Intermediately operator preserves normalizer-closedness

Statement
Suppose $$p$$ is a fact about::normalizer-closed subgroup property and $$q$$ is a subgroup property obtained by applying the fact about::intermediately operator to $$p$$. Then, $$q$$ is also a normalizer-closed subgroup property.

Proof idea
The key idea is that if $$N_G(H) \le K \le G$$, then $$N_K(H) = N_G(H)$$.

Proof details
Given: A normalizer-closed property $$p$$. A group $$G$$ with a subgroup $$H$$ such that $$H$$ satisfies property $$p$$ in every intermediate subgroup of $$G$$.

To prove: $$N_G(H)$$ satisfies property $$p$$ in every intermediate subgroup of $$G$$.

Proof: Let $$K$$ be a subgroup of $$G$$ containing $$N_G(H)$$. Our goal is to show that $$N_G(H)$$ satisfies property $$p$$ in $$K$$.

By assumption, $$H$$ satisfies property $$p$$ in $$K$$. Further, $$N_K(H) = K \cap N_G(H) = N_G(H)$$ (since $$N_G(H) \le K$$). Thus, since $$p$$ is normalizer-closed, $$N_G(H) = N_K(H)$$ also satisfies property $$p$$ in $$K$$, completing the proof.