Let $G$ act 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 $\beta\in A$ 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.
 A: 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.
A: *

*The stabilizers of the points of one same orbit are conjugate. So, if one is $\{1\}$, all the others are $\{1\}$ either.


*If the action is transitive, then the whole $A$ is one same orbit.
From 1 and 2 your claim follows.
A: Hint:
There exists $h\in G$ such that $h\alpha=\beta$.
Now, if $g\beta=\beta$, what can you deduce for $\alpha$?
A: Let us do this explicitly, let us break down all the definitions, since $G$ is a transitive action, we have that there is a single orbit, that is given $x,y\in A$ there is some $g\in G$ such that $gx=y$. Let us also recall the definition of the stabilizer, we have that $G_x=\{g\in G:gx=x\}$.
Now say that we have $G_\alpha=\{1\}$, now let $\beta\in A$, let us show that $G_\beta=\{1\}$. Thus, let $g\in G_\beta$ (our goal will be to show that $g=1$). Thus, we have that $g\beta=\beta$. Now we have that there is some $h\in G$ such that $h\alpha=\beta$ (equivalently we will have that $h^{-1}\beta=\alpha$), so we will have that $gh\alpha=\beta$, and then multiplying by $h^{-1}$ on the left again, we conclude that $h^{-1}gh\alpha=h^{-1}\beta=\alpha$. Thus, we will have that $h^{-1}gh\in G_\alpha=\{1\}$. Thus, we conclude that $h^{-1}gh=1$, and from this we conclude that $g=1$.
