CorollaryStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-11

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The order of a permutation is the least positive common multiple of its nontrivial cycle lengths, with value $1$ for the identity

Statement

Let the nontrivial cycles in the disjoint-cycle decomposition of a permutation $σ$ have lengths $d_{1}, \dots, d_{r}$ . The order of $σ$ is the least positive natural number divisible by every $d_{i}$ . For the identity, where $r = 0$ , the order is $1$ .

Facts & Assumptions

Given: A permutation $σ$ of a finite set and its order as the least positive exponent giving the identity.

[L1]

Every finite permutation has a disjoint-cycle decomposition, unique up to reordering and cyclic rotation (Every permutation of a finite set is a product of pairwise disjoint cycles, uniquely up to reordering and cyclic rotation).

[L2]

Cycles with disjoint supports commute (Cycles with disjoint supports commute).

[L3]

The order of an element is the least positive natural exponent giving the identity (The order $∣ G ∣$ of a finite group and the order $ord (g)$ of an element, with $ord (g) = \infty$ when no positive power of $g$ is the identity).

Proof

technique · direct

1.1

Write $σ = γ_{1} \dots γ_{r}$ as in [L1]. Since the factors commute by [L2], $σ^{k} = γ_{1}^{k} \dots γ_{r}^{k}$ for every natural $k$ .

givenL1L2

2.1

The $k$ -th power of a $d_{i}$ -cycle shifts its displayed entries by $k$ positions, so it is the identity exactly when $k \equiv 0 (m o d d_{i})$ , equivalently when $d_{i}$ divides $k$ . Because the supports are disjoint, $σ^{k}$ is the identity exactly when every $γ_{i}^{k}$ is the identity.

step 1.1L1L2

3.1

Thus the positive exponents giving the identity are precisely the positive common multiples of $d_{1}, \dots, d_{r}$ , so their least element is the order of $σ$ by [L3]. If $r = 0$ , then $σ$ is the identity and its order is $1$ .

step 2.1L1L3∎

Depends on

Used by

For prime p, a transitive subgroup of Sₚ containing a transposition is all of Sₚ Theorem

Dependency tree · two levels

19 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

J. S. Milne, Group Theory, §4, cycle decompositions (standard reference, not scraped)

The order of a permutation is the least positive common multiple of its nontrivial cycle lengths, with value 11 for the identity