Show that $|X^G|\equiv |X| \bmod p$

Question

Question:When a finite group $G$ acts on a finite set $X$, $X^G$ is defined as $$ X^G=\{x\in X \mid \forall g\in G,~~gx=x \} $$ If $|G|$ is power of $p$ ($p$ is a prime number), show that $$ |X^G|\equiv |X| \mod p$$ where $|A|$ is number of elements of $A$.

Answer 1

Using orbit-stabilizer theorem we have, $X= X^G \bigcup_{x \notin X^G} \mathbb{O}_x$. Also, this theorems says (In finite setup) $\#\mathbb{O}_x \mid \#G $. Since $G$ is $p$-group and $\#\mathbb{O}_x \neq 1$, Implies $\#\mathbb{O}_x $ is $p$-powers. hence divisible by $p$. This implies your result.