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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:

2) $T$ 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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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 $(1,p,p^2,\ldots)$? –  S. Carnahan Apr 4 '11 at 8:40
    
In fact, by the same argument, the torsion subgroup seems to be uncountable, so very far from $\bigoplus\mathbb{Z}/p^n\mathbb{Z}$. –  Alex B. Apr 4 '11 at 10:55
    
@S, @Alex: thanks; I changed this to what I really meant. –  Pete L. Clark Apr 4 '11 at 11:41
    
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. –  Kevin Buzzard Apr 4 '11 at 18:36
    
[the bit I'm missing is "what does the Pont. dual of a torsion-free discrete group look like?"] –  Kevin Buzzard Apr 4 '11 at 18:37
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1 Answer

up vote 13 down vote accepted

These are the (torsion) cotorsion groups. By a theorem of Baer all torsion abelian groups that are cotorsion are direct sums of a divisible group and a group of bounded order.

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.

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

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@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...) –  Pete L. Clark Apr 4 '11 at 11:46
    
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. –  Pete L. Clark Apr 4 '11 at 11:51
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