# Weakly closed implies normalizer-relatively normal

This article gives the statement and possibly, proof, of an implication relation between two subgroup-of-subgroup properties. That is, it states that every subgroup-of-subgroup satisfying the first subgroup-of-subgroup property (i.e., weakly closed subgroup) must also satisfy the second subgroup-of-subgroup property (i.e., normalizer-relatively normal subgroup)
View all subgroup-of-subgroup property implications | View all subgroup-of-subgroup property non-implications
Get more facts about weakly closed subgroup|Get more facts about normalizer-relatively normal subgroup

## Statement

Suppose $H \le K \le G$ are groups such that $H$ is a Weakly closed subgroup (?) of $K$ relative to $G$. Then, $H$ is a Normalizer-relatively normal subgroup (?) of $K$ relative to $G$: in other words, $H$ is a normal subgroup of $N_G(K)$.

## Proof

Given: $H \le K \le G$ such that $H$ is weakly closed in $K$ relative to $G$.

To prove: $H$ is normal in $N_G(K)$.

Proof: For $g \in N_G(K)$, $gHg^{-1} \le gKg^{-1} = K$. Thus, $gHg^{-1} \le K$, so by the condition of being weakly closed, $gHg^{-1} \le H$. Thus, $H$ is normal in $N_G(K)$.