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Why is that if $g:\mathbb{Z_{10}}$$\rightarrow$$U_{20}$ is a group homomorphism, then the order of $g(1)$ is either $1$ or $2$?

Also, $g$ is a function, $\mathbb{Z_{10}}$ is the group of integers modulo $10$, and $U_{20}$ is the group of units modulo $20$.

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closed as off-topic by Shaun, Gibbs, José Carlos Santos, John B, Brahadeesh Dec 10 '18 at 12:07

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    $\begingroup$ $1$ has order $10$ in $\mathbb{Z}_{10}$. $\endgroup$ – Randall Dec 10 '18 at 5:18
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Hint: $\operatorname{Im}(g)$ is a subgroup of $U_{20}$ and $1$ (as an element of $\Bbb Z_{10}$) generates $\Bbb Z_{10}$.

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In fact, $U(20)=\mathbb Z_4\oplus \mathbb Z_2$. So $\operatorname{Im}g=\langle g(1)\rangle $ has order dividing $10$ and $4$.

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