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This question already has an answer here:

All group homomorphism from $ \mathbb{Z} _m $ to $ \mathbb{Z}_n $

How could I find every group homomorphism?

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marked as duplicate by Mikko Korhonen, The Chaz 2.0, Alexander Gruber, rschwieb, Davide Giraudo Apr 25 '13 at 14:28

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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The map $\phi : Z_m \to Z_n$ given by $\phi(r\mod m) = k r\mod n $ is a well-defined homomorphism if and only if $n$ divides $km$. Every homomorphism from $Z_m \to Z_n$ is of this form.

A homomorphism is determined by $\phi(1)$ in this case. (Of course, if the group has more generators, you should check all the generators.)

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A homomorphism from the cyclic group $\mathbb{Z}_m$ into any other group is determined by where it sends a generator. The generator must be sent to an element whose order divides $m$.

In the case of this problem, let $d=\operatorname{gcd}(m,n)$. For every $d'|d$, there exists a unique subgroup $\mathbb{Z}_{d'}\le\mathbb{Z}_n$. Map a generator of $\mathbb{Z}_m$ to each generator of $\mathbb{Z}_{d'}$ for all $d'|d$ to obtain all possible homomorphisms. There are $\varphi(d')$ generators for $\mathbb{Z}_{d'}$, so the total number group homomorphisms is $$\sum_{d'\mid d}\varphi(d')=d$$ Here $\varphi$ is Euler's totient function. The above formula is a standard property of Euler's function.

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  • $\begingroup$ Why $ d ' $ is considered as $ d'|d $? $\mathbb{Z}_k$ where $ k|n $ can be a subgroup of $ \mathbb{Z}_n$. Then I suppose that $ d' $ would satisfy the condition of $ d'|n $ $\endgroup$ – user73309 Apr 25 '13 at 14:06
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Hint: each homomorpism $f$ is uniquely determined by $f(1)$.

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