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TheoremStatement: Literature-sourcedProof: Literature-sourcedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-13
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Two elements of Sn are conjugate if and only if they have the same cycle type

Statement

For σ,τSn, there is a gSn with τ=gσg1 if and only if σ and τ have the same cycle type, including their numbers of fixed points.

Facts & Assumptions

Given: Permutations σ,τSn.

[F1]

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

[F2]

Cycle type records the number ck of cycles of each length k, including 1-cycles (Support, fixed points, disjoint cycles, cycle length, disjoint-cycle decompositions, and cycle type).

Proof

technique · iff
1.1

Suppose τ=gσg1. Conjugate every factor in the disjoint-cycle decomposition of σ; [F3] preserves each cycle length and relabels fixed points.

F1F3
1.2

Conversely, suppose the cycle types agree. By [F1] and [F2], pair every cycle of σ with a cycle of τ of the same length, including the fixed points, and define g by sending entries positionwise in each pair.

F1F2
2.1

Hence σ and τ have the same multiplicities ck and the same cycle type.

F2step 1.1
2.2

The paired cycles partition the underlying set, so the map in step 1.2 is a bijection and hence an element of Sn.

step 1.2algebra
3.1

Applying [F3] to each paired cycle gives gσg1=τ.

F3step 1.2step 2.2

Depends on

Used by

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 34 results over 9 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources