Pronormal implies weakly closed in intermediate nilpotent

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


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

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

Related facts

 * Pronormal implies intermediately subnormal-to-normal

Stronger facts

 * Paranormal implies weakly closed in intermediate nilpotent: Paranormality is a somewhat weaker assumption than pronormality, and hence the corresponding result is stronger. The proof is almost the same.

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).

Pronormal subgroup
A subgroup $$H$$ of a group $$G$$ is termed paranormal in $$G$$ if for any $$g \in G$$, there exists $$x \in \langle H, H^g \rangle$$ such that $$H^x = H^g$$.

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::Pronormal implies intermediately subnormal-to-normal
 * 2) uses::Nilpotent implies every subgroup is subnormal
 * 3) uses::Normality satisfies intermediate subgroup condition

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 $$K$$.

Now suppose $$g \in G$$ is such that $$H^g \le K$$. Then, $$\langle H, H^g \rangle \le K$$. By fact (3), $$H$$ is normal in $$\langle H, H^g \rangle$$. Thus, for any $$x \in \langle H, H^g \rangle$$, we have $$H^x \le H$$.

By the definition of pronormality, we also have $$x \in \langle H, H^g \rangle$$ such that $$H^x = H^g \rangle$$. Since $$H^x \le H$$, we get $$H^g \le H$$, completing the proof.

Textbook references

 * , Page 266, Exercise 9, Chapter 7