# For the additive group modulo $n$, what's the proof that the order of each element is given by $\frac{|G|}{\text{ GCD}(x,|G|}$

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

Outline:

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

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

• 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 $6$ – john fowles Sep 24 '18 at 6:02
• You're right. I should have said $x\cdot \mid x\mid=\operatorname{lcm}(x,\mid G\mid)$... – Chris Custer Sep 24 '18 at 6:13
• Ok. is there a proof for the first equality? – john fowles Sep 24 '18 at 7:11
• Yes. It's pretty obvious because it's the smallest multiple of $x$ that is a multiple of $\mid G\mid$... – Chris Custer Sep 24 '18 at 16:06
• Thanks! I was also wondering which groups this property, $\frac{|G|}{\text{ GCD}(x,|G|)}$ also holds for? – 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 $$\text{lcm}(x,n)$$.