§ Geometric characterization of normal subgroups

Stab(Orb(x))=Stab(x)  ⟺  Stab(x) is normalStab(Orb(x)) = Stab(x) \iff Stab(x) \text{ is normal}
∀x′∈Orb(x),Stab(x′)=Stab(x)  ⟺  Stab(x) 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 GG act on a set XX with action (  ˙):G×X→X(~\dot~) : G \times X \rightarrow X. Let H⊆GH \subseteq G be the stabilizer of a point x∈Xx \in X. Now, let K=kHk−1K = kHk^{-1}, a conjugacy class of HH. Clearly, the element (k⋅x)(k \cdot x) in the orbit of xx is stabilized by KK. If the group HH is normal, then K=HK = H. So every element in the orbit of xx is stabilized by HH.

§ Interaction of stablizer and the orbit:

Stab(g⋅x)=gStab(x)g−1Stab(g \cdot x) = g Stab(x) g^{-1}
g−1Stab(g⋅x)g=Stab(x)g^{-1} Stab(g \cdot x) g = Stab(x)
  • Proof of s∈Stab(x)  ⟹  gsg−1∈Stab(g⋅x)s \in Stab(x) \implies gsg^{-1} \in Stab(g \cdot x): The action of gsg−1gsg^{-1} on g⋅xg \cdot x is: (g⋅x→g−1x→sx→gg⋅x)(g \cdot x \rightarrow_{g^-1} x \rightarrow_s x \rightarrow_g g \cdot x).
  • Proof of s′∈Stab(g⋅x)  ⟹  g−1s′g∈Stab(x)s' \in Stab(g \cdot x) \implies g^{-1}s'g \in Stab(x): The action of g−1s′gg^{-1}s'g on xx is: (x→gg⋅x→s′g⋅x→g−1x)(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⋅x)=gStab(x)g−1Stab(g \cdot x) = g Stab(x) g^{-1}. If the entire orbit has the same stabilizer, Stab(g⋅x)=Stab(x)Stab (g \cdot x) = Stab(x). Hence, we get Stab(x)=gStab(x)g−1Stab(x) = g Stab(x) g^{-1}, proving that it's normal.