I was reading Atiyah's and Wall's section on group cohomology (on Casslels-Frölich) and they went quite quickly on the homology section, and I wanted to understand if the following argument is correct.
We want to define the group homology of a group $G$ with values in a $\mathbb{Z}[G]$-module $M$ as the derived functor
\begin{equation} \operatorname{H}^i(G, M) = \operatorname{Tor}_i^{\mathbb{Z}[G]}(\mathbb{Z}, M). \end{equation}
This is a balanced functor which means we can calculate this from a projective resolution of $\mathbb{Z}$ (and not of $M$). For this consider the standard resolution
\begin{equation} \dots \to P_i \to P_{i-1} \to \dots \to P \to \mathbb{Z} \to 0 \end{equation}
where $P_i = \mathbb{Z}[G^{i+1}]$ and the differentials are defined via the alternating sum $$d(\sigma_0, \dots, \sigma_i) = \sum_{k \in [n]}(-1)^k(\sigma_0, \dots, \hat\sigma_k, \dotsm \sigma_i).$$ Finally, we obtain the homology by applying the right exact functor $-\otimes_GM$ and take the homology of the complex.
So far so good. Now, the $P_i$ are actually free $\mathbb{Z}[G]$-modules of rank $i+1$ right? So this means that \begin{equation} P_i \otimes_G M = M^{i+1}? \end{equation}
If this is correct then the differential operator now how do we describe the differntial operators? Of course it's not the same formula as the above mutatis mutandis (I believe this would imply that the homology is always zero). I'm having trouble opening up the isomorphisms and writing it down, but the image of $M^2$ in $M$ needs to be $I_GM$ for this to work (where $I_G$ is the augmentation ideal of $\mathbb{Z}[G]$, kernel of $\mathbb{Z}[G] \to \mathbb{Z}$ sending every group element $\sigma \mapsto 1$).