= Finite cyclic group
A <finite cyclic group> of order $n$ is isomorphic to the additive <group> $C_n=\mathbb Z/n\mathbb Z$. The <group isomorphism> sends the <congruence class> of $k$ to $g^k$, where $g$ is a chosen <generator of a group>. For $n=6$, the <identity element> is $0$, $4+5=3$ modulo six, and the <inverse element> of $4$ is $2$.
One <finite cyclic group> can have several choices of <generator of a group>. Adding $1$ visits all six elements before returning to zero; adding $5$ also visits all six, in the reverse order. Adding $2$ reaches only $\{0,2,4\}$ from zero, and adding $3$ reaches only $\{0,3\}$. The following <permutations> of the six elements show what each repeated step can reach.
\Image[/cyclic-group-generator-cycles.png]
{title=Repeated addition modulo six: one cycle, two cycles, or three cycles}
{description=Each arrow applies $x\mapsto x+k$ modulo six. The panels use $k=1,2,3$. Colors distinguish the separate cycles; only the cycle containing zero is the <cyclic subgroup> generated by $k$. For $k=2$, the other cycle is the <coset> $\{1,3,5\}$; for $k=3$, the other two cycles are its other <cosets>.}
This gives a precise version of a familiar scheduling question: does a fixed step visit every position, or repeatedly miss some? The <generators of a finite cyclic group> are exactly the steps <coprime> to the total number of positions. Changing the step changes the orbit through the same underlying <group>; the diagram does not replace its <group operation> with a different one.
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