Fully invariant not implies abelian-potentially verbal in abelian group

Statement
It is possible to have an abelian group $$A$$ and a fully invariant subgroup $$B$$ of $$A$$ such that $$A$$ is not an abelian-potentially verbal subgroup of $$B$$, i.e., there is no abelian group $$C$$ containing $$B$$ such that $$A$$ is a verbal subgroup of $$C$$.

Facts used

 * 1) uses::Divisible abelian subgroup of abelian group contains no proper nontrivial verbal subgroup

Proof
Let $$B$$ be the $$p$$-quasicyclic group, i.e., the group of all $$(p^k)^{th}$$ roots of unity for all $$k$$, under multiplication. Let $$A$$ be the subgroup of $$B$$ comprising the group of $$p^{th}$$ roots of unity. Thus, $$A$$ is a fully invariant subgroup of $$B$$ that is neither trivial nor the whole group.

Since $$B$$ is a divisible abelian group, fact (1) tells us that for any abelian group $$C$$ containing $$B$$, any verbal subgroup of $$C$$ contained in $$B$$ is equal to either the trivial subgroup or $$B$$. In particular, $$A$$ is not a verbal subgroup of $$C$$.