Proof of Burnside's Theorem

proofgroup-theory

Theorem

The number of distinct orbits of _____ is equal to,

$$\#orbits=\frac{1}{|G|}\sum _{g\in G}|{\mathrm{fix}}_x(g)|.$$

Proof

Let $I=\{(g,x)\in G\times X:g\bullet x=x\}$.

$$\begin{array}{rl}\#I& =\sum _{g\in G}|{\mathrm{fix}}_x(g)|\\ & =\sum _{x\in X}|{\mathrm{stab}}_G(x)|\end{array}$$

$\frac{1}{|G|}{\sum }_{g\in G}^{}|{\mathrm{fix}}_x(g)|=\frac{1}{|G|}{\sum }_{x\in X}^{}|{\mathrm{stab}}_G(x)|$.

If $x^{\prime }\in \mathrm{orb}(x)⟹x^{\prime }=g\bullet x$ for some $g\in G⟹\mathrm{stab}(x^{\prime })=g\cdot \mathrm{stab}(x)\cdot g^{-1}$. $h\cdot x=x⟹(h{g}^{-1}g)\cdot x=x$ $(h{g}^{-1})\bullet (g\cdot x)=x$ $(gh{g}^{-1})\bullet (g\cdot x)=g\bullet x$ $⟹|\mathrm{stab}(x^{\prime })|=|\mathrm{stab}(x)|$

Then rearranging $X$ by orbits, ${\sum }_{x\in X}^{}|{\mathrm{stab}}_G(x)|={\sum }_{\text{distinct orbits}}^{}{\sum }_{x\in \text{ such orbit}}^{}|\mathrm{stab}(x)|$ $={\sum }_{\text{distinct orbits}}^{}|G|=\#\text{ distinct orbits}$