Let $X$ be a transitive $G$-set. ($G$ acts on $X$ transitively.) If $X$ is finite and has at least two elements, show that there is some element $g$ $\in$ G which does not have any fixed points; that is, such that $g$$($$x$$)$ $\ne$ $x$ for all $x$ $\in$ $X$
I am trying to use contradiction, but it is not very clear to me why this is true.
Best Answer
The action has only one orbit, and since (this is Burnside's lemma)
$$|X/G||G|= \sum_{g\in G} |X^g|$$
where $X^g$ is the set of fixed points of $g$, it is then clear that if every $|X^g|\geqslant 1$ we obtain a contradiction. Namely, that $|G| > |G|$.