Reasoning for this particular orbit

Question

In my lecture today, my class came to the following conclusion regarding the $D_n$-orbits of size 1: that they are the set of all two neighboring points (when viewing $D_n$ has the symmetries on an n-gon). This was given as an example to the orbit definition, so I'm just a little confused on the reasoning for these orbits.

Edit: of size 2, not 1.

Answer 1

Let's start with the definition of orbit.

Consider a group $G$ acting on a set $X$. The orbit of an element $x$ in $X$ is the set of elements in $X$ to which $x$ can be moved by the elements of $G$. The orbit of $x$ is denoted by $G\cdot x$: $$G\cdot x=\left\{g\cdot x\mid g\in G\right\}$$

In your case the group $G$ is $D_n$, the symmetries of a regular polygon with $n$ points. And the set $X$ consists of the points of the polygon.

If we pick any particular point $x$ of the polygon, and apply each of the successive rotations to it, we rotate the point through each of the other points until we are back at the original point. So the corresponding orbit $D_n\cdot x$ consists of all $n$ points of the polygon.

In other words, I'm afraid that there is no $D_n$-orbit of size 1 in a polygon with at least 3 points. The only case that we can have an orbit of size 1 is if the polygon consists of a single point.

Edit: for the same reason an orbit of size 2 means that the polygon consists of 2 points.