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 $20$.


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

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Shaun, Gibbs, José Carlos Santos, John B, Brahadeesh
If this question can be reworded to fit the rules in the help center, please edit the question.

  • 1
    $\begingroup$ $1$ has order $10$ in $\mathbb{Z}_{10}$. $\endgroup$ – Randall Dec 10 '18 at 5:18

Hint: $\operatorname{Im}(g)$ is a subgroup of $U_{20}$ and $1$ (as an element of $\Bbb Z_{10}$) generates $\Bbb Z_{10}$.


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


Not the answer you're looking for? Browse other questions tagged or ask your own question.