Difficulty showing that a group $G$ applied on $X$, $G_x$ ($x \in X$) & $G_y$, w/ $y \in G(x)$ are same iff $G_x$ is a normal subgroup of $G$. I am difficulty proving that with a group $G$ acting on $X$, stabilizers $G_x$ ($x \in X$) and $G_y$, with $y \in G(x)$ ($x$'s orbit) are same if and only if $G_x$ is a normal subgroup of $G$.
This is what I have so far:
Let $g \in G$. Then, by the definition of stabilizers, ${g_x}x = x$ (with $g_x \in G_x$), and $y = gx$, with $g_ygx = gx$ ($g_y \in G_y$).
However, I am not too sure where to go from here. I believe that if $G_x = G_y$, then for the same $x$ value, $g_x = g_yg$. I'm troubled as how to use the properties of the normal subgroup (https://brilliant.org/wiki/normal-subgroup/) to further my proof.
 A: Suppose $G_x$ is a normal subgroup. Then $g^{-1}G_xg=G_x$, from which we see that if $g_y\in G_y$ iff $g^{-1}g_y g x = x$. But $g^{-1}g_y g$ is thus in $G_x$ and thus also $gg^{-1}g_yg g^{-1}=g_y\in G_x$. Also $g_x\in G_x$ if and only if for each $g$ we get $g_x g x = g x$ (again because $g^{-1}g_xg x=x$), so $g_x\in G_y$.
Now suppose $G_x=G_y$. Then we get $g_ygx = gx$ and $g_y x= x$. Thus we get $g_ygx=g g_y x$ and thus $g^{-1}g_y g x= x$. Thus for each $g_y\in G_y=G_x$ we get $g^{-1}g_yg\in G_x = G_y$.
But I think this is as far as this goes now. We can get normality if we demand that $G_x=G_y$ for all $y\in G(x)$.
EDIT: In fact we need this. Suppose $g=e$, then $y=x$. As $G_x=G_x$ it is easy to see that the given premisses are not suffiecient to get normality.
A: Let $G\curvearrowright X$ be an action of a group on a set. For a point $x\in X$ the stabilizer of $x$ is defined as $G_x:=\{g\in G: gx=x\}$. It is trivial that $G_x$ are subgroups of $G$: indeed, if $g,h\in G_x$ then since $hx=x$ we have $x=h^{-1}x$ so $gh^{-1}x=gx=x$, and thus $gh^{-1}\in G_x$.
Now let $x\in X$ be a point and assume that for any $y\in\text{orb}(x)$ we have $G_x=G_y$. This of course means that $G_x=G_{gx}$ for all $g\in G$. We show that $G_x$ is normal; let $g\in G_x$ and $h\in G$. We have to show that $hgh^{-1}\in G_x$. But, note that $G_{h^{-1}x}=G_x$, so $g\in G_{h^{-1}x}$. We compute: $hgh^{-1}x=h(g(h^{-1}x))=h(h^{-1}x)=x$. so $hgh^{-1}\in G_x$, proving normality.
Conversely, assume that $G_x$ is normal and let $y\in\text{orb}(x)$, say $y=gx$ for some fixed $g\in G$. We need to show that $G_x=G_{gx}$. Let $h\in G_x$. Then $g^{-1}hg\in G_x$ by normality, so $g^{-1}hgx=x$, so $hgx=gx$, so $h\in G_{gx}$, showing that $G_x\subset G_{gx}$. Now let $h\in G_{gx}$; then $hgx=gx$, so $g^{-1}hgx=x$, so $g^{-1}hg\in G_x$. But since $G_x$ is normal, we also have $g(g^{-1}hg)g^{-1}\in G_x$, i.e. $h\in G_x$.
