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If $G$ is a topological group and $H$ is a closed subgroup, is it the case the $H$ is either discrete or else $H=G$? I see this is true for $G=\mathbb{R}^d$ in Subgroup of $\mathbb{R}$ either dense or has a least positive element?

Does the same hold for general $G$? I'm willing to assume $G$ is locally compact, second-countable, Hausdorff (i.e. a Polish group).

nullUser
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    The question you linked to seems to be about dense subgroups? – bof Oct 28 '16 at 01:14
  • @bof Yes I meant to ask something slightly different, but you guys are too good at answering quickly! Stay tuned for my revised question... – nullUser Oct 28 '16 at 01:18
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    Very no. Closed subgroups needn't be discrete. Examples with Lie groups abound and are very easy to make. For instance, $\Bbb R^k\subset\Bbb R^n$ for $1<k<n$, or $$SO(n)\subset O(n)\subset SL(n, \Bbb R)\subset GL(n,\Bbb R),\textrm{ or}$$ $$SU(n)\subset U(n)\subset GL(n,\Bbb C),\textrm{ or}$$ $$\textrm{unitriangulars}\subset\textrm{upper triangular}\subset GL,$$ or $H\subset H\times K$ where $K$ is another positive dimension Lie group, etc. etc. – anon Oct 28 '16 at 01:25

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$R$, the real line, is a closed subgroup of $R\times R$, but is not discrete.

Tsemo Aristide
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Consider the multiplicative group $(\mathbb{C} \setminus \{0\}, \cdot )$, which is locally compact and Hausdorff. Then the torus $( \mathbb{T}, \cdot)$ is a non-discrete proper closed subgroup of $(\mathbb{C} \setminus \{0\}, \cdot )$.