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I am wondering if there is a good method to write down a finite equational basis for a finite group. Especially I am wondering if there is a good method in following situations:

  1. We can write a group as a direct or semidirect product of two groups which finite basis we already know how to write, is there any relationship between the identities of the factors and the big group?

  2. Can we do it if we have a given finite presentation (generators and relations)

  3. Group is small, in this case I am wondering if something like gap, magma or sage has a good way of dealing with this problem? I tried looking it up but I didn't find anything and I am not really good at these languages.

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  • $\begingroup$ I'm no expert but my feeling is no way way to get a finite basis is known $\endgroup$
    – Benjamin Steinberg
    Commented May 17, 2023 at 21:31
  • $\begingroup$ Have you tried following through the proof in Oates and Powell, "Identical relations in finite groups", J Algebra 1 (1964), 11-39? It's also worth looking in MR for papers referring to this one, although most of them aren't about groups. I don't think there are any algorithms implemented in the usual algebra packages, and it's probably a hard problem in general. $\endgroup$
    – Dave Benson
    Commented May 18, 2023 at 10:05
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    $\begingroup$ @DaveBenson Unfortunately Oates and Powell don't provide a general construction which is to be expected. It is hard to do in general, I was hoping that given a sufficiently nice description of a group, like finite presentation or as a product of two groups whose finite basis is already known, one would be able to construct the basis for the group at hand. This doesn't seem to be the case however. $\endgroup$
    – Todor Antic
    Commented May 18, 2023 at 10:12
  • $\begingroup$ I believe all the known proofs show that finite groups generate a Cross variety and then there is some general argument that cross varieties are finitely based so I think the proof doesn't really give something concrete and algorithmic. I think the nicest proof is in Kovacs and Newman, which can also be found in Hanna Neumann's book. $\endgroup$
    – Benjamin Steinberg
    Commented May 18, 2023 at 10:29
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    $\begingroup$ The identities satisfied by a direct product consists of those identities satisfied by both factors. But it is hard to see how you can extract a basis for this from a basis for each factor $\endgroup$
    – Benjamin Steinberg
    Commented May 18, 2023 at 10:31

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