Exercise 2147Algebra › GroupsM1Added 3 Nov 2026

Burnside's lemma

#Burnside lemma#group action#counting

Statement

Let $G$ (finite) act on a finite set $X$ . Show that the number of orbits equals
$\frac{1}{|G|}\sum_{g\in G}|X^g|,$
where $X^g=\{x\in X:gx=x\}$ .

Hint

Count the set $\{(g,x):gx=x\}$ in two ways.

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Solution

Double count. Let $S=\{(g,x)\in G\times X:gx=x\}$ . Then
$|S|=\sum_{g\in G}|X^g|=\sum_{x\in X}|\operatorname{Stab}(x)|.$

Orbit-stabilizer. For each orbit $O$ , all stabilizers have the same size $|G|/|O|$ (since $[G:\operatorname{Stab}(x)]=|O|$ ). So
$\sum_{x\in O}|\operatorname{Stab}(x)|=|O|\cdot\frac{|G|}{|O|}=|G|.$
Summing over orbits: $\sum_{x\in X}|\operatorname{Stab}(x)|=|G|\cdot\#\text{orbits}$ .

Conclusion. $\#\text{orbits}=\frac{1}{|G|}\sum_g|X^g|$ .

Burnside's lemma

Statement

Solution

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