# Permutation is Cyclic iff At Most One Non-Trivial Orbit

## Theorem

Let $S$ be a set.

Let $\rho: S \to S$ be a permutation on $S$.

Then:

$\rho$ is a cyclic permutation
$S$ has no more than one orbit under $\rho$ with more than one element.

## Proof

### Sufficient Condition

Recall the definition of cyclic permutation.

Let:

$\rho = \begin {bmatrix} i & \map \rho i & \ldots & \map {\rho^{k - 1} } i \end{bmatrix}$

Note that the orbit $\set {i, \map \rho i, \ldots, \map {\rho^{k - 1} } i}$ is the only non-trivial orbit.

(If $k = 1$, then the above-mentioned orbit is also trivial.)

$\Box$

## Also see

Some sources use this result as a definition for a cyclic permutation.