Free Abelian Group - Subgroup Closure

Subgroup Closure

Every subgroup of a free abelian group is itself a free abelian group. This is similar to the Nielsen–Schreier theorem that a subgroup of a free group is free.

Theorem: Let be a free abelian group generated by the set and let be a subgroup. Then is a free abelian group.

Proof: If, the statement holds, so we can assume that is nontrivial. First we shall prove this for finite by induction. When, is isomorphic to (being nontrivial) and clearly free. Assume that if a group is generated by a set of size, then every subgroup of it is free. Let, the free group generated by and a subgroup. Let be the projection If, then is a subset of and free by the induction hypothesis. Thus we can assume that the range is nontrivial. Let be the least such that and choose some such that . It is standard to verify that and if, then, where and . Hence . By the induction hypothesis and are free: first is isomorphic to a subgroup of and the second to .

Assume now that is arbitrary. For each subset of let be the free group generated by, thus is a free subgroup and denote .

Now set

Formally is an injective (one-to-one) map

such that generates .

Clearly is nonempty: Let us have an element in . Then and thus the free group generated by contains and the intersection is a nontrivial subgroup of a finitely generated free abelian group and thus free by the induction above.

If, define order if and only if and the basis is an extension of ; formally if and, then and .

If is a -chain ( is some linear order) of elements of, then obviously

,

so we can apply Zorn's lemma and conclude that there exists a maximal . Since, it is enough to prove now that . Assume on contrary that there is .

Put . If then it means that, but they are not equal, so is bigger, which contradicts maximality of . Otherwise there is an element such that and .

The set of for which there exists such that forms a subgroup of . Let be a generator of this group and let with . Now if, then for some, where .

On the other hand clearly, so is a basis of, so contradicting the maximality again.