§ Geometric characterization of normal subgroups

$Stab(Orb(x)) = Stab(x) \iff Stab(x) \text{ is normal}$

$\forall x' \in Orb(x), Stab(x') = Stab(x) \iff Stab(x) \text{ is normal}$

§ Forward: if the stabilizer is normal, then all elements in the orbit have the same stabilizer

Let a group $G$ act on a set $X$ with action $(~\dot~) : G \times X \rightarrow X$. Let $H \subseteq G$ be the stabilizer of a point $x \in X$. Now, let $K = kHk^{-1}$, a conjugacy class of $H$. Clearly, the element $(k \cdot x)$ in the orbit of $x$ is stabilized by $K$.
If the group $H$ is normal, then $K = H$. So every element in the orbit of $x$ is stabilized by $H$.

§ Interaction of stablizer and the orbit:

$Stab(g \cdot x) = g Stab(x) g^{-1}$

$g^{-1} Stab(g \cdot x) g = Stab(x)$

• Proof of $s \in Stab(x) \implies gsg^{-1} \in Stab(g \cdot x)$: The action of $gsg^{-1}$ on $g \cdot x$ is: $(g \cdot x \rightarrow_{g^-1} x \rightarrow_s x \rightarrow_g g \cdot x)$.

• Proof of $s' \in Stab(g \cdot x) \implies g^{-1}s'g \in Stab(x)$: The action of $g^{-1}s'g$ on $x$ is: $(x \rightarrow_{g} g \cdot x \rightarrow_{s'} g \cdot x \rightarrow_{g^{-1}} x)$.

Hence, both containments are proved.

§ Backward: if all elements in the orbit have the same orbit, then the stabilizer is normal.

From the above equation $Stab(g \cdot x) = g Stab(x) g^{-1}$. If the entire orbit has the same stabilizer, $Stab (g \cdot x) = Stab(x)$. Hence, we get $Stab(x) = g Stab(x) g^{-1}$, proving that it's normal.