1. **Problem statement:** Show that $\mathrm{Hom}_{\mathbb{Z}}(\mathbb{Z}/n\mathbb{Z}, \mathbb{Q}) = \mathbb{Q}$. 2. **Recall 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 (i.e., group homomorphisms). 3. **Key fact:** Any homomorphism $f: \mathbb{Z}/n\mathbb{Z} \to \mathbb{Q}$ is determined by the image of the generator $1 + n\mathbb{Z}$. 4. **Condition on the image:** Since $f$ is a group homomorphism, for the generator $1 + n\mathbb{Z}$, we have $$f(n \cdot (1 + n\mathbb{Z})) = f(0) = 0,$$ which implies $$n f(1 + n\mathbb{Z}) = 0.$$ 5. **Since $\mathbb{Q}$ is torsion-free:** The only element $q \in \mathbb{Q}$ satisfying $nq=0$ is $q=0$. 6. **Therefore:** $$f(1 + n\mathbb{Z}) = 0,$$ which means the only homomorphism is the zero map. 7. **Conclusion:** $$\mathrm{Hom}_{\mathbb{Z}}(\mathbb{Z}/n\mathbb{Z}, \mathbb{Q}) = \{0\}.$$ **Note:** The original claim $\mathrm{Hom}_{\mathbb{Z}}(\mathbb{Z}/n\mathbb{Z}, \mathbb{Q}) = \mathbb{Q}$ is false. The correct result is the zero group. If the problem intended $\mathrm{Hom}_{\mathbb{Z}}(\mathbb{Q}, \mathbb{Z}/n\mathbb{Z})$, that is also zero since $\mathbb{Q}$ is divisible and $\mathbb{Z}/n\mathbb{Z}$ is torsion. Hence, the hom group is trivial, not $\mathbb{Q}$.