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Let $G$ be a topological group, $U$ is a neighborhood of $e$ which is the unit element of $G$.

My question is does there exist a neighborhood $H \subseteq U$ of $e$ s.t.

  1. $H$ is a subgroup of $G$?
  2. $H$ is a normal subgroup of $G$?

If 1 is true, then it seems can be concluded that there is a symmetric neighborhood $H$ of $e$ such that $H \cdot H \subseteq U$.


Unfortunately I have found both 1 and 2 are falsifiable, e.g. $U=(e^{-i\frac{\pi}{6}},e^{i\frac{\pi}{6}})$ in $S^1$

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up vote 5 down vote accepted

Yes; let $H = e$.

To see that the answer is no to all of your questions in general if $H$ is required to be nontrivial, let $G = S^1$. I don't understand what you mean by "topological subgroup." Every subgroup automatically inherits the subspace topology and is a topological group with respect to this topology.

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Thanks, however, I feel sorry that I have lost a condition that $H$ need to be a neighborhood of $e$ too. So would you please reconsider it? – Popopo Nov 4 '12 at 9:40
@Popopo: the counterexample holds. The statement you want to prove is much weaker than the statement that you can find a neighborhood that is a subgroup (which is false). – Qiaochu Yuan Nov 4 '12 at 17:43
(Non sequitur: Congratulations!) – Brian M. Scott Nov 6 '12 at 0:28
@Brian: thanks! – Qiaochu Yuan Nov 6 '12 at 1:23

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