$(X/H)/(G/H)=X/G$? Here $G$ is a topological group acting continuously on $X$, $H$ is a closed normal subgroup of $G$.

Question

Let $G$ be a topological group acting continuously on a topological space $X$ (on the left, denoted by $g\cdot x$, $g\in G, x\in X$), $H$ be a closed normal subgroup of $G$. Then

1. The quotient group $G/H = \{gH\mid g\in G\}$ becomes a topological group when given the quotient topology.

2. The continuous group action of $G$ on $X$ induces a quotient topology on the orbit space $X/G:=\{G\cdot x \mid x\in X\}$. The quotient space $X/H$ is defined in the same way.

3. The continuous group action of $G$ on $X$ induces a (left) continuous group action of $G/H$ on $X/H$ by: $$gH\cdot (H\cdot x) := H\cdot(g\cdot x),\quad x\in X, g\in G.$$ This action also induces a quotient topological space $(X/H)/(G/H)$.

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The natural question is, what relations are there between $(X/H)/(G/H)$ and $X/G$? It's easy to verify that they are the same set, and the quotient topology on $(X/H)/(G/H)$ is coarser than that on $X/G$ (by the universal property of quotient topology).

So what I'm wondering is whether the two quotient topologies are the same...

Answer 1

Actually the two quotient topologies are the same. Just using the more detailed universal property of quotient topology presented in Munkres' topology book, Theorem 22.2.

PS: This conclusion may be referred to as the third isomorphism theorem.