# 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

## 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

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