# Characteristic equals fully invariant in odd-order abelian group

This article gives the statement and possibly, proof, of an implication relation between two subgroup properties, when the big group is a odd-order abelian group. That is, it states that in a Odd-order abelian group (?), every subgroup satisfying the first subgroup property (i.e., Characteristic subgroup (?)) must also satisfy the second subgroup property (i.e., Fully invariant subgroup (?)). In other words, every characteristic subgroup of odd-order abelian group is a fully invariant subgroup of odd-order abelian group.

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## Contents

## Statement

In an odd-order abelian group, a subgroup is characteristic if and only if it is fully invariant.

## Related facts

### Opposite facts

- Characteristic not implies fully invariant
- Characteristic not implies fully invariant in finitely generated abelian group
- Characteristic not implies fully invariant in finite abelian group
- Characteristic not implies fully invariant in class three maximal class p-group
- Characteristic not implies fully invariant in odd-order class two p-group

## Proof for a group of prime power order

**Given**: An abelian group of odd prime power order.

.

where we have:

.

such that .

is a characteristic subgroup of .

**To prove**: is fully invariant in .

**Proof**:

### It is a direct sum of its intersections with the direct summands

We first prove that is a direct sum of the intersections .

For this, consider the projection map that sends an element to the element . is an endomorphism of . Consider the map:

.

We claim that is an automorphism. For this, note that acts as the identity map on all coordinates other than , and as the doubling, or square, map on the coordinate. Since the group has odd order, the doubling map is an automorphism of the coordinate, so is an automorphism. Thus, we have:

.

Since is a subgroup, we get:

.

Thus, the projections of any element of are in . Thus, is a direct sum of its projections.

### A characteristic subgroup satisfies the conditions for being fully invariant

For , there is an injective homomorphism:

that sends the generator on the left to times the generator on the right.

There is also a surjective homomorphism:

that sends the generator to the generator.

Recall that we have:

.

For , define as the endomorphism of that sends to via the map:

.

and is zero elsewhere. Define as:

.

Clearly, is an automorphism of .

Similarly, define as the endomorphism of that sends to via the map and is zero elsewhere.

Define as:

.

Clearly is an automorphism of .

Thus, for to be characteristic, it must be invariant under both and . This forces to be invariant under and . If the order of is , we obtain the following:

- The endomorphism sends to itself: Thus, injects into via , so .
- The endomorphism sends to itself: Thus, induces a surjective endomorphism from to , forcing .

Now, we show that if satisfies the conditions described above, then is fully invarant in .

Suppose is an endomorphism. Since is a direct sum of , it suffices to show that . For this, in turn, it suffices to show that the coordinate of is contained in . This is easily done in three cases:

- : In this case, it is direct since is a fully invariant subgroup of the cyclic group (all subgroups of cyclic groups are fully invariant).
- : In this case, . Consider the homomorphism from to obtained by composing the projection with . This map must send to a subgroup of size at most in . But since , this subgroup of size is contained in the subgroup that has order .
- : In this case, . Consider the homomorphism from to obtained by composing the projection with . The index of the image of is at least the index of in , which is . Thus, the image has size at most because . Thus, it is in .

## Proof for an odd-order abelian group

This is direct because the group splits up as a direct sum of its -Sylow subgroups, and characteristic or fully invariant subgroups also split up as direct sums.