TITLE: Every countable subgroup of $\mathbb{Z}^\mathbb{N}$ is free.
QUESTION [4 upvotes]: This is a guided exercise in Rotman's Homological Algebra. Is stated as follows


(Pontrjagin) If $A$ is a countable torsion-free abelian group each of whose subgroups $S$ of finite rank is free, prove that $A$ is free.
Prove that any subgroup of finite rank in $\mathbb{Z}^\mathbb{N}$ (the product of countable many copies of $\mathbb{Z}$) is free.
Prove that any countable subgroup of $\mathbb{Z}^\mathbb{N}$ is free.


The only easy part is $1)+2)\implies 3)$. But I can't solve $1)$ or $2)$.
The first part has as hint to see a technique used in the proof of Kaplansky theorem wich said that any projective module is a direct sum of countably generated modules. 
Using this hint my attempt is to find a chain of submodules $S_n$ of finite rank such that $\bigcup_{n\in\mathbb{N}}S_n=A$ and every $S_n$ is a direct summand of $S_{n+1}$ because in that case is not difficult to prove that $A\cong \bigoplus_{n\in \mathbb{N}} S_{n+1}/{S_n}$ with each $S_{n+1}/{S_n}$ isomorphic to a submodule of $A$ and hence free (because they are of finite rank). But I can't construct such a chain.
For the second part there is already an answer in this site but I don't understand the solution there.
Any help would be appreciated. Thanks in advance.

REPLY [1 votes]: First note that if $B$ is a torsion free abelian group then the natural map $B \hookrightarrow B \otimes \mathbb Q$ is an injection, and that tensoring with $\mathbb Q$ is exact, hence preserves inclusions.

Following your plan, let $\{g_1,\ldots, g_n, \ldots\}$ be a countable set (assumed infinite - the finite case is trivial) of generators for $A$. Let $G_n = \langle g_1, \ldots, g_n \rangle$ be the subgroup generated by the first $n$ generators and set $S_n = G_n \otimes \mathbb Q \cap A$ (intersection inside $A \otimes \mathbb Q$).
We may assume for simplicity that $S_n$ has rank $n$ (in general we would set $S_n =G_m \otimes \mathbb Q \cap A$ where $m$ is the first index such that $G_m$ has rank $n$). As you indicate, it is enough to show that the inclusion of $S_n \cong \mathbb Z^n$  in $S_{n+1} \cong \mathbb Z^{n+1}$ splits. If this were not the case then the rank $1$ (finitely generated) abelian group $S_{n+1}/S_n$ would have torsion, i.e. there would exist an element $g \in S_{n+1}\setminus S_n$ and an integer $k$ such that $kg \in S_{n}$. Writing $g=\alpha_1 g_1 + \ldots + \alpha_{n+1} g_{n+1}$ with $\alpha_i \in \mathbb Q$ we see that $kg \in S_n \Rightarrow k\alpha_{n+1}=0 \Rightarrow \alpha_{n+1}=0$. This means that $g\in S_n$, which is a contradiction.

Let $A \subset \mathbb Z^{\mathbb N}$ be a subgroup of finite rank. It is enough to check that, for $N$ large enough, the composite $A \hookrightarrow \mathbb Z^{\mathbb N} \xrightarrow{\pi} \mathbb Z^N$ (with $\pi$ the projection onto the first $N$ coordinates) is injective. In view of the commutative diagram
$\require{AMScd}$
\begin{CD}
A  @>>> {\mathbb Z^{\mathbb N}} @>>>  {\mathbb Z^N} \\
@VVV @VVV @VVV \\
{A\otimes \mathbb Q} @>>> {\mathbb Z^{\mathbb N}\otimes \mathbb Q} @>>> {\mathbb Q^N}
\end{CD}
it is enough to show that the composite along the bottom row, denoted $\phi_N$, is an injection for some $N$. The rank of $\phi_M$ increases with $M$ and, since $A$ has finite rank, must eventually stabilize. Suppose that $\textrm{rank } \phi_M =r$ for $M\geq N$. Then for all $k\geq 1$, the projections $\pi^{N+k}_N \colon \mathbb Q^{N+k} \to \mathbb Q^N$ restrict to isomorphisms from the image of $\phi_{N+k}$ to the image of $\phi_N$. This means that, for each $a\in A \otimes \mathbb Q$, the finite sequence $\phi_N(a)$ has a unique completion to a sequence $\phi_{N+k}(a)$ for all $k\geq 1$. It follows that there is a unique element of $\mathbb Z^{\mathbb N} \otimes \mathbb Q \subset \mathbb Q^{\mathbb N}$ lifting $\phi_N(a)$. Hence $\mathbb Z^{\mathbb N} \otimes \mathbb Q \to \mathbb Q^N$ is injective on the image of $A\otimes \mathbb Q$, as required (and of course $r$ is the rank of $A$).
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