Fully invariant not implies verbal in finite abelian group

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

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 $$\Omega_1(G)$$.