Verbal not implies iterated agemo

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., iterated agemo subgroup of group of prime power order) 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 iterated agemo subgroup of group of prime power order|Get more facts about verbal subgroup

EXPLORE EXAMPLES YOURSELF: View examples of subgroups satisfying property iterated agemo subgroup of group of prime power order but not verbal subgroup|View examples of subgroups satisfying property iterated agemo subgroup of group of prime power order and verbal subgroup

Statement

It is possible to have a group of prime power order $G$ and a verbal subgroup (hence, a Verbal subgroup of group of prime power order (?)) $H$ of $G$ that is not an iterated agemo subgroup of $G$.

Proof

Further information: prime-cube order group:U(3,p)

Suppose $p$ is an odd prime. Consider the group $G := U(3,p)$, which is the unique (up to isomorphism) non-abelian group of order $p^3$ and exponent $p$.

We define $H := [G,G]$, so $H$ is the commutator subgroup of $G$. Clearly, $H$ is a verbal subgroup of $G$. On the other hand, $H$ is not an iterated agemo subgroup of $G$, because $\mho^1(G)$ is trivial, so the only agemo subgroups (and hence the only iterated agemo subgroups) are the whole group and the trivial subgroup.