Group of order $p^n$ acting on a set of set of order $pm$

Question

Let $G$ be a finite group of order $p^n$ where p is prime. Suppose $G$ acts on a set $X$ of order $pm$ for some non negative integer $m$. Prove or disprove that if $F=\{x\in X : g.x=x \ \forall g\in G\}\neq \emptyset $, then $p$ divides $|F|$.

I guess it is true. I tried to prove it by using some properties that the sizes of the orbits and their sum have, but failed.

Any help will be appreciated.

Answer 1

• Orbits have lengths dividing $p^n$, so either $1$ or a multiple of $p$.
• The sum of all orbit lengths must be $pm$.
• $F$ collects precisely those elements of orbit length $1$.