get $H^g \le H^G \cap N_G(H) = H$, so $H^G \le H$.