Let $A$ be a subgroup of a group $G$. Then since $A$ is a subgroup of the fundamental group $\pi_1(K(G,1))=G$, there is a covering space $p\colon Y\to K(G,1)$ with $p_*(\pi_1(Y))=A$. So the homology of $Y$ should be completely determined by $A$ and $G$. Suppose that $A$ and $G$ is known, how can one compute $H_*(Y)$, the homology groups of $Y$?
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Short answer: it is not so easy (especially for infinite index $A$) Long answer: Read Ken Brown's "Cohomology of groups". 


This is not a trivial problem. A favorite example of mine is the case of a knot complement, $S^3\setminus K$. (It is known that these are EilenbergMaclane spaces.) If you pick $A$ to be the commutator subgroup of $\pi_1(S^3\setminus K)=G$, then $Y$ is called the universal Abelian cover, and $H_1(Y;\mathbb Q)$ turns out to be a torsion $\mathbb Q[t,t^{1}]$module, called the Alexander module of the knot. The order of the Alexander module is called the Alexander polynomial. One can calculate the Alexander module from the fundamental group using Fox calculus. 

