A homology of p-covering space 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: 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.
A: Here is an example of a $p$-sheeted covering space $X\to Y$ such that $H_1(X)$ has an element of order $p$ but $H_1(Y)$ is free. The space $X$ is the union of two pieces: a torus $A$ and surface $B$ obtained from a torus by deleting the interiors of $p$ disjoint disks that are equally spaced along a longitude of the torus. Both $A$ and $B$ have a free action of a cyclic group of order $p$ rotating the tori in the longitudinal direction. We form $X$ by attaching $B$ to $A$ by identifying its $p$ boundary circles with $p$ meridian circles of $A$ so that the $p$-fold rotational symmetries of $A$ and $B$ match up to give a free action of the cyclic group of order $p$ on $X$. Factoring out this action gives the covering space $X\to Y$, where $Y$ is a torus (the quotient of $A$) with a torus-minus-a-disk (the quotient of $B$) attached.
To compute $H_1(X)$ we can first deform $X$ to a homotopy equivalent space $Z$ by deforming the way $B$ attaches to $A$ so that all $p$ boundary circles of $B$ attach to the same meridian circle $C$ of $A$. The image $B'$ of $B$ in $Z$ is thus $B$ with all its boundary circles identified to the single circle $C$. It's easy to compute $H_1(B')$ from a cell structure on $B'$, and one finds that it is the direct sum of a cyclic group of order $p$ (generated by the circle $C$) and a free abelian group of rank $p+1$.  From the Mayer-Vietoris sequence for $Z = A \cup B'$ (or from a cell structure on $Z$) we then get that $H_1(Z)=H_1(X)$ is the direct sum of a cyclic group of order $p$ and a free abelian group of rank $p+2$.  For the quotient space $Y$ it's easy to check that $H_1$ is free abelian of rank $3$.
