Non-vanishing of the Tate-Shafarevich kernel in group cohomology

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 (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, 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 $Ш(G,M)=0$.

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).

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 $Ш(G,M)=0$.

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));

The output 1, 1, 1 confirms the computation above.

• How do you calculate this with a computer? – user19475 Jun 13 '16 at 7:38
• I used Magma, but the example is small enough that one can do it by hand. A reference for the relevant functions in Magma is magma.maths.usyd.edu.au/magma/handbook/… It is also possible to use GAP, see gap-system.org/Manuals/doc/ref/chap39.html#X7CA0B6A27E0BE6B8 – Kasper Andersen Jun 13 '16 at 13:10
• Could you please explain, how one can do it hand? The group $A_4$ is of order 12, and $M$ is of dimension 5, so maybe brute force is not sufficient.... What tricks can I use? Please kindly add this to your answer (you can edit your answer). – Mikhail Borovoi Jun 13 '16 at 18:08
• Also, could you please add your Magma code to your answer? I have never used computer algebra, but I am a former programmer, and I would be happy to learn using Magma by this example, and then to compute other examples! – Mikhail Borovoi Jun 13 '16 at 18:15
• Dear Mikhail! I added a sketch of how to do the computation by hand and the Magma code I used. Let me know if I need to add more detail. – Kasper Andersen Jun 14 '16 at 19:22