Fully invariant not implies verbal in finite abelian group

This article gives the statement and possibly, proof, of a non-implication relation between two subgroup properties, when the big group is a finite abelian group. That is, it states that in a finite abelian group, every subgroup satisfying the first subgroup property (i.e., fully invariant subgroup) need not satisfy the second subgroup property (i.e., verbal subgroup)
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Statement

A fully characteristic subgroup of a finite abelian group need not be a verbal subgroup.

Related facts

• Fully invariant equals verbal in homocyclic group
• Fully invariant equals verbal in free abelian group
• Characteristic not implies fully invariant in finite abelian group
• Characteristic equals fully invariant in odd-order abelian group

Proof

Example of a group of prime-cube order

Let $C_{p}$ and $C_{p^{2}}$ denote the cyclic groups of order $p$ and $p^{2}$ respectively. Let $G=C_{p}\times C_{p^{2}}$. Consider the omega-1 subgroup:

$\Omega _{1}(G):=\{x\in G\mid px=0\}$.

In other words, it is the subgroup of $G$ comprising the elements of order dividing $p$. Then, $\Omega _{1}(G)$ is fully characteristic: any endomorphism of $G$ preserves the condition. On the other hand, $\Omega _{1}(G)$ is not verbal: the only possible verbal subgroups of $G$ are the agemo subgroups: the whole group $G$, the set of multiples of $p$, and the trivial subgroup. None of these equals $\Omega _{1}(G)$.