Pronormal not implies join with any distinct conjugate is the whole group

Statement
We can have a pronormal subgroup $$H$$ of a group $$G$$, and a conjugate subgroup $$K \ne H$$ of $$H$$ in $$G$$ such that $$\langle H, K \rangle$$ is not the whole group $$G$$.

Converse
Join with any distinct conjugate is the whole group implies pronormal

Facts used

 * 1) uses::Sylow implies pronormal

Example of the symmetric group of degree four
Suppose $$G$$ is the symmetric group on $$\{ 1,2,3,4 \}$$. Let $$H$$ be the $$3$$-Sylow subgroup generated by $$\{ (1,2,3) \}$$. Then, by fact (1), we get that $$H$$ is pronormal in $$G$$. On the other hand, the subgroup generated by $$H$$ and its conjugate subgroup $$\langle (1,3,4) \rangle$$ is the alternating group on $$\{ 1,2,3,4 \}$$, which is a proper subgroup of $$G$$.