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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.

Pete L. Clark
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