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The above claim was made at the very beginning of a proof of the structure theorem for finitely-generated abelian groups and brushed off as easy. However, I think the problem is easy if the torsion subgroup is finitely-generated, but this does not seem to necessarily be true, or at least not obviously. Must the torsion subgroup be finitely-generated? And is the claim in the title true?

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  • $\begingroup$ An abelian broup is a $\mathbf Z$-module, and $\mathbf Z$ is a noetherian ring, so that a submodule of a finitely generated module is finitely generated. $\endgroup$ – Bernard Oct 3 '18 at 8:14
  • $\begingroup$ How would I show something like this without knowing what a Noetherian ring is? Is there an easy way to see this by just using basic group theory language? $\endgroup$ – Jon Hillery Oct 3 '18 at 8:17
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Any subgroup of a finitely generated abelian group (not true for non abelian groups) is finitely generated. So yes, if you have a finitely generated abelian group then its torsion subgroup is also finitely generated, and hence must be isomorphic to a group of the form $\mathbb{Z_{m_1}}\times\mathbb{Z_{m_2}}\times...\times\mathbb{Z_{m_k}}\times \mathbb{Z^r}$ when $m_1|m_2|...|m_k$. Now, all the elements of the torsion subgroup must have finite order so $r=0$ and we really get that it is finite.

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  • $\begingroup$ But can you show that subgroups of finitely-generated abelian groups are finitely-generated without resorting to saying that $\mathbb{Z}$ is Noetherian? $\endgroup$ – Jon Hillery Oct 3 '18 at 8:27
  • $\begingroup$ Are you familiar with free abelian groups? And with the fact that if $A$ is a free abelian group of rank $n$ then any of its subgroups is also free abelian of rank $k\leq n$? If you do then I can show you an easy proof of that a subgroup of a finitely generated abelian group is finitely generated. $\endgroup$ – Mark Oct 3 '18 at 8:29
  • $\begingroup$ Ah, okay I've seen this result before. Forgot about that. Thanks! $\endgroup$ – Jon Hillery Oct 3 '18 at 8:31
  • $\begingroup$ Surely you should prove the first sentence in your answer, or at least edit it to include your comment! I mean, this is where the OP's issue lies, and he said that he knows how to get from finite generation to finite. $\endgroup$ – user1729 Oct 3 '18 at 9:32
  • $\begingroup$ @user1729, well, it's easy to prove it if you know the theorem about subgroups of free abelian groups, and as you can see OP wrote in the comment that he knows this proof. I didn't write a proof in my answer from the beginning because if OP didn't know about free abelian groups then I would have no idea how to prove it in another way. $\endgroup$ – Mark Oct 3 '18 at 11:37

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