1. **Problem Statement:** Show that $\mathrm{Hom}_{\mathbb{Z}}(\mathbb{Z}/n\mathbb{Z}, \mathbb{Q}) = 0$. 2. **Recall the definitions:** - $\mathbb{Z}/n\mathbb{Z}$ is the cyclic group of integers modulo $n$. - $\mathbb{Q}$ is the additive group of rational numbers. - $\mathrm{Hom}_{\mathbb{Z}}(A,B)$ denotes the group of group homomorphisms from $A$ to $B$ that respect the $\mathbb{Z}$-module structure. 3. **Key fact:** Any homomorphism $f : \mathbb{Z}/n\mathbb{Z} \to \mathbb{Q}$ is determined by the image of $1 + n\mathbb{Z}$. 4. **Use the property of the domain:** Since $1 + n\mathbb{Z}$ has order $n$ in $\mathbb{Z}/n\mathbb{Z}$, we have $$n \cdot (1 + n\mathbb{Z}) = 0.$$ 5. **Apply the homomorphism:** For $f$ to be a group homomorphism, $$f(n \cdot (1 + n\mathbb{Z})) = n \cdot f(1 + n\mathbb{Z}) = f(0) = 0.$$ 6. **Implication:** This means $$n \cdot f(1 + n\mathbb{Z}) = 0$$ in $\mathbb{Q}$. 7. **Since $\mathbb{Q}$ is torsion-free:** The only element annihilated by a nonzero integer $n$ is $0$. 8. **Therefore:** $$f(1 + n\mathbb{Z}) = 0,$$ which implies $f$ is the zero homomorphism. 9. **Conclusion:** The only homomorphism from $\mathbb{Z}/n\mathbb{Z}$ to $\mathbb{Q}$ is the zero map, so $$\mathrm{Hom}_{\mathbb{Z}}(\mathbb{Z}/n\mathbb{Z}, \mathbb{Q}) = 0.$$