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I saw this statement given in a solution that listed the orders of each element of the group of elements of the additive modulo group $\mathbb{Z}/12\mathbb{Z}$ where $G=\{0,1,2,...,11\}$ and the order for each corresponding element is given by $1,12,6,...$. How could I prove this?

Thanks

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Outline:

$• x\cdot\mid x\mid=\operatorname{lcm}(x,\mid G\mid)$

$•$ in general, $\operatorname{lcm} (m,n)\cdot\operatorname{gcd}(m,n)=m\cdot n$

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  • $\begingroup$ But, on your first point, in $\mathbb{Z}/12\mathbb{Z}$ under addition, $|2|\neq \text{ lcm} (2,12)$, since the order of $2$ should be $6$ $\endgroup$ – john fowles Sep 24 '18 at 6:02
  • $\begingroup$ You're right. I should have said $x\cdot \mid x\mid=\operatorname{lcm}(x,\mid G\mid) $... $\endgroup$ – Chris Custer Sep 24 '18 at 6:13
  • $\begingroup$ Ok. is there a proof for the first equality? $\endgroup$ – john fowles Sep 24 '18 at 7:11
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    $\begingroup$ Yes. It's pretty obvious because it's the smallest multiple of $x$ that is a multiple of $\mid G\mid$... $\endgroup$ – Chris Custer Sep 24 '18 at 16:06
  • $\begingroup$ Thanks! I was also wondering which groups this property, $\frac{|G|}{\text{ GCD}(x,|G|)}$ also holds for? $\endgroup$ – john fowles Sep 24 '18 at 22:15
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The order of an element $x$ in (the additive group) $\mathbb{Z}/n\mathbb{Z}$ is given by the least positive integer $m$ satisfying $mx = 0 \pmod n$. In other words, the smallest $m > 0$ for which there exists $y \in \mathbb{Z}$ satisfying $$mx = ny$$

To prove that $m = \dfrac{n}{\gcd(x,n)}$, you can first verify that $nx = \gcd(x,n)\cdot \text{lcm}(x,n)$, and that for any other $m$ satisfying the above equality, $mx$ must be a multiple of $\text{lcm}(x,n)$.

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