Solution to Abstract Algebra by Dummit & Foote 3rd edition Chapter 4.1 Exercise 4.1.2
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 $\bigcap_{\sigma \in G} \sigma \mathsf{stab}_G(a) \sigma^{-1} = 1$.
Solution: 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 $\bigcap_{\sigma \in G} \sigma \mathsf{stab}_G(a) \sigma^{-1} = 1$.
