Suppose we have a projective resolution of $G$ modules (consider the standard resolution or example)
$$\cdots \rightarrow P_2 \rightarrow P_1 \rightarrow P_o \rightarrow \mathbb Z \rightarrow 0 $$
We can dualize it as
$$0 \rightarrow\mathbb Z \rightarrow Hom(P_0, \mathbb Z) \rightarrow Hom(P_1,\mathbb Z) \rightarrow Hom(P_2,\mathbb Z) \rightarrow \cdots $$
I want to prove that this later sequence is exact.
As $Hom(\_, \mathbb Z)$ is left exact, we only know, prima facie,
$$0 \rightarrow\mathbb Z \rightarrow Hom(P_0, \mathbb Z) \rightarrow Hom(P_1,\mathbb Z)$$ is exact.
So I want to prove that if $ P_{i+1} \rightarrow P_{i} \rightarrow P_{i-1} $ is exact then so is its $\mathbb Z$ dual.
I know $P$ being projective over $ \mathbb Z [G]$ should be free over $\mathbb Z$ of finite rank as $G$ is finite.
Motivation: I want to understand how complete resolutions are constructed in the context of Tate cohomology. (Ref: Cassels and Frohlich page 103)
These are $\Bbb ZG$-modules. The point here is that your resolution is split when considered as a sequence of $\Bbb Z$-modules. Let $K_i$ be the kernel of $P_i\to P_{i-1}$. Your sequence is then really a bunch of short exact sequences $$0\to K_i\to P_i\to K_{i-1}\to0.$$ As $K_{i-1}$ is a $\Bbb Z$-submodule of the free $\Bbb Z$-module $P_{i-1}$ this sequence splits as a sequence of $\Bbb Z$-modules. Therefore $$0\to\text{Hom}(K_{i-1},\Bbb Z)\to\text{Hom}(P_i,\Bbb Z)\to\text{Hom}(K_{i},\Bbb Z)\to0$$ also splits and then the sequence $$\cdots\to\text{Hom}(P_{i-1},\Bbb Z)\to\text{Hom}(P_i,\Bbb Z)\to\text{Hom}(P_{i+1},\Bbb Z)\to\cdots$$ is exact over $\Bbb Z$ and so exact over $\Bbb ZG$.