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 $\pi_1(G/H,e) = H?$.

Related: https://math.stackexchange.com/questions/287155/ – Watson Aug 03 '17 at 19:25 — Watson, Aug 03 '17 at 19:25

score 1 · Accepted Answer · answered Jul 18 '16 at 17:19

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 $\pi_1(G/H)=H$.

answered Jul 18 '16 at 17:19

Tsemo Aristide

87,475

thanks! Do you know any simple proof? I mean, that does not use Deck transformations and universal covering? – L.F. Cavenaghi Jul 18 '16 at 17:23

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

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