# Burnside's Lemma

(Redirected from Burnside's Counting Theorem)

## Theorem

Let $G$ be a finite group acting on a set $X$.

Let $X / G$ be the set of orbits under this action.

For $x \in X$, let $\Stab x$ be the stabilizer of $x$ by $G$.

For $g \in G$, let $X^g$ denotes the set of all elements in $X$ which is fixed by $g$, that is:

$X^g := \set {x \in X: g x = x}$

Then:

$\ds \size {X / G} = \frac 1 {\order G} \sum_{g \mathop \in G} \size {X^g}$

In words, the number of orbits equals the average number of fixed elements.

## Proof

 $\ds \frac 1 {\order G} \sum_{g \mathop \in G} \size {X^g}$ $=$ $\ds \frac 1 {\order G} \sum_{g \mathop \in G} \size {\set {x \in X: g x = x} }$ by definition $\ds$ $=$ $\ds \frac 1 {\order G} \sum_{x \mathop \in X} \size {\set {g \in G: g x = x} }$ Same summation, different indexing $\ds$ $=$ $\ds \frac 1 {\order G} \sum_{x \mathop \in X} \order {\Stab x}$ Definition of Stabilizer $\ds$ $=$ $\ds \frac 1 {\order G} \sum_{x \mathop \in X} \frac {\order G} {\order {\Orb x} }$ Orbit-Stabilizer Theorem $\ds$ $=$ $\ds \sum_{x \mathop \in X} \frac 1 {\order {\Orb x} }$ $\ds$ $=$ $\ds \sum_{\Orb x \mathop \in X / G} \paren {\sum_{x \mathop \in \Orb x} \frac 1 {\order {\Orb x} } }$ $\ds$ $=$ $\ds \sum_{\Orb x \mathop \in X / G} 1$ $\ds$ $=$ $\ds \order {X / G}$

$\blacksquare$

## Also known as

This theorem is also known as Burnside's Counting Theorem.

## Source of Name

This entry was named for William Burnside.