I was wondering what the condition is for the restriction map (in group cohomology) $H^i(G,A)\to H^i(H,A)$ to be surjective.


I am a little confused about when maps between cohomology groups are considered to be $G$-, $H$- or $G/H$-module homomorphisms. In particular, in the exact sequence of low degrees


$0\to H^1(G/H,A^H)\to H^1(G,A)\to H^1(H,A)^{G/H}\to H^2(G/H,A^H)\to H^2(G,A)$


are the maps all meant to be $G$-module homomorphisms?


EDIT: Does someone have an example when $\mathrm{res}$ is not surjective?



Conditions for the restriction $H^i(G,A)\to H^i(H,A)$ being surjective
You asked for examples when $\text{res}^G\_H\colon H^i(G;A)\to H^i(H;A)$ is not surjective. Here are some examples for $i=1$.


* Let $G=\text{SL}\_2\mathbb{Z}$ and $H=\mathbb{Z}$ generated by $\begin{bmatrix}1&1\\\\0&1\end{bmatrix}$. Then $H^1(G;\mathbb{Z})=0$ but $H^1(H;\mathbb{Z})=\mathbb{Z}$ so $\text{res}$ is not surjective.
* Let $G=D\_\infty=\text{Isom}(\mathbb{Z})=\mathbb{Z}/2\ltimes\mathbb{Z}$ and $H=\text{Isom}^+(\mathbb{Z})=\mathbb{Z}$. Now $[G:H]=2<\infty$ but still $H^1(G;\mathbb{Z})=0$ and $H^1(H;\mathbb{Z})=\mathbb{Z}$ so $\text{res}$ is not surjective.
* Let $G$ be the rank 2 free group $G=F\_2=\langle a,b\rangle$, and let $H$ be any subgroup of index $n-1$. Then $H$ is a free group of rank $n$ so $H^1(H;\mathbb{Z})=\mathbb{Z}^n$. But $H^1(G;\mathbb{Z})=\mathbb{Z}^2$ so $\text{res}$ is not surjective for $n>2$.
* For a finite example, let $G$ be the group of upper-triangular matrices in $\text{SL}\_3\mathbb{F}\_p$, and let $H$ be its center: $H=\begin{bmatrix}1&0&\ast\\\\0&1&0\\\\0&0&1\end{bmatrix}$. Then the map from $H^1(G;\mathbb{F}\_p)=\mathbb{F}\_p^2$ to $H^1(H;\mathbb{F}\_p)=\mathbb{F}\_p$ is the zero map (this is a good exercise), so $\text{res}$ is not surjective. (For a related topological example, let $T^1\Sigma\_g$ be the unit tangent bundle of an orientable surface of genus $g>1$, let $G=\pi\_1(T^1\Sigma\_g)$, and let $H\approx \mathbb{Z}$ be its center.)
* More generally, let $G$ be any group and let $H$ be its commutator subgroup $[G,G]$. As long as your coefficients are untwisted (trivial as a $G$-module), the map $H^1(G;A)\to H^1([G,G];A)$ is always the zero map. Thus if $G$ is any group for which $H^1(G;A)\neq 0$ and $H^1([G,G];A)\neq 0$, $\text{res}$ is not surjective. Examples of such $G$ include free groups, surface groups, non-abelian nilpotent groups (either with $A=\mathbb{Z}$ or $A=\mathbb{F}\_p$), etc.


You can also get examples for $i=0$ when $A$ is a nontrivial $G$-module, since $H^0(G;A)$ is the $G$-invariant elements of $A$, while $H^0(H;A)$ is the $H$-invariant elements of $A$. Thus whenever $A$ contains an $H$-invariant element not fixed by $G$, the restriction $\text{res}$ will not be surjective. For example, for $[G:H]<\infty$ you could take $A=\mathbb{Q}[G/H]$.


