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.
Definition of topological group
2 Answers
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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0Thank you very much, Alex. – 2011-08-19
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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0Yes, you're right. The multiplication by $h$ is an open map because it is a homeomorphism. – 2011-08-19