An Equivalence class in a group of permutation

We know that every equivalence relation, induced by a partition on a set for example $X$ , make some equivalence classes. Now, if a group $G$ acts on $X$ then the associated equivalence classes are exactly the orbits of $X$ under the action of $G$. In fact, if $x\in X$ then we know $[x]$ in which $[x]=\{x^g\mid g\in G\}$ as an equivalence classes for $X$. Here, I am facing an example in permutation groups as following

Let $G=\operatorname{Aut}(\mathbb Q,\le)$ be the group of permutations of $\mathbb Q$ which preserves the usual ordering $\le$ . Also, suppose that $G$ acts on $\mathbb Q^2$ transitively, so we have two orbits of this action; $\Delta:=\{(\alpha,\beta)\in \mathbb Q^2\mid \alpha<\beta\}$ and $\Delta^*:=\{(\beta,\alpha)\mid (\alpha,\beta)\in ∆\}$ clearly.

I am misunderstanding of what would be that representing $x$ above for one of these two orbits in the example? Could every element of $\Delta$ be taken as a representing element to show the equivalance class? Indeed, I am baffling of the way we denote $\mathbb Z_p=\{[0],[1],[2],\dots,[p-1]\}$ every time we need it. Thanks for sharing the thoughts.

-
I would expect three orbits. I really don't see how the diagonal $$\overline{\Delta}=\{(\alpha,\alpha)\in\mathbb{Q}^2\mid \alpha\in\mathbb{Q}\}$$ could fail to be stable under the action of $G$. In other words, the action of $G$ is not transitive. It is transitive on $\mathbb{Q}$ though. – Jyrki Lahtonen May 14 '12 at 8:05
@JyrkiLahtonen: Yes Exactly. I made a mistake to note that these orbits above are nondiagonal. – Babak S. May 14 '12 at 8:13

Yes, if $\langle\alpha,\beta\rangle\in\Bbb Q^2$ is such that $\alpha<\beta$, then $[\langle\alpha,\beta\rangle]=\Delta$. To see this, you just have to show that if $\langle\alpha,\beta\rangle$ and $\langle\gamma,\delta\rangle$ are any members of $\Bbb Q^2$ with $\alpha<\beta$ and $\gamma<\delta$, there is some $g\in G$ such that $$\langle\alpha,\beta\rangle^g=\langle g(\alpha),g(\beta)\rangle=\langle\gamma,\delta\rangle\;.$$ In other words, you have to show that there is some $g\in G$ such that $g(\alpha)=\gamma$ and $g(\beta)=\delta$.

This isn’t too hard, if you define $g$ in three pieces. If $\xi\le\alpha$, let $g(\xi)=\xi+(\gamma-\alpha)$; this maps the ray $(\leftarrow,\alpha]$ in an order-preserving fashion onto $(\leftarrow,\gamma]$ by simply translating it $\gamma-\alpha$ units. Similarly, if $\xi\ge\beta$, let $g(\xi)=\xi+(\delta-\beta)$; this translates the ray $[\beta,\to)$ onto the ray $[\delta,to)$. To finish the job, you need only find a way to map the open interval $(\alpha,\beta)$ in order-preserving fashion onto the interval $(\gamma,\delta)$. This can be done easily with a linear function, but I’ll leave it to you for now to find one that works.

-
Thank you for saving me. – Babak S. May 14 '12 at 7:50
Is the bijective $f(x)$ = $\left(\frac{δ-γ}{β-α} \right)(x-α)+γ$ does what you left for me? – Babak S. May 16 '12 at 15:10
@Babak: Yes, that function will do nicely – Brian M. Scott May 16 '12 at 18:58

In general, an equivalence class can be represented by any of its elements. We simply pick the ones it is most convenient to work with. In the case of $\mathbb{Z}_p$ we choose $0,1,\dots, p-1$ simply because we usually define arithmetic on $\mathbb{Z}_p$ and it's much easier to do arithmetic with the representative $3$ then with the representative $432534421423$.

In the case of the example you gave, I think $(0,1)$ and $(1,0)$ are nice representatives.

-