Stabilisers of group action open imply the action is continuous Let $\mu \colon X \times G \longrightarrow X$ be the action of a topological group on a set $X$. We consider $X$ to be a topological space with the discrete topology. Suppose that for all $x \in X$, the stabiliser $I_x =\{ g \in G \mid xg = x \}$ is open in $G$. Why does this mean that the function $\mu$ is continuous?
 A: Because for a given $x \in X$ we have
$$
\mu^{-1}(\{ x \}) = \{ (y, g) : y g = x \}
= \bigcup \{ z \} \times I_{z} g_{zx},
$$
where the union ranges over all $z$ in the orbit of $x$, and for each such $z$, we choose $g_{zx}$ to be a fixed element such that $z g_{zx} = x$.
Clearly the translate $I_{z} g_{zx}$ of $I_{z}$ is open in $G$, and so is $\{ z \}$ in $X$, so their product is open in $X \times G$.
A: Here I have written the action on the left and change the notation of the stabilizer. I think it will be easy to translate the idea into your formalism...
To show that $\mu$ is continuous, you need to show that $\mu^{-1}(U)$ is open in $X\times G$ for the product topology. Because $X$ is discrete a topological base for $X$ is given by the $\{x\}$'s for $x\in X$, thus it suffices to show that $\mu^{-1}(\{x\})$ is open. Then :
$$\mu^{-1}(\{x\})=\{(y,g)\in X\times G|g.y=x\} $$
$$\mu^{-1}(\{x\})=\bigcup_{y\in X}\{(y,g)|g\in G \text{ and } g.y=x\} $$
Now if $y\notin G.x$ then $\{(y,g)|g\in G \text{ and } g.y=x\}=\emptyset$ so :
$$\mu^{-1}(\{x\})=\bigcup_{y\in G.x}\{(y,g)|g\in G \text{ and } g.y=x\} $$
For each $y\in G.x$ take $g_y\in G$ such that $g_y.x=y$. Then :
$$\mu^{-1}(\{x\})=\bigcup_{y\in G.x}\{(y,g)|g\in G \text{ and } g.g_y^{-1}x=x\} $$
Now $(y,g)\in \{(y,g)|g\in G \text{ and } g.g_y^{-1}x=x\}$ if and only if $gg_y^{-1}\in Stab_G(x)$ if and only if $g\in Stab_G(x)g_y$. Finally :
$$\mu^{-1}(\{x\})=\bigcup_{y\in G.x}\{y\}\times (Stab_G(x)g_y) $$
Because $X$ is endowed with the discrete topology $\{y\}$ is open, because $G$ is a topological group and $Stab_G(x)$ is open by assumption, $ Stab_G(x)g_y$ is open as well. The product of both is open and then $\mu^{-1}(\{x\})$ is open.
