# Why is every finite subset of an $\aleph_{1}$-free abelian group $M$ contained in a finitely-generated pure subgroup of $M$?

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?

• 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). Aug 26, 2020 at 20:47
• 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 $A$. Aug 26, 2020 at 22:59
• @AndreasBlass, thank you. Could you add this as an answer so I can close de question?
– frch
Aug 27, 2020 at 9:21

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.