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Let $U$ be open in a topological group, G. Why then is it necessarily true that $UH$ where $H$ is some subgroup of $G$ open in $G$? (I think I don't quite get the idea of a topological group even after reading its definition on Wiki. Grateful if someone could explain the idea.) Thanks.

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As you know, in any topological space, arbitrary unions of open sets are open. $UH$ is a union of cosets, $UH=\bigcup_{h\in H} Uh$. Now, you should check your understanding by proving for yourself that each of the cosets $Uh$ is open. If you can't, then you should try to identify your difficulty with the definitions and ask about that. Just repeating the Wikipedia definition here is not a very worthwhile thing to do.

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Thanks. Unfortunately, I can't quite locate the gaping hole in my knowledge, but I am guessing that $Uh$ is open because we are composing the open set $U$ with a constant element. But I can't recall what theorem or rule it is that allows this to be true. –  brian Aug 19 '11 at 9:04
    
@brian All you need to know are the definitions of topology, continuous maps, and topological groups. From your comment, my guess is that it's the second of the three that you are having trouble with, but it's only a guess. It is your job to form a well-posed question that can be answered and that you will benefit from. –  Alex B. Aug 19 '11 at 9:10
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Ah! Would it be that the map of multiplying a set by $h^{-1}$ is continuous, hence the the preimage of $U$ under such a map is open? –  brian Aug 19 '11 at 9:12
    
@brian A nitpick: you don't multiply a set by $h^{-1}$. You multiply elements. That gives a function from one set to another (namely $G\rightarrow G$), and you can look at images and preimages of sets under this map. The rest of what you say is correct. –  Alex B. Aug 19 '11 at 11:08
    
Thank you very much, Alex. –  brian Aug 19 '11 at 12:26
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This is not a full answer, but just a comment. Let $h$ in $H$ and consider the set $U_h = h\cdot U$ in $G$. Suppose that we have shown that $U_h$ is open. Then

$$HU = \bigcup_{h\in H} U_h$$ is open, because the union of open subsets is open.

Thus, we have to show that $U_h$ is open in $G$ for any $h$ in $H$. This probably follows from the definition of a topological group. That is, by definition, the multiplication by $h$ is a continuous map from $G$ to itself. Thus, you have a continuous map from $U$ to $U_h$. Is this map open?

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Ow, I see Alex B. posted a similar answer. –  Oen Aug 19 '11 at 8:44
    
Thanks anyway, Oen. Like my comment to Alex B's post, I can't quite prove the openness of the $U \to U_h$ map. –  brian Aug 19 '11 at 9:06
    
In your comment on Alex B.'s answer you suggest a solution to your own problem. The continuous map given by multiplication by h has an inverse which is continuous. That should do it, right? –  Oen Aug 19 '11 at 9:26
    
Thanks, Oen, I wasn't sure that I was right... So I am? –  brian Aug 19 '11 at 9:29
    
Yes, you're right. The multiplication by $h$ is an open map because it is a homeomorphism. –  Oen Aug 19 '11 at 9:32
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