Let \(p\) be prime, \(G\) be a group of order \(p^n > 1\), and \(X\) be a finite \(G\)-set whose size is not divisible by \(p\). Prove that \(G\) has a fixed point in \(X\): i.e., an \(x \in X\) such that \(gx = x\) for all \(g \in G\).
Solution
1. Every orbit has cardinality a power of \(p\), and a singleton orbitis exactly a point fixed by all of \(G\).
For \(x\in X\), let \(G_x=\{g\in G:gx=x\}\) be its stabilizer. The map \(G/G_x\to Gx\) sending \(gG_x\) to \(gx\) is a bijection: two representatives give the same point exactly when they lie in the same left coset. Hence \[ |Gx|=[G:G_x]=\frac{p^n}{|G_x|}, \] which is a power of \(p\). The orbit is a singleton precisely when \(gx=x\) for every \(g\in G\). Every other orbit has size divisible by \(p\).
2. The global fixed-point set \(X^G\) is nonempty.
The orbits partition the finite set \(X\). The singleton orbits contribute exactly \(|X^G|\), and all other contributions are divisible by \(p\). Therefore \[ |X|\equiv |X^G|\pmod p. \] The left side is nonzero modulo \(p\) by hypothesis, so \(|X^G|\) cannot be zero. Any \(x\in X^G\) is fixed by every element of \(G\), as required.