A nonzero free abelian group has a subgroup of index n for every positive integer n.
Proof: Consider the free abelian group F(S), where S is the set of generators. Let x be some element of S, and let xn be some element of S'. Then F(S') is a subgroup of F(S), and F(S)/F(S') is isomorphic to Z/nZ.
So showing this answers my question right. By saying its isomorphic we see that the abelian group has a subgroup of index n?