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Consider a short exact sequence of (finite, or pro-finite) groups $$ 1 \to H \to G \xrightarrow{\pi} G/H \to 1 $$ and suppose for simplicity that $H$ lies in the centre of $G$. Such short exact sequence corresponds to a 2-cocycle in $H^2(G/H,H)$ under the correspodence between central extensions and second group cohomology. The cocycle is calculated as follows: take a section $j$ of $\pi$ and let $$ h_{\sigma,\tau}=j(\sigma)+\sigma j(\tau) - j(\sigma\tau) $$

My question is about a map $$ f: H^1(G,A^H) \to H^2(G/H,A^H) $$ which is defined by a somewhat similar formula.

Take $\{g_\sigma\} \in H^1(G,A^H)$, take a section $j$ as above, and define the 2-cocycle as follows $$ h_{\sigma,\tau}=g_{j(\sigma)}+j(\sigma)g_{j(\tau)}-g_{j(\sigma\tau)} $$ One can check that the class of $h$ does not depend neither on a particular choice of a cocycle $g$ in a cohomology class, nor on the section $j$, the map is also functorial in $A$.

What is known about this map? Is it a part of some standard exact equence, or a composition of some well-known functors (like restriction/corestriction) ?

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This is more of a guess but have you looked into papers on equivariant group cohomology? –  Simon Markett May 24 '12 at 12:24

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up vote 3 down vote accepted

Consider the 1-cochain $\phi_g$ for the bar resolution of $G/H$ with values in $A^H$ given on the free generators $[x]: x \in G/H$ by $[x]\mapsto g(j(x))$. Then $$\partial \phi_g [x|y] = \phi_g([x]+x[y]-[xy])=g(j(x)) + xg(j(y)) - g(j(xy))$$ (note the action of $x \in G/H$ is well-defined on $A^H$) which is your (second) map $h$. Thus $h$ is zero in cohomology.

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oh, you are right, of course. the context in which this map has arised made me believe that it might be non-zero. thanks! –  Dima Sustretov May 24 '12 at 13:58
    
Do you know about the 5-term exact sequence associated to the Lyndon-Hochschild-Serre spectral sequence for your group extension? The cohomology groups you are talking about show up there. It is in Benson - Representations and Cohomology vol 1 for example. –  mt_ May 24 '12 at 14:12
    
yes, in this sequence there is a term in between the two groups. If the composition of the two maps in this sequence were my map, consequently my map would have to be zero, but as I supposed that my map might be non-zero, I excluded this possibility. –  Dima Sustretov May 24 '12 at 14:16

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