Divisible abelian subgroup of abelian group contains no proper nontrivial verbal subgroup

Statement
Suppose $$G$$ is an abelian group, and $$H$$ is a subgroup of $$G$$ that is a fact about::divisible abelian group. Then, if $$K$$ is a subgroup of $$H$$ that is a verbal subgroup of $$G$$, then either $$K$$ is trivial or $$K = H$$.

Facts used

 * 1) uses::verbal subgroup equals power subgroup in abelian group

Proof
Given: An abelian group $$G$$, a subgroup $$H$$ of $$G$$ that is divisible, and a subgroup $$K$$ of $$H$$ that is verbal in $$G$$.

To prove: $$K = H$$ or $$K$$ is trivial.

Proof: By fact (1), there exists an integer $$n$$ such that $$K$$ is the set of $$n^{th}$$ powers in $$G$$. If $$n = 0$$, $$K$$ is trivial. If, on the other hand, $$n$$ is nonzero, then every element of $$H$$ is a $$n^{th}$$ power, so $$H \le K$$. Since $$K \le H$$ by assumption, we get $$H = K$$.