Let \(n\) be odd, and let \(G\le S_n\) have order a power of \(2\). Prove that there is some \[ i\in\{1,\dots,n\} \] such that \[ \sigma(i)=i \] for every \(\sigma\in G\).
Solution
Let \[ \Omega\coloneqq\{1,\ldots,n\}. \]
1. Every orbit of the action of \(G\) on \(\Omega\) has size a powerof \(2\).
For \(i\in\Omega\), the orbit-stabilizer theorem gives \[ \abs{G\cdot i}=[G:G_i], \] which divides \(\abs G\). Since \(\abs G\) is a power of \(2\), so is every orbit size.
2. At least one orbit has size \(1\).
If every orbit had size greater than \(1\), then step 1 would make every orbit size even. The disjoint union of the orbits is \(\Omega\), so \[ n=\abs\Omega \] would be a sum of even integers and hence even. This contradicts the hypothesis that \(n\) is odd.
3. There is \(i\in\Omega\) fixed by every element of \(G\).
2.Then\[ G\cdot i=\{i\}, \]so \(\sigma(i)=i\) for every \(\sigma\in G\).
4. Q.E.D.
Step 3 is the required common fixed point.