Homology of 5-manifold

Question

Let $M$ be a closed connected 5-manifold such that $\pi_1(M)=\mathbb{Z}_7$ and $H_2(M)=\mathbb{Z}^2$. I would like to compute its homology and cohomology.

Standard computations involving Poincaré duality and the universal coefficient theorem for cohomology settle the orientable case. In summary:

1. $H_0(M)=\mathbb{Z}$ by connectedness
2. $H_1(M)=Ab(\pi_1(M))=\mathbb{Z}_7$
3. $H_5(M)=\mathbb{Z}$ by orientability
4. $H_3(M)=\mathbb{Z}^2\oplus \mathbb{Z}_7$
5. $H_4(M)=0$ (here one can either use the universal coefficients theorem twice, or use it once and remember that if $M$ is an orientable n-manifold, then $H_{n-1}(M)$ has no torsion.

I am having come trouble figuring out the non-orientable case. From start we know

1. $H_0(M)=\mathbb{Z}$ by connectedness
2. $H_1(M)=Ab(\pi_1(M))=\mathbb{Z}_7$
3. $H_5(M)=0$ since $M$ is not orientable.
4. $H_4(M)$ has $\mathbb{Z}_2$-torsion, since $M$ is non-orientable, so $H_4(M)=\mathbb{Z}^{k_4}\oplus \mathbb{Z}_2$.
5. The Euler characteristic is zero since $M$ has odd dimension, so in particular $k_4=k_3-1$, and further $k_3>0$.

So I still need to figure out $k_3$ and the torsion part of $H_3(M)$. What ingredient am I missing?

Answer 1

$M$ is orientable because its oriented double cover should have fundamental group which is a subgroup of $\pi_1(M)$ with index $2$ but $\pi_1(M)\cong\mathbb{Z}_7$ which has no proper non-trivial subgroups.