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?




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

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

  • $\begingroup$ 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
  • 1
    $\begingroup$ 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

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 $\text{lcm}(x,n)$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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