Stabilizer commutes with conjugation

Let G be a permutation group on the set A (i.e. G \leq S_A), let \sigma \in G, and let a \in A. Prove that \sigma \mathsf{stab}_G(a) \sigma^{-1} = \mathsf{stab}_G(\sigma(a)). Deduce that if G acts transitively on A then \bigcap_{\sigma \in G} \sigma \mathsf{stab}_G(a) \sigma^{-1} = 1.


First we show that \sigma \mathsf{stab}_G(a) \sigma^{-1} = \mathsf{stab}_G(\sigma(a)).

(\subseteq) Let x \in \sigma \mathsf{stab}_G(a) \sigma^{-1}. Then x = \sigma y \sigma^{-1} for some y \in \mathsf{stab}_G(a). Now x \cdot \sigma(a) = \sigma y \sigma^{-1} \sigma \cdot a = \sigma y \cdot a = \sigma \cdot a = \sigma(a); hence x \in \mathsf{stab}_G(\sigma(a)).

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

Now if G acts transitively on A, the kernel of the action is \bigcap_{\sigma \in G} \sigma \mathsf{stab}_G(a) \sigma^{-1} by the previous exercise. Moreover, because G \leq S_A, the homomorphism \iota : G \rightarrow S_A producing this action is injective, and thus has a trivial kernel. Thus \bigcap_{\sigma \in G} \sigma \mathsf{stab}_G(a) \sigma^{-1} = 1.

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