TITLE: First group homology with general coefficients
QUESTION [13 upvotes]: When $G$ acts trivially on $M$, the first homology group is just the abelianisation of $G$ tensored with $M$, i.e. $H_1(G;M)=(G/[G,G])\otimes_\mathbb Z M$.
Is there any similar statement when $G$ acts not trivially on $M$? Where does the abelianisation come from?

REPLY [7 votes]: Using the short exact sequence $0\to I\to \mathbb Z G \to \mathbb Z \to 0$ and the associated long exact sequence for ${\rm Tor}_*^{\mathbb ZG}(-,M),$ we get the following exact sequence
$$0 \longrightarrow H_1(G,M) \longrightarrow I \otimes_{\mathbb ZG} M \longrightarrow M \longrightarrow H_0(G,M) \longrightarrow 0.$$
In particular, we obtain 
$$H_1(G,M)={\rm Ker}(I \otimes_{\mathbb ZG} M \longrightarrow M).$$ 
If $G$ acts trivially on $M,$ the map $I \otimes_{\mathbb ZG} M \longrightarrow M$ is trivial and we get $H_1(G,M)=I\otimes_{\mathbb ZG}M=I/I^2 \otimes M=G_{\rm ab}\otimes M.$ Hence, it is a generalisation of your formula.