Let $G$ be a finite group, and suppose it is acting on the finite set $\Omega$. Suppose $N$ is a normal subgroup of $G$ and $N$ acts transitively on $\Omega$. Let $\omega \in \Omega$, and let $G_\omega$ be the stabilizer in $G$ of $\omega$. Prove that $|G : N| = |G_\omega: G_\omega \cap N|$.
$\textit{My Attempt:}$ $\ \ $As $N \unlhd G$, $G/N$ forms a well-defined group with coset multiplication. Let $$\mu: G_\omega \longrightarrow G/N$$ be defined by, for $f \in G_\omega$, $\mu(f) = fN$. Suppose $z \in \ker\mu$, so that $zN = \mu(z) = N$, and $\mu \in N$. Thus, $z \in G_\omega \cap N$. If $z \in G_\omega \cap N$, then $z \in N$ so that $\mu(z) = zN = N$, and thus $z \in \ker \mu$. Therefore $\ker\mu = G_\omega \cap N$. By the First Isomorphism Theorem, it suffices to show that $\mu$ is surjective, which is where I am stuck. Suppose $f N \in G/N$. Thus $f \in G$, and $f \cdot \omega \in \Omega$. Since $N$ acts transitively, there is $n \in N$ for which $f \cdot \omega = n \cdot \omega$. I don't know where to go from here. Can I conclude that $f^{-1} \cdot \omega = n^{-1}\cdot \omega$? If so, then I can conclude that $$(fn^{-1}) \cdot \omega = f \cdot (n^{-1} \cdot \omega) = f \cdot (f^{-1} \cdot \omega) = (f f^{-1}) \cdot \omega = e \cdot \omega = \omega$$ so that $fn^{-1} \in G_\omega$ and $\mu(fn^{-1}) = fn^{-1}N = fN$, as $n \in N$ and $N \leq G$ implies $n^{-1} \in N$. Finally, $$G_\omega/ \ker\mu = G_\omega/G_\omega \cap N \cong \mu[G_\omega] = G/N$$ but I doubt I could make the assumption I did just before. I never use the fact that $G$ and $\Omega$ are finite either. I considered perhaps using an argument through Lagrange's Theorem to conclude surjectivity of $\mu$ but didn't know how it would play out. It's likely that my entire approach all together here isn't the right way. Thanks in advance for any response.
$G / G_w$ and $N / N_w$ are both bijetive with $\Omega$ because the action is transitive. The bijection comes from the action: send a coset $gG_w$ to $gw$.
($N_w$ is the stabilizer in $w$.)
Now, $N_w = N \cap G_w$, and the result follows by re-arranging these equalities:
$| G : N | = |G| / |N|$, $|G : G_w| = |G| / |G_w|= |N : N_w|= |N|/|N_w|$ and $| G_w : N_w | = |G_w| / |N_w|$.