Let $G$ be a connected Lie group with a simple Lie algebra $\mathfrak{g}$. Show that any proper normal Lie subgroup $H$ of $G$ is contained in the centre of $G$.
I know that if $H$ is discrete. Then the result holds by discrete normal subgroup of a connected group. One approach is to show that any proper normal Lie subgroup is discrete. But to show this it involves the use of manifolds, which I'm trying to avoid.
Another approach I have in mind is to use that $Z(G)=\ker(ad)$ from Centre of a connected Lie group.. And show that $H$ is in $ker(ad)$. Since $H$ is a normal Lie subgroup, it must be the kernel of some Lie homomorphism map. But I'm not sure how to show that $H\subset Ker(ad)$.
Any help will be appreciated!
Best Answer
There is a correspondence between normal subgroups of $G$ and ideals of $\mathfrak g$. Since $H$ is normal in $G$, then $\mathfrak h$ is an ideal of $\mathfrak g$. But $\mathfrak g$ is simple, so either $\mathfrak h = \{0\}$ or $\mathfrak h = \mathfrak g$. In the former case, $H$ must be discrete, and you know how to conclude. In the other case, $G$ connected implies $H = G$. But $H$ is a proper subgroup, so this case does not occur.