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I hope that my question is appropriate for MO, since it might turn out te be mainly a question about GAP or other group theory software.

Is there an algorithm to produce all non-nilpotent groups of odd order (up to some given upper bound)?

All groups of odd order are solvable by the famous Feit-Thompson theorem; I guess that this fact could be useful in enumerating all these groups, but I don't know how.

Any general ideas to produce many such groups (not necessarily all of them) would also be greatly appreciated.

I already checked the SmallGroups GAP database, and it turns out that there are only 1016 such groups of order $\leq 2015$.

EDIT: The following question probably makes more sense:

Is there an algorithm to produce all groups of odd order (up to some given upper bound) not admitting a non-cyclic nilpotent quotient?

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    $\begingroup$ Just a remark: if $G$ is nilpotent of odd order then something like $G\times(C_7\rtimes C_3)$ will be in your list, so non-nilpotency is perhaps not really restrictive $\endgroup$
    – Tim Dokchitser
    Mar 10, 2011 at 12:22
  • $\begingroup$ That's true; I guess the new question (edited) makes more sense. $\endgroup$
    – Tom De Medts
    Mar 10, 2011 at 14:28
  • $\begingroup$ Look at the largest nilpotent quotient. It must be cyclic of order, say, k. The next chief factor below it must have order coprime to k, and so be of order q^n where n is the order of q mod k. Now you are just looking for downwards extensions of a fixed (solvable) group. In GAP you might use "Extensions" to find all the extensions for a given module. $\endgroup$
    – Jack Schmidt
    Mar 10, 2011 at 15:24
  • $\begingroup$ Usually when I do this, I have specific ideas of what sort of modules I want to allow. I think as posed, you need to find all modules (reducible, decomposable, etc.) up to a certain dimension. There are probably too many to make that a sane undertaking. If the Sylow p-subgroup is cyclic, it might be more reasonable. If you want the Fitting series to be a chief series then it should be relatively easy. $\endgroup$
    – Jack Schmidt
    Mar 10, 2011 at 15:25
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    $\begingroup$ I think your low number (1016) is just due to not having enough room for the p-groups to get out of control by order 2015. You have to waste a little space with that top k*q^n, and then maybe the chief factors below it are not order p, but rather p^2 or p^3 or something, but you still get the ridiculous growth of p-groups. $\endgroup$
    – Jack Schmidt
    Mar 10, 2011 at 15:30

1 Answer 1

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Hi Tom,

the answer (at least to your second, refined question) is "Yes! or at least "Yes, soon!" :). I first wanted to post this as a comment, but since it is rather lengthy, I figured it made more sense to give this as an answer, even though it might not be completely satisfying.

There are algorithms that can generate all groups up to a given order; those were used to create the database of small groups. Indeed, I am currently working on a refined set of such algorithms. We plan to use this to extend the database to small groups to orders up to 10,000 (excluding multiples of 1024 and $3^7$ or $3^8$). As part of this, I am working on algorithms that allow constructing all extensions of a group $A$ by another group $B$; but also allow restriction to say all metabelian groups of a given order; etc.

Of course you can just generate all groups up to a given order, and then remove all you don't need, but that's very wasteful. A first refinement is to restrict to generating all groups of odd order, that's already considerably better.

But you can do more: Say a group $G$ has property * if it has odd order, is solvable and has no non-cyclic nilpotent quotients. To find all these groups up to order $n$, it suffices to compute all extensions $E$ (up to isomorphism) of a solvable group $N$ by a cyclic group $Q$, both of odd order, for which $[E,N]=N$.

This is sufficient because the quotients of $E$ by its lower central series are all nilpotent, so must all be cyclic if property * is to hold. But then we can assume $N$ to be the last term of the lower central series (last here means: the term from which on the series becomes stable). And have that $E/N$ is cyclic, and $[E,N]=N$.

The condition that $[E,N]=N$ translates into a restriction on the action of $Q$ on $N/N'$. For it implies (with some handwaving) that $N/N' = [E,N]/N' = [Q,Q] [Q,N/N'] [N/N', N/N'] = [Q,N/N']$ (as $Q$ and $N/N'$ are abelian). This can now be used to effectively cut down on what groups and couplings between them are possible for $Q$ and $N$.

This is indeed a special case of an algorithm we (Bettina Eick and me) are planning to include in our new GAP package. As of now, though, I have not yet turned to working on this algorithm, but it'll happen in the forseeable future.

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