Orientable manifold with finite first homology group. I've started going through old comps and have hit a road block, the following: Let $M$ be a closed, connected, orientable $n$-manifold, with $n \geq 3$. Show if $H_1 (M)$ is finite then $H_{n-1} (M) =0$. I have tried looking at Euler characteristics and the long exact sequence of relative homology but every route leads me to thinking the supposition of $H_1 (M)$ finite is not enough to go on, which in turn tells me there is something about the properties of the manifold that either I am not using or do not know to use. A proof of the statement would be appreciated and references that expound on manifolds for which the supposition holds would be even better.

If $H_{1}(M)$ is finite then by the universal coefficient theorem, $$H^1(M) \cong \hom_{\mathbb{Z}}(H_1(M), \mathbb{Z})$$ is zero (because $\mathbb{Z}$ has no nonzero elements of finite order).

And by Poincaré duality (the manifold is connected, closed, and orientable), there is an isomorphism $H_{n-1}(M) \cong H^1(M)$.