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

if and only if:

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

Proof

Necessary Condition

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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.)

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$\Box$

Also see

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

Sources

• 1966: Richard A. Dean: Elements of Abstract Algebra ... (previous) ... (next): $\S 1.6$