Existentially bound-word not implies verbal

This article gives the statement and possibly, proof, of a non-implication relation between two subgroup properties. That is, it states that every subgroup satisfying the first subgroup property (i.e., existentially bound-word subgroup) need not satisfy the second subgroup property (i.e., verbal subgroup)
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Statement

It is possible to have a group $G$ and an existentially bound-word subgroup $H$ of $G$ that is not a verbal subgroup of $G$.

Facts used

1. verbal subgroup equals power subgroup in abelian group

Proof

Let $G$ be the direct product of cyclic group of prime-square order and cyclic group of prime order for a prime $p$, and let $H=\Omega _{1}(G)$ be the subgroup of elements of order dividing $p$. Then, $H$ is an existentially bound-word subgroup of $G$, in the sense that it is the set of solutions to a system of equations, but it is not a verbal subgroup of $G$, something we can see, for instance, from fact (1).

For instance, $G$ could be the direct product of Z4 and Z2 or the direct product of Z9 and Z3.