# Classification of fully characteristic subgroups in finitely generated Abelian groups

## Contents

## Statement

### Case of groups of prime power order

Suppose is an Abelian group whose order is a power of a prime . Then, we can write, by the structure theorem for finitely generated Abelian groups:

where:

.

where .

Then a subgroup of is a fully characteristic subgroup of , and this happens if and only if the following are satisfied:

- is the direct sum of the intersections .
- If the orders of are , then and .

### Case of finitely generated groups

Suppose is a finitely generated Abelian group. Then, we can write:

where is a torsion-free Abelian group, and each is an Abelian group of prime power order for different primes. Then, a subgroup of is a fully characteristic subgroup, if and only if the following are satisfied:

- is the direct sum of the intersections .
- Each is characteristic (equivalently, fully characteristic) in .

## Proof

### Case of prime power order

**Given**: is an Abelian group whose order is a power of a prime . Then, we can write, by the structure theorem for finitely generated Abelian groups:

where:

.

where .

A subgroup of .

**To prove**: is characteristic if and only if it is fully characteristic in , if and only if is the direct sum of , and if the orders of the intersections are , then and .

**Proof**: The proof relies on two important homomorphisms. 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.

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

We are now in a position to prove the main result.

Suppose is a fully characteristic subgroup of . We first show that is a direct sum of . For this, suppose the element:

.

Since is fully characteristic, it is invariant under the projections to the direct factors, which are endomorphisms. Thus, each is in . Thus, every element of can be expressed as a sum of elements in , and we get that is a direct sum of the s.

Finally, we need to show the conditions on the s. For this, observe that for , we have:

- 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 characteristic 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 characteristic subgroup of the cyclic group (all subgroups of cyclic groups are fully characteristic).
- : 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 .

### General case

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