Kernel of Group Action is Normal Subgroup

Theorem
Let $G$ be a group.

Let $X$ be a set.

Let $\phi: G \times X \to X$ be a group action.

Let $G_0$ denote the kernel of $\phi$.

Then $G_0$ is a normal subgroup of $G$.

Proof
Let $h \in G_0$.

From Stabilizer is Subgroup:
 * $\Stab x \le G$

Thus $G_0$ is the intersection of subgroups.

By Intersection of Subgroups is Subgroup:
 * $G_0 \le G$

To prove normality we need to show:
 * $\forall g \in G: g G_0 g^{-1} = G_0$

Let $h \in G_0, g \in G$ be 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:
 * $h = g h' g^{-1} \in g G_0 g^{-1}$

and so:
 * $G_0 \subseteq g G_0 g^{-1}$

This concludes the proof.