$\DeclareMathOperator{\Res}{Res}
\DeclareMathOperator{\Cor}{Cor}$
This question was [asked in MSE](https://math.stackexchange.com/q/4533187/37763).
It got no answers or comments, and so I post it here.


Let $H$ be a subgroup of a finite group $G$, and let $M$ be a $G$-module,
that is, an abelian group on which $G$ acts. Write $n=[G:H]$.
Consider the restriction and corestriction homomorphisms in Tate cohomology:
\begin{align\*}
\Res\colon\, &H^{-1}(G,M) \to H^{-1}(H,M),\\
\Cor\colon\, &H^{-1}(H,M) \to H^{-1}(G,M).
\end{align\*}
Since
$$ \Cor \circ \Res =n,$$
we know that for $\xi \in H^{-1}(G, M)$,
$$\text{if}\ \, \Res \xi=0,\ \text{then}\ \, n\xi=0.$$



> 
> **Question:**
> What is an example of $(G,H,M,\xi)$ such that
> $$n\xi=0,\ \ \text{but}\ \ \Res \xi \neq 0\ ?$$
> 
> 
> 



Restriction vs. multiplication by $n$ in Tate cohomology
Let $G = \Bbb Z/4$, acting on the Gaussian integers $M = \Bbb Z[i]$ via multiplication by $i$. The transfer $M\_G \to M^G$ is given by multiplication by $1 + i + (-1) + (-i) = 0$, so
$$H^{-1}(G,M) = M\_G = \Bbb Z[i] / (1-i) \cong \Bbb Z/2.$$
In particular, all elements are 2-torsion.


Similarly, if $H$ is the index-2 subgroup $\Bbb Z/2$, then
$$H^{-1} = M\_H = \Bbb Z[i] / (2).$$
Under this identification, the restriction $\Bbb Z[i]/(1-i) \to \Bbb Z[i]/(2)$ is multiplication by $(1+i)$. The restriction is therefore injective.


