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
Best Answer
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":