# Show that $x∼y ⇔∃g∈G : g.x=y$ is a equivalence relation on $X$.

To put this in context, this is my first day on working with group actions.

Let $G$ a group and $X$ a $G$-set. Let the group action of $G$ on $X$ be:

$G×X→X:(g,x)\mapsto g.x$

Show that $x∼y ⇔∃g∈G : g.x=y$ is a equivalence relation on $X$.

This is what I got:
Let $x∈X$, then $e.x=x$ by definition, therefore $x∼x$.
Let $x,y∈G$ and $x∼y$, then $∃g:g.x=y$....

Now I don't know what to do, I know the rule $g_1.(g_2.x)=(g_1 \circ g_2).x$. And I thought about using $λ_{g^{-1}}$ on both side, but my problem is that $g.x ∈X$ and not in the group $G$, so I don't know how I can deal with this.

-

If $g\cdot x=y$, where $\cdot$ represents the action, then $g^{-1}\cdot(g\cdot x)=g^{-1}\cdot y$. Now use $(gg')\cdot x=g\cdot (g'\cdot x)$.

For transitivity, suppose $g\cdot x=y$ and $g'\cdot y =z$ and use $(gg')\cdot x=g\cdot (g'\cdot x)$ yet again.

You can equivalently show that the action of $G$ on $X$ partitions $X$ into $G$-orbits of the form $$Gx=\{gx:g\in G\}$$

for each $x\in X$, so that the partition gives rise to an equivalence relation.

-
Aaaah thanks ! I see, so you use some kind of similar function as $λ_{g^{-1}} : G→G :h\mapsto g^{-1}\circ h$, but now in the form of: $G×X → X : (g^{-1},x) \mapsto g^{-1}.x$ Right ? – Kasper Mar 28 '13 at 0:07
@Kasper You needn't think about it in terms of functions. Just use the action of $G$ on $X$. You know you can perform $g^{-1}\cdot y$ for any $y\in X$. And in particular if $x\in X$ and $g\in G$, $y=g\cdot x\in X$. That's all there is to it. – Pedro Tamaroff Mar 28 '13 at 0:09
Cool, thanks ! Things are becoming to get clear ! – Kasper Mar 28 '13 at 0:12

If $g\cdot x=y$, then $x=g^{-1}\cdot (g \cdot x)=g^{-1}\cdot y$.

If $g\cdot x = y$ and $g'\cdot y = z$, then $(g'g)\cdot x = z$.

-