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:

$\rho = \begin{bmatrix} i & \rho \left({i}\right) & \ldots & \rho^{k-1} \left({i}\right) \end{bmatrix}$

Note that the orbit $\left\{{i \quad \rho \left({i}\right) \quad \ldots \quad \rho^{k - 1} \left({i}\right)}\right\}$ 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.