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Let $G_1$ and $G_2$ be groups and let $M$ be a vector space equipped with actions of $G_1$ and $G_2$. The free product $G_1 \ast G_2$ thus acts on $M$. How can one compute the twisted group homology $H_{\ast}(G_1 \ast G_2;M)$ in terms of $H_{\ast}(G_1;M)$ and $H_{\ast}(G_2;M)$? Since I doubt that $H_{\ast}(G_1;M)$ and $H_{\ast}(G_2;M)$ are enough to completely determine $H_{\ast}(G_1 \ast G_2;M)$, I guess my question more precisely is what additional information is needed?

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@Lewis: You have Meyer-Vietoris sequence, see Ken Brown's book "Cohomology of groups". Situation is the same as with topological spaces since $H_*(G,M)=H_*(K(G,1), {\mathbb M})$, where ${\mathbb M}$ is an appropriate sheaf on $K(G,1)$. – Misha Mar 27 '12 at 23:39

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