Let $M$ be a finitely generated $ZG$-module for a finite group G. I want to show that there exists a $ZG$-resolution $F$ of $M$ such that each $P_n$ is a finitely generated free $ZG$-module.
Since $M$ is finitely generated we find a finitely generated free module $A = \oplus_{i=1}^n ZG $ and a surjection $A\rightarrow M$ (just map the basis of $A$ onto the generators of $M$). Let $B$ be the kernel of this surjection. Is $B$ also finitely generated and free? If this is true then we are done.
Best Answer
Since $G$ is finite, $\mathbb{Z}G$ is noetherian, i.e. any submodule of a finitely generated $\mathbb{Z}G$ module is again finitely generated. That is because a $\mathbb{Z}G$ module $M$ is a finitely generated $\mathbb{Z}G$ module if and only if $M$ is finitely generated as a $\mathbb{Z}$ module. This is where we need $G$ to be finite. Now if $N \subset M$ is any $\mathbb{Z}G$ submodule it is in particular a $\mathbb{Z}$ submodule and hence a finitely generated $\mathbb{Z}$ module since so is $M$ and $\mathbb{Z}$ is a noetherian ring. But then $N$ is a finitely generated $\mathbb{Z}G$ module by our above observation.
With this in mind we might build a free $\mathbb{Z}G$ resolution of $M$ as usual, i.e. pick a finite generating set $S$ of $M$ and surject $\mathbb{Z}G\{S\} \xrightarrow{d_0}M$ (where $\mathbb{Z}G\{S\}$ is the free $\mathbb{Z}G$ module generated by $S$). Proceed inductively and suppose we are given a partial resolution
$$F_n \xrightarrow{d_n} F_{n-1} \xrightarrow{d_{n-1}} F_{n-2} \rightarrow \cdots \rightarrow F_0 \rightarrow M \rightarrow 0 $$
of $M$ by finitely generated free $\mathbb{Z}G$ modules $F_i$. By picking a finite generating set $S'$ for ker $d_n$ (which can be done since $F_n$ is finitely generated and $\mathbb{Z}G$ is noetherian) we obtain a partial resolution
$$ F_{n+1} \xrightarrow{d_{n+1}} F_n \xrightarrow{d_n} F_{n-1} \rightarrow \cdots \rightarrow F_0 \rightarrow M \rightarrow 0 $$
of $M$ by finitely generated free $\mathbb{Z}G$ modules where $d_{n+1} : F_{n+1} = \mathbb{Z}G\{S'\} \twoheadrightarrow \text{ker }d_n \hookrightarrow F_n$.Hence we can construct a resolution of $M$ by finitely generated free $\mathbb{Z}G$ modules.