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 $Ш^1_\omega(G,M)\ne 0$.

 We ask our question with a hope to answer  [this hard question](http://mathoverflow.net/q/240439/4149) (in positive or negative). The relation is the following.
Let $L/k$ be a Galois extension of number fields with Galois group $G$. Then we may regard $M$ as a $\mathrm{Gal}(\bar{k}/k)$-module. In this case
$$  Ш^1(k,M)\subset Ш^1_\omega(G,M).$$
Moreover, if  the decomposition groups of *all* the places of $k$ in $\mathrm{Gal}(L/k)=G$ are *cyclic*, then 
$$  Ш^1(k,M)= Ш^1_\omega(G,M).$$



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 $Ш^1_\omega(G,M)$.
If $G$ acts trivially on $M$, then clearly  $Ш(G,M)=0$ (because then $H^1(G,M)=\mathrm{Hom}(G,M)$ ).
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, we 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 $Ш(G,M)=0$.

*Proof.* Let $P$ be a Sylow $p$-subgroup of $G$, then the map
$$ \mathrm{Cor}\circ\mathrm{Res}\colon H^1(G,M)\to H^1(P,M)\to H^1(G,M)$$
is the multiplication by $[G:P]$. 
Since $P$ is cyclic, this map is 0 on $Ш(G,M)$.
Since $pM=0$, the multiplication by $p$ on $Ш(G,M)$ is 0 as well.
Since the numbers $p$ and $[G:P]$ are coprime, we conclude that $Ш(G,M)=0$. 


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. (For $p=2$, this is the Galois module $T[2]$ from [this answer](http://mathoverflow.net/a/240449/4149)).

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

I would be especially interested in a counter-example with $p=2$.

The proposition above shows that for a counter-example $M(G,H,p)$, the group $G$ must have a noncyclic Sylow $p$-subgroup. 

> **Proposition.** If $p\nmid [G:H]$, then for $M=M(G,H,p)$ we have $Ш(G,M)=0$.

*Proof* (due to user nfdc23). Write $X=G/H$, then $\# X$ is prime to $p$.
It follows that $M=M(G,H,p)$ is isomorphic (as a $G$-module) to a direct summand of the $G$-module $\mathrm{Maps}(X,\mathbb{F}_p)$.
Since $Ш(G,\mathrm{Maps}(X,\mathbb{F}_p))=0$, we conclude that $Ш(G,M)=0$.