Prove that $\Bbb Q$, as a $\Bbb Z$-module, is a direct summand in a direct product of copies of $\Bbb Q/\Bbb Z$.
This is a problem from Hilton and Stammbach's Homological Algebra. If this is true, then there exists a homomorphism from the direct product of some copies of Q/Z to Q. But if the number of the copies is infinite, it isn't necessary that only a finite number of components have non-zero values. So how to define this homomorphism to Q?
On
We know every abelian group sits inside a direct product of copies of $\mathbb{Q}/\mathbb{Z}$, hence we have an exact sequence $$0 \to \mathbb{Q} \to \prod_{i \in I} (\mathbb{Q}/\mathbb{Z})_{i} \to ( \ \prod_{i \in I}(\mathbb{Q}/\mathbb{Z})_{i}) /\mathbb{Q} \to 0.$$ Now since $\mathbb{Q}$ is divisible, hence injective, the sequence splits. We are done.
Consider the direct product $G=(\mathbb{Q}/\mathbb{Z})^{\mathbb{N}}$; this group is not torsion, because you can take the element $$ x=\left(\frac{1}{n+1}+\mathbb{Z}\right)_{n\in\mathbb{N}} $$ which is not annihilated by any integer. Thus $G/t(G)\ne0$, where $t(G)$ denotes the torsion subgroup of $G$. But $G/t(G)$ is a torsion free divisible group, so it is a direct sum of copies of $\mathbb{Q}$. So $\mathbb{Q}$ is a direct summand of $G/t(G)$. Since $t(G)$ is divisible, it splits, so $G=t(G)\oplus H$, where $H\cong G/t(G)$.