Properties of set $\mathrm {orb} (x)$

Question

Properties of set $\mathrm {orb} (x)$:

${\displaystyle \bigcup_{x\in X}\mathrm{orb}(x)=X}$;

$\mathrm{orb}(x)\cap\mathrm{orb}(y)=\emptyset$ for all $x,y\in X, x\neq y$

How to prove it? Please help.

Appedix: Let $\phi: G \times X \longrightarrow X$ - action of the group G on the non-empty set $X$. The set $\mathrm {orb} (x) = \{ \phi (g,x) \in X: g \in G \}$ called orbit of $x \in X$

Answer 1

This statement is false:

$\text{orb}(x)\cap \text{orb}(y)=\emptyset$ for all $x,y\in X$ such that $x\not= y$

Instead, I think that you mean to say that $X$ is partitioned into disjoint orbits under the action of $G$. This is given by the fact that the relation $x \sim y \Leftrightarrow y\in \text{orb}(x)$ is an equivalence relation. Here is a proof of that fact.

Answer 2

And may not be as?:

Suppose that there is $a \in X$ such that $a \in \mathrm {orb} (x) \cap \mathrm {orb} (y)$ for all $x, y \in X$, where $x \neq y$. This means that $a \in \mathrm {orb} (x)$ and $a \in \mathrm {orb} (y)$ for all $x, y \in X$, that is $\mathrm {orb} (x)$  and $\mathrm {orb} (y)$ are two orbits having a common element. By definition, exists $g \in G$, such that $a = \phi (g, x)$. Hence, for some $g \in G$

$\mathrm {orb} (a) = \mathrm {orb} (\phi (g, x)) = \{ \phi (h, \phi (g, x) ) \in X : h \in G \} = \{ \phi (hg, x) \in X: h \in G \} = \mathrm {orb} (x)$.

Similarly, $\mathrm {orb} (a) = \mathrm {orb} (y)$. Thus, $\mathrm {orb} (x) = \mathrm {orb} (y)$. Hence, we obtain that if one element belongs to $\mathrm {orb} (x) \cap \mathrm {orb} (y)$, these orbits must be equal.

We obtain, that every two orbits are either disjoint or equal, so $X$ is the sum of disjoint orbits.