TheoremStatement: Literature-sourcedProof: Literature-sourcedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-13

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

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.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

For $n \geq 2$ , an $S_{n}$ -class of an even permutation splits in $A_{n}$ exactly when all cycle lengths, including $1$ -cycles, are odd and distinct

Statement

Let $n \geq 2$ and let $σ \in A_{n}$ . Its $S_{n}$ -conjugacy class splits into two $A_{n}$ -conjugacy classes of equal size if and only if the lengths of all cycles in its cycle decomposition, including $1$ -cycles for fixed points, are odd and no two lengths are equal.

Facts & Assumptions

Given: An integer $n \geq 2$ and an even permutation $σ \in A_{n}$ ; for such $n$ , $A_{n}$ has index $2$ in $S_{n}$ , which is the hypothesis [F3] requires and which fails for $n \leq 1$ , where $A_{n} = S_{n}$ .

[F1]

$A_{n}$ is the kernel of sign (The alternating group $A_{n} = \ker (sgn)$ of even permutations).

[F2]

A $k$ -cycle has sign $(- 1)^{k - 1}$ (A $k$ -cycle has sign $(- 1)^{k - 1}$ , and $sgn (σ) = (- 1)^{n - c (σ)}$ when fixed points are counted as cycles), and sign is multiplicative (The sign is a homomorphism $S_{n} \to {+ 1, - 1}$ , surjective exactly when $n \geq 2$ ).

[F3]

An $S_{n}$ -class splits in the index-two subgroup $A_{n}$ exactly when its centralizer in $S_{n}$ is contained in $A_{n}$ (A conjugacy class in an index-two normal subgroup remains one class or splits into two equal classes, with a centralizer criterion).

[F4]

Every 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).

[F5]

If a permutation has $c_{k}$ cycles of length $k$ , including fixed points, then its centralizer has cardinality $\prod_{k = 1}^{n} k^{c_{k}} c_{k}!$ (If $σ \in S_{n}$ has $c_{k}$ cycles of length $k$ , then $∣ C_{S_{n}} (σ) ∣ = \prod_{k = 1}^{n} k^{c_{k}} c_{k}!$ ).

Proof

technique · iff

1.1

If $σ$ has an even-length cycle, that cycle commutes with every disjoint cycle of $σ$ , so it lies in $C_{S_{n}} (σ)$ ; [F2] makes it odd.

F2F4algebra

1.2

If two cycles have the same odd length $k$ , including $k = 1$ , the permutation swapping their entries positionwise is a product of $k$ transpositions. It centralizes $σ$ and is odd by [F2].

F2F4algebra

1.3

Conversely, suppose all cycle lengths are odd and distinct. Products of independent powers of the disjoint cycles form a subgroup $P \subseteq C_{S_{n}} (σ)$ of cardinality $\prod_{k} k$ . Here each $c_{k}$ is $0$ or $1$ , so [F5] gives $∣ C_{S_{n}} (σ) ∣ = \prod_{k} k = ∣ P ∣$ . Hence $C_{S_{n}} (σ) = P$ : every centralizing permutation preserves each cycle and restricts to a power of it.

F4F5algebra

2.1

In either case the centralizer is not contained in $A_{n}$ , so [F3] says the class does not split.

F1F3step 1.1step 1.2

2.2

Each cycle has odd length, so [F2] makes it and all its powers even. By step 1.3 every centralizing permutation is a product of such powers and is therefore even.

F2step 1.3

3.1

Hence $C_{S_{n}} (σ) \subseteq A_{n}$ , so [F3] gives two equal $A_{n}$ -classes.

F1F3step 2.2∎

Depends on

Used by

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 69 results over 17 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

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

For n≥2n≥2, an SnSn​-class of an even permutation splits in AnAn​ exactly when all cycle lengths, including 11-cycles, are odd and distinct