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How do I compute the group cohomology $H^2(G,A)$ if G is a finite abelian group acting nontrivially on a finite abelian group A?

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Type it into a computer. Seriously. Magma will definitely do it. – Kevin Buzzard May 30 '11 at 19:44
GAP too........ – Fernando Muro Sep 14 '11 at 12:10

If $G$ is any group and $A$ is any $G$-module, then $H^2(G,A)$ can be identified with the set of the equivalence classes of extensions $$1\to A\to H\to G\to 1$$

such that the action of $G$ on $A$ is the given action. Two extensions $H_1,H_2$ are said to be equivalent if there is an isomorphism $H_1\to H_2$ that makes the extension exact sequences commute. See K. Brown, Group cohomology, chapter 4.

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One can do the calculation using Kunneth theorem and the cohomology of cyclic group.

See eqn J18 and appendix J.6 and J.7 in a physics paper

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I don't think this works so easily with non-trivial coefficients. – Fernando Muro Sep 14 '11 at 12:12
Dear Fernando: Eqn. J60 - eqn. J70 in the above paper give some explicit results for non-trivial coefficients, for some simple Abelian groups. But do your suggest that I cannot use Kunneth theorem for non-trivial coefficients? – Xiao-Gang Wen Sep 14 '11 at 16:02
I agree with Fernando. At least the derivation of (J54) is doubtful: It's based on (J43), but in (J43) one has $H^i(G_1;M)\otimes_M H^{n-i}(G_2;M)$ while by setting $G_1 := Z_2^T, G_2 := Z_n$, (J54) reads: $H^i(G_1;Z_T) \otimes_Z H^{d-i}(Z_n;Z)$, i.e. in both components of the tensor product in (J43) the coefficients are equal, while in (J54) they differ! Also be aware of the wikipedia references for Kuenneth-formulars: They require trivial coefficients! (otherwise wikipedia had to use (co)homology with local coefficients, what they don't). – Ralph Sep 14 '11 at 20:58
Dear Ralph: Thanks for the comments. I agrees with you that eqn. J43 from a webpage is intended for trivial coefficient. But if Kuenneth formula only depends on the cohomological structure algebraically, should it also apply to non-trivial coefficients, provided that the group action "splits" in some way? Here $Z_T$ is the same as $Z$. Just that $G_1$ has a non-trivial action on $Z_T$. In fact, $G1\times G_2$ acts "naturally" on $H^i(G_1,Z_T)\otimes_Z H^{d−i}(G2,Z)$, Let $a\in H^i(G_1,Z_T)$ and $b\in H^{d−i}(G_2,Z)$. We have a group action $(g1,g2) \cdot (a\otimes b)=(g1\cdot a)\otimes b$. – Xiao-Gang Wen Sep 14 '11 at 22:31
In the case that I am interested in, the action does not split in any way at all. To make things more precise, in my case $G$ is a transitive abelian subgroup of $S_n$ acting on $A=(Q/Z)\times\ldots \times (Q/Z)$ by permuting factors (in the concrete problem I actually have the multiplicative group of complex numbers without $0$ insead of $Q/Z$). – Mitja Sep 15 '11 at 12:27

You can compute it using the Bar resolution, see [Weibel, H-book].

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I know exactly how it is related to extensions and how to compute it explicitly if the actions is trivial (using Kunneth formula to translate the problem to cyclic groups). I was just trying to find out if there is some relatively easy procedure that gives you an explicit answer for nontrivial actions (perhaps by somehow using the long exact sequence to translate the problem to that of trivial actions). – Mitja May 30 '11 at 21:06
I don't think it's a good idea to use the bar resolution here since it's way to large. You'll get a projective resolution of mimimal complexity by tensoring the periodic resolutions of the cyclic summands of $G$. – Ralph May 31 '11 at 4:14

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