Retract implies local divisibility-invariant

Statement
Suppose $$G$$ is a group and $$H$$ is a retract of $$G$$. Then, $$H$$ is a local divisibility-invariant subgroup of $$G$$. In other words: suppose $$n$$ is a natural number. Suppose $$h \in H$$ is such that there exists $$x \in G$$ such that $$x^n = h$$. Then, there exists $$y \in H$$ (possibly equal to $$x$$ such that $$y^n = h$$.