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suppose $X$ is a topological space and $G$ and $H$ are groups acting on it.

1) if $G$ is isomorphic to $H$ do we have necessarely $X/G$ is homeomorphic to $X/H$

2) suppose $G$ and $H$ are two conjugate subgroups of a group $K$ acting on $X$, in this case do we have $X/G$ is homeomorphic to $X/H$

3) Conversely, if $X/G$ is homeomorphic to $X/H$ does this implie that $G$ is isomorphic to $H$?

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Regarding your question, it should be intuitively clear that the actions of $G$ and $H$ on $X$ must be connected in some way if we wish to deduce (1). Otherwise, we can certainly have something like $G=H$ and consider two totally different actions of $G$ on $X$. The question looks very much like homework; please do correct me if I am wrong. Also, please explain what you have tried; firing questions at us will help nobody. – Amitesh Datta Jun 21 '11 at 8:31
up vote 1 down vote accepted

The following are hints:

(1) Consider the case $G=H$ and consider one action of $G$ on $X$ to be defined by $g\cdot x=x$ for all $x\in X$. (Prove that this is indeed an action of $G$ on $X$ if you have not done so already.) The quotient space $X/G$ is naturally homeomorphic to $X$. Find an example of a non-trivial topological group $X$ and an action of $G$ on $X$ such that $X/G$ has exactly one orbit (for example). Conclude that $X/G$ is not homeomorphic to $X/H$.

(2) Let $k\in K$ be such that $k^{-1}Gk=H$. Prove that $x\mapsto k\cdot x$ induces a homeomorphism $X/G\to X/H$. (In particular, you need to prove that if two elements of $X$ lie in the same $G$-orbit, then their images under this map lie in the same $H$-orbit. For this, use $k^{-1}Gk=H$.)

(3) Consider two non-isomorphic groups acting trivially on $X$.

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the: the counter example in 3) is clear . thanks. what about 2) i think it is true but i don't knowa why conjugacy implies homeomorphism – palio Jun 21 '11 at 8:48
Dear palio, did you understand the counterexample in (1) (which should also (hopefully) be clear)? Regarding (2), I really do think you should try to work this out on your own. I have given a hint (in brackets). Basically, you should be able to prove that the map $x\mapsto k\cdot x$ is a homeomorphism $X\to X$. In order to prove that it induces a homeomorphism $X/G\to X/H$, you really only need to show that if two elements of $X$ lie in the same $G$-orbit, then their images under the map $x\mapsto k\cdot x$ lie in the same $H$-orbit. Can you see how to prove this? – Amitesh Datta Jun 21 '11 at 8:52
it's all clear now. thanks alot!!! – palio Jun 21 '11 at 9:01

First a question. When you talk about `action of a group on a topological space' is it assumed that the action is continuous?

  1. No. You could let $\mathbf{Z}$ act on $\mathbf{R}$ by translation. Then $\mathbf{R}/\mathbf{Z}$ is homeo to the circle. If we let $\mathbf{Z}$ act on $\mathbf{R}$ trivially, we just get $\mathbf{R}/\mathbf{Z} = \mathbf{R}$.

So the point is that the same group can give many different actions and, therefore, many different quotients.

  1. No. (I'm not sure if the following example works.) Let G be the cyclic group of order two acting on the circle S^1 via multiplication with -1. Let H be the trivial group acting trivially on the circle. Note that S^1/H = S^1 and S^1/G$ is (also?) homeomorphic to the circle.
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Dear Luffy, the question refers to the action of a group $G$ on a topological group $X$. Therefore, it seems reasonable to believe that each element of the group $G$ should induce an isomorphism of $X$ (via the map $x\mapsto g\cdot x$ for $x\in X$) in the category of topological groups. – Amitesh Datta Jun 21 '11 at 8:45
@ Amitesh Datta : $X$ is not a topological group but it is only a topological space – palio Jun 21 '11 at 8:53

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