# Finite implies local powering-invariant

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 $x \in H$.