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?

  • 1
    $\begingroup$ The cohomology groups are not $G$-modules. $\endgroup$
    – Angelo
    Apr 24, 2012 at 16:04
  • $\begingroup$ At the very least $H^1(H,A)$ should be a $G/H$-module. $\endgroup$
    – Earthliŋ
    Apr 24, 2012 at 16:10
  • 3
    $\begingroup$ @Angelo, if $A$ is a $G$-module, and if $H$ is a normal subgroup of $G$, then the conjugation action of $G$ on $H$ together with the given action of $G$ on $A$ induces an action of $G$ on the cohomology group $H^i(H,A)$. It turns out that this action factors through the quotient group $G/H$. You also obtain an action of $G$ on $H^i(G,A)$, though this action turns out to be trivial. Similarly, $G/H$ (hence also $G$) acts trivially on $H^i(G/H,A^H)$. $\endgroup$ Apr 24, 2012 at 16:41
  • $\begingroup$ @s.barmeier, along the lines of my previous comment, the maps are all $G$-module homomorphisms, though all of the spaces in the exact sequence are trivial $G$-modules, so this isn't saying very much. $\endgroup$ Apr 24, 2012 at 16:42
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    $\begingroup$ To Christopher: ok, I should have said that those cohomology groups are not $G$-modules in a non-trivial way. This reminds me of a talk I gave when I was young, with a great mathematician in the audience. I said something like "homology of singular spaces does not have a natural ring structure"; this really upset him, and objected very loudly "sure it does: you could define all products to be 0". $\endgroup$
    – Angelo
    Apr 24, 2012 at 17:22

2 Answers 2


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 $A=\mathbb{Q}[G/H]$.

  • $\begingroup$ Thanks, these examples are really helpful. Although $C^1(G,A)\to C^1(H,A)$ is always surjective the induced $\mathrm{res}$ is not... (I think that is where I got confused before.) Now, looking at your 4th example, $\mathrm{res}\colon H^1(G,A)\to H^1(H,A)^{G/H}$ is not even surjective in this case. I tried your exercise in the next comment. $\endgroup$
    – Earthliŋ
    Apr 25, 2012 at 4:09
  • $\begingroup$ For $[x]\in H^1(H,A)$ define a normalized 1-cochain in $C^1(G,A)$ by $y(\sigma)=y(g)+gx(h)$, where $\sigma=gh$ and $g$ is a coset rep. Using that $H$ is central, the cocycle condition on $C^1(G,A)$ gives $\partial^1 y+g_1g_2(g_2^{-1}x(h_1)-x(h_1h_2)+x(h_2))=0$. I interpret this to mean that $\mathrm{res}[y]=[x]$ as long as $G$ acts trivially on $A$. But it does, whence my conclusion seems to be that $\mathrm{res}$ is surjective. Except that it isn't. $\endgroup$
    – Earthliŋ
    Apr 25, 2012 at 4:11
  • $\begingroup$ I must admit that I didn't follow your argument, but the approach I would take is to show that $H^1(G;\mathbb{F}_p)\approx \text{Hom}(G,\mathbb{F}_p)$ for any group $G$ (step 1). Since here $H$ lies in the commutator subgroup of $G$ (step 2), any homomorphisms $G\to \mathbb{F}_p$ restricts trivially to $H$. Thus the restriction map $\text{Hom}(G,\mathbb{F}_p)\to \text{Hom}(H,\mathbb{F}_p)$ is the zero map (step 3). For step 1 you can quote the universal coefficient thm, but I recommend proving it directly with cochains. (This argument only needs $H\subset [G,G]$, but here in fact $H=[G,G]$.) $\endgroup$
    – Tom Church
    Apr 25, 2012 at 23:59

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.

  • $\begingroup$ Well, I'm interested in the surjectivity of the restriction, because I am interested in the kernel of the transgression $H^1(H,A)^{G/H}\to H^2(G/H,A^H)$ and the exact sequence in low degrees tells me $\mathrm{im} \mathrm{res}=\ker \mathrm{tg}$. But in all the examples I can think of the restriction map is always surjective (unless there was some $G$-module structure I overlooked, which doesn't seem to be the case). ... which just seems too strange to be true. Hence, I wondered under what precise conditions the restriction is surjective. $\endgroup$
    – Earthliŋ
    Apr 24, 2012 at 18:16
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    $\begingroup$ What are your examples? $\endgroup$ Apr 24, 2012 at 18:43

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