Orbit-Stabilizer Theorem/Proof 2

Proof
Let $x \in X$.

Let $\phi: \operatorname{Orb} \left({x}\right) \to G / H^l$ be a mapping from the orbit of $x$ to the left coset space of $\operatorname{Stab} \left({x}\right)$ defined as:
 * $\forall g \in \operatorname{Orb} \left({x}\right): \phi \left({g * x}\right) = g \operatorname{Stab} \left({x}\right)$

where $*$ is the group action.

Note: this is not a homomorphism because $\operatorname{Orb} \left({x}\right)$ is not a group.

Suppose $g * x = h * x$.

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

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

Thus:
 * $h^{-1} g \in \operatorname{Stab} \left({x}\right)$

so by Coset Spaces form Partition:
 * $g \operatorname{Stab} \left({x}\right) = h \operatorname{Stab} \left({x}\right)$

demonstrating that $\phi$ is well-defined.

Let $\phi \left({g_1 * x}\right) = \phi \left({g_2 * x}\right)$ for some $g_1, g_2 \in G$.

Then:
 * $g_1 \operatorname{Stab} \left({x}\right) = g_2 \operatorname{Stab} \left({x}\right)$

and so by Coset Spaces form Partition:
 * $g_2^{-1} g_1 \in \operatorname{Stab} \left({x}\right)$

So by definition of $\operatorname{Stab} \left({x}\right)$:
 * $x = g_2^{-1} g_1 * x$

Thus:

thus demonstrating that $\phi$ is injective.

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

The result follows.