TheoremStatement: Literature-sourcedProof: Literature-sourcedaudited 2026-08-28

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The cycle index of the dihedral group D_{2n}

Statement

Let $D_{2 n}$ act on the vertices of a labelled $n$ -gon, with $n \geq 1$ .

If $n$ is odd, then

$Z (D_{2 n}) = \frac{1}{2} Z (C_{n}) + \frac{1}{2} s_{1} s_{2}^{(n - 1) / 2} .$

If $n$ is even, then

$Z (D_{2 n}) = \frac{1}{2} Z (C_{n}) + \frac{1}{4} (s_{1}^{2} s_{2}^{(n - 2) / 2} + s_{2}^{n / 2}) .$

Facts & Assumptions

Given: an integer $n \geq 1$ and the full symmetry action of $D_{2 n}$ on a labelled $n$ -gon.

[L1]

The rotations contribute the cyclic-group cycle index $Z (C_{n})$ (The cycle index of the cyclic group C_n).

Proof

technique · cases

1.1L1

The subgroup of rotations has $n$ elements. Since $D_{2 n}$ has $2 n$ elements, the total rotational contribution to $Z (D_{2 n})$ is $(1 / 2) Z (C_{n})$ by [L1].

2.1step 1.1algebra

Suppose $n$ is odd. Every reflection fixes exactly one vertex and swaps the remaining $n - 1$ vertices in $(n - 1) / 2$ transpositions. Thus each reflection contributes the monomial $s_{1} s_{2}^{(n - 1) / 2}$ . There are $n$ reflections, so after division by $2 n$ their total contribution is $(1 / 2) s_{1} s_{2}^{(n - 1) / 2}$ .

2.2step 1.1algebra

Suppose $n$ is even. Then there are two reflection types. The $n / 2$ reflections through opposite vertices fix two vertices and swap the remaining $n - 2$ vertices in $(n - 2) / 2$ transpositions, so they contribute $s_{1}^{2} s_{2}^{(n - 2) / 2}$ . The $n / 2$ reflections through opposite edges fix no vertex and consist of $n / 2$ transpositions, so they contribute $s_{2}^{n / 2}$ . Dividing the sum of these $n$ reflection monomials by $2 n$ gives the contribution $\frac{1}{4} (s_{1}^{2} s_{2}^{(n - 2) / 2} + s_{2}^{n / 2})$ .

3.1step 1.1step 2.1step 2.2∎

Combine step 1.1 with step 2.1 in the odd case and with step 2.2 in the even case. This yields the two displayed formulas.

Depends on

Used by

Bracelet count from the dihedral-group cycle index Corollary
The cycle index of D₈ Example
Two-colour bracelets of length 6 Example

Dependency tree · two levels

7 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

Ben Lynn, Polya Theory: The Cycle Index Polynomial (standard reference, not scraped)
Applied Combinatorics, Section 15.5: Applications of Pólya's Enumeration Formula (standard reference, not scraped)