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?
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 $$ hUh^{-1}U^{-1}\cap H=\{e\}. $$
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.