Finite implies local powering-invariant

Statement
Suppose $$G$$ is a group and $$H$$ is a finite subgroup of $$G$$. Then, $$H$$ is a local powering-invariant subgroup of $$G$$: for any $$g \in H$$ and $$n \in \mathbb{N}$$ such that the solution to $$x^n = g$$ is unique for $$x \in G$$, we must have $$x \in H$$.

Related facts

 * Finite implies powering-invariant
 * Finite not implies divisibility-closed