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Let $n \in \mathbb{N}$, $\mathbb{Z}_n = \mathbb{Z} / n\mathbb{Z}$.
For $m \in \mathbb{Z}$, what is the order of $[m]$?

To my understanding, quiet naive question, but I cannot really find out how to phrase it.
This is what I tried:
Suppose the order is k, then $k[m] = [km] = [0]$ by definition of order, which forces either $k=0$ or $m=0$. The latter gives order of $[0]$ (is this even a valid statement?), and the former is just the order doesn't exists, i.e. infinity.
Any suggestions?

Bill Dubuque
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Eva Lin
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  • It is not true that either $k=0$ or $m=0$. The element $[km]=[0]$ just means that $km$ is a multiple of $n$. For example, when $n=6$ and $m=3$, we get $k=2$. I hope that makes sense! – Coherent Sheaf Oct 02 '23 at 04:25
  • @CoherentSheaf That make sense...I totally overlooked that, my bad. – Eva Lin Oct 02 '23 at 04:27
  • By the linked dupe $,km\equiv0\pmod{!n}\iff n\mid km\iff n/\gcd(n,m)\mid k \iff {\rm lcm}(n,m)/m\mid k.\ $ (this is presuming you seek the additive order, as written). – Bill Dubuque Oct 02 '23 at 04:32
  • Well, you first have to ensure that $m$ and $n$ are coprime. You can then prove that $m$ must possess an order by noting that there are finitely many congruence classes and appealing to the pigeonhole principle. But the question itself is a little strange, as it asks you to find the order of $m$ for any $m$ (and presumably any $n$), but of course they will not all have the same order for any given value of $n$... – H. sapiens rex Oct 02 '23 at 04:36
  • @H.sap The question as written concerns the additive order, i.e. the order in the additive group $\Bbb Z.\ \ $ – Bill Dubuque Oct 02 '23 at 04:37
  • @BillDubuque thanks a lot for the suggestion! I get it now. I actually have to solve quiet similar in another part of the question (to find all generator of $\mathbb{Z}_n$). I was just very confused about the wording originally. – Eva Lin Oct 02 '23 at 04:48

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