Warner Ex. 3.12; Prove Fundamental Group of a Lie Group is Abelian Using "Discrete normal subgroups are central"

Question

Specifically, Warner says to prove that if a Lie hom $\phi:G\to H$ has a discrete kernel then the kernel is actually contained in the center. (I can do this part.) The exercise then says to use this fact to prove that the fundamental group of a Lie group is abelian. (I'm familiar with other proofs of this, e.g. Hilton-Eckmann.) I suppose the idea here is to exhibit the fundamental group as the kernel of some hom, but I just don't have any ideas on how to do that.

Answer 1

Let $\rho:\widetilde{G}\rightarrow G$ be the universal covering homomorphism. You already know the kernel $K$ is a discrete central subgroup of $\widetilde{G}$. We have the following

Lemma: For every $k\in K$, the map $\phi_k:\widetilde{G}\rightarrow\widetilde{G}$ given by $g\mapsto gk$ is a deck transformation. The map $K\rightarrow\operatorname{Aut}(\rho)$ is a group isomorphism.

Your question is solved by this, because $\operatorname{Aut}(\rho)$ is isomorphic to $\pi_1(G)$.

Proof of Lemma: $\phi_k$ is clearly a homeomorphism, and it is a deck transformation because $\rho(gk)=\rho(g)$. The map $k\mapsto\phi_k$ is injective, because $\phi_k(1)=\phi_h(1)$ implies $k=h$. It is surjective because if $\phi$ is any deck transformation, then $\phi(1)=k\in K$ since $\rho\phi(1)=1$. But then $\phi\equiv\phi_k$, because both are deck transformations, agreeing at a point.

Finally, the map is a homomorphism because \begin{align*} \phi_{kh}(g) &= gkh\\ &= ghk\\ &= \phi_k\phi_h(g) \end{align*} where the second equality follows because $K$ is central.

Answer 2

I assume that $f:G\rightarrow H$ is a morphism of Lie groups and $G, H$ are connected. Suppose that $Kerf$ is discrete and let $g\in Kerf$. Consider $f_g(x) =xgx^{-1}$ it is continous this implies that the image of $f_g$ is connected since $G$ is connected. Since $Kerf$ is discrete and the image of $f_g$ is contained in $Kerf$ since $Kerf$ is normal, we deduce that the image of $Kerf$ contains only one element $1g1^{-1}=g$. This implies that $Kerf$ is central.

If $H$ is the universal cover of $G$, and $f$ the covering map, the kernel of $f$ is discrete since it is a local homeomorphism. The first part implies that $Kerf=\pi_1(H)$ is abelian.