Let $|G|=16$ which acts on a set $|X|=31$, show that it exist an element of $X$ which remains stable under the group action.
I denote $X^G=\{x\in X\mid \forall g\in G, gx=x \}$, we have $|G|=16=2^4$ , $2$ is prime
then by the theorem we get $|X|\equiv |X^G|\mod{2}$
$2$ doesn't divide $31$, so it must be $|X^G|>0$
Can I somehow show that $|X^G|=1$ ?
If $|X|\equiv |X^G|\mod 2$, then $|X^G|$ must be odd just like $|X|$. Therefore $|X^G|\not=0$, done. But you cannot show $|X^G|=1$. For example, $|X^G|=31$ is possible for the trivial action of $G$. (In fact, one can show that $|X^G|$ can possibly be any odd value $\le 31$.)