Homotopy groups of quotient groups.

Question

I'd like to ask how to compute homotopy groups of quotient groups, whose homotopy groups I already know. I found this answer, but I don't understand how to derive the homotopy group of $\pi_n (G/H)$ using the long exact sequence.

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In general, if I know $\pi_n(G)$ and $\pi_n(H)$, can I compute $\pi_n(G/H)$?

• What if $\pi_n(G)$ is trivial?
• Or $\pi_n(H)= \mathbb{Z}_N$?
• What about a combination of the two?
• Any special cases that are easy?
• What about $n=1$?

I know these are many questions so to clarify, in principal I'd like the most general answer but if one is not known, then any examples where a computation can be done is acceptable.

Answer 1

It turns out that the long exact sequence associated to the fibration $H \rightarrow G \rightarrow G/H$ is actually the same as the long exact sequence of relative homotopy groups associated to the pair $(G,H)$. On top of being somewhat surprising (since we usually want to compare the relative homotopy groups to the topological, not group quotient), this actually can be useful in certain situations.

For example, we want to show that $G/H$ is weakly equivalent to the classifying space of some functor, we can show that the relative homotopy groups of $(G,H)$ classify the functor applied to the spheres.

Answer 2

For the cases you're interested in $H$ is discrete which simplifies the situation a lot. In this case most of the long exact sequence in homotopy vanishes and we get that $G$ and $G/H$ have the same $\pi_n, n \ge 2$, so the only thing left to do is analyze $\pi_1$. Here the long exact sequence ends

$$1 \to \pi_1(G) \to \pi_1(G/H) \to H \to 1$$

so we get that $\pi_1(G/H)$ is an extension of $H$ by $\pi_1(G)$. Furthermore in all the examples you've named $\pi_1(G)$ vanishes so we just get that $\pi_1(G/H)$ is $H$.