How to prove simple transitivity from faithful transitivity immediately from stabilizers? The question comes from first few pages of Marcel Berger's textbook "Geometry I". Below is an excerpt in page 8:

1.5.3. One always has
$$G_{g(x)}=gG_xg^{-1}.$$
In other words, $G_{g(x)}$ and $G_x$ are conjugate subgroups of $G$,
and in particular are always isomorphic. Example 1.4.4.1 is an
$\underline{\textrm{immediate consequence}}$ of this remark.

where Example 1.4.4.1 is a proposition

1.4.4.1. Proposition. If $G$ is an abelian group, any faithful transitive action is simply transitive.

I have no difficulty in proving the claims in 1.5.3 and proposition 1.4.4.1 per se, but the last sentence of 1.5.3 (underlined in black) frustrated me. When $G$ is abelian, the two stabilizers merge into one, i.e., $G_x=G_{g(x)}$. But then I don't know how to proceed to establish simple transitivity. How to get a contradiction against faithfulness if there are another element in $G$ that sends $x$ to $g(x)$? I believe it will greatly deepen my understanding of concepts involved, so I greatly appreciate it if you can help me figure out how to prove the simple transitivity (immediately) using results of 1.5.3. Thank you.
 A: This is a general fact of group actions: if $g$ and $h$ do the same thing to $x$, that is, if $gx=hx$, then $h^{-1}g\in G_x$. This follows by applying $h^{-1}$ to both $gx$ and to $hx$, and using that in a group action $\sigma(\tau x) = (\sigma\tau)x$, and that $ex = e$.
This is a general fact about group actions: the intersection of the point stabilizers, $\cap_{x\in X} G_x$ is the kernel of the action; that is, the set (actually, normal subgroup) of all elements $g\in G$ such that $gx=x$ for all $x\in X$.
Now, from 1.5.3 you immediately get that if $x$ and $y$ are in the same orbit of the action of $G$, then $G_x$ and $G_y$ are conjugate.
In 1.4.4.1 you have the following hypotheses: the group $G$ is abelian; the action is transitive; the action is faithful.
Since the group is abelian, as you observe, if $x$ and $y$ are in the same orbit of $G$, then $G_x=G_y$. Okay: so, since the action is transitive there is only one orbit, so $G_x=G_y$ for all $x,y\in X$. Since the action is faithful, the kernel of the action is trivial. From the general fact in the second paragraph above, that means that the intersection of all point stabilizers is trivial; but that intersection is just equal to a single point stabilizer (because they are all the same subgroup!) So the one point stabilizers are all trivial.
But that means that if $gx=hx$ for any $x$, then $g=h$, by the “general fact” in the first paragraph.
This yields that the action is simply transitive (you already knew it was transitive).
So this follows pretty immediately from the assumptions that $G$ is abelian, and the action is faithful and transitive, since the two general facts I outline are pretty standard and basic about group actions.
