Existentially bound-word not implies verbal

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) uses::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 particular example::direct product of Z4 and Z2 or the particular example::direct product of Z9 and Z3.