Proving that $0 \rightarrow \Bbb Z \rightarrow \Bbb Q \rightarrow \Bbb Q / \Bbb Z \rightarrow 0$ does not split.

Question

Can I prove that the $\Bbb Z$-module exact sequence $0 \rightarrow \Bbb Z \rightarrow \Bbb Q \rightarrow \Bbb Q / \Bbb Z \rightarrow 0$ is non-split exact by proving that $\Bbb{Z} \bigoplus \Bbb{Q} / \Bbb{Z}$ is not isomorphic to $\Bbb{Q}$? The way I prove this is by noticing that in $\Bbb{Z} \bigoplus \Bbb{Q} / \Bbb{Z}$, there exist elements of order $a$ in which $a$ is the smallest natural number that satisfy $ar \in \Bbb Z$, where $r \in \Bbb Q$ (e.g. order of $(0, \frac{1}{2} + \Bbb Z)$ is $2$), whereas the order of elements in $\Bbb Q$ is either $1$ or $\infty$, showing that they are not isomorphic to each other. Is this correct?

Answer 1

There are many equivalent criterion that determine when a sequence splits. I think the most-common definition is the existence of a splitting map — in this case a map $f\colon \Bbb Q/\Bbb Z\to\Bbb Q$ such that $g\circ f=\operatorname{Id}_{\Bbb Q/\Bbb Z}$, where $g\colon \Bbb Q\to\Bbb Q/\Bbb Z$ is the quotient map.

In this case, for any $x\in \Bbb Q/\Bbb Z$, we know that $nx=0$ for some integer $n$. So if $f$ is a $\Bbb Z$-module map, then $nf(x)=f(nx)=0$. In $\Bbb Q$, this means $f(x)=0$, so $f$ must be the zero map.

As you already noted, though, the existence of a splitting map implies that $\Bbb Z\oplus \Bbb Q/\Bbb Z\cong\Bbb Q$, which is of course absurd as $\Bbb Q$ is torsion-free. You can see how both proofs capture the same idea.