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I have the following question: given that $f: \mathbb{Z}_{28},+ \to \mathbb{Z}_{10},+$ is a morphism (here $\mathbb{Z}_n$ is the set of integers modulo $n$), then what is the value of $f(\overline{16})$?

In some previous exercise, I have proven that the order of $f(g)$ divides the order of $g$, so I computed the order of $\overline{16}$ in $\mathbb{Z}_{28}$ and found that it should be 7. This would imply that the only possibility for $f(\overline{16})$ is zero, since $7$ can not be the order of an element in $\mathbb{Z}_{10}$... However, if I define the morphism $h$ mapping $\overline{1}$ to $\overline{3}$, I find that this is also a morphism.

This makes me doubt my solution, but also the statement of the order of $f(g)$. Where did I make a mistake?

$\textbf{EDIT:}$ I think I have made a mistake in my 'morphisme' mapping $\overline{1}$ onto $\overline{3}$, since $28 \cdot \overline{1} = \overline{0}$, but $28\cdot \overline{3} = \overline{4}$ (modulo 10)...

So is the answer then $f(\overline{16})$ equals zero?

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    Yes, it has to be $\bar{0}$. – 2017-02-12
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    Ah thank you! Do you know another way of achieving this result (whitout using that the order of the image divides the original order?) – 2017-02-12
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    No, I don't. I think this is the simplest way to justify this answer. – 2017-02-12
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    Really appreciate your quick answer, thanks! – 2017-02-12

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