It is known that a discrete normal subgroup $N$ of the connected group $G$ is contained in the center of $G$. But we also know that if a group is discrete then its Lie algebra is $\{0\}$, and we know that if $G$ is connected then $\exp(\mathfrak{g})$ generates $G$, where $\mathfrak{g}$ is its Lie algebra.
So consider the Lie algebra of the subgroup $N$: this has to be $\{0\}$, and so $N$ has to be $\{1\}$, which means $N$ must be in the center. However this seems false somehow. Where exactly am I going wrong?
From the comments above.
A discrete topological space is connected if and only if it is a singleton. In particular, the hypothesis that $G$ is connected does not imply that the discrete normal subgroup $N$ is also connected. Hence, it is not true that $N = \langle \exp(X) : X \in \mathfrak{n} \rangle$.
Thus, $\mathfrak{n} = \{0\}$ does not imply that $N = \{e\}$.