Group action with a single orbit has a $g \in G$ with no fixed points Let $G$ be a group acting on a set $S$ where $|S| \geq 2$.  If this action has one orbit, show that there exists a $g \in G$ which has no fixed points.
I feel I am having trouble proving this because I am not sure exactly what it means to have one orbit.   I am taking the set $Gs = \{g\cdot s : g \in G\}$ to be the orbit of $s \in S$.  For every $s \in S$, we have $e \cdot s = s$.  In this case doesn't 'one orbit' mean that the orbit would be all of $S$?  Any help is appreciated.
 A: It is possible for a group acting on a finite set to have more than one orbit. For instance:


*

*Consider the trivial group acting on any finite set: then every singleton subset is a distinct orbit, and so there are as many orbits as there are elements. 

*Or, suppose $G$ acts transitively on $X$ (i.e. all of $X$ is a single orbit) and $Y$ is a set with more than one element; then $G$ acts on $X\times Y$ via $g\cdot(x,y):=(gx,y)$, in which case the orbits are the subsets $X\times\{y\}$ for all $y\in Y$, in which case there is more than one orbit.

*Suppose $G$ acts on $X$ and $y\not\in X$. Then $G$ acts on $X\sqcup\{y\}$ by doing its usual thing to elements of $X$, and fixes $y$ (i.e. $gy=y$ for all $g\in G$). Then $\{y\}$ is its own orbit distinct from all of the orbits in the original set $X$.


Having one orbit is equivalent to either of the following equivalent conditions:


*

*for any $x_1,x_2\in X$ there exists a $g\in G$ such that $x_2=gx_1$

*there exists a $x\in X$ such that (for all $x'\in X$ there exists a $g\in G$ such that $gx=x'$)


If the set is just one orbit, the action is called transitive.

One possible route: the reason the condition that $G$ is finite might be important is that we may be able to leverage a counting argument. Do we have any combinatorial identity that involves counting fixed points of individual elements? Yes: Burnside's Lemma.
