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Let $G$ be a discrete group and $M$ be an $RG$-module for some ring $R$ (I'm happy to assume that $R = \mathbb{Q}$). Define $R[M]$ to be the set of $R$-linear combinations of formal symbols of the form $[m]$ with $m \in M$ (this is sort of like a group-ring; there are no relations between the symbols, so for instance it is not true that $[m_1+m_2]$ and $[m_1] + [m_2]$ are related in any way). Clearly the group $G$ acts on $R[M]$.

Question : Is there any relationship between $H_{\ast}(G;M)$ and $H_{\ast}(G;R[M])$? More generally, what can one say about $H_{\ast}(G;R[M])$?

The specific example I'm interested in is as follows. Let $\mathbb{Q}^n$ be the usual representation of $\text{GL}(n,\mathbb{Z})$. It is easy to see that $H_{k}(\text{GL}(n,\mathbb{Z});\mathbb{Q}^n) = 0$ for $k \geq 1$, basically because the center acts as $-1$. Is there any way to calculate $H_{\ast}(\text{GL}(n,\mathbb{Z});\mathbb{Q}[\mathbb{Q}^n])$? Does it vanish?

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$R[M]$ is just a permutation module, so, by Shapiro's Lemma, $H_{\ast}(G;R[M])$ is the direct sum, over a set of representatives $m$ of orbits of $G$ acting on $M$, of the homology $H_{\ast}(G_m;R)$ with constant coefficients of the point stabilizer $G_m$. In particular ($m=0$) it contains $H_{\ast}(G;R)$ as a direct summand, and so in your special case, $H_{\ast}(\text{GL}(n,\mathbb{Z});\mathbb{Q}[\mathbb{Q}^n])$ won't vanish, as it contains $H_{\ast}(\text{GL}(n,\mathbb{Z});\mathbb{Q})$.

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