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?

  • $\begingroup$ This is not because the Lie algebra of $N$ is 0 that $N=1$... $\endgroup$
    – user598294
    Oct 7, 2018 at 20:38
  • $\begingroup$ @AlexL I realize that, I'm wondering where I'm going wrong... $\endgroup$ Oct 7, 2018 at 20:58
  • $\begingroup$ 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, $\endgroup$ Oct 8, 2018 at 4:00
  • $\begingroup$ @JyrkiLahtonen I see...N is not connected, even though G is. This is the flaw $\endgroup$ Oct 8, 2018 at 4:43

1 Answer 1


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 $N = \{e\}$.


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