Let $G=\langle g\rangle$ be a cyclic group of order $n$, let $t=1+g+\cdots+g^{n-1}$ be the trace of $G$ and let $M$ be a finite right $G$-module. Applying the functor ${\rm Hom}_G(-, M)$ to the right $\mathbb{Z}G$-resolution $$\cdots\rightarrow\mathbb{Z}G \overset{t}\rightarrow \mathbb{Z}G \overset{g-1}{\rightarrow} \mathbb{Z}G \overset{t}{\rightarrow} \mathbb{Z}G \overset{g-1}{\rightarrow} \mathbb{Z}G \rightarrow \mathbb{Z}$$ we get the new sequence $$0{\rightarrow}M \overset{g-1}{\rightarrow}M \overset{t}\rightarrow M\overset{g-1}{\rightarrow}M \overset{t}{\to}M\rightarrow \cdots.$$ Then, if we let $\alpha$ and $\beta$ be the right multiplication by $g-1$ and $t$ on $M$, respectively, one has by definition $$H^i(G,M)\simeq\frac{\text{Ker }\alpha}{\text{Im }\beta}$$ for even $i>0$ and $$H^i(G,M)\simeq\frac{\text{Ker }\beta}{\text{Im }\alpha}$$ for odd $i>0$.
Now, my question is the following:
Why is it true that if $H^i(G,M)=0$ for some positive $i$, then it is zero for every $i>0$?
One should clearly use the finiteness of $M$ in deriving "$\text{Ker }\beta=$ Im $\alpha$" from "Ker $\alpha=$ Im $\beta$" and a hint could be the existence of an integer $r$ for which $M=\text{Ker }\alpha^r\oplus\text{Im }\alpha^r=\text{Ker }\beta^r\oplus\text{Im }\beta^r$.
Let $m$, $\alpha_i$, $\alpha_k$, $\beta_i$, and $\beta_k$ be the cardinalities of $M$, the image of $\alpha$, the kernel of $\alpha$, the image of $\beta$, and the kernel of $\beta$ respectively. Then from the first isomorphism theorem,
$$ m/\alpha_k = \alpha_i $$ and $$ m/\beta_k = \beta_i, $$ so $$ \beta_i\beta_k = \alpha_i\alpha_k $$ and therefore $\alpha_k / \beta_i = \beta_k / \alpha_i$.