Pronormality is normalizing join-closed

Statement
Suppose $$H, K \le G$$ are pronormal subgroups such that $$K \le N_G(H)$$: in other words, $$K$$ normalizes $$H$$. Then the join of subgroups $$\langle H, K \rangle$$ (also the same as $$HK$$) is also a pronormal subgroup.

Similar facts: statements about normalizing join-closedness
Some facts with very similar proofs:


 * Weak pronormality is normalizing join-closed
 * Intermediate isomorph-conjugacy is normalizing join-closed
 * Intermediate automorph-conjugacy is normalizing join-closed

Other facts about normalizing joins, but with a different kind of proof:


 * Subnormality is normalizing join-closed

Related facts about pronormality

 * Nilpotent join of pronormal subgroups is pronormal
 * Pronormality is not join-closed
 * Pronormality is not intersection-closed

Join-closedness of some related properties

 * Paranormality is strongly join-closed
 * Polynormality is strongly join-closed

Proof
Given: A group $$G$$, pronormal subgroups $$H,K \le G$$ such that $$K \le N_G(H)$$.

To prove: $$HK$$ is pronormal in $$G$$. Specifically, for any $$g \in G$$, our goal is to find a $$z \in \langle HK, (HK)^g \rangle$$ such that $$(HK)^z = (HK)^g$$.

Proof: The overall problem in showing that there exists such an element $$z$$ is to find something that simultaneously works for both $$H$$ and $$K$$ for the same element $$g$$. The solution to that is to first find an element (here called $$x$$) that works for $$H$$. This original element $$x$$ need not work for $$K$$. However, since $$K$$ normalizes $$H$$, we can now tweak this element to fix the behavior on $$K$$ without altering the behavior on $$H$$. The tweaking happens via multiplication by a suitable element in $$N_G(H)$$.

The proof has the flavor of constructing an upper triangular transformation.

Steps (6) and (7) clinch the proof.