Let $G$ be a finite group, and let $F$ be a complete resolution for $G$. In other words, $F$ is an acyclic chain complex of projective $\mathbb{Z}G$-modules together with a map $\varepsilon:F_0\to\mathbb{Z}$ such that $\varepsilon:F_+\to\mathbb{Z}$ is a projective resolution of $\mathbb{Z}$ over $\mathbb{Z}G$.
It can be shown that we may construct a complete resolution for $G$ in such a way that each $F_n$ is finitely generated. Such a resolution is said to be of finite type.
If $M$ is a $\mathbb{Z}G$-module, we can define the Tate cohomology groups of $G$ with coefficients in $M$ as follows:
$$\hat{H}\,^n(G,M):=H^n(\mathrm{Hom}_G(F,M))$$
This definition is independent of choice of resolution $F$ because any two choices are homotopy equivalent.
Now, let $M=\mathbb{Z}$ be the trivial $\mathbb{Z}G$-module. Using what I've written down, I'd like to see why $\hat{H}\,^n(G,\mathbb{Z})$ is a finitely generated abelian group for all $n$. I know that in this case $\mathrm{Hom}_G(F_n,\mathbb{Z})\cong F_n^G$ is the group of invariants of $F_n$, so that we are considering the abelian groups $H^n(F^G)$, but how do we show that they are finitely generated?
$F_n$ is a finitely-generated $G$-module, so it is also finitely-generated as a $\mathbf Z$-module because $\mathbf Z[G]$ is finitely generated as a $\mathbf Z$-module (since $G$ is finite).
Now $F_n^G$ is a subgroup of a finitely-generated abelian group, so it is also finitely-generated. Thus $H^n(F^G)$ is a subquotient of a finitely-generated abelian group, so it is also finitely-generated.