TITLE: Prove that there is no homomorphism $f:\mathbb { Z } / n \mathbb { Z }\rightarrow \mathbb { Z }$
QUESTION [5 upvotes]: Problem Prove that there is no homomorphism $f:\mathbb { Z } / n \mathbb { Z } \rightarrow \mathbb { Z }$

My attempt: By contradiction, let's suppose that there exists as such. We then have $0 = f(0) = f(n \cdot \overline { 1 })=n\cdot f(\overline { 1 })= n$. Absurd. 
What do you think of my solution?

REPLY [6 votes]: A ring homomorphism  $f:\mathbb { Z } / n \mathbb { Z }\rightarrow \mathbb { Z }$ induces a homomorphism of additive groups $\mathbb { Z } / n \mathbb { Z }\rightarrow \mathbb { Z }$.
A slight generalization of your argument proves that this group homomorphism is the zero homomorphism and so $f$ is the zero map. Whether the zero map qualifies as a ring homomorphism depends on your definitions.