Stabilizer commutes with conjugation

Let G be a group and A a nonempty set. Let G act on A. Prove that if a,b \in A and b = g \cdot a for some g \in G, then \mathsf{stab}_G(b) = g \mathsf{stab}_G(a) g^{-1}. Deduce that if G acts transitively on A, then the kernel K of the action is \bigcap_{g \in G} g \mathsf{stab}_G(a) g^{-1}.

First we prove that \mathsf{stab}_G(b) = g \mathsf{stab}_G(a) g^{-1}.

(\subseteq) If x \in \mathsf{stab}_G(b), then x \cdot b = b. Now g^{-1}xg \cdot a = g^{-1}x \cdot b = g^{-1} \cdot b = a, so that g^{-1}xg \in \mathsf{stab}_G(a). Hence x \in g \mathsf{stab}_G(a) g^{-1}.

(\supseteq) Let x \in g \mathsf{stab}_G(a) g^{-1}. Then gxg^{-1} \cdot b = gx \cdot a = g \cdot a = b, so that gxg^{-1} \in \mathsf{stab}_G(b). Hence g \mathsf{stab}_G(a) g^{-1} \subseteq \mathsf{stab}_G(b).

Suppose now that the action of G on A is transitive; that is, for all a,b \in A, there exists g \in G such that b = g \cdot a. We show that, if K is the kernel of the action, K = \bigcap_{g \in G} g \mathsf{stab}_G(a) g^{-1}.

(\subseteq) Let x \in K. Then x \cdot a = a for all a \in A. Let g \in G. Then g^{-1}xg \cdot a = g^{-1} \cdot (x \cdot (g \cdot a)) = g^{-1} \cdot (g \cdot a) = g^{-1}g \cdot a = 1 \cdot a = a, so that g^{-1}xg \in \mathsf{stab}_G(a) for all g \in G. Thus x \in g \mathsf{stab}_G(a) g^{-1} for all g \in G, hence x \in \bigcap_{g \in G} g \mathsf{stab}_G(a) g^{-1}.

(\supseteq) Let x \in \bigcap_{g \in G} g \mathsf{stab}_G(a) g^{-1}, and let b \in A. Because the action of G on A is transitive, we have b = h \cdot a for some h \in G. Now x = hyh^{-1} for some y \in \mathsf{stab}_G(a), thus x \cdot b = hyh^{-1} \cdot b = hy \cdot a = h \cdot a = b. Hence x stabilizes b; since b \in A is arbitrary, x \in K.

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