Orbits of Group Actions

Question

What is an orbit of a group action? What is an example of a group action on a set with 10 elements withe exactly two orbits?

I'm not entirely sure what an orbit is. This is a definition I found online:

Let $S$ be a $G$-set and $s\in S$. The orbit of $s$ is the set $G\cdot s =\{g\cdot s \mid g\in G\}$, the full set of objects that $s$ is sent to under the action of $G$.

This definition doesn't make sense to me, but is it a correct definition?

How would I use this definition to find an example?

Answer 1

This is correct. The idea of a group action is that you have a set (with no additional structure), and a group $G$ which acts on that set $S$ by permutations.

For a simple example, let $S$ be the letters $\{a,b,c,d,e\}$, and let $G$ be the cyclic group of order $3$. An action of $G$ on $S$ is essentially just a way to think of each element of $G$ as a function $S\to S$.

Write $G=\{1,\sigma,\sigma^2\}$, where $1$ is the identity. One possible action is to associate $1$ to the identity map $S\to S$ and $\sigma$ to the map defined as

\begin{align} \sigma(a)&=b\\ \sigma(b)&=c\\ \sigma(c)&=a\\ \sigma(d)&=d\\ \sigma(e)&=e. \end{align}

Then necessarily, this means

\begin{align} \sigma^2(a)&=c\\ \sigma^2(b)&=a\\ \sigma^2(c)&=b\\ \sigma^2(d)&=d\\ \sigma^2(e)&=e. \end{align}

You can see that $G$ can carry $a$ to $b$ and $c$, but never to $d$ or $e$. We say that the orbit of $a$ is the set $\{a,b,c\}=Ga$, and that the orbit of $d$ is $\{d\}=Gd$.

I am not being 100% precise in my definition of a group action, but this is the main idea. The "orbit" is meant to have sort of a physical interpretation in the sense that the orbit $Gx$ of $x$ is the set of points in $S$ which $x$ "visits" under the action by $G$.

Answer 2

The notion of (abstract) group action arises as soon as you move from the original notion of "group" as group of permutations (which natively fulfills the conditions $\iota(s)=s, \forall s\in S$, and $\sigma(\tau(s))=(\sigma\tau)(s),\forall s\in S, \forall \sigma,\tau\in \operatorname{Sym}(S)$) to the one of abstract group. Like a group of permutations acts (colloquially) on $S$ by natively fulfilling the above conditions, so an abstract group $G$ acts (technically) on $S$ if there is a map $G\times S\to S$ (whose image elements we denote by $g\cdot s$) which fulfils the "abstract version" of the above conditions, namely:

1. $\space\space e\cdot s=s, \space\forall s\in S$;
2. $g\cdot(h\cdot s)=(gh)\cdot s, \space\forall g,h\in G, \forall s\in S$.

Now, 1 and 2 are precisely the conditions which make of $(s\sim t \stackrel{(def.)}{\iff} \exists g\in G\mid t=g\cdot s)$ an equivalence relation in $S$. The equivalence class of $s\in S$ is then:

\begin{alignat}{1} [s]_\sim &= \{t\in S\mid t\sim s\} \\ &= \{g\cdot s, g\in G\} \\ &= G\cdot s \\ \end{alignat}

So, the orbits are equivalence classes induced on the acted set by the acting group.

As an example (this goes back to a group of permutations), the natural action of $\langle\sigma\rangle$ on $X:=\{1,\dots,n\}$, where $\sigma\in S_n$ has cycle structure $(m,n-m)$ (for some $m$, $1\le m\le n-1$), partitions $X$ into precisely $2$ orbits, namely:

$$O(i)=\{i,\sigma(i),\sigma^2(i),\dots,\sigma^{m-1}(i)\}$$ $$O(j)=\{j,\sigma(j),\sigma^2(j),\dots,\sigma^{n-m-1}(j)\}$$

where $j\notin O(i)$.