Let $G$ be a group (not necessarily abelian) acting on an abelian group $M$. Consider the group ring $\Bbb Z[G]$ (which is in fact an algebra). The action of $G$ on $M$ makes $M$ as a $\Bbb Z[G]$-module in an obvious way, namely, $$ (\sum _i n_i g_i) a:=\sum_i n_i (g_i a).$$
Suppose $M$ is a free $\Bbb Z[G]$-module. Viewing $M$ as an abelian group, let $N$ be the subgroup generated by all elements of the form $ga-a$, for $g \in G$ and $a\in M$, and consider the quotient group $M/N$. Is there something that we can say about the resulting group $M/N$? It seems like it would be a free abelian group, but I'm not sure.
Thanks in advance.
Let us first consider the case that $M = \mathbb{Z}[G]$ is free of rank $1$. The map $$ \varepsilon : \mathbb{Z}[G] \to \mathbb{Z}, g \mapsto 1 $$ is a $\mathbb{Z}[G]$-module homomorphism if we give $\mathbb{Z}$ the trivial $\mathbb{Z}[G]$-module structure and its kernel is $N = \left< g a - a \:|\: g \in G, a \in M \right>$. In fact, $N \subseteq \ker(\varepsilon)$ is clear and conversely, if we have $a = \sum_{g \in G} a_g g \in \ker(\varphi)$, then $\sum_{g \in G} a_g = 0$ and so $$ a = \sum_{g \in G} a_g g - \sum_{g \in G} a_g e = \sum_{g \in G} a_g (g e - e) \in N $$ where $e \in G$ denotes the identity element. It follows that $\varepsilon$ induces an isomorphism $M/N \cong \mathbb{Z}$ and thus the quotient is a free abelian group of rank $1$.
In general, with $M$ being a free $\mathbb{Z}[G]$-module, $M$ is isomorphic to a direct sum of copies of $\mathbb{Z}[G]$, say $M \cong \mathbb{Z}[G]^{\oplus T}$ and under such an isomorphism $N$ will correspond to a direct sum of copies of $\ker(\varepsilon)$, i.e. $N \cong \ker(\varepsilon)^{\oplus T}$ and this will yield $M/N \cong \mathbb{Z}^{\oplus T}$, a free abelian group.