Kernel of Group Action is Normal Subgroup

Theorem
Let $G$ be a group of transformations of a set $X$.

The set $G_0$ defined as $G_0 = \left\{{g \in G: \forall x \in X: g \cdot x = x}\right\}$ is a normal subgroup of $G$.

Proof
Let $h \in G_0$.

Thus $G_0$ is the intersection of subgroups ($\operatorname{Stab} \left({x}\right) \le G$ from Stabilizer is Subgroup) and by Intersection of Subgroups: Generalized Result, $G_0 \le G$.

To prove normality we need to show that for all $g \in G$, $g G_0 g^{-1} = G_0$.

First suppose that $h \in G_0$, $g \in G$ are arbitrary.

Then

Therefore $ g h g^{-1} \in G_0$, so $g G_0 g^{-1} \subseteq G_0$.

Conversely suppose that $h \in G_0$.

Then by the above, $h' = g^{-1} h g \in G_0$.

Therefore $g = g h' g^{-1} \in g G_0 g^{-1}$, and $G_0 \subseteq g G_0 g^{-1}$.

This concludes the proof.