# The second homologygroup of a orientable three manifold

If a short question: I have a closed orientable 3-Manifold, which has a perfect fundamental group.

therfore $H_1(M)=0,H_0(M)=H_3(M)=\mathbb{Z}$ somehow it seems like i can conclude that $H_2(M)=0$, but i don't know how to conclude this. by Poincaire duality I could say that $b_2=0$ but why cant there be Torsion.

• You can also use the universal coefficient theorem. – Angina Seng May 7 '17 at 16:08

The torsion parts of $H^3(M) = \mathbb{Z}$ and $H_2(M)$ coincide by the universal coefficient theorem.
• More generally, if $M$ is a closed $n$-dimensional orientable manifold, $H_{n-1}(M; \mathbb{Z})$ is free. – Michael Albanese May 7 '17 at 17:16