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For an abelian group $G$, let $G[\operatorname{tors}]$ be its torsion subgroup.

Consider the torsion sequence:

$0 \rightarrow G[\operatorname{tors}] \rightarrow G \rightarrow G/G[\operatorname{tors}] \rightarrow 0$.

For which torsion abelian groups $T$ is it the case that for all abelian groups $G$ with $G[\operatorname{tors}] \cong T$, the torsion sequence splits?

I know some sufficient conditions:

  1. $T$ is divisible. Indeed, this holds iff $T$ is injective as a $\mathbb{Z}$-module iff any short exact sequence $0 \rightarrow T \rightarrow G \rightarrow G/T \rightarrow 0$ splits.

Thus divisibility is necessary and sufficient if one considers arbitrary short exact sequences, but in the special case $T = G[\operatorname{tors}]$ divisibility is not necessary. The torsion sequence also splits if:

  1. $T$ has bounded order: $T = T[n]$ for some $n \in \mathbb{Z}^+$. (For this see e.g. see Corollary 20.14 of these notes of K. Igusa.)

I do know some examples where the torsion sequence does not split, e.g., when $G = \prod_{n=1}^{\infty} \mathbb{Z}/p^n \mathbb{Z}$.

But in fact I am interested in the case in which $T$ has "cofinite type", i.e., $T$ can be injected into $(\mathbb{Q}/\mathbb{Z})^n$ for some $n \in \mathbb{Z}^+$. (I am making up the terminology here; if I ever knew what the infinite abelian group people call this, it's not coming to mind at the moment.)

So for instance the simplest case that I don't know at the moment would be something like $T = \mathbb{Z}/p\mathbb{Z} \oplus \mathbb{Q}_p/\mathbb{Z}_p$.

Not that it makes any difference as to what the answer is, but I would be very pleased to hear that the torsion sequence splits whenever $G[\operatorname{tors}]$ has "cofinite type". If you care why, see Theorem 5 here.

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    $\begingroup$ I am having difficulty seeing why the subgroup in your nonsplit extension is in fact the torsion subgroup. Can't you have an element of order $p$ of the form $(1,p,p^2,\ldots)$? $\endgroup$
    – S. Carnahan ♦
    Commented Apr 4, 2011 at 8:40
  • $\begingroup$ In fact, by the same argument, the torsion subgroup seems to be uncountable, so very far from $\bigoplus\mathbb{Z}/p^n\mathbb{Z}$. $\endgroup$
    – Alex B.
    Commented Apr 4, 2011 at 10:55
  • $\begingroup$ @S, @Alex: thanks; I changed this to what I really meant. $\endgroup$
    – Pete L. Clark
    Commented Apr 4, 2011 at 11:41
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    $\begingroup$ Pete: you can translate your question into an equivalent one by taking Pontrjagin duals. Your discrete group $G$ becomes a compact group $C$, the torsion subgroup of cofinite type becomes a topologically finitely generated profinite group which is a quotient of $C$, and the torsion-free part is...umm...some sub of $C$ which I don't understand very well but perhaps google could help...maybe at least it gives you another way of thinking about the problem. $\endgroup$
    – Kevin Buzzard
    Commented Apr 4, 2011 at 18:36
  • $\begingroup$ [the bit I'm missing is "what does the Pont. dual of a torsion-free discrete group look like?"] $\endgroup$
    – Kevin Buzzard
    Commented Apr 4, 2011 at 18:37

1 Answer 1

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These are the (torsion) cotorsion groups. The following follows from a theorem of Baer:

A torsion abelian group is cotorsion if and only it is direct sum of a divisible torsion abelian group and an abelian group of bounded exponent.

The original paper of R. Baer is "The subgroup of the elements of finite order of an abelian group", Ann. of Math. 37 (1936), 766-781. (See in particular Theorem 8.1.)

[I have made Baer's paper available here. --PLC]

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  • $\begingroup$ @Gjergji: thanks, this is very helpful. Do you happen to know whether every "cofinite type" torsion group is a cotorsion group? (I guess I will learn the answer to this by reading the relevant parts of Fuchs' book...) $\endgroup$
    – Pete L. Clark
    Commented Apr 4, 2011 at 11:46
  • $\begingroup$ Wait, never mind -- I guess the answer is obviously no: the group $\bigoplus_{p \in \mathcal{P}} \mathbb{Z}/p\mathbb{Z}$ (where $\mathcal{P}$ is the set of all prime numbers) is a counterexample. $\endgroup$
    – Pete L. Clark
    Commented Apr 4, 2011 at 11:51
  • $\begingroup$ The link to eom.springer.de is broken, but the article can now be found at encyclopediaofmath.org/wiki/Cotorsion_group. $\endgroup$
    – The Amplitwist
    Commented Jul 24, 2022 at 11:51

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