Why does $G$ act trivially on group of coinvariants? Given a $\Bbb{Z}G$-module $M$, define $M_g$ to be the quotient of $M$ by the subgroup $\langle m - gm \rangle$, $m \in M$, $g \in G$.   My question is why does $G$ act trivially on $M_g$?

Taking the quotient by what is generated by all $m-gm$ means that in the quotient (the image of) $m-gm$ becomes zero, i.e. $g\bar m=\bar m$ holds for all $g$ and $\bar m$ in this quotient.
This is a standard trick and often the very reason to pass to a quotient. In a similar fashion, $\mathbb R[X]/(X^2+1)$ is a ring where "suddenly" the equation $x^2+1=0$ has a solution (namely one solution is the image of $X$, the other is $-X$) and thus $\mathbb C$ is obtained.
