# Torsion subgroup of a finitely-generated abelian group is finite?

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?

• 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. – Bernard Oct 3 '18 at 8:14
• 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? – Jon Hillery Oct 3 '18 at 8:17

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.
• But can you show that subgroups of finitely-generated abelian groups are finitely-generated without resorting to saying that $\mathbb{Z}$ is Noetherian? – Jon Hillery Oct 3 '18 at 8:27
• 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. – Mark Oct 3 '18 at 8:29