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!
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.