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I need to prove that the third (co)homology group of a certain finitely presented group is not finitely generated. The group is not an amalgamated product or an HNN extension, and it does not act nicely on a CAT(0) cube complex. It does have torsion and its presentation is very non-aspherical. What do I do? That is what are the methods of proving it?

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  • $\begingroup$ I keep looking at this question, thinking "Surely the OP has tried Lyndon-Hochschild-Serre spectral sequence methods," but you didn't mention trying that, and nobody else has asked--so I apologize for suggesting something very naive, but have you looked for filtrations of your group by normal subgroups, such that you have some cohomological understanding of the filtration quotients? It may be possible to use the LHSSS of each resulting group extension, iteratively, to get some kind of hold on H^3 of your group. Sorry again for suggesting something more naive than what you already mentioned :/ $\endgroup$
    – user126192
    Jul 24, 2018 at 1:11
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    $\begingroup$ Oh, that's a good point: if H_2 of the group (i.e., its Schur multiplier) turns out to be sufficiently "interesting," you can automatically get that H^3 of the group is huge. For example, if H_2(G; Z) is Q, then the universal coefficient sequence gives you a copy of Ext^1_Z(Q, Z) inside H^3(G; Z), so H^3(G; Z) is uncountable, so certainly not finitely generated. $\endgroup$
    – user126192
    Jul 24, 2018 at 4:45
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    $\begingroup$ @aaaaaaaaaaaaaaa — yes, but we are told that the group is finitely presented, which is much stronger than finitely generated. $\endgroup$
    – HJRW
    Jul 24, 2018 at 15:43
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    $\begingroup$ In general, lacunary hyperbolic groups indeed have infinitely generated Schur multiplyers. This includes Tarski monsters of unbounded exponents. But these are not finitely presented. $\endgroup$
    – user6976
    Jul 24, 2018 at 16:02
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    $\begingroup$ I'd say best call here would be using homological shifts (concretely, $H_*(F/R, R_{ab}) = H_{*+1}(F/R, \Delta) = H_{*+2}(F/R, \Bbb Z)$) and trying some explicit computations with Lyndon's partial resolvent. This approach is, of course, much better suited for proving that some homology is finitely generated, but may help anyway. $\endgroup$
    – Denis T
    Jul 24, 2018 at 20:03

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