Question: Let $G$ be a group acting transitively on a set $A$. Show that if there is some $\alpha\in A$ such that $G_\alpha=\{1\}$, then $G_\beta=\{1\}$ for all $\beta\in A$.
My thoughts: Since $G$ acts transitively on $A$, there exist $a,b\in A$ such that $g\cdot a=b$, by definition of transitive actions. As $G_\alpha=\{1\}$, we have that $g\cdot\alpha=\alpha$, so I need to try and see if I replace $\alpha$ with any element of $A$, then the result still holds. I'm assuming that this comes from us only having a single orbit, since the action is transitive, but I am having some issues "rigorously" proving it. Any help is greatly appreciated! Thank you.
Best Answer
Here are two approaches:
Fancy approach: Prove $G_{g\cdot\alpha}=gG_{\alpha}g^{-1}$. Then in your scenario you get $G_\beta=G_{g\cdot\alpha}=gG_\alpha g^{-1}=g\{1\}g^{-1}=\{1\}$.
Direct approach: You just need to show if $g'\in G_\beta$ then $g'=1$. To this end, if $g'\cdot\beta=\beta$ then $g'\cdot(g\cdot\alpha)=g\cdot\alpha$, and try to conclude $g^{-1}g'g=1$ then conclude the result from here.