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](https://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 
$$ Ш^1(k,M)= Ш^1\_\omega(G,M).$$
Here $ Ш^1(k,M)$ is the "honest" Tate-Shafarevich group of the $\mathrm{Gal}(\bar{k}/k)$-module $M$.


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 $p^n M=0$ for some $n\ge 1$, and if a Sylow $p$-subgroup of $G$ is *cyclic*, then $Ш(G,M)=0$.
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*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 $p^n M=0$, the multiplication by $p^n$ on $Ш(G,M)$ is 0 as well.
Since the numbers $p^n$ 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](https://mathoverflow.net/a/240449/4149)).



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



Non-vanishing of the Tate-Shafarevich kernel in group cohomology
I think the following is an example of $Ш(G,M(G,H,\Bbb{F}\_2))\neq 0$: Take $G=A\_4$ and $H$ of order $2$. Then $M$ has dimension $5$ and a (computer) calculation shows that $Ш(G,M(G,H,\Bbb{F}\_2))$ has dimension $1$.


The following is a sketch of how to do the computation by hand. First of all $A\_4$ has two conjugacy classes of cyclic subgroups namely $H=\langle (1,2)(3,4)\rangle$ and $K=\langle (1,2,3)\rangle$. Since $\lvert K\rvert=3$ it suffices to prove that the restriction map $H^1(G,M)\rightarrow H^1(H,M)$ is not injective. Let $V\_4=\langle (1,2)(3,4),(1,3)(2,4)\rangle$. The spectral sequence associated the sequence $1\rightarrow V\_4\rightarrow A\_4\rightarrow A\_4/V\_4\cong C\_3\rightarrow 1$ collapses to give $H^\*(A\_4,M)\cong H^\*(V\_4,M)^{C\_3}$. From this it is a straightforward (but tedious) computation to see that both $H^1(G,M)$ and $H^1(H,M)$ have dimension $1$ and that the restriction map is the zero map.


Here is the Magma code I used to verify the example:



> 
> `G:=Alt(4);
>  H:=sub<G|(1,2)(3,4)>;
>  M:=PermutationModule(G,H,GF(2));
>  M:=M/Fix(M);
>  XG:=CohomologyModule(G,M);
>  H1G:=CohomologyGroup(XG,1);
>  XH:=Restriction(XG,H);
>  H1H:=CohomologyGroup(XH,1);
>  ims:=[IdentifyOneCocycle(XH,OneCocycle(XG,H1G.i)) : i in [1..Dimension(H1G)]];
>  res:=hom<H1G->H1H|ims>;
>  print Dimension(H1G),Dimension(H1H),Dimension(Kernel(res));`
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The output `1, 1, 1` confirms the computation above.


