TheoremStatement: Literature-sourcedProof: Literature-sourcedSession-authored (Fable 5 assisted)audited 2026-08-28

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The cycle index of A_n is the parity-filtered symmetric-group sum

Statement

For every integer $n \geq 2$ ,

$Z (A_{n}) = 2 \sum_{\begin{array}{c} m_{1}, \dots, m_{n} \geq 0 \\ \sum_{d = 1}^{n} d m_{d} = n \\ \sum_{d = 1}^{n} (d - 1) m_{d} even \end{array}} \frac{1}{\prod_{d = 1}^{n} d^{m_{d}} m_{d}!} \prod_{d = 1}^{n} s_{d}^{m_{d}} .$

Equivalently,

$Z (A_{n}) = \sum_{\begin{array}{c} m_{1}, \dots, m_{n} \geq 0 \\ \sum_{d = 1}^{n} d m_{d} = n \end{array}} \frac{1 + (- 1)^{\sum_{d} (d - 1) m_{d}}}{\prod_{d = 1}^{n} d^{m_{d}} m_{d}!} \prod_{d = 1}^{n} s_{d}^{m_{d}} .$

Facts & Assumptions

Given: an integer $n \geq 2$ .

[L1]

A permutation lies in $A_{n}$ exactly when it is even (The alternating group $A_{n} = \ker (sgn)$ of even permutations).

[L2]

A permutation with cycle counts $m_{d}$ has sign $(- 1)^{\sum_{d} (d - 1) m_{d}}$ (A $k$ -cycle has sign $(- 1)^{k - 1}$ , and $sgn (σ) = (- 1)^{n - c (σ)}$ when fixed points are counted as cycles).

[L3]

For $n \geq 2$ , the alternating group has half the elements of $S_{n}$ , so $∣ A_{n} ∣ = n! / 2$ ( $A_{n}$ is normal in $S_{n}$ ; for $n \geq 2$ , $2 ∣ A_{n} ∣ = n!$ , while $A_{n} = S_{n}$ for $n = 0, 1$ ).

[L4]

The symmetric-group cycle index is the sum over cycle types (The cycle index of S_n is the sum over cycle types).

Proof

technique · direct

1.1

By [L1] and [L2], a permutation of cycle type $(m_{1}, \dots, m_{n})$ belongs to $A_{n}$ exactly when $\sum_{d} (d - 1) m_{d}$ is even.

L1L2

2.1

For such an even cycle type, the number of permutations in $A_{n}$ with that type is the same as the number in $S_{n}$ , namely $n! / \prod_{d} d^{m_{d}} m_{d}!$ , because an entire cycle type is either even or odd. Therefore the unnormalized cycle-index sum over $A_{n}$ is $\sum_{\begin{array}{c} m_{1}, \dots, m_{n} \geq 0 \\ \sum_{d} d m_{d} = n \\ \sum_{d} (d - 1) m_{d} even \end{array}} \frac{n!}{\prod_{d} d^{m_{d}} m_{d}!} \prod_{d} s_{d}^{m_{d}}$ .

step 1.1L4

3.1

Divide step 2.1 by $∣ A_{n} ∣ = n! / 2$ from [L3]. This multiplies the cycle-type coefficients by $2 / n!$ , leaving exactly $2 \sum_{\begin{array}{c} m_{1}, \dots, m_{n} \geq 0 \\ \sum_{d} d m_{d} = n \\ \sum_{d} (d - 1) m_{d} even \end{array}} \frac{1}{\prod_{d} d^{m_{d}} m_{d}!} \prod_{d} s_{d}^{m_{d}}$ . The equivalent filtered formula follows because $1 + (- 1)^{e}$ is $2$ when $e$ is even and $0$ when $e$ is odd.

step 2.1L3∎

Depends on

Used by

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Sources

Eric W. Weisstein, Cycle Index, Wolfram MathWorld (standard reference, not scraped)