non-split sequence of torsion-free groups

Question

Can anyone give an example of a non-split short exact sequence of torsion-free abelian groups? Better if they are countable. If there is no such a thing, please explain why.

Context: I heard someone say that exact sequences determine the group in the middle up to torsion, so I started wondering to which extent this statement is true.

My best attempt was to include $\mathbb{Q}$ into its algebraic closure (to make things countable) and then project onto the quotient. This doesn't work because, indeed, as groups, $\overline{\mathbb{Q}}$ is the direct sum of the rationals and the quotient.

Answer 1

Another way of getting examples is to take any torsion-free, but not free, abelian group $G$, write it as a quotient of a free abelian group $F$, and let $K$ be the kernel of the quotient map $q:F\to G$. Then $$0\to K\to F\stackrel{q}{\to} G\to0$$ is a non-split short exact sequence.

For example, take $G=\mathbb{Q}$, $F$ the group of finite sequences $(a_n)_{n\geq1}$ of integers, with $$q\left((a_n)\right)=\sum_n\frac{a_n}{n},$$ so $$K=\left\{(a_n): \sum_n\frac{a_n}{n}=0\right\}.$$

Answer 2

Not an explicit example, but a proof that you can find plenty of examples.

Consider the exact sequence $\def\Q{\mathbb{Q}}\def\Z{\mathbb{Z}}0\to\Z\to\Q\to\Q/\Z\to0$ and apply to it the functor $\DeclareMathOperator\H{Hom}\H(-,\Z)$, getting the exact sequence $$\DeclareMathOperator\E{Ext} 0\to\H(\Z,\Z)\to\E(\Q/\Z,\Z)\to\E(\Q,\Z)\to0\tag{1} $$ You can also apply the functor $\H(\Q/\Z,-)$, getting the exact sequence $$ 0\to\H(\Q/\Z,\Q/\Z)\to\E(\Q/\Z,\Z)\to\E(\Q/\Z,\Q)\to0\tag{2} $$ Since $\H(\Q/\Z,\Q/\Z)$ is uncountable, also $\E(\Q/\Z,\Q)$ is uncountable, from $(2)$. On the other hand $\H(\Z,\Z)$ is countable, so we get from $(1)$ that $\E(\Q,\Z)\ne0$ (actually it is uncountable). In particular, there are nonsplit exact sequences $0\to\Z\to G\to\Q\to0$ where, of course, $G$ is torsion-free and countable.