Intersect of Stabilizers is a Normal Group I am looking for guidance for two problems on group action, one of them is here and the other one will be posted in another page:

Assume that $G$ operates on a set $\Omega.$ Show that $\bigcap_{\omega \in \Omega} G_\omega$ is a normal subgroup of $G.$

Here $G_\omega$ is stabilizer defined as $G_\omega : = \{g \in G : \omega g = \omega \} $, and here are what I have managed so far: $$\Omega = \{\omega_1, \omega_2, \omega_3 ...\},$$
$$\bigcap_{\omega \in \Omega}G_\omega = \{G_{\omega_1} \cap G_{\omega_2} \cap G_{\omega_3} ... \}$$
$$\Rightarrow \{g \in G : (\omega_1 g = \omega_1) \land (\omega_2 g = \omega_2) \land (\omega_3 g = \omega_3) ...\}$$
$$\Rightarrow g \in G : \Omega g = \Omega$$
$$ ... $$
$$ ... $$
But then I am stuck here because, I think, my understanding of group action is not fluid enough. I would appreciate any help to give me good understanding of group action. Thanks for your time and help.
PS. My class note defines group action as follow: Let $G$ and $\Omega$ be a group and a non-empty set, we say that $G$ acts (or operates) on $\Omega$ if there exists a homomorphism $\phi$ from $G$ to $Sym(\Omega).$ Thanks again.
 A: The intersection of subgroups is always a subgroup, so we only need to show that $\bigcap_{\omega \in \Omega} G_\omega$ is normal.
Let $g \in \bigcap_{\omega \in \Omega} G_\omega$, and let $h \in G$. We want to show $hgh^{-1} \in \bigcap_{\omega \in \Omega} G_\omega$.
Indeed,
$$\omega (hgh^{-1})=(\omega h) g h^{-1} = (\omega h)h^{-1} = \omega(hh^{-1})=\omega.$$
A: A group action of $G$ on $\Omega$ can be seen as homomorphism from $G$ to $S_\Omega$. Under this view if $g$ is in the intersection of all the stabilizers then $g$ maps to the identity of $S_\Omega$. so what you have is the kernel of a homomorphism.
A: You are almost done. You know that $g$ fixes every element of $\Omega$ (although saying $\Omega g = \Omega$ does not really mean that: it means that $g$ sends $\Omega$ to itself).
$\bigcap_{\omega \in \Omega} G_\omega$ is a subgroup of $G$, since it is the intersection of a bunch of subgroups. And from your analysis, any $g\in \bigcap_{\omega \in \Omega} G_\omega$ is such that $\phi(g)$ is the identity in $Sym(\Omega)$. Thus this is just the kernel of the map $G\to Sym(\Omega)$, which is a normal subgroup.
