Pronormality is not finite-upper join-closed

Statement
We can have a subgroup $$H \le G$$ such that $$H$$ is a pronormal subgroup in two intermediate subgroups $$K$$ and $$L$$, but $$H$$ is not pronormal in the join $$\langle K, L \rangle$$.

Related facts about pronormality

 * Pronormality satisfies intermediate subgroup condition: If $$H \le K \le G$$ and $$H$$ is pronormal in $$G$$, then $$H$$ is also pronormal in $$K$$.

Related facts about upper join-closedness

 * 2-subnormality is not upper join-closed
 * Subnormality is not upper join-closed
 * Normality is upper join-closed
 * Characteristicity is not upper join-closed

Example
Let $$G$$ be the symmetric group on the set $$\{ 1,2,3,4 \}$$, $$K$$ be the subgroup comprising permutations on $$\{ 1,2,3 \}$$, $$L$$ be the subgroup comprising permutations on $$\{ 1,2,4 \}$$, and $$H = K \cap L$$ be the two-element subgroup generated by $$(1,2)$$. Observe that:


 * $$H$$ is a pronormal subgroup in both $$K$$ and $$L$$. In fact, it is a maximal subgroup in both, and maximal implies pronormal.
 * $$\langle K, L \rangle = G$$.
 * $$H$$ is not pronormal in $$G$$. That's because the subgroup generated by $$(3,4)$$ is a conjugate of $$H$$, and these two subgroups are not conjugate in the subgroup they generate.