$\hom_{\mathbb{Z}}(\mathbb{Q}, C) = 0$ for every cyclic group $C$

Question

This is part of an exercise I'm doing, exercise 2.22 Rotman, Introduction to homological algebra.

Prove that $$\hom_{\mathbb{Z}}(\mathbb{Q}, C) = 0$$ for every cyclic group $C$.

Any hint ?

Answer 1

$\mathbf{Q}$ is a divisible group.

Answer 2

The book already shows that $\textrm {Hom}_{\mathbb Z}(\mathbb Q, \mathbb Z) = 0$, so assume that $C=\mathbb Z/n\mathbb Z$. Let $f: \mathbb Q\to C$ be a morphism and compute \begin{align*} f(a/b) = f(na/nb) = nf(a/nb) = 0 \end{align*} for any rational $a/b$.