Permutation is Cyclic iff At Most One Non-Trivial Orbit

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Theorem

Let $S$ be a set.

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


Then:

$\rho$ is a cyclic permutation

if and only if:

$S$ has no more than one orbit under $\rho$ with more than one element.


Proof

Necessary Condition


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.


Sources