Problem Suppose that $G$ is a finite group whose abelianization is trivial. Suppose also that $G$ acts freely on $S^3$. Compute the homology groups (with integer coeffcients) of the orbit space $M=S^3/G$.

This is an algebraic-topology problem of Harvard qualifying exam. There is a solution in the website:

Solution: Note that $M$ is a smooth manifold, and that $\pi_1 M = G$. By Poincare's [sic] theorem $H_1 S^3/G = 0$, as is $H^1(S^3/G; A) = \hom(\pi_1 M, A)$ for any abelian group $A$. This implies that $M$ is orientable. It then follows from Poincare [sic] duality that $H_2(M;A) = 0$ for any abelian group $A$ and that $H_3(M;A) = A$.

I think, before using Poincaré duality, we should first check it's orientable.
For example, $M$ is orientable if and only if the action $G$ on $S^3$ preserve the orientation; so, I guess this solution manual is not correct. Also, perhaps I made some stupid mistake.

Please help! Thank you!

  • 2
    $\begingroup$ The first result of Poincare's mentioned is not Poincare duality; it's the result that $H_1$ is the abelianization of $\pi_1$. (I don't think I've seen that result attributed to Poincare, or at least referred to as 'Poincare's theorem', though. It's part of, for example, Hurewicz's theorem.) $\endgroup$
    – anomaly
    Jul 6, 2015 at 13:35
  • $\begingroup$ Sorry,I made a typo-mistake here. $\endgroup$
    – Hang
    Jul 7, 2015 at 4:53

1 Answer 1


If $S^3/G$ was not orientable, then the orientation double cover would be non-trivial. From the correspondence between covers of $S^3/G$ and subgroups of $\pi_1(S^3/G)\cong G$, the cover would correspond to an index two subgroup of $G$ (call this subgroup $H$). But now $G/H \cong \Bbb Z/2\Bbb Z$ is a non-trivial abelian quotient of $G$, so $G$ cannot have trivial abelianization.

Alternatively, one can use the universal coefficient theorem to show $H^1(S^3/G,\Bbb Z/2\Bbb Z)=0$, so every vector bundle over $S^3/G$ has vanishing 1st Stiefel-Whitney class. Equivalently, every vector bundle (including the tangent bundle) over $S^3/G$ is orientable.

  • 1
    $\begingroup$ Here's another way of saying this: by definition if $S^3/G$ is non-orientable, $\pi_1$ acts nontrivially on the local homology $H_3(M, M-x_0) \cong \mathbb Z$. So we have a nontrivial homomorphism $G = \pi_1(S^3/G) \to \mathbb Z/2 = \text{Aut}(\mathbb Z)$. $\endgroup$
    – user98602
    Jul 6, 2015 at 18:17

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