Image-closed fully invariant not implies verbal

Statement
It is possible to have a group $$G$$ and a subgroup $$H$$ of $$G$$ such that:


 * $$H$$ is an image-closed fully invariant subgroup of $$G$$: for any surjective homomorphism $$\rho:G \to K$$, $$\rho(H)$$ is a fully invariant subgroup of $$K$$.
 * $$H$$ is not a verbal subgroup of $$G$$.

Facts used

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

Example of the quasicyclic group
For any prime $$p$$, let $$G$$ be the quasicyclic group for the prime $$p$$. This is defined as the inverse limit of the sequence:

$$ \dots \to \mathbb{Z}/p^n\mathbb{Z} \to \mathbb{Z}/p^{n-1}\mathbb{Z} \to \dots \to \mathbb{Z}/p\mathbb{Z} \to 0$$

It can also be defined as the group of all $$(p^k)^{th}$$ roots of unity for all $$k$$.

Define $$H$$ as the subgroup comprising the elements of order $$p$$. Then:


 * Any nontrivial normal subgroup of $$G$$ contains $$H$$. Thus, for any surjective homomorphism $$\rho:G \to K$$, either the map is an isomorphism or we have that $$\rho(H)$$ is trivial. In either case, $$\rho(H)$$ is fully invariant in $$K$$.
 * $$H$$ is not verbal in $$G$$: A verbal subgroup of an abelian group is a power subgroup -- it is the set of $$m^{th}$$ powers for some $$m$$. But in this group $$G$$, the set of $$m^{th}$$ powers is equal to $$G$$ for all $$m$$.