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This is one of the Exercise problem in Hatcher's AT.

(Exercise 3.3.18) Show that a direct limit $\varinjlim G_\alpha$ of torsionfree abelian group $G_\alpha$ is torsionfree. More generally, show that any finitely generated subgroup of $\varinjlim G_\alpha$ is realized as a subgroup of some $G_\alpha$.

I've already proved the first statement (torsionfree). For the second statement, I want to use the torsionfree result but I don't know how to connected these two. There's a post especially for the second statement. But I don't want to use finitely presented concept for this (Hatcher didn't introduce such concept). Could you help?

one potato two potato
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  • The other post to which you refer is discussing a direct limit of arbitrary groups, which is not the same as the second statement: both of Hatcher's statements only refers to a direct limit of torsion-free abelian groups. Another relevant post: https://math.stackexchange.com/questions/1399573/finite-generated-abelian-torsion-free-group-is-a-free-abelian-group/1399615. – John Palmieri Nov 05 '21 at 17:50
  • @JohnPalmieri So in the second statement, $G_\alpha$ are torsion free abelian groups? It seems the post you linked is not very relevant. – one potato two potato Nov 06 '21 at 04:53
  • Any finitely generated subgroup of a direct limit is the image of a subgroup of one of the $G_{\alpha}$ (but if the structure maps are not injections, then it need not be isomorphic to a subgroup). For given finitely many elements of the direct limit, you can find finitely many indices $\alpha_1,\ldots,\alpha_n$ and elements in the corresponding $G_{\alpha_i}$ mapping to them; then picking $\alpha\geq \alpha_i$ for all $i$ you find a subgroup of $G_{\alpha}$ that maps onto the subgroup you have. – Arturo Magidin Nov 07 '21 at 00:31
  • @ArturoMagidin Thank you for your comment. As you said, generally the image need not be injective. So in the problem, freeness is necessary to conclude the image of finitely generated subgroup is isomorphic to some subgroup of $G_\alpha$. – one potato two potato Nov 07 '21 at 00:44

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$G_{\alpha}$ is described in the first sentence as torsion free abelian, and that persists for the second sentence. In the second part you are considering a finitely generated subgroup $H$ of the direct limit, and you know that the direct limit is torsion free by part one. Subgroups of torsion free abelian groups must be torsion free, so the subgroup is finitely generated torsion free abelian, and therefore (for example by the link I posted earlier) must be free abelian. Choose generators $\{a_1, ..., a_n\}$ for the subgroup $H$. Now you could imitate the proof in the post that you cited: each generator $a_i$ must lie in the image (in the direct limit) of some $G_\alpha$, and since there are finitely many of them, they must all lie in the image of a common $G_\alpha$, and therefore the subgroup itself is in the image of that $G_\alpha$.

John Palmieri
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  • I can see know. But why the second statement implies the first one? – one potato two potato Nov 06 '21 at 10:02
  • For any non-identity element $x$ in the direct limit, the subgroup generated by $x$ must be torsion-free, according to the second statement. Therefore $x$ is not a torsion element. – John Palmieri Nov 06 '21 at 16:40
  • I think there's one issue : Once $a_i$'s are mapped to $G_\alpha$ by structure map, isn't it possible that two $a_i$'s mapped to same element? i.e., how can I guarantee the injectivity? – one potato two potato Nov 06 '21 at 23:54
  • The $a_i$'s lie in $H$ which is in the direct limit, and you could choose them to be distinct (for example choose a free basis, not just generators). So the $a_i$'s are not mapped anywhere — they are the images of other elements. Since the $a_i$'s are distinct, so are the elements mapping to them. – John Palmieri Nov 06 '21 at 23:59