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This question is very simple, but I don't get the right idea.

Assume $H$ be a locally closed dense subgroup of a topological group $G$. Prove that $H=G$.

I need to prove that $gH\cap H\ne\emptyset$ for all $g\in G$. I know that $H$ is open in $\overline{H}$, but I don't know of it is important.

LBJFS
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    Definitions are your friends. What does it mean to be "locally closed'? – hardmath Mar 19 '19 at 14:23
  • Means that (i) $H=U\cap F$ for some open $U$ and closed $F$; (ii) $H$ open in $\overline{H}$; (iii) for every $y\in H$ there is nhbd $N$ in $G$ s.t. $N\cap H$ is closed in $N$. – LBJFS Mar 19 '19 at 14:32
  • Wait! I apologize, but I want to prove that $H$ equals $G$. Now I edit – LBJFS Mar 19 '19 at 14:34

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Actually you don't need the two assumptions on $H$, namely that $H$ is locally closed AND that it is open, because each one of these assumptions, separately, imply that $H$ is closed. Proofs of both these assertions can be found here. Now since $H$ is also dense in $G$, its closure coincides with $G$, hence $H=G$.

uniquesolution
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