# Characteristic not implies fully invariant 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., characteristic subgroup) neednotsatisfy the second subgroup property (i.e., fully invariant subgroup)

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## Contents

## Statement

In a finite abelian group, a characteristic subgroup need not be a fully invariant subgroup.

## Related facts

- Characteristic equals fully invariant in odd-order abelian group
- Characteristic not implies fully invariant in finitely generated abelian group
- Characteristic equals verbal in free abelian group

## Proof

Let be the direct sum of the infinite cyclic group and the cyclic group of order two:

.

Let be the cyclic subgroup of generated by .

### The subgroup is characteristic

Set:

and

.

and:

.

Thus, we have:

.

Clearly, any automorphism of sends to itself, sends to itself, and sends to itself. Thus, any automorphism of sends to itself. Thus, any automorphism of sends to itself. Note that comprises *precisely* those elements of that have the second coordinate equal to : in particular, , so the subgroup generated by equals . Thus, any automorphism of preserves .

### The subgroup is not fully invariant

Consider the map:

.

This map is an endomorphism of , but the image of under this map is , which is not an element of . Thus, is not fully invariant in .