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: 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.
A: 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 $H$.
A: Hint,
See: Lecture V - Topological Groups, Theorem 5.5.
