Let $X$ be a Hausdorff space, and $G$ a group acting on $X$ by homeomorphisms. Let $H$ be a normal subgroup of $G$. Is it true that $X/G$ is homeomorphic to $(X/H)/(G/H)$ ?
If so, can you please provide a proof or a reference, and if not, are there additional informations on the type of action of $G$ (discrete, proper, discontinuous...) needed to make the statement true ?
Thanks in advance.
Consider the natural map between $(X/H)$ and $(X/G)$ given by sending the orbit of a point $x$ under $H$ to the orbit of $x$ under $G$. This map is continuous, because the preimages of open sets in $(X/G)$ are unions of open sets in $(X/H)$. It is constant on each $G$-orbit (which is a $G/H$-orbit), so it induces a continuous bijection between the quotient $(X/H)/(G/H)$ and $(X/G)$.
Now do the same thing, sending each point in $X$ to its image in $(X/H)$ and then to its image in $(X/H)/(G/H)$. The pre image of an open set in $(X/H)/(G/H)$ is the union of open sets in $(X/H)$ and the primates of these are open sets in $X$. Also, this map is constant on $G$-orbits. So we get a continuous bijection between $(X/H)/(G/H)$ and $(X/G)$ which is the inverse of the previous map, giving a homeomorphism.