- Prove Burnside’s theorem.
Solution
Let a finite group \(G\) act on a finite set \(X\), and let \[ S=\{(g,x)\in G\times X:gx=x\}. \] Counting first by \(g\) gives \[ |S|=\sum_{g\in G}|X^g|. \] Counting first by \(x\) gives \[ |S|=\sum_{x\in X}|G_x|. \] By orbit-stabilizer, \[ |G_x|=\frac{|G|}{|Gx|}. \] For each orbit \(O\), summing this quantity over \(x\in O\) contributes \[ |O|\frac{|G|}{|O|}=|G|. \] Therefore \[ \sum_{g\in G}|X^g| =|G|\cdot |X/G|, \] and hence \[ |X/G|=\frac1{|G|}\sum_{g\in G}|X^g|. \] This is Burnside’s lemma.