Order of an element of a finite cyclic group ($\mathrm{ord}(k)=n/\gcd (n,k)$)

For a positive integer $n$ and a congruence class $k$ in the additive finite cyclic group $\mathbb{Z}/n\mathbb{Z}$, the order of a group element isIndeed, write $d=\gcd (n,k)$, $n=d{n}^{\prime }$ and $k=d{k}^{\prime }$. Returning to zero after $m$ steps means $n\mid mk$, equivalently $n^{\prime }\mid m{k}^{\prime }$. Since $n^{\prime }$ and $k^{\prime }$ are coprime, this holds exactly when $n^{\prime }\mid m$; the least positive such $m$ is $n^{\prime }$. This argument also covers $k=0$, since $\gcd (n,0)=n$.

For $n=6$, the orders of $0,1,2,3,4,5$ are respectively $1,6,3,2,3,6$. The translation in a group $x\mapsto x+k$ has the same period from every starting element. Consequently its cycles each have length $n/d$ and there are $d$ of them. They are exactly the cosets of the cyclic subgroup generated by $k$, which explains both the lengths and the number of components in the picture.