1. ## Explanation of orbits

I have been reading in my book I don't quite understand orbit.

$G(x)=\{gx\in X:g\in G\}$

Can someone explain this and show some easy examples?

2. ## Re: Explanation of orbits

Originally Posted by dwsmith
I have been reading in my book I don't quite understand orbit.

$G(x)=\{gx\in X:g\in G\}$

Can someone explain this and show some easy examples?
To create an orbit you just take an element $x\in X$ and apply every group element to it. For example, if $G$ is a finite group and $X=\left\{(g_1,\cdots,g_n)\in G^n:g_1\cdots g_n=1\right\}$ then $\mathbb{Z}_n$ acts on $X$ via cyclic shifting backward of the tuples (i.e. $k\cdot(g_1,\cdots,g_n)=(g_{k+1\text{ mod }n},\cdots,g_{k+n\text{ mod }n})$). Then, the orbit of $(g_1,\cdots,g_n)$ consists of $0\cdot(g_1,\cdots,g_n)=(g_1,\cdots,g_n)$, $1\cdot(g_1,\cdots,g_n)=(g_2,g_3,\cdots,g_n,g_1)$, etc. Make sense?

3. ## Re: Explanation of orbits

Originally Posted by Drexel28
To create an orbit you just take an element $x\in X$ and apply every group element to it. For example, if $G$ is a finite group and $X=\left\{(g_1,\cdots,g_n)\in G^n:g_1\cdots g_n=1\right\}$ then $\mathbb{Z}_n$ acts on $X$ via cyclic shifting backward of the tuples (i.e. $k\cdot(g_1,\cdots,g_n)=(g_{k+1\text{ mod }n},\cdots,g_{k+n\text{ mod }n})$). Then, the orbit of $(g_1,\cdots,g_n)$ consists of $0\cdot(g_1,\cdots,g_n)=(g_1,\cdots,g_n)$, $1\cdot(g_1,\cdots,g_n)=(g_2,g_3,\cdots,g_n,g_1)$, etc. Make sense?
Nope

4. ## Re: Explanation of orbits

the motivating example of a group acting on a set is G = Sn, where X is the set {1,2,...,n}. this action is transitive, which means for any element x of X, G(x) = X.

so let's consider a subgroup of G, say H = <(1 2 3)>. what is the orbit H(1)? well, if h = (1 2 3), h(1) = 2, so 2 is in the orbit of 1. similarly h^2 = (1 3 2), so h^2(1) = 3,

so 3 is in the orbit of 1. it should be pretty clear that if x is not 1,2 or 3, then h(x) = x. in particular none of {4,....,n} are in the orbit of 1. obviously the identity map sends 1-->1,

so 1 is in the orbit of 1. so the orbit of 1 under H, H(1) = {1,2,3}.

we can also consider an action of G on itself, by defining for g in G and x in X (=G), g(x) = gx, the product in G. similarly, we can also let a subgroup H act on G the same way:

define h(g) = hg (the product in G). let's take G = Z10, H = Z5, and look at H(4).

the elements of H are: {0,5}. if h = 0, h(4) = 0+4 = 4, so 4 is in the orbit of 4. if h = 5, then h(4) = 5+4 = 9, so H(4) = {4,9}. note that in this case, H(n) is the coset H+n.

note also that H(9) = {0+9,5+9} = {9,4} = H(4), perhaps this makes clear the term "orbit"...the action of H on G partitions G into disjoint sets, and on each of these orbits,

H sends an element of that orbit to another element of that orbit.