I was reading Brown's Cohomology Theory of Finite groups and was wondering whether there's an example of the following.


Let $n\in \mathbb{N}$. Does there exist a $\mathbb{Z}\_n$ module $M$ which is cohomologically trivial (i.e. $\smash{\hat{H}}^i(\mathbb{Z}\_n,M)$ is trivial for each $i$) but such that there exists some $p,i\in \mathbb{N}$ such that $\smash{\hat{H}}^i(\mathbb{Z}\_n,M/pM)$ is non-trivial?


In the case that $M$ has no $n$ torsion this is not possible by Lemma 8.6 of Brown's Cohomology of groups, I was wondering what goes wrong in the sense of a counterexample when $M$ has $n$ torsion.



Cohomologically trivial module $M$ such that $M/pM$ is not cohomologically trivial for some $p\in\mathbb{N}$
For $n=p$ an odd prime, $M=\mathbb Z/p^2$, where a generator of $\mathbb Z/p$ acts by multiplication by $1+p$, is an example.


$M/pM$ is $\mathbb Z/p$ with the trivial action of $\mathbb Z/p$, which has cohomology groups $\mathbb Z/p$ in each degree.


But for $M$ itself, the invariants are generated by $p$ and the coinvariants by $1$, and the norm map from invariants to coinvariants sends $1$ to $\sum\_{i=1}^{p-1} (1+p)^i = p + p \binom{p}{2} + \dots$, which is $p$ times a unit since $\binom{p}{2}$ is divisible by $p$, so the norm map is an isomorphism, and thus the Tate cohomology in degrees $0$ and $-1$ vanish. By periodicity, they vanish in all degrees.


