1

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.

AMD
  • 311

1 Answers1

0

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.

anon
  • 155,259
  • I can't see any finiteness assumptions in the question. – Derek Holt Jun 20 '17 at 08:22
  • @DerekHolt I assumed the condition was in the context that OP is quoting from (lecture notes or textbook). The exercise is false otherwise (e.g. in $SO(3)$ acting on $S^2$ every rotation has a fixed point). – anon Jun 20 '17 at 12:18