# Fundamental group of prime closed 3-manifold and it's cover

Let $M$ be prime closed compact orientable 3-manifold with infinite fundamental group $G=\pi_1(M)$. Can there be finite order elements in $G$ ? What are the possibilities for $H_1(M)$ - can it be any abelian finitely generated group ?

If we have finite cover $N\to M$. In this case there is monomorphism $\pi_1(N)\to\pi_1(M)$. Is there also monomorphism $H_1(N)\to H_1(M)$ ?

From the "virtually Haken conjecture" I conclude that there exists finite cover $N$ which is Haken. It means it contains incompressible surface. It means $H_2(N)\neq0$.

If $M$ is closed compact orientable 3-manifold with finite fundamental group then $M$ is spherical. It means it is quotient of $S_3$ by finite subgroup of $SO_4$. This subgroup is fundamental group of $M$.

It cannot be $S^1\times S^2$, so in fact it is irreducible. So $\pi_1$ infinite implies its universal cover is contractible and moreover is $\mathbb R^3$ [Geometrization Theorem]. To see it's universal cover is contractible, observe that irreducibility implies $\pi_2$ is zero. So infact it's universal cover has $\pi_2$ zero. Now $\pi_1$ infinite implies it's universal cover is non-compact, so $H_3=0$ and thus by Hurewich theorem $\pi_3=0$. And with similar arguments it's all homotopy groups are zero. So it is contractible. And Thurston-Parelman's geometrization says that it is infact diffeomorphic with $\mathbb R^3$
Now if it has a finite order element in $\pi_1$ then infact we can have a finite free group action on $\mathbb R^3$ (WHY?). But it is not possible. So contradiction.
• Thank you. How do you deduce that universal cover is contractible ? Then $H_1$ is just sum of few copies of $Z$... or maybe not . I am not familiar with infinite group properties. Jun 30, 2018 at 13:44