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

Question

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

Answer 1

It's fairly easy to see that in a cyclic group,$\langle g\rangle$, the order of $g^k$ is given by $|g^k|=\dfrac{|g|}{\gcd(|g|,k)}$.

But $\Bbb Z_n=\langle1\rangle$. Thus $x=1^x$ for any $x\in\{0,\dots,n-1\}$.

Substituting, we get $|x|=|1^x|=\dfrac n{\gcd(n,x)}$.

Don't let the multiplicative notation throw you.

Answer 2

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