# Homomorphism from $\mathbb{Z}/n\mathbb{Z}$

Does there exist a non-zero homomorphism from $\mathbb{Z}/n\mathbb{Z}$ to $\mathbb{Z}$ ? If yes, state the mapping. How is this map exactly?

• This is a strange question, because for any two groups $G$ and $H$, there always exists a homomorphism from $G$ to $H$. Are you sure you have the question right? – Derek Holt May 14 '13 at 12:36

Suppose there is a homomorphism $f: \mathbb{Z}/n\mathbb{Z} \to \mathbb{Z}$. What could $f(1)$ be? Let's call it $a$. Then $f(2) = f(1+1) = f(1)+f(1) = a+a = 2a$, and likewise $f(3) = 3a$, $\dots,$ $f(\underbrace{1+\cdots+1}_{n\text{ times}}) = na$. But $\underbrace{1+\cdots+1}_{n\text{ times}} = 0$, and $f(0) = 0$. So what can $a$ be?
Once you've worked out which value(s) of $a$ is/are allowed, can you check that the associated map $f$ is a homomorphism? (We've already done this, really, but make sure you're sure of that.)
Hint: Every element of $\Bbb{Z/nZ}$ has finite order.
Let $\varphi : \mathbb{Z}_n \to \mathbb{Z}$ be a homomorphism. Using isomorphism theorem, $\mathbb{Z}_n / \text{ker}(\varphi)$ is isomorphic to $\text{Im}(\varphi)$. Because $\text{Im}(\varphi) \leq\mathbb{Z}$, either $\text{Im}(\varphi)= \{0\}$ or $\text{Im}(\varphi) \simeq \mathbb{Z}$; on the other hand, $\mathbb{Z}_n/ \text{ker}(\varphi)$ is necessarily a finite group. Therefore, $\varphi =0$.