Let $G$ a simple connected topological group and $H$ a normal discrete subgroup, then $\pi_1(G/H,e) = H.$

Question

I know that $G$ is a covering space for $G/H$ and there is a injection between the fundamental group of $G$ and $G/H.$ How to proceed to show that $\pi_1(G/H,e) = H?$.

Answer 1

$G$ is simply connected and $H$ is discrete implies the action of $H$ on $G$ induced by right translations is proper and free. This implies that $G$ is the universal cover of $G/H$ and $H$ is the group of Deck transformations of $G/H$ and $\pi_1(G/H)=H$.