Conditions for the restriction $H^i(G,A)\to H^i(H,A)$ being surjective 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?
 A: 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 $A=\mathbb{Q}[G/H]$.
A: The following is a small collection of examples.  I am not sure why you are interested in the restriction map being surjective, but I know a big use of restriction maps deals with their kernel, and in fact the intersection of its kernels to all subgroups (called Essential Cohomology).
But first, for example by Frobenius reciprocity, the corestriction map $\text{cor}^G_H$ is a homomorphism of $\widehat{H}^\ast(G,A)$-modules. This is better than having the trivial actions of the groups themselves on their cohomology.
Also, if $H$ is a normal subgroup of $G$ and $A$ is a $G$-module, then the conjugation action of $G$ on $(H,A)$ induces an action of $G/H$ on $H^\ast(H,A)$. But yes, $G$ acts trivially on $H^*(G,A)$.  I refer you to Ken Brown's Cohomology of Groups for more information.
1) By Mislin's theorem, the restriction map from $G$ to $H$ (a subgroup) on mod-p cohomology is an isomorphism if and only if $H$ controls p-fusion in $G$.
2) A rather trivial example: The inclusion $G\hookrightarrow G\times\mathbb{Z}_p$ induces a surjective restriction map (mod-p coefficients) by functoriality.
3) You can also go the other route and write down all the vanishing and nilpotence results on $H^\ast(H,A)$; there is a ton.
[[Edit]]:  Now the response to the restriction map not being surjective is much heavier, but you can easily just look some of this stuff up in Ken Brown's textbook, so I don't think it should be asked on this forum.
1) If $G$ is finite and $H$ is an abelian sylow-p subgroup, and $A$ is a trivial $G$-module, then the image of the restriction map is $H^\ast(H,A)^{N_G(H)}$.
2) If $|G:H|<\infty$ and $A$ is a field of characteristic $p$ relatively prime to $|G:H|$, then the restriction map is injective, because $\text{cor}\circ\text{res}(z)=|G:H|z$.
3) If $|G:H|<\infty$ and is invertible in $A$ (a $G$-module), then the restriction map sends $H^\ast(G,A)$ isomorphically onto $H^\ast(H,A)^G$.
4) Let $G$ be a $2$-group.  It turns out that $\text{Ker}(\text{res}^G_H)$ is the principal ideal $(x)$, where $|G:H|=2$ and $x\in H^1(G,\mathbb{Z}_2)$ is a homomorphism $x:G\rightarrow \mathbb{Z}_2$ such that $\text{Ker}x=H$.  Since the maximal subgroups of a $p$-group are the subgroups of index $p$, we see that every nontrivial element $x$ corresponds to some maximal subgroup $M\subset G$ (which has index $2$) with $\text{Ker}x=M$.
Note, for injective restriction maps, couple it to Mislin's theorem to see when it is not surjective.
You can also take a look at this short paper I wrote of explicit calculations of restriction maps and their kernels: http://arxiv.org/abs/1006.4836.
** In fact, given a collection $\mathcal{H}$ of subgroups of $G$, the ring $H^\ast(G,A)$ is said to be detected by $\mathcal{H}$ if all the restriction maps associated to $\mathcal{H}$ are injective.  As an example, $H^\ast(G)$ is detected on the set of Sylow subgroups.
