Let a finite group \(G\) act transitively on a set \(X\) with \(|X|\ge2\). Prove that some element \(g\in G\) acts on \(X\) without fixed points.
Solution
For \(g\in G\), let \[ \operatorname{Fix}(g)=\{x\in X:g x=x\}. \]
1.\[ \sum_{g\in G}\abs{\operatorname{Fix}(g)}=\abs{G}. \]
Count the set \[ S=\{(g,x)\in G\times X:g x=x\} \] in two ways. Counting first by \(g\) gives \[ \abs{S}=\sum_{g\in G}\abs{\operatorname{Fix}(g)}. \] Counting first by \(x\) gives \[ \abs{S}=\sum_{x\in X}\abs{G_x}, \] where \(G_x\) is the stabilizer of \(x\). Transitivity implies that every orbit has size \(\abs{X}\), so orbit-stabilizer gives \[ \abs{G_x}=\frac{\abs{G}}{\abs{X}} \] for every \(x\in X\). Therefore \[ \abs{S} = \abs{X}\frac{\abs{G}}{\abs{X}} = \abs{G}. \]
2. Some element \(g\in G\) satisfies\[ \abs{\operatorname{Fix}(g)}=0. \]
1. Thus at least one element has no fixed point.
3. Q.E.D.
The element supplied by step 2 is the required derangement.