Fully invariant not implies abelian-potentially verbal in abelian group

This article gives the statement and possibly, proof, of a non-implication relation between two subgroup properties, when the big group is a abelian group. That is, it states that in a abelian group, every subgroup satisfying the first subgroup property (i.e., fully invariant subgroup) need not satisfy the second subgroup property (i.e., abelian-potentially verbal subgroup)
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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. 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 $C$.