# When is the torsion subgroup of an abelian group a direct summand?

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.

• 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 $(1,p,p^2,\ldots)$? Apr 4, 2011 at 8:40
• In fact, by the same argument, the torsion subgroup seems to be uncountable, so very far from $\bigoplus\mathbb{Z}/p^n\mathbb{Z}$. Apr 4, 2011 at 10:55
• @S, @Alex: thanks; I changed this to what I really meant. Apr 4, 2011 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. Apr 4, 2011 at 18:36
• [the bit I'm missing is "what does the Pont. dual of a torsion-free discrete group look like?"] Apr 4, 2011 at 18:37

• 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. Apr 4, 2011 at 11:51
• The link to eom.springer.de is broken, but the article can now be found at encyclopediaofmath.org/wiki/Cotorsion_group. Jul 24 at 11:51