Orbit-Stabilizer Theorem/Proof 2

Proof
Let $x \in X$.

Let $\phi: \Orb x \to G \mathbin / \Stab x$ be a mapping from the orbit of $x$ to the left coset space of $\Stab x$ defined as:
 * $\forall g \in G: \map \phi {g * x} = g \, \Stab x$

where $*$ is the group action.

Note: this is not a homomorphism because $\Orb x$ is not a group.

Suppose $g * x = h * x$ for some $g, h \in G$.

Then:
 * $h^{-1} g * x = h^{-1} h * x$

and so:
 * $h^{-1} g * x = x$

Thus:
 * $h^{-1} g \in \Stab x$

so by Left Coset Space forms Partition:
 * $g \, \Stab x = h \, \Stab x$

demonstrating that $\phi$ is well-defined.

Let $\map \phi {g_1 * x} = \map \phi {g_2 * x}$ for some $g_1, g_2 \in G$.

Then:
 * $g_1 \, \Stab x = g_2 \, \Stab x$

and so by Left Coset Space forms Partition:
 * $g_2^{-1} g_1 \in \Stab x$

So by definition of $\Stab x$:
 * $x = g_2^{-1} g_1 * x$

Thus:

thus demonstrating that $\phi$ is injective.

As the left coset $g \, \Stab x$ is $\map \phi {g * x}$ by definition of $\phi$, it follows that $\phi$ is a surjection.

The result follows.