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Let $G$ be a finite group with trivial action on $\mathbb C^\times$. And given a 2-cocycle $\alpha$ in its Schur multiplier group $H^2(G,\mathbb C^\times)$, as an explicit map from $G\times G\to \mathbb C^\times$. I am wondering if there is an algorithm to determine if $\alpha$ is $0$ in $H^2(G,\mathbb C^\times)$? I vaguely remember there is such an algorithm mentioned in some paper but cannot find it now. Also I am wondering if there is such an implementation in software like GAP or Magma.

Edit: I want to work with $k=\mathbb C$.

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  • $\begingroup$ It depends on how $G$ and $k^{\times}$ are given. But if $G$ is given by a list of its elements, and $k^{\times} $ as an explicit finitely generated abelian group, you have a presentation of $C^1(G, k^{\times}) $ and can compute whether your cocycle is in the image of the coboundary map using Smith normal form. $\endgroup$
    – Achim Krause
    May 3, 2023 at 6:36

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Have you looked at IdentifyTwoCocycle and IsTwoCoboundary here: http://magma.maths.usyd.edu.au/magma/handbook/text/812 ?

You may have to reduce down to a finite subgroup of $k^\times$ and put it in explicity as a $\mathbb{Z}G$-module first.

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  • $\begingroup$ Yes I have looked at it and failed because magma cannot compute second Cohomology with coefficient in $\mathbb C^*$ ( I want to work with $k=\mathbb C$). Will try reducing to finite subgroup though! $\endgroup$
    – JKDASF
    May 3, 2023 at 7:32
  • $\begingroup$ You can always reduce to the $|G|$th roots of unity, and often much smaller. $\endgroup$
    – Dave Benson
    May 3, 2023 at 7:47
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    $\begingroup$ Computationally, you should never be working with complex numbers. Since most of them don't have names, they are represented by approximations, and that drastically changes whether something is a coboundary. $\endgroup$
    – Dave Benson
    May 3, 2023 at 9:31

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