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?

  • $\begingroup$ 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? $\endgroup$ – 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.

  • 1
    $\begingroup$ Is that really relevant? The answer would be yes anyway. $\endgroup$ – Derek Holt May 14 '13 at 8:04
  • $\begingroup$ @Derek: What question would you answer "yes"? Will you marry me? I think I didn't ask that here. Homomorphisms [of abelian groups] map torsion elements to torsion elements. In torsion-free groups this means just the zero element. $\endgroup$ – Asaf Karagila May 14 '13 at 16:10
  • $\begingroup$ The question asked was "does there exist a homomorphism from some group to some other group", and the answer to that would be yes irrespective of what the groups are. I know that your answer is providing more information, which is likely to be helpful, but I am a great believer in answering the question asked - so my comment should really have been aimed at the question asked rather than at your answer. $\endgroup$ – Derek Holt May 14 '13 at 17:55

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, $\varphi =0$.


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