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)
View a complete list of subgroup property non-implications | View a complete list of subgroup property implications
Get more facts about image-closed fully invariant subgroup|Get more facts about verbal subgroup

EXPLORE EXAMPLES YOURSELF: View examples of subgroups satisfying property image-closed fully invariant subgroup but not verbal subgroup|View examples of subgroups satisfying property image-closed fully invariant subgroup and verbal subgroup

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 $ρ : G \to K$ , $ρ (H)$ is a fully invariant subgroup of $K$ .
$H$ is not a verbal subgroup of $G$ .

Facts used

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:

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

It can also be defined as the group of all $(p^{k})^{t h}$ 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 $ρ : G \to K$ , either the map is an isomorphism or we have that $ρ (H)$ is trivial. In either case, $ρ (H)$ 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 $m^{t h}$ powers for some $m$ . But in this group $G$ , the set of $m^{t h}$ powers is equal to $G$ for all $m$ .