compact group transitive implies fundamental group has finite commutator subgroup Suppose that M is a manifold admitting a smooth transitive action by a compact Lie group. Prove that the fundamental group of M fits into the SES
$$
1 \to \Gamma \to \pi_1(M) \to A \to 1
$$
Where $ \Gamma $ is finite and $ A $ is abelian.
This is equivalent to the commutator subgroup of $ \pi_1(M) $ (the kernel of the first Hurewicz map) being finite.
For example this criterion shows that the klein bottle does not admit a transitive action by any compact group.
 A: I think this is very similar to what tkf suggested! See theorem 2.2 of
https://www.uni-muenster.de/imperia/md/content/theoretische_mathematik/diffgeo/mr1783960.pdf
Suppose the compact group G acts transitively on M. Then the universal cover
$$
\widetilde{G}\cong K \times \mathbb{R}^n
$$
also acts transitively (here $ K $ is compact and simply connected). So for some closed stabilizer group $ H $ we have
$$
M \cong \widetilde{G}/H
$$
Consider the projection p of H onto the second factor $ \mathbb{R}^n $. Then we have a SES
$$
1 \to ker(p) \to H \to  p(H) \to 1
$$
Observe that p(H) is abelian since it is a subgroup of $ \mathbb{R}^n $ and $ ker(p) $ is compact because it is a subgroup of $ K \times 0 $. Take $ \pi_0 $ of this SES then we have another SES of discrete groups
$$
1\to \pi_0(ker(p)) \to \pi_0(H) \to \pi_0(p(H)) \to 1
$$
The first component group is compact and discrete so it must be finite. The third component group is a discrete subgroup $ \mathbb{R}^n $ so just $ \mathbb{Z}^k $. This proves that $\pi_0(H) $ is finite by abelian. Since $ \widetilde{G} $ is simply connected then $ \pi_1(M)=\pi_0(H) $ is finite by abelian.
Related fun facts from this same reference:
If G compact acts transitively then M admits a nonnegative curvature metric with respect to which G acts by isometries.
