# Orbit of an element of the group $G$ as opposed to the orbit of an element of a set acted upon by the group $G.$

At this point of this lecture by Professor Macauley at Clemson University the following statement appears:

Every element in a group traces out an orbit.

I am trying to familiarize with group theory concepts a bit (beginner), and find this confusing because I thought orbits applied to group actions so that considering the action of group $G$ on a set $X,$ and taking an element of $x \in X,$ its orbit would be all the results of the binary operation of this action of $G$ on $x:$

$$G.x=\{g.x\;|\;g\in G\}$$

However, if $x$ is not an element of some set $X,$ separate from the set $G$ that defines the group, but rather is an element of $G,$ i.e. $x\in G,$ wouldn't then, $G.x=G?$ In other words, the result of all the elements of the set constituting $G,$ acting on any given element $x\in G,$ would reproduce the entire set $G?$ The orbit of any given element would then be the entire group, resulting in a single partition without much interest...

Am I confusing "acting on" with "composed with" as in $Gx =\{g\;\circ\; x\;|\;g\in G\},$ where $\circ$ denotes the group operation, and where every element of the group appears once and only once?

• He is calling orbit of $g$ to $\{e,g,g^2,g^3,...\}$
– arts
Nov 24, 2017 at 22:21

## 1 Answer

There is a way to make the "orbit" as he calls it the actual orbit of an action (the "orbit" of $g \in G$ is the orbit of $e$ under the left multiplication action of $\langle g\rangle$ on $G$) but this isn't necessary. I don't think he really means for you to be thinking about actions at this point, he just means the path you get through the Cayley diagram by multiplying by $g$ repeatedly. Since this path path forms a loop starting at $e$ he's just informally calling this loop an orbit.