Join with any distinct conjugate is the whole group implies pronormal

Statement
Suppose $$G$$ is a group and $$H$$ is a subgroup of $$G$$ such that for any $$g \in G$$, the conjugate subgroup $$H^g = g^{-1}Hg$$ is either equal to $$H$$, or satisfies $$\langle H, H^g \rangle = G$$.

Then, $$H$$ is a pronormal subgroup of $$G$$.

Converse
The converse is false. For instance, Sylow subgroups are pronormal, but need not satisfy the condition that the join with any distinct conjugate is the whole group.

Proof
Given: A group $$G$$ with a subgroup $$H$$ such that for every $$g \in G$$, $$H^g = H$$ or $$\langle H, H^g \rangle = G$$.

To prove: $$H$$ is pronormal in $$G$$.

Proof: Suppose $$g \in G$$. We need to show that $$H$$ and $$H^g$$ are conjugates in $$\langle H, H^g \rangle$$. We consider both cases:


 * $$H = H^g$$: In this case, $$H$$ and $$H^g$$ are conjugate by the identity element, which is certainly in $$H$$.
 * $$\langle H, H^g \rangle$$: In this case, $$g \in G = \langle H, H^g \rangle$$, so $$H$$ and $$H^g$$ are conjugate in $$\langle H, H^g \rangle$$.