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I need to determine how many homomorphisms there are from $\Bbb{Z}_6 \to \Bbb{Z}_{18}$.

I have never solved that kind of question. I do know that orders are preserved and that some elements can be exchanged with other elements with the same properties.

I am a bit confused. A thorough explanation is more than welcome, as well as references and links.

Thank you.

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  • $\begingroup$ Do you know about group actions? $\endgroup$ – flawr Jan 25 '15 at 20:57
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    $\begingroup$ Allegedly, yes. $\endgroup$ – Donna Jan 25 '15 at 20:57
  • $\begingroup$ What do you suggest? $\endgroup$ – Donna Jan 25 '15 at 20:58
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HInt: We have that $\mathbb{Z}_6 = \langle \overline{1}\rangle$ and $\mathbb{Z}_{18} = \{\overline{1}, \ldots, \overline{17}\}$. then there exists a homomorphism $f: \langle \overline{1}\rangle \to \mathbb{Z}_{8}$ such that $f(a) = b$, if and only if, $\mathcal{O}(b)$ divides $\mathcal{O}(a)$.

Note: $\mathcal{O}$ stand for order of an element.

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  • $\begingroup$ I always see that noting $\overline{a}$. What does that mean? $\endgroup$ – Donna Jan 25 '15 at 21:06
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    $\begingroup$ It means it's a class whose representant is $a$. For example $\overline{2}$ in $\mathbb{Z}_6$ then it means that you're taking the class of all numbers in $\mathbb{Z}$ such that its remainder when divided by $6$ equals $2$. $\endgroup$ – Aaron Maroja Jan 25 '15 at 21:07
  • $\begingroup$ Okay, so just so I don't get confused and, both solutions are equivalent, right? $\endgroup$ – Donna Jan 25 '15 at 21:09
  • $\begingroup$ You mean the answers? Yes. $\endgroup$ – Aaron Maroja Jan 25 '15 at 21:11
  • $\begingroup$ Feel free to ask, if you have any questions. $\endgroup$ – Aaron Maroja Jan 25 '15 at 21:12
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Since $\Bbb Z_6$ is cyclic and generated by $1$, the value $f(1)$ determines the homomorphism. On the other hand, $f(1)$ must have order $6$ or a divisor of $6$, so $f(1)\in\{0,3,6,9,12,15\}$. This yields six homomorphisms.

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