$|G|<\infty$ and $|X|<\infty$ a $G$-set $\Rightarrow \bigcap_{g \in G}Fix(g)\not= \emptyset$

Question

Question

Let $G$ a finite group with $|G|=p^n$ for a prime $p$,and $|X|<\infty$ a $G$-set.If $p\not| |X|$ then show that $$\bigcap_{g \in G}Fix(g)\not= \emptyset$$

Attempt

I know that the number of orbits of $G$ in $X$ are $$N=\dfrac{1}{|G|}\sum_{g\in G}|Fix(g)|$$

Also that $|G|=p^n \Rightarrow Z(G)\not=\{1_G\}$ but can't put them together.Any hints?

Answer 1

So,thanks to @Tobias Kildetoft I'll give it a try: $$\bigcap_{g\in G}Fix(g)\not=\emptyset \Leftarrow \exists x\in X:|O(x)|=1$$

because $\exists x\in X:|O(x)|=1 \Rightarrow \exists x\in X:O(x)=\{x\} \Rightarrow x\in\bigcap_{g\in G}Fix(g) \Rightarrow \bigcap_{g\in G}Fix(g)\not=\emptyset$

So I'll so that $\exists x\in X:|O(x)|=1$:

We know that $|O(y)| | |G|=p^n\Rightarrow |O(y)|=p^m$ for $0\leq m\leq n $

If $m>0\forall y\in X$ then $p|\sum_{y\in S}|O(y)|=|X|$ contradiction!

If I missed something please let me know!