Transitive action of the group implies isomorphism with the quotient by stabilizer Let $\Omega$ be a set and $G$ a subgroup of the group $Sym(\Omega)$ of permutations of $\Omega$. Let $\omega \in \Omega$ and
let $G_{\omega}$ denote the stabilizer of $\omega$ in $G$.
If $G$ acts transitively on $\Omega$, show that $G/G_ω \simeq Ω$ as G-sets.
I am unable to think what map should I define. Please give me solution. Thank you
 A: Define a function $f:G\to \Omega$ by $f(g)=g\omega$. The function $f$ is surjective by the assumption that $\Omega$ is a transitive $G$-set. Transitivity is required for $f$ to be surjective.
Observe that
$$\begin{align*}
f(g_1)=f(g_2) &\iff g_1\omega = g_2\omega\\
&\iff (g_2^{-1}g_1)\omega=\omega\\
&\iff g_2^{-1}g_1\in G_\omega &\text{(by definition of $G_\omega$)}\\
&\iff g_1G_\omega = g_2G_\omega &\text{(these are cosets)}
\end{align*}$$
Because $f(g_1)=f(g_2) \impliedby g_1G_\omega=g_2G_\omega$, the function $f$ descends to (a.k.a., factors through) a function $\phi:G/G_\omega\to\Omega$ defined by $\phi(gG_\omega):=g\omega=f(g)$. (This is often phrased as "proving $\phi$ is well-defined";  see Timothy Gowers's post about this.)
Because $\phi(g_1 G_\omega)=f(g_1)=f(g_2)=\phi(g_2 G_\omega) \implies g_1G_\omega=g_2G_\omega$, the function $\phi$ is injective.
Because $f$ is surjective, so is $\phi$.
Thus, $\phi$ is an isomorphism of $G$-sets $\phi:G/G_\omega\xrightarrow{\;\cong\;} \Omega$. While the particular isomorphism $\phi$ defined this way does depend on the choice of $\omega$, the fact that $\Omega$ is isomorphic to a set of left cosets of $G$ does not. For, from transitivity, it follows that stabilizers $G_\omega$ for points $\omega\in\Omega$ are conjugates.

More generally, there is a "first isomorphism theorem of sets":

For any function of sets $f:A\to B$, define an equivalence relation on $A$ by
  $$a_1\sim a_2\;\text{ when }\;f(a_1)=f(a_2)$$
  Then $f$ factors through a surjection, a bijection, and an injection as follows:
  $$\large A\; \xrightarrow[\text{surjective}]{\;\large a\;\mapsto\;[a]\;}\;A/{\sim}\; \xrightarrow[\text{bijective}]{\;\large [a]\;\mapsto\;f(a)\;} \;f(A)\;\xrightarrow[\text{injective}]{\;\large f(a)\;\mapsto\;f(a)\;}\;B$$

