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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 $V_4=\lvert (1,2)(3,4),(1,3)(2,4)\rvert$. 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.

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.

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 $V_4=\lvert (1,2)(3,4),(1,3)(2,4)\rvert$. 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 Dimnsion(H1G),Dimension(H1H),Dimension(Kernel(res));

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

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 $V_4=\lvert (1,2)(3,4),(1,3)(2,4)\rvert$. 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.

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 $1$.

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 H1G,H1H,Kernel(res);

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 $V_4=\lvert (1,2)(3,4),(1,3)(2,4)\rvert$. 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 Dimnsion(H1G),Dimension(H1H),Dimension(Kernel(res));

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