Group homology

Remarks

The groups do not depend on the choice of projective resolution, up to canonical isomorphism. In degree $0$, $H_0(G, M) = M_G$ is the module of coinvariants, the quotient of $M$ by the subgroup generated by all $gm - m$. In degree $1$, $H_1(G, \Z) \cong G / [G, G]$.

Example: $G = \Z$ written multiplicatively with generator $t$, so that $\Z[G] = \Z[t, t^{-1}]$. The sequence $0 \to \Z[G] \xrightarrow{\ t - 1\ } \Z[G] \to \Z \to 0$ is a free resolution of $\Z$. Tensoring with the trivial module $\Z$ gives the complex $0 \to \Z \xrightarrow{0} \Z \to 0$. So $H_0(\Z, \Z) = \Z$, $H_1(\Z, \Z) = \Z$, and $H_n(\Z, \Z) = 0$ for $n \ge 2$.

Non-example: the groups $\mathrm{Tor}^{\Z}_n(\Z, M)$, taken over $\Z$ instead of $\Z[G]$, forget the action. They vanish for $n \ge 1$, while $H_1(\Z, \Z) = \Z$.