Image-closed fully invariant 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., image-closed fully invariant 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 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. verbal subgroup equals power subgroup in abelian group

Proof

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 $m$.