Proof of the fixed point congruence for p-group actions

Proof

We split $X$ into orbits and reduce their sizes modulo $p$. Let $e$ be the identity of $G$.

1. First, $\abs{G}$ is a power of $p$. Let $q$ be a prime that divides $\abs{G}$. By Cauchy's theorem, some $g \in G$ has order $q$.
2. Since $G$ is a $p$-group, the order of $g$ is $p^s$ for some $s \geq 0$. So $q = p^s$, and as $q$ is prime, $q = p$.
3. So $p$ is the only prime factor of $\abs{G}$. By the fundamental theorem of arithmetic, $\abs{G} = p^r$ for some $r \geq 0$.
4. Let $d$ be a positive divisor of $p^r$ and $q$ a prime dividing $d$. Then $q \mid p^r$, so $q \mid p$ by Euclid's lemma, and $q = p$. As in step 3, $d$ is a power of $p$.
5. By the orbit size formula, the size of each orbit divides $\abs{G} = p^r$. By step 4, each orbit has size $p^j$ for some $j \geq 0$.
6. An orbit $G \cdot x$ has size $1$ if and only if $G \cdot x = \set{x}$, since $x = e \cdot x$ lies in it. This means $g \cdot x = x$ for all $g$, that is, $x$ is a fixed point.
7. So the orbits of size $1$ are the sets $\set{x}$ with $x \in X^G$. Every other orbit has size $p^j$ with $j \geq 1$, which is divisible by $p$.
8. The orbits partition $X$, and $X$ is finite. So $\abs{X}$ is the sum of the orbit sizes, which is $\abs{X^G}$ plus a sum of multiples of $p$.
9. Hence $\abs{X} \equiv \abs{X^G} \pmod p$.