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could any one give me hint for this one?

$G$ be a connected group, and let $H$ be a discrete normal subgroup of $G$, then we need to show $H$ is contained in the center of $G$

first of all, I have no clear idea what is meant by discrete subgroup and its any special properties?

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  • $\begingroup$ A discrete subgroup is just a subgroup such that the subspace topology it gets as a subset of $G$ is the same as the discrete topology. $\endgroup$ – Zev Chonoles Jan 26 '13 at 6:59
  • $\begingroup$ an example will be appreciated $\endgroup$ – Marso Jan 26 '13 at 7:00
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    $\begingroup$ An example: circle group and $n$-th root of unity. $\endgroup$ – user27126 Jan 26 '13 at 7:07
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    $\begingroup$ Hint: Fix $h \in H$. What must the set $ghg^{-1}$ be, using connectedness? $\endgroup$ – user27126 Jan 26 '13 at 7:08
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Hint, See: Lecture V - Topological Groups, Theorem 5.5.

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    $\begingroup$ The link is broken but this one seems to work. $\endgroup$ – Watson Jul 3 '17 at 17:59
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Suppose $h \in H$, $g\in G $ and $ghg^{-1}\not=h$, then since $G$ is connected manifold, hence path connected, one can find a path $g(t)$ in $G$ going from $e$ to $g$. Notice $a(t):=g(t)hg(t)^{-1}$ realizes a path lying entirely in $H$ from $h$ to $ghg^{-1}$ which contracts the discreteness of $H$.

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  • $\begingroup$ I would agree if $G$ is assumed to be a Lie group, but it may happen that $G$ is a connected topological group which is not path-connected. $\endgroup$ – Watson Aug 3 '17 at 19:37
  • $\begingroup$ Why must it be a normal subgroup??? $\endgroup$ – swedishfished Oct 7 '18 at 19:55
  • $\begingroup$ H should be a normal subgroup so that we have $g(t) h g(t)^{-1} \in H \ (\forall \ t)$. $\endgroup$ – gpr1 Dec 2 '18 at 16:46

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