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In looking up this question, I found this site: Physics Forums. In it, someone claims that $f(x) = kx$ is a homomorphism from the group $\mathbb{Z}_{m}$ to $\mathbb{Z}_{n}$ if $m$ divides $kn$. I assume this is all with the group operation being addition. However, in the case of $m=10, n=15$ it seems that $k=2$ doesn't work. I found that $f(1+9) = f(0) = 0\ne f(1) + f(9) = 2 + 18 = 2+3 = 5$. Am I misunderstanding the claim or not computing something correctly?

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I think you've misunderstood the claim. $f([x]) = [kx]$ is a group homomorphism from $\mathbb{Z}_n$ to $\mathbb{Z}_m$ if $m | kn$. You've got $n$ and $m$ switched in the claim, although not in your question title.

In your example, $f([x]_{15}) = [2x]_{10}$ is indeed well-defined and a group homomorphism $\mathbb{Z}_{15} \rightarrow \mathbb{Z}_{10}$ (not the other way around).

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