= Order of an element of a finite cyclic group
{title2=$\operatorname{ord}(k)=n/\gcd(n,k)$}
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> is
$$
\operatorname{ord}(k)=\frac{n}{\gcd(n,k)}.
$$
Indeed, write $d=\gcd(n,k)$, $n=dn'$ and $k=dk'$. Returning to zero after $m$ steps means $n\mid mk$, equivalently $n'\mid mk'$. Since $n'$ and $k'$ are <coprime>, this holds exactly when $n'\mid m$; the least positive such $m$ is $n'$. 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.
Back to article page
