CorollaryStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-13

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The conjugacy classes of $S_{n}$ are indexed by the tuples $(c_{1}, \dots, c_{n})$ with $\sum k c_{k} = n$

Statement

The conjugacy classes of $S_{n}$ are in bijection with the tuples of nonnegative integers $(c_{1}, \dots, c_{n})$ satisfying $\sum_{k = 1}^{n} k c_{k} = n .$ For $n = 0$ , the unique empty tuple indexes the identity class of $S_{0}$ .

Facts & Assumptions

Given: The symmetric group $S_{n}$ for $n \geq 0$ .

[F1]

The cycle type of a permutation is the tuple $(c_{1}, \dots, c_{n})$ counting its orbits of each length, with fixed points counted as $1$ -cycles (Support, fixed points, disjoint cycles, cycle length, disjoint-cycle decompositions, and cycle type).

[F2]

Two permutations in $S_{n}$ are conjugate exactly when they have the same cycle type (Two elements of $S_{n}$ are conjugate if and only if they have the same cycle type).

Proof

technique · bijection

1.1

Assign to a conjugacy class the cycle type of any representative; [F2] makes this well-defined and injective.

1.2

Conversely, for any such tuple, partition $n$ symbols into $c_{k}$ blocks of size $k$ and put a $k$ -cycle on each block; their product has that cycle type. For $n = 0$ , use the empty product on the empty set.

F1algebra

2.1

The orbits counted in [F1] partition the $n$ symbols, so counting their points gives $\sum_{k} k c_{k} = n$ . Thus the image consists of tuples satisfying the displayed equation.

F1step 1.1algebra

3.1

Thus the assignment is surjective and hence a bijection.

step 1.1step 2.1step 1.2∎

Depends on

Used by

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 21 results over 10 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)

The conjugacy classes of SnSn​ are indexed by the tuples (c1,…,cn)(c1​,…,cn​) with ∑kck=n∑kck​=n