It's well-known that that [a subgroup of a free group is free](http://ncatlab.org/nlab/show/Nielsen-Schreier+theorem). Is a subgroup of a free *abelian* group (may not be finitely generated) always a free abelian group?



Is a subgroup of a free abelian group free abelian?
[Yes](https://en.wikipedia.org/w/index.php?title=Free_abelian_group&oldid=317358888#Subgroup_closure).


(EDIT: If you don't like following links, this is the Wikipedia article on Free abelian groups which, uncharacteristically, contains a complete (and correct) proof of precisely that statement).


