Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $H$?

share|cite|improve this question
    
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 $X$.

share|cite|improve this answer
    
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
1  
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.
share|cite|improve this answer
    
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.