Let \(S_n\) act on \(n\) objects and let \(G\le S_n\) have order \(p^k\), where the prime \(p\) does not divide \(n\). Show that \(G\) has a fixed point.
Solution
Let \(X\) be the set of the \(n\) objects on which \(S_n\), and hence \(G\), acts. Decompose \(X\) into \(G\)-orbits: \[ X=\mathcal O_1\sqcup\cdots\sqcup\mathcal O_r. \]
1. Every orbit has cardinality a power of \(p\).
For \(x\in X\), the orbit-stabilizer theorem gives \[ |Gx|=[G:G_x]. \] Since \[ |G|=p^k, \] the index of every subgroup of \(G\) is a power of \(p\). Thus each orbit size is \[ |Gx|=p^j \] for some \(0\le j\le k\).
2. At least one orbit has size one.
Suppose no point were fixed by all of \(G\). Then no orbit would have size \(1\), so every orbit size would be divisible by \(p\). Therefore \[ n=|X|=\sum_{i=1}^r|\mathcal O_i| \] would be divisible by \(p\). This contradicts the hypothesis \[ p\nmid n. \] Hence some orbit has size \(1\). Its unique point is fixed by every element of \(G\).