CorollaryStatement: AI-adaptedProof: AI-adaptedprecheck passaudited 2026-08-13

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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.

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The class equation of $S_{n}$ is $n! = \sum_{\sum k c_{k} = n} n! / \prod_{k} k^{c_{k}} c_{k}!$

Statement

For every $n \geq 0$ , $n! = \sum_{\begin{array}{c} c_{1}, \dots, c_{n} \geq 0 \\ \sum_{k = 1}^{n} k c_{k} = n \end{array}} \frac{n!}{\prod_{k = 1}^{n} k^{c_{k}} c_{k}!} .$ When $n = 0$ , the one empty tuple contributes $1$ .

Facts & Assumptions

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

[F1]

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

[F2]

A permutation of type $(c_{k})$ has centralizer cardinality $\prod k^{c_{k}} c_{k}!$ (If $σ \in S_{n}$ has $c_{k}$ cycles of length $k$ , then $∣ C_{S_{n}} (σ) ∣ = \prod_{k = 1}^{n} k^{c_{k}} c_{k}!$ ).

[F3]

A conjugacy class has cardinality $∣ {Cl}_{G} (x) ∣ = [G : C_{G} (x)]$ ( $G / C_{G} (x) \to {Cl}_{G} (x)$ is a bijection, so $∣ {Cl}_{G} (x) ∣ = [G : C_{G} (x)]$ whenever these cardinalities are finite), and $∣ G ∣ = [G : H] ∣ H ∣$ for $H \leq G$ (Lagrange's theorem: $∣ G ∣ = [G : H] ∣ H ∣$ for every subgroup $H$ of a finite group $G$ ), so the class size is $∣ G ∣ / ∣ C_{G} (x) ∣$ .

[F4]

If $x_{1}, \dots, x_{r}$ represent the non-singleton conjugacy classes of a finite group $G$ , then $∣ G ∣ = ∣ Z (G) ∣ + \sum_{i} [G : C_{G} (x_{i})]$ (The class equation $∣ G ∣ = ∣ Z (G) ∣ + \sum_{i} [G : C_{G} (x_{i})]$ for a finite group).

[F5]

The symmetric group on $n$ symbols has $n!$ elements (A finite set $A$ with $∣ A ∣ = n$ has exactly $n!$ bijections onto itself, and $n!$ bijections onto any set of the same cardinality).

Proof

technique · counting

1.1

By [F1], index the conjugacy classes by the displayed tuples.

2.1

For a tuple $(c_{k})$ , [F2], [F3], and [F5] give class size $n! / \prod k^{c_{k}} c_{k}!$ .

F2F3F5step 1.1

3.1

In [F4], the central term counts the singleton conjugacy classes and the sum counts every remaining class. Thus summing the sizes from step 2.1 and using [F5] for $∣ S_{n} ∣$ gives the displayed identity.

F4F5step 1.1step 2.1

4.1

At $n = 0$ , [F1] gives the one empty type and all empty products and $0!$ equal $1$ , so the same formula reads $1 = 1$ .

F1F2F5∎

Depends on

Used by

Dependency tree · two levels

33 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)

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