Pronormal implies self-conjugate-permutable

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) uses::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 $$HH^g = H^gH$$ 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, $$H = H^g$$.