# Free abelian group has a subgroup of index n

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?

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The way is the good one, but your answer is not correct. But since you're really near to the solution I gonna give you some step to complete the proof.

• Try to write down explicitly what are $S$ and $S'$;

• motivate formally why $F(S)/F(S')$ should be $\mathbb Z/n \mathbb Z$.

If you complete all the step above you should be able to build up an good proof of your claim.

Anyway I would suggest to avoid using the notation $F(S')$ to denote the subgroup of $F(S)$ generated by some subset $S'$ of $F(S)$. It would be better if you denote that subgroup with $\langle S' \rangle$, or even $\langle x_1,x_2,x_2,\dots,x_n\rangle$ (when $S'=\{x_1,\dots,x-n\}$).

Feel free to ask if you have any other problem.

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