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I want to prove that every subgroup of finite rank of $\mathbb Z^\mathbb N$ is free abelian. A group $G$ has finite rank $N$ if there exists a subset $ \{e_1,...,e_N\}$ of $G$ such that:

$(1)$ If $a_i \in \mathbb Z$, $\sum_{i=1}^{N} a_i e_i=0 $ implies $a_i=0$ for $i=1,2,...,N$.

$(2)$ For every $g\in G$, exist $k \in \mathbb Z ,k\neq 0$ and $a_i \in \mathbb Z$, $i=1,2,...,N$ such that $kg=\sum_{i=1}^{N} a_i e_i$.

So let $G_n=\{a\in \mathbb Z^\mathbb N:\exists k\in \mathbb Z, k\neq 0 $ such that $ka=\sum_{i=1}^{n} a_i e_i\}$. I was trying to find a free $\mathbb Z$-module $M$ such that $G_n$ is a submodule of $M$ and use that every submodule of a free $R$-module, where $R$ is a PID, is free. But I'm having troubles finding this module.

Any idea would be appreciated, thanks.

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    @OlivierBégassat does the condition (2) imply that it is finitely generated as a $\mathbb Z$-submodule? Maybe I'm being too formal but instead of $kg=\sum a_i e_i$ shouldn't it be $g=\sum a_i e_i$? – 2012-12-18

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So the subgroup $H$ of $G_n$ generated by the $e_i$ is free abelian of rank $n$. If $G_n$ were not finitely generated, then it would have a subgroup $K$ with $n+1$ generators $f_1,f_2,\ldots,f_{n+1}$, and since $K$ is torsion-free, it would be free abelian with the $f_i$ as free basis. For each $i$ there exists $0 \ne k_i \in {\mathbb Z}$ with $k_if_i \in H$. Since $H$ has rank $n$, the elements $k_if_i$ must be linearly dependent in $H$, which implies that the $f_i$ are linearly dependent, contradiction.