# discrete subgroup of complex Lie group is normal automatically?

This is in relation to Kodaira's Complex Manifolds and Deformation Complex Structures Chpt 2, Sec 2.

$$W$$ is a complex Lie group. A discrete subgroup $$G\leq W$$ gives properly discontinuous and fixed point free action on $$W$$. Thus $$W/G$$ makes sense as a complex manifold. Now the book says $$W/G$$ is a complex Lie group as well without mentioning $$G$$ normal.

$$\textbf{Q:}$$ Why does $$W/G$$ inherits a group structure? Note that I need $$W\times W\xrightarrow{\cdot} W$$ descends to the quotient level map. From standard group theory, $$W/G$$ is group iff $$G$$ is normal by considering $$wG\cdot 1G=wG$$. The natural procedure is to assume $$G$$ is normal which forces $$G\leq Z(W)$$ if $$W$$ is connected where $$Z(W)$$ is the center. However, the book did not mention $$G$$ being normal. Where does normality coming from then or have I missed something here?

• In a connected group every discrete normal subgroup is central. Hence discrete subgroups are non-normal as soon as they are not contained in the center. So there are plenty of examples of non-normal discrete subgroups. – YCor Jun 2 at 20:07
• I checked Kodaira's book, the statement occurs on page 48 and is indeed plain false. Moreover, if $G$ is non-abelian discrete and $W$ is simply-connected then $W/G$ cannot even be homeomorphic to a Lie groups, as Lie groups have abelian fundamental groups. As Yves said, there are many examples of nonabelian discrete subgroups of complex Lie groups. For instance, take the permutation group $S_n$ as a subgroup of $GL(n, {\mathbb C})$. – Moishe Kohan Jun 2 at 22:39
• @MoisheKohan That is unfortunate. Thank. – user45765 Jun 2 at 23:10

I can't comment on the book, since I don't have access to it now. But I can tell you that the statement is false. Just take, for instance$$\left\{\begin{bmatrix}1&0\\0&1\end{bmatrix},\begin{bmatrix}1&0\\0&-1\end{bmatrix}\right\},$$which is a non-normal subgroup of $$GL(2,\mathbb C)$$.

• So you mean without normality, it is not group. I just checked your example. Denote off diagonal matrix as $A=a_{ij}$ with $a_{ii}=0$ and $a_{12}=1,a_{21}=1$ for $A$. Similarly for $B=b_{ij}$ with $b_{ii}=0,b_{12}=-1,b_{21}=1$. Set $G$ the group to be quotient out in your example. Then clearly $AG=BG$. So $A^2G=B^2G$ if there is group structure. One compute to get $A^2G=IG, B^2G=-IG$ which is impossible. – user45765 Jun 2 at 18:51
• I don't really understand what you were computing. But I know this: if $\left[\begin{smallmatrix}a&b\\c&d\end{smallmatrix}\right]\in GL(2,\mathbb C)$, then$$\begin{bmatrix}a&b\\c&d\end{bmatrix}.\begin{bmatrix}1&0\\0&-1\end{bmatrix}.\begin{bmatrix}a&b\\c&d\end{bmatrix}^{-1}=\begin{bmatrix}\frac{b c+a d}{a d-b c} & \frac{2 a b}{b c-a d} \\ \frac{2 c d}{a d-b c} & \frac{b c+a d}{b c-a d}\end{bmatrix},$$which will seldom be equal to $\left[\begin{smallmatrix}1&0\\0&\pm1\end{smallmatrix}\right]$. Therefore, my subgroup is not normal. – José Carlos Santos Jun 2 at 19:09
• I was trying to check whether there is group structure endowed on $GL(2,C)/G$. And I checked the specific elements for multiplication property on the coset to see the quotient not being group which deduces non-normality. – user45765 Jun 2 at 19:16