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.