Weakly normal implies weakly closed in intermediate nilpotent

Statement with symbols
Suppose $$H \le K \le G$$ are groups such that:


 * $$H$$ is a fact about::weakly normal subgroup of $$G$$.
 * $$K$$ is a fact about::nilpotent group.

Then, $$H$$ is a fact about::weakly closed subgroup of $$K$$.

Related facts

 * Weakly normal implies intermediately subnormal-to-normal

Corollaries

 * Paranormal implies weakly closed in intermediate nilpotent
 * Pronormal implies weakly closed in intermediate nilpotent

Definitions used
For these definitions, $$H^g = g^{-1}Hg$$ denotes the conjugate subgroup by $$g \in G$$. (This is the right-action convention; however, adopting a left-action convention does not alter any of the proof details).

Paranormal subgroup
A subgroup $$H$$ of a group $$G$$ is termed paranormal in $$G$$ if for any $$g \in G$$, $$H^g \le N_G(H)$$ implies $$H^g \le H$$. In other words, it is weakly closed in its normalizer.

Weakly closed subgroup
Suppose $$H \le K \le G$$ are groups. We say $$H$$ is weakly closed in $$K$$ with respect to $$G$$ if, for any $$g \in G$$ such that $$H^g \le K$$, we have $$H^g \le H$$.

Facts used

 * 1) uses::Weakly normal implies intermediately subnormal-to-normal
 * 2) uses::Nilpotent implies every subgroup is subnormal

Proof
Given: $$H \le K \le G$$ with $$H$$ a paranormal subgroup of $$G$$ and $$K$$ a nilpotent group.

To prove: For any $$g \in G$$ such that $$H^g \le K$$, we have $$H^g \le H$$.

Proof: By fact (2), $$H$$ is a subnormal subgroup of $$K$$. By fact (1), $$H$$ is therefore normal in $$K$$. Thus, $$K \le N_G(H)$$.

Since $$H$$ is weakly closed in $$N_G(H)$$ by definition, whenever $$H^g \le N_G(H)$$, we get $$H^g \le H$$. In particular, whenever $$H^g \le K$$, we get $$H^g \le H$$, so $$H$$ is weakly closed in $$K$$.