# Orbit of permutaton and orbit of group action

1) Suppose $$G$$ is a group and $$S$$ is a set and we consider the action of the group $$G$$ on a set $$S$$. Then the orbit of $$x\in S$$ is the following subset of $$S$$, $$\text{Orb}_x=\{g\cdot x|g\in G\}.$$

2) But let's take some set $$S$$ and it's symmetric group $$A(S)$$ and some $$\theta\in A(S)$$. We say that $$a\sim b$$ iff $$\theta^{i}(a)=b$$ for some integer $$i$$. We can easily check that this is an equivalence relation. We call the equivalence class of an element $$s\in S$$ the orbit of $$s$$ under $$\theta$$; thus the orbit of $$s$$ under $$\theta$$ consists of all the elements $$\{\theta^i(s): i\in \mathbb{Z}\}$$.

I would like to ask the following question: Are the orbit notions in 1) and 2) are the same? If yes, can anyone explain what is the group action in 2)?

P.S. I have found a couple of topics in MSE which contain the same question but did not find an answer to my question. So please do not duplicate. Moreover, it could be interested to people who have the same question.

• The equivalence relation in $2$ is on some set $S$ : is it $S_n$, or is it $\{1,2,...,n\}$? Because both give equivalence classes : one is the orbits given by the cyclic group generated by $\theta$, and the other is the cycle decomposition of $\theta$. That is, are $a,b \in S_n$ or $a,b$ are between $1$ and $n$? – астон вілла олоф мэллбэрг Oct 26 '18 at 16:16
• @астонвіллаолофмэллбэрг, I have edited please take a look – ZFR Oct 26 '18 at 16:20
• Thank you for the edit. – астон вілла олоф мэллбэрг Oct 26 '18 at 16:21

Yes, they are the same, in that 2) is a special case of 1).

How so ? Well consider the set $$S=\{1,...,n\}$$ and $$G=\langle \theta \rangle$$, the subgroup of $$S_n$$ generated by $$\theta$$. Then $$G$$ acts on $$S$$, and the orbit of $$a$$ under this action is the equivalence class of $$a$$ under your equivalence relation

• Indeed, this makes sense! – ZFR Oct 26 '18 at 16:22
• As far as I understand, we just need to consider the action of subgroup on the set $S$, where the subgroup is the cyclic group generated by our $\theta$, right? – ZFR Oct 26 '18 at 16:23
• Yes, that's it ! – Max Oct 26 '18 at 17:30
• Thanks for help! I appreciate it! – ZFR Oct 26 '18 at 19:12

Consider the natural action of $$\mathfrak S_n$$ on the set $$\{1,\ldots ,n\}$$ defined by $$\forall \sigma \in \mathfrak S_n, \forall x \in \{1,\ldots ,n\}, \quad\sigma \cdot x := \sigma (x)$$

The orbit of an element $$x$$ under this action is the set $$\{\sigma(x) | \sigma \in \mathfrak S_n\}$$.

Now, fix $$\theta \in \mathfrak S_n$$ and consider the restriction of this action to the subgroup $$\langle \theta \rangle$$ generated by $$\theta$$. More precisely, if $$\rho: \mathfrak S_n \rightarrow \mathfrak S_n$$ is the morphism of groups describing the action, consider its restriction $$\left.\rho\right|_{\langle \theta \rangle}$$.

Then, the orbit of $$x$$ under this restricted action is given by $$\text{Orb}_x = \{\theta^i(x): i\in \mathbb{Z}\}$$, as $$\langle \theta \rangle = \{\theta^i: i\in \mathbb{Z}\}$$.

For a fixed permutation $$\theta$$, the equivalence relation $$a\sim b$$ iff $$\theta^{i}(a)=b$$, means that $$a$$ and $$b$$ are on the same cycle of the permutation $$\theta$$. In the last defintion you apply the same permutation several times to a

Notice that when you define the orbit of an element $$x$$ the permutations in $$G=S_{n}$$ are variable (this is your $$g$$ running in $$S_n$$) while your element $$x$$ is fixed.