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Two elements of $S_{n}$ are conjugate if and only if they have the same cycle type

Statement

For $σ, τ \in S_{n}$ , there is a $g \in S_{n}$ with $τ = g σ g^{- 1}$ if and only if $σ$ and $τ$ have the same cycle type, including their numbers of fixed points.

Facts & Assumptions

Given: Permutations $σ, τ \in S_{n}$ .

[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 $c_{k}$ of cycles of each length $k$ , including $1$ -cycles (Support, fixed points, disjoint cycles, cycle length, disjoint-cycle decompositions, and cycle type).

[F3]

Conjugating a cycle by $g$ relabels all its entries by $g$ (Conjugating a cycle relabels each entry: $g (a_{1} \dots a_{k}) g^{- 1} = (g (a_{1}) \dots g (a_{k}))$ ).

Proof

technique · iff

1.1

Suppose $τ = g σ g^{- 1}$ . 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 $c_{k}$ 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 $S_{n}$ .

step 1.2algebra

3.1

Applying [F3] to each paired cycle gives $g σ g^{- 1} = τ$ .

F3step 1.2step 2.2∎

Depends on

Used by

All 3-cycles are conjugate in Aₙ for n≥5 Corollary
The conjugacy classes of Sₙ are indexed by the tuples (c₁,…,cₙ) with ∑ k cₖ=n Corollary
Z(Sₙ) is trivial for n≥3 Corollary
A₄ has a normal Klein four subgroup and four conjugacy classes Example
Conjugating (1 4)(2 5 3) by an explicit permutation in S₅ Example
The Sylow subgroups of S₄ Example
False statement: all subgroups of the same p-power order are conjugate False statement

Dependency tree · two levels

11 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

K. Conrad, Conjugacy Classes (standard reference, not scraped)

Two elements of SnSn​ are conjugate if and only if they have the same cycle type