I have some questions.


I want to know informations of a (first) homology group of covering spaces from the homology groups of the base space.
If the first homology group of the base space is free, does that of the p-covering space have no p-torsion?
And more generally, is that of the finite abelian cover free?
If not, please let me know some counterexamples. 


And if there are some results or calculation methods about the (finite abelian or p-)covering's homology group, please introduce me.



A homology of p-covering space
This is certainly false! One should look at the corresponding group theory problem: any group $G$ is the fundamental group of a manifold $M$, and the first homology of $M$ is $\pi\_1(M)^{ab} = G^{ab}$. Thus the problem becomes: given a group $G$ and a finite index subgroup $H$ of index $p$, does $G^{ab}$ torsion free imply that $H^{ab}$ is torsion free ( **edit**  $p$-torsion free)? (It is unclear whether you are insisting that the $p$-covers be Galois, which would correspond to insisting that $H$ is normal, but both variations have negative answers.) 


For an explicit example, let $G = \langle x,y \ | \ [x,y]^2 \rangle$. Then
$G^{ab} = \mathbf{Z}^2$, but the homomorphism $G \rightarrow \mathbf{Z}/2$ sending
$x$ and $y$ to $1$ has kernel $H = \langle a,b,c \ | \ (cb^{-1}a^{-1})^2, (c^{-1}ba)^2 \rangle$, where $a = yx^{-1}$, $b = x^2$, and $c = xy$. In particular, we see that $H^{ab} = \mathbf{Z}^2 \oplus \mathbf{Z}/2$.


 **A previous version discussed knot complements and the Alexander polynomial, but I had missed read "torsion free" for "$p$-torsion free", and so the example does not apply.** 


One positive remark in the direction of your question: If you are assuming that $H$ is normal, then $H^{ab}$ is $p$-torsion free if the rank of $G^{ab}$ is less than two. This is because the latter condition implies that the $p$-completion of $G$ is cyclic, a condition which is inherited by a normal subgroup of index $p$.


The methods used to compute homology vary considerably depending on what information you have. If you have a presentation for the group, you can use the Reidemeister--Schreier algorithm to compute a presentation of $H$, from which it is easy to compute $H^{ab}$. The more you understand the geometry of the situation, however, the better.


