Join with any distinct conjugate is the whole group implies pronormal

This article gives the statement and possibly, proof, of an implication relation between two subgroup properties. That is, it states that every subgroup satisfying the first subgroup property (i.e., subgroup whose join with any distinct conjugate is the whole group) must also satisfy the second subgroup property (i.e., pronormal subgroup)
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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 $⟨H,H^g⟩=G$.

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

Related facts

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.

Further information: Pronormal not implies 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 $⟨H,H^g⟩=G$.

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

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

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