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Let $G$ be a discrete group and let $M$ be a $G$-module. Assume that I have a resolution $$\cdots \rightarrow M_1 \rightarrow M_0 \rightarrow M \rightarrow 0$$ of $M$ by $G$-modules (with no further assumptions on the $M_k$; certainly they do not need to be free or acyclic or anything). What is the relationship between $H_{\ast}(G;M)$ and the abelian groups $H_{\ast}(G;M_k)$? It feels like there should be some kind of spectral sequence here, but I'm having trouble sorting out the relevant homological algebra.

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This is an example of the hyperhomology spectral sequences in the case of chain complexes of $G$-modules. See chapter 5.7 (especially proposition 5.7.6) of Weibel's book "An introduction to homological algebra".

More precisely, let $P_{\ast\ast}$ be a Cartan–Eilenberg resolution of the chain complex of $G$-modules $M_\ast$. Then each of the two spectral sequences (coming from the horizontal and vertical filtrations) associated with the double chain complex $P_{\ast\ast}/G$ will converge to the homology of the total complex of $P_{\ast\ast}/G$. The $E^2_{p,q}$-term of the first spectral sequence is clearly $H_p(H_q(G;M_\ast))$, i.e. the $p$-th homology of the chain complex $H_q(G;M_\ast)$. On the other hand, the $E^2_{p,q}$-term of the second spectral sequence is $H_p(G;H_q(M_\ast))$. Since $H_q(M_\ast)$ is $M$ if $q=0$ and is trivial otherwise, we conclude that the second spectral sequence degenerates at the $E^2$ page, and its limit is canonically identified with $H_\ast(G;M)$. Therefore, the first spectral sequence converges to $H_\ast(G;M)$: $$ H_p(H_q(G;M_\ast)) \Longrightarrow H_{p+q}(G;M) $$

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Great, thanks!! –  Richard Jun 19 '13 at 3:45
    
@Richard: I am glad to be of help. –  Ricardo Andrade Jun 19 '13 at 5:16

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