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$$)$ $\ne$ $x$ for all $x$ $\in$ $X$
I am trying to use contradiction, but it is not very clear to me why this is true.
I'm assuming $G$ and hence $X$ are finite.
For $g \in G, x \in X$, let $\delta_{gx} = 1$ if $g$ fixes $x$, and $0$ if not. Suppose by way of contradiction that for every $g \in G$, there exists an $x \in X$ for which $\delta_{gx} = 1$. Then $$\sum\limits_{g \in G} \sum\limits_{x \in X} \delta_{gx} \geq |G|$$ In fact, that inequality $\geq$ is actually a strict inequality $>$, because if we consider $g = 1_G$, then $\delta_{1_Gx} = 1$ for all $x \in X$, and $X$ has at least two elements.
For each $x \in X$, let $\textrm{Stab } x$ be the subgroup of $G$ consisting of elements which fix $x$. By the orbit stabilizer theorem, and the fact that the action of $G$ on $X$ is transitive, we have $|\textrm{Stab } x | = \frac{|G|}{|X|}$ for all $x \in X$. Now $$\sum\limits_{g \in G} \sum\limits_{x \in X} \delta_{gx} = \sum\limits_{x \in X} \sum\limits_{g \in G} \delta_{gx} = \sum\limits_{x \in X} |\textrm{Stab } x| = |G|$$ So $|G| > |G|$, absurd.
The action has only one orbit, and since (this is Burnside's lemma)
$$|X/G||G|= \sum_{g\in G} |X^g|$$
where $X^g$ is the set of fixed points of $g$, it is then clear that if every $|X^g|\geqslant 1$ we obtain a contradiction. Namely, that $|G| > |G|$.
By Burnside's lemma, $$ 1=|X/G|=\frac{1}{|G|}\sum_{g\in G}|X^g|, $$ the first equality holding since the action is transitive, i.e., there is one $G$-orbit.
Now try to prove $|X^g|=0$ for some $g\in G$.
Since $e$ fixes $X$, $|X^e|=|X|\geq 2$. Then there must be some other $g\in G$ such that $|X^g|=0$, for if not, $\sum_{g\in G}|X^g|\geq |G|+1$.
