[proofplan]
We prove that the orbit relation is exactly the [equivalence relation](/page/Equivalence%20Relation) generated by the [group action](/page/Group%20Action). First, every point lies in its own orbit because the identity element acts as the identity on $X$. Then we prove the key dichotomy: if two orbits intersect, then each orbit is contained in the other, using inverses in the group action. This shows that distinct orbits cannot overlap, and hence the family of all orbits partitions $X$.
[/proofplan]
[step:Show that every point lies in its own orbit]
Let $e \in G$ denote the identity element of the group $G$. Since the given map is a group action, $e \cdot x = x$ for every $x \in X$. Therefore, for every $x \in X$,
\begin{align*}
x = e \cdot x \in G \cdot x.
\end{align*}
Thus every element of $X$ lies in at least one member of the collection $X/G$.
[/step]
[step:Prove that intersecting orbits are equal]
Let $x, y \in X$, and suppose that the two orbits $G \cdot x$ and $G \cdot y$ intersect. Choose an element $z \in (G \cdot x) \cap (G \cdot y)$. By the definition of orbit, there exist elements $g, h \in G$ such that
\begin{align*}
z = g \cdot x
\end{align*}
and
\begin{align*}
z = h \cdot y.
\end{align*}
Hence
\begin{align*}
g \cdot x = h \cdot y.
\end{align*}
Applying $h^{-1}$ to both sides and using the compatibility law for the group action gives
\begin{align*}
y = h^{-1} \cdot (g \cdot x) = (h^{-1}g) \cdot x.
\end{align*}
Thus $y \in G \cdot x$.
We now prove $G \cdot y \subset G \cdot x$. Let $w \in G \cdot y$. Then there exists $a \in G$ such that
\begin{align*}
w = a \cdot y.
\end{align*}
Since $y = (h^{-1}g) \cdot x$, the action law gives
\begin{align*}
w = a \cdot ((h^{-1}g) \cdot x) = (a h^{-1} g) \cdot x.
\end{align*}
Because $a h^{-1} g \in G$, this shows $w \in G \cdot x$. Therefore $G \cdot y \subset G \cdot x$.
Similarly, applying $g^{-1}$ to the equality $g \cdot x = h \cdot y$ gives
\begin{align*}
x = (g^{-1}h) \cdot y.
\end{align*}
The same containment argument, with $x$ and $y$ interchanged, gives $G \cdot x \subset G \cdot y$. Hence
\begin{align*}
G \cdot x = G \cdot y.
\end{align*}
[guided]
The main point is that a group action lets us move both forward and backward along an orbit, because every group element has an inverse. Suppose $G \cdot x$ and $G \cdot y$ have a common element. This means there is some point $z \in X$ that can be reached from $x$ and also from $y$. By the definition of orbit, there exist $g, h \in G$ such that
\begin{align*}
z = g \cdot x
\end{align*}
and
\begin{align*}
z = h \cdot y.
\end{align*}
Therefore
\begin{align*}
g \cdot x = h \cdot y.
\end{align*}
To compare the two orbits, we first express $y$ in terms of $x$. Apply $h^{-1}$ to both sides. Since this is a group action, the identity element acts as the identity and the action satisfies $a \cdot (b \cdot u) = (ab) \cdot u$ for all $a,b \in G$ and $u \in X$. Thus
\begin{align*}
h^{-1} \cdot (g \cdot x) = h^{-1} \cdot (h \cdot y).
\end{align*}
Using the action law on both sides gives
\begin{align*}
(h^{-1}g) \cdot x = (h^{-1}h) \cdot y = e \cdot y = y.
\end{align*}
So
\begin{align*}
y = (h^{-1}g) \cdot x.
\end{align*}
This says precisely that $y \in G \cdot x$.
Now take any element $w \in G \cdot y$. By definition of the orbit of $y$, there exists $a \in G$ such that
\begin{align*}
w = a \cdot y.
\end{align*}
Substitute the expression for $y$ in terms of $x$:
\begin{align*}
w = a \cdot ((h^{-1}g) \cdot x).
\end{align*}
Using the action law again,
\begin{align*}
w = (a h^{-1} g) \cdot x.
\end{align*}
Since $a h^{-1} g$ is an element of $G$, this shows $w \in G \cdot x$. Because $w$ was arbitrary, we have proved
\begin{align*}
G \cdot y \subset G \cdot x.
\end{align*}
The reverse containment is obtained by the same explicit argument starting from
\begin{align*}
g \cdot x = h \cdot y.
\end{align*}
Apply $g^{-1}$ to both sides:
\begin{align*}
g^{-1} \cdot (g \cdot x) = g^{-1} \cdot (h \cdot y).
\end{align*}
Using the action law and the identity law gives
\begin{align*}
x = (g^{-1}h) \cdot y.
\end{align*}
If $v \in G \cdot x$, then $v = b \cdot x$ for some $b \in G$, and therefore
\begin{align*}
v = b \cdot ((g^{-1}h) \cdot y) = (b g^{-1} h) \cdot y.
\end{align*}
Since $b g^{-1} h \in G$, we get $v \in G \cdot y$. Hence
\begin{align*}
G \cdot x \subset G \cdot y.
\end{align*}
The two containments imply
\begin{align*}
G \cdot x = G \cdot y.
\end{align*}
[/guided]
[/step]
[step:Conclude that the collection of orbits is a partition of $X$]
We have shown that every $x \in X$ belongs to at least one orbit, namely $G \cdot x$. We have also shown that whenever two orbits $G \cdot x$ and $G \cdot y$ intersect, they are equal. Equivalently, if two orbits are not equal, then their intersection is empty. Therefore the collection
\begin{align*}
X/G = \{G \cdot x : x \in X\}
\end{align*}
covers $X$ and consists of pairwise disjoint nonempty subsets of $X$. Hence the orbits of the action form a partition of $X$.
[/step]
