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According to 2.3 Theorem (p. 98) from Eklof and Mekler's Almost free modules, an abelian group $M$ is $\aleph_{1}$-free, that is, all its countable subgroups are abelian free, if and only if $M$ is torsion-free and every finite subset of $M$ is contained in a finitely-generated pure subgroup of $M$ (B is a pure subgroup of $M$ if $M/B$ is torsion-free). The left to right implication goes as follows.

$M$ is torsion-free since otherwise $M$ contains a finite torsion group, and thus cannot be $\aleph_{1}$-free. Let $S$ be a finite subset of $M$. If $\langle S\rangle_{\ast}$ is not finitely-generated, then there is a countably generated subgroup $N$ of $\langle S\rangle_{\ast}$ containing $S$ which is not finitely-generated. But then $N$ is not free, since it has finite rank but is not finitely-generated. This contradicts the assumption that $M$ is $\aleph_{1}$-free.

$\langle S\rangle$ denotes the group generated by $S$ and $\langle S\rangle_{\ast}$ denotes the pure closure of $\langle S\rangle$, that is, the smallest pure subgroup of $M$ containing $\langle S\rangle$, which has the form $\{x\in M:nx\in \langle S\rangle \text{ for some } n\in\mathbb{N}\}$.

What definition of rank are we using to conclude that $N$ has finite rank?

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  • $\begingroup$ I am currently reading through the same book and am a bit stuck on this part also. I believe that the notion of "rank" that is applicable is the basis number for the group. $S$ is a finite set, so by construction $\langle S \rangle$ is at most countable. And I BELIEVE that the pure-closure of $\langle S \rangle$ is also countable. Which means that because $M$ is $\aleph_1$-free, $\langle S \rangle_*$ is free. I may be misunderstanding this though (hence my reply as a comment). $\endgroup$
    – dbossaller
    Aug 26, 2020 at 20:47
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    $\begingroup$ The rank of a torsion-free abelian group $A$ is the dimension of the rational vector space $A\otimes_{\mathbb Z}\mathbb Q$. Equivalently, it is the maximum size of a linearly independent (over $\mathbb Z$ or over $\mathbb Q$, they're equivalent) subset of $A$. $\endgroup$ Aug 26, 2020 at 22:59
  • $\begingroup$ @AndreasBlass, thank you. Could you add this as an answer so I can close de question? $\endgroup$
    – frch
    Aug 27, 2020 at 9:21

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As suggested by OP, I'm promoting my comment to an answer.

The rank of a torsion-free abelian group 𝐴 is the dimension of the rational vector space $A\otimes_{\mathbb Z}\mathbb Q$. Equivalently, it is the maximum size of a linearly independent (over ℤ or over ℚ, they're equivalent) subset of 𝐴.

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