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We say that a group $G$ is of type $FL$ if there exists a resolution $L_\bullet \to \mathbb{Z}$ of finite length of finitely generated, free $\mathbb{Z}G$-modules. Now, un unproven proposition in Brown's book "Cohomology of groups" says that if there exists a finite complex $X$ which is a $K(G,1)$, then $G$ is of type $FL$. I thought immediately to the cellular homology chain $$\ldots\to H_{n+1}(X_{n+1},X_n)\to H_n(X_n,X_{n-1})\to\ldots$$ where $X_n$ is the $n$-skeleton of $X$, but I am not sure about it and I don't know how to equip these free groups with a $G$-module structure. I suppose that I should use that $G\cong\pi_1(X,x_0)$ but I would like to know, at least, if I'm on the right way or not.

Thank you in advance, bye

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The universal cover of a $K(G, 1)$ is a contractible space with a free action of $G$, so that's a more natural place to look for $G$-modules. – Qiaochu Yuan Jun 15 '11 at 11:00
up vote 1 down vote accepted

On one hand, $H_n(X_n,X_{n-1})$ is just the abelian group generated by $n$-cells of $X$. Since there is no natural action of $G$ on $BG=K(G,1)$ (let alone a free one), it has no natural structure of $G$-module.

On the other hand, it's universal cover, $EG$, has a free action of $G$ (and $H(EG_n,EG_{n-1})$ is finitely generated as $G$-module, since it's generated exactly by $H(BG_n,BG_{n-1})$). So one can take $L_n=H_n(EG_n,EG_{n-1})=H_n(BG_n,BG_{n-1};\mathbb Z[G])$ (where $\mathbb Z[G]$ is the local system on $K(G,1)$ given by natural action of $\pi_1=G$ on $\mathbb Z[G]$)

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