Let $G$ be a finite group and $M$ be a finitely generated $G$-module,
that is, a finitely generated abelian group on which $G$ acts.
Consider the functor
$$ (G,M)\rightsquigarrow F(G,M):= (M\_G)\_{\rm Tors}$$
the torsion subgroup of the group of coinvariants $M\_G$ of $M$.
For a cyclic subgroup $i\colon C\hookrightarrow G$, we have a natural homomorphism
$$i\_\*\colon F(C,M)\to F(G,M),$$
which might be non-surjective when $M$ is infinite.
Consider the finite abelian group
$$S(G,M)={\rm coker}\left[ \bigoplus\_{C\subseteq G {\rm \ cyclic}}\!\!\!\! F(C,M)\,\longrightarrow\, F(G,M)\right].$$
If $G$ is cyclic or $M$ is finite, then $S(G,M)=0$.



> 
> **Question.** What are $G$ and $M$ such that $S(G,M)\neq 0$ ?
> 
> 
> 


I would be happy to get an example with $G={\Bbb Z}/2{\Bbb Z}\oplus {\Bbb Z}/2{\Bbb Z}$ and $M$ torsion free.


**Motivation:** When $M$ is torsion free, $F(G,M)=H^{-1}(G,M)$ (Tate's cohomology).
Cyclic subgroups suggest a relation with Chebotarev's density theorem.


I asked this seemingly elementary question in math.stackexchange.com and got no answers or comments, so I ask it here.



The torsion subgroup of the coinvariants for a $G$-module
Let $\mathbb Z[\mathbb Z/2\times \mathbb Z/2]$ be the free module of rank one over $G=\mathbb Z/2\times\mathbb Z/2$. Let $\mathbb Z$ be the trivial $G$-module. There is an obvious diagonal inclusion $\mathbb Z\hookrightarrow \mathbb Z[\mathbb Z\_2\times \mathbb Z\_2]$. Let $M$ be the quotient of this inclusion. As an abelian group, $M\cong \mathbb Z^3$. For every sugbroup $H$ of $G$ there is an exact sequence


$$\mathbb Z\_H \to \mathbb Z[\mathbb Z\_2\times \mathbb Z\_2]\_H\to M\_H \to 0.$$


For $H=G$ this sequence becomes


$$\mathbb Z\xrightarrow{\cdot 4}\mathbb Z\to \mathbb Z/4\to 0.$$
It follows that $F(G, M)=\mathbb Z/4$. On the other hand, if $H\cong\mathbb Z/2$ is a non-trivial cyclic subgroup of $G$ then the sequence has the form


$$\mathbb Z\xrightarrow{(2,2)} \mathbb Z\oplus\mathbb Z \to \mathbb Z\oplus \mathbb Z/2 \to 0$$
and it follows that $F(H, M)=\mathbb Z/2$. If $H$ is the trivial group then clearly $F(H, M)=0$. It follows that the cokernel of the homomorphism $\bigoplus F(C, M)\to F(G, M)$, where $C$ ranges over cyclic subgroups, is $\mathbb Z/2$.


