# Why is group homomorphism determined by $\phi(1)$?

A lot of solutions to problems say that for a cyclic group, such as $\mathbb{Z}/\mathbb{Z}_3$, $\mathbb{Z}/\mathbb{Z}_{10}$, etc., a group homomorphism $\phi$ from $\mathbb{Z}/\mathbb{Z}_m$ to $\mathbb{Z}/\mathbb{Z}_n$ is determined by $\phi(1)$, but I never really understood why... can someone help me? Thanks so much in advance!

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## 2 Answers

Because $\Bbb Z/n\Bbb Z$ is generated by $1$. Thus, $\varphi(2)=\varphi(1+1)=\varphi(1)+\varphi(1)$, and similarly for every other element of $\Bbb Z/n\Bbb Z$.

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Thanks, this all makes sense but I guess my question is more what exactly does φ(1) determine and tell us specifically about this specific homomorphism? I am so lost =/ – arcastar Oct 11 '12 at 5:36
@arcastar: $\varphi(1)$ completely determines the homomorphism. This simply means that if you know $\varphi(1)$, then you know $\varphi(a)$ for every $a\in\Bbb Z/n\Bbb Z$; in other words, you know the whole function. That one value pins down every value, leaving no wiggle room. – Brian M. Scott Oct 11 '12 at 5:38
I just got the notification for this new post, so postponed. But thank you so much! It was really helpful! – arcastar Oct 11 '12 at 6:52

Hint: For example $$\varphi(2)=\varphi(1+1)=\varphi(1)+\varphi(1)$$

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