TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-13

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Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

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If $σ \in S_{n}$ has $c_{k}$ cycles of length $k$ , then $∣ C_{S_{n}} (σ) ∣ = \prod_{k = 1}^{n} k^{c_{k}} c_{k}!$

Statement

If $σ \in S_{n}$ has $c_{k}$ cycles of length $k$ , including fixed points when $k = 1$ , then $∣ C_{S_{n}} (σ) ∣ = \prod_{k = 1}^{n} k^{c_{k}} c_{k}! .$ The formula uses the empty product $1$ when $n = 0$ .

Facts & Assumptions

Given: A permutation $σ \in S_{n}$ of cycle type $(c_{1}, \dots, c_{n})$ .

[F1]

The centralizer consists of the permutations commuting with $σ$ (The conjugacy class ${Cl}_{G} (x)$ and centralizer $C_{G} (x)$ of an element).

[F2]

Every permutation has a disjoint-cycle decomposition unique up to reordering and cyclic rotation, and its cycle type counts the orbits of each length, including fixed points as $1$ -cycles (Every permutation of a finite set is a product of pairwise disjoint cycles, uniquely up to reordering and cyclic rotation, Support, fixed points, disjoint cycles, cycle length, disjoint-cycle decompositions, and cycle type).

[F3]

There are $c_{k}!$ bijections of a $c_{k}$ -element set (A finite set $A$ with $∣ A ∣ = n$ has exactly $n!$ bijections onto itself, and $n!$ bijections onto any set of the same cardinality).

[F4]

The cardinality of a finite product of finite choice sets is the product of their cardinalities, with the empty product having cardinality $1$ (The product rule: $∣ A \times B ∣ = ∣ A ∣ ∣ B ∣$ , and $∣ \prod_{i < m} A_{i} ∣ = \prod_{i < m} ∣ A_{i} ∣$ ).

Proof

technique · counting

1.1

If $g σ = σ g$ and $O$ is a $⟨ σ ⟩$ -orbit, then $g (O)$ is another orbit of the same size.

F1F2algebra

2.1

For the $c_{k}$ orbits of size $k$ , $g$ may permute those orbits in $c_{k}!$ ways by [F3]. Once a target orbit is chosen, the image of one marked point has $k$ choices, and commutation forces all other images; hence there are $k^{c_{k}} c_{k}!$ choices at length $k$ .

F1F2F3step 1.1

3.1

Conversely, arbitrary orbit permutations and cyclic offsets from step 2.1 assemble on the disjoint orbits to a unique bijection $g$ , and the forced-image rule makes $g σ = σ g$ .

F2step 2.1algebra

4.1

Choices for distinct lengths are independent, so [F4] and steps 2.1--3.1 give the displayed product. If $c_{k} = 0$ its factor is $k^{0} 0! = 1$ ; for the identity the result is $n!$ , and for $n = 0$ it is the empty product $1$ .

F3F4step 2.1step 3.1∎

Depends on

Used by

Dependency tree · two levels

27 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

D. A. Craven, Groups, Geometry and Representation Theory (standard reference, not scraped)
K. Conrad, Conjugacy Classes (standard reference, not scraped)

If σ∈Snσ∈Sn​ has ckck​ cycles of length kk, then ∣CSn(σ)∣=∏k=1nkckck!∣CSn​​(σ)∣=∏k=1n​kck​ck​!