# Discrete normal subgroup of a connected linear lie group

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?

• This is not because the Lie algebra of $N$ is 0 that $N=1$...
– user598294
Oct 7, 2018 at 20:38
• @AlexL I realize that, I'm wondering where I'm going wrong... Oct 7, 2018 at 20:58
• The Lie algebra of a discrete subgroup $N$ is $\{0\}$. A connected Lie group is generated by the exponentials of its Lie algebra. The only explanation is that a discrete Lie group is not necessarily connected. Hence not necessarily generated by those exponentials. Wait! A discrete topological space is connected if and only if it is a singleton, Oct 8, 2018 at 4:00
• @JyrkiLahtonen I see...N is not connected, even though G is. This is the flaw Oct 8, 2018 at 4:43

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 $$N = \{e\}$$.