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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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Since $Y$ is a covering space of $K(G,1)$, the two spaces share the same higher homotopy groups, which are trivial in this case. Then, $Y = K(A,1)$. In that case, $H_i(Y)=H_i(K(A,1))$, which are the homology groups of the group $A$. – Aru Ray Apr 18 '12 at 21:42
@Aru Ray, so you mean that the result has nothing to do with $G$? – Zuriel Apr 18 '12 at 21:48
@Zuriel I'm not exactly an expert, but I think so. Presumably there is some relationship between $K(A,1)$ and $K(G,1)$ when $A\subseteq G$ - as Igor says in his answer, one can possibly find such a result in "Cohomology of groups" – Aru Ray Apr 18 '12 at 22:04
Many thanks @Aru Ray! I will find the book and look for related results. – Zuriel Apr 18 '12 at 22:16

Short answer: it is not so easy (especially for infinite index $A$) Long answer: Read Ken Brown's "Cohomology of groups".

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@Igor Rivin, for your long answer, may I know which part(s) of the book should I read to find answers to my question? – Zuriel Apr 19 '12 at 11:05
I would say chapter 7, but you would do well to read the whole book up to that point... – Igor Rivin Apr 19 '12 at 16:29

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 Eilenberg-Maclane 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.

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