For an action of $G$ on a set, every point of some orbit has the same stabilizer if and only if this stabilizer is a normal subgroup.

Here is my proof of the first part.

(=>). Let $G$ act on a set $X$ and for some $x\in X$. In order to show that $G_x$ is a normal subgroup of $G$, we will show that for all $g \in G$ we have $gG_x = G_xg$. Let $g$ be any element of $G$ and $y = gx$, an element in the orbit of $G(x)$. We know from a previous result that points in the same orbit have conjugate stablizers, i.e. $gG_xg^{-1}=G_y$, and since $G_x=G_y$, we have $gG_xg^{-1}=G_x$ or $gG_x = G_xg$. The stabilizer is normal.

(<=). Suppose $G_x \lhd G$ for some $x \in X$ and let $y \in G(x)$. I have no idea where to go from here.

Solution 1:

($<=$)

Since $y\in G(x)$ we have $y=g.x$ for some $g\in G$.

Then $$h\in G_y\iff g^{-1}hg\in G_x\tag1$$ or equivalently:$$G_y=gG_xg^{-1}\tag2$$

Then the normality of $G_x$ tells us that $G_x=G_y$.

Solution 2:

If $y=g\cdot x$ is an element of the orbit of $x$ (with $g\in G$), then $h\in G$ fixes $y$ if $hg\cdot x=y=g\cdot x$ which happens if and only if $g^{-1}hg\cdot x=x$, in other words $g^{-1}hg$ is in the stabiliser subgroup $G_x$, which happens if and only if $h\in gG_xg^{-1}$, so $G_y=gG_xg^{-1}$. Then $G_y=G_x$ means $gG_xg^{-1}=G_x$, and this holds for all $g\in G$ if and only if $G_x$ is a normal subgroup.

Solution 3:

The proper answer to your question has been given by drhab (since it gives the converse part for your own proof). So I will give an alternative answer with a different proof.

Elements $y \in G(x)$ are exactly the ones of the form $y = gx$ for some element $g \in G$. Then \begin{align*} h \in G_y & \iff hy =y\\ & \iff h(gx) = (hg)x = gx\\ & \iff (g^{-1}hg) x = x\\ & \iff g^{-1}hg \in G_x\\ & \iff h \in g G_x g^{-1}. \end{align*} Thus, the stabilizer of $y$ is $G_y = g G_x g^{-1}$, for every $y = gx \in G(x)$, the orbit of $x$.

Therefore, $G_x = G_y$ for every $y = gx \in G(x)$ if and only if $g G_x g^{-1} = G_x$, for every $g \in G$, i.e., $G_x \unlhd G$.

Note that both parts of the result are proved "in one go".