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?
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 put the proof in the url in M.Sina’s answer for convenient.
Theorem. Let $G$ be a connected topological group. Let $H$ be a discrete normal subgroup of $G$. Then $H$ is contained in the center of $G$.
Proof. Let $h\in H$. Put $$ K=\{g\in G\mid hg=gh\}, $$ which is a subgroup of $G$. It suffices to show that $K=G$.
Since $H$ is discrete, we can choose an open neighborhood $U$ of the unit $e$ such that $$ hUh^{-1}U^{-1}\cap H=\{e\}. $$
In fact, by the discreteness, choose an open neighborhood $V$ of $e$ such that $V\cap H=\{e\}$. Consider the continuous map $$ f\colon G\times G\to G,\quad(g,g’)\mapsto hgh^{-1}(g’)^{-1}. $$ Then we can choose open neighborhoods $U_1$, $U_2$ of $e$ such that $$ U_1\times U_2\subseteq f^{-1}(V), \quad\text{thus}\quad hU_1h^{-1}U_2^{-1}\cap H=\{e\}. $$ Now we get a neighborhood $U=U_1\cap U_2$ as desired.
Now it suffices to show that $U\subseteq K$, since $U$ generates the group $G$ (see Lemma below). Let $u\in U$. Then since $H$ is a normal subgroup, $$ huh^{-1} u^{-1} \in hUh^{-1} U^{-1}\cap H=\{e\},\quad\text{i.e.,}\quad hu=uh. $$ This shows that $U\subseteq K$ and hence the theorem.
Lemma. Let $G$ be a connected topological group. Let $U$ be an open neighborhood of $e$. Then $U$ generates $G$, i.e., $\langle U\rangle=G$.
Proof of Lemma. Since $G$ is connected, it suffices to show that $\langle U\rangle$ is open and closed.
Since the set $U\cup U^{-1}$ is open and $$ \langle U\rangle =\bigcup_{n\ge0} (U\cup U^{-1})^n, $$ we see that $\langle U\rangle$ is open.
Moreover $\langle U\rangle$ is closed since $$ G\setminus\langle U\rangle= \bigcup_{x\in G\setminus\langle U\rangle}xU $$ is open.
