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There’s a pretty simple classification of finitely generated Abelian groups, and there’s a relatively understandable classification of countably infinite Abelian $p$-groups for any prime $p$. But my question is, what is the general classification upto isomorphism of countably infinite Abelian groups?

How complicated is it to state? Is there a sentence we can write, like “Two countably infinite Abelian groups are isomorphic if and only if ...”?

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    $\begingroup$ Even classifying, for prime $p$, subgroups of $\mathbf{Z}[1/p]^2$ up to isomorphism, is quite complicated. Note that there are continuum many. $\endgroup$ – YCor Nov 29 '20 at 23:01
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There is no simple classification theorem for countably infinite Abelian groups.

There are many different theorems in the literature that make the sentence above precise in various ways, but here's one that I think is pretty compelling: Hjorth proved that the isomorphism relation for countable abelian groups (even torsion-free ones) is non-Borel.

Another result along the same line is a result of Thomas, which says that if you look at torsion-free countable abelian groups of finite rank, there is no Borel map that takes invariants that work for rank $n$ groups and produces invariants that work for rank $n+1$ groups. Thus the complexity problems gets strictly harder as the rank increases.

The results I've mentioned here show that the classification problem is hard even for torsion-free countable abelian groups, so you can imagine the situation certainly isn't any better for arbitrary ones.

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