Pronormal implies self-conjugate-permutable

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., pronormal subgroup) must also satisfy the second subgroup property (i.e., self-conjugate-permutable subgroup)
View all subgroup property implications | View all subgroup property non-implications
Get more facts about pronormal subgroup|Get more facts about self-conjugate-permutable subgroup

Statement

Property-theoretic statement

The subgroup property of being pronormal is stronger than the subgroup property of being self-conjugate-permutable.

Verbal statement

Any pronormal subgroup is self-conjugate-permutable.

Facts used

1. Product of conjugates is proper: If $G$ is a group and $H$ is a subgroup such that there exists $g\in G$ for which $HH^{g}=G$, then $H=G$.

Proof

Given: A group $G$, a pronormal subgroup $H$.

To prove: If $HH^{g}=H^{g}H$ for some $g\in G$, then $H^{g}=H$.

Proof: Let $K=\langle H,H^{g}\rangle =HH^{g}$.

1. (Given data used: $H$ is pronormal in $G$): There exists $x\in K$ such that $H^{g}=H^{x}$. Thus, $K=HH^{x}$.
2. (Fact used: fact (1), product of conjugates is proper): $K$ is the product of two conjugate subgroups, so fact (1) forces that $K=H$. Since $x\in K$, we also get $K=H^{x}$. Thus, $H=H^{g}$.