# 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) neednotsatisfy the second subgroup property (i.e., verbal subgroup)

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## Contents

## Statement

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

## Related facts

## Proof

### Example of a group of prime-cube order

Let and denote the cyclic groups of order and respectively. Let . Consider the omega-1 subgroup:

.

In other words, it is the subgroup of comprising the elements of order dividing . Then, is fully characteristic: any endomorphism of preserves the condition. On the other hand, is not verbal: the only possible verbal subgroups of are the agemo subgroups: the whole group , the set of multiples of , and the trivial subgroup. None of these equals .