Prove: There is a $g \in G$ such that $\forall$ $x \in X: g \circ x \neq x$

Question

I have to prove this theorem for my math study:

Let $G$ be a finite group, and $X$ a set with #$X \geqslant 2$. Let the action of $G$ on $X$ be transitive.

Prove: There is a $g \in G$ such that $\forall$ $x \in X: g \circ x \neq x$

So far, I'm not getting any further than this:

For any $x \in X$, the orbit of x is given by $Gx = ${$g \circ x | g \in G$}. So from the transitive action of $G$ on $X$, we can conclude that #$Gx = 1$

I don't have any idea on how to complete the proof, or if it's actually right what I'm doing. Could you please explain me how to complete the proof?

Thanks in advance!

Answer 1

since $G$ acts transitively on $X$ so $|X/G|=1$... again if we assume there is no such $g \in G$ then $|X^g| \geq 1$ and $|X^e|=|X|$...then by burnside's lemma (http://en.wikipedia.org/wiki/Burnside%27s_lemma) $|X/G| >1$ which is a contradiction.