Let $G$ be a finite group. Let $M$ be a finite $G$-module (a finite abelian group with an action of $G$).
We consider a special kind of $G$-modules; in particular, our $M$ is a finite dimensional representation of $G$ over $\mathbb{F}_p$. For our $G$-modules $M$, we ask whether it is possible that $Ш^1_\omega(G,M)\ne 0$.
 Our question is inspired by [this hard question](http://mathoverflow.net/q/240439/4149).

Following [Sansuc](http://www.digizeitschriften.de/dms/img/?PID=GDZPPN002198746), we define
$$Ш^1_\omega(G,M)=\mathrm{ker}\left[H^1(G,M)\to\prod_C H^1(C,M)\right],$$
where $C$ runs over the set of cyclic subgroups of $G$.
We write $Ш(G,M)$ for $Ш^1_\omega(G,M)$.
Sansuc proves that if all the Sylow subgroups of $G$ are *cyclic*, then $Ш(G,M)=0$ for any $G$-module $M$.
Using his method, one can prove the following proposition.

> **Proposition.** Let $p$ be a prime number. If $M$ is a $G$-module such that  $pM=0$ and if a Sylow $p$-subgroup of $G$ is *cyclic*, then $Ш(G,M)=0$.


Let $H$ be a subgroup of $G$ (e.g., $H=\{1\}$). We consider the $G$-set $X:=G/H$. We embed $\mathbb{F}_p$ into $\mathrm{Maps}(X,\mathbb{F}_p)$ as the subspace of constant maps, and  we set 
$$M(G,H,p):=\mathrm{Maps}(X,\mathbb{F}_p)/\mathbb{F}_p.$$
 Then $M(G,H,p)$ is a finite dimensional representation of $G$ over $\mathbb{F}_p$, hence a $G$-module.

> **Question.** Do there exist $G$, $H$, and $p$ such that for $M=M(G,H,p)$ we have $Ш(G,M)\ne 0$?

The proposition above shows that for such $M(G,H,p)$, the group $G$ should have a noncyclic Sylow $p$-subgroup. One could try to construct a desired counter-example with $G=(\mathbb{Z}/p\mathbb{Z})^m$, $H=0$.