Why is there no compact manifold without boundary with the following homology groups?

Solution 1:

Denote our space by $M$.

Note that the fundamental group cannot have any subgroup of index two since this would constitute a nontrivial map $\pi_1M \rightarrow \mathbb{Z}/2$ which must factor through the abelianization of $\pi_1M$, i.e. $\mathbb{Z}/3$. But every map $\mathbb{Z}/3 \rightarrow \mathbb{Z}/2$ is trivial.

Whence $M$ has no nontrivial double covers. If $M$ were a manifold, this would imply that the orientation double cover was trivial, whence $M$ would be orientable- but this is clearly false because the homology groups do not satisfy Poincaré duality for any possible dimension for the manifold.