Group and orbit question.

Question

Suppose group $G$ acts on a set $A$.

a) If $x$ and $y$ are in the same orbit, show that there exists some $g \in G$ such that $gG_x g^{-1} = G_y$.

b) Show that if $|G.x|$ is finite, then $|G.x| = |G| / |G_x|$.

I'm beginning to get used to my first abstract, proof-based math course (still have a lot to do though). Here is what I have so far. Thanks for the help.

a) Since $x,y$ are in the same orbit, $G_x = G_y$. Pick an element $g$ such that $g^m = x$ and $g^n = y$ for some $m,n \in \mathbb{Z}$. Then I got stuck.

b) Note: $G.x$ where $x \in A$ denotes $G$ acting on $A$. $|G.x|$ is finite so each equivalence class in $G$ is finite. The right hand side also denotes an arbitrary equivalence class of $G$.

Answer 1

Hints:

(1) If $\,x,y\,$ are in the same orbit, then $\,\exists\,g\in G\,\;s.t.\;\;x=g\cdot y\,$ , and from here

$$\forall\,h\in G_x:\;\;\;g^{-1}hg\cdot y=g^{-1}h\cdot x =g^{-1}\cdot x=\ldots$$

(Note: the element $\,g\,$ above is the same in both lines)

(b) Denote by $\,G_x/G\,$ the set of all left cosets of the group $\,G_x\,$ in $\,G\,$ , and by $\,\mathcal O(x)\,$ the orbit of an element $\,x\in G\,$ , and define

$$f: \mathcal O(x)\longrightarrow G_x/G\;,\;\;f(g\cdot x):=gG_x$$

Prove this function is well defined and bijective.

Answer 2

Concerning (b), it might be worth remarking that this follows from what is sometimes called the first isomorphism theorem for sets.

This says that if you have a surjective map $f : A \to C$ between sets, then the relation $R$ on $A$ defined by $a R b$ if and only if $f(a) = f(b)$ is an equivalence relation. And if $A/R = \{ [a] : a \in A \}$ is the set of the equivalence classes (so $[a] = \{ y \in A : y R a \}$), then the map $f' : A/R \to C$ that sends $[a] \mapsto f(a)$ is well-defined, and it is a bijection. (One could also characterize $f'$ as the unique map that makes a certain diagram commute.)

In your case, $A = G$, $C = G.x$, and $f(g) = gx$. The relation $R$ becomes $g R h$ if and only if $gx = hx$ if and only if $(g^{-1} h) x = x$ if and only if $g^{-1} h \in G_{x}$, So $[g] = g G_{x}$, and you obtain the bijection already stated by @DonAntonio (actually, its inverse) between the orbit $G.x$, and the set of left cosets of $G_{x}$ in $G$, which has cardinality $\lvert G \rvert / \lvert G_{x} \rvert$.

This might appear involved in this particular case, but it is often useful to construct bijections starting from arbitrary maps.