Connected topological group $G$ with discrete normal subgroup $H$. Prove that $\pi_1(G/H)$ is commutative.

Question

I'm trying to prove the second part of exercise 4.6 of Greenberg and Harper. So far, I have proved that $H$ is central.

I thought of using the fact that in this situation $G/H$ is locally isomorphic to $G$ but I am stuck on how to proceed so some hints or advice would be greatly appreciated.

You might want to read this: http://math.stackexchange.com/questions/686496/the-fundamental-group-of-a-topological-group-is-abelian — freakish, Mar 22 '17 at 21:54

score 1 · Accepted Answer · answered Mar 22 '17 at 20:20

1

Note that the fundamental group of a a topological group is abelian (presumably in G&H, if not, easy to prove by explicit homotopy).

Now, prove that $G/H$ is a topological group.

answered Mar 22 '17 at 20:20

Igor Rivin

26,372

Connected topological group $G$ with discrete normal subgroup $H$. Prove that $\pi_1(G/H)$ is commutative.

1 Answers1