# discrete normal subgroup of a connected group

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?

• 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. – Zev Chonoles Jan 26 '13 at 6:59
• an example will be appreciated – Marso Jan 26 '13 at 7:00
• An example: circle group and $n$-th root of unity. – user27126 Jan 26 '13 at 7:07
• Hint: Fix $h \in H$. What must the set $ghg^{-1}$ be, using connectedness? – user27126 Jan 26 '13 at 7:08

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 $H$.
• 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. – Watson Aug 3 '17 at 19:37
• H should be a normal subgroup so that we have $g(t) h g(t)^{-1} \in H \ (\forall \ t)$. – gpr1 Dec 2 '18 at 16:46