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I tried to calculate the number of groups of order $2058=2\times3\times 7^3$ and aborted after more than an hour. I used the (apparently slow) function $ConstructAllGroups$ because $NrSmallGroups$ did not give a result.

The number $n=2058$ is (besides $2048$) the smallest number $n$, for which I do not know $gnu(n)$

The highest exponent is $3$, so it should be possible to calculate $gnu(2058)$ in a reasonable time.

What is $gnu(2058)$. If a result is too difficult, is it smaller than ,larger than or equal to $2058$ ?

Olexandr Konovalov
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Peter
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    I never realized this was A000001 in OEIS! OEIS lists a reference going up to...2047. D; – GPhys Jan 02 '16 at 20:10
  • I would also have expected other sequences at place one : The natural numbers or the prime numbers. – Peter Jan 02 '16 at 20:46
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    If you will call `SetInfoLevel(InfoGrpCon,4);` before calling `ConstructAllGroups` then you will be able to see info messages about the progress of the calculation. Apparently it starts the step described as `compute ff groups with socle 343 and size 2058` where things are starting to be computationally hard. – Olexandr Konovalov Jan 02 '16 at 20:50
  • Hi, Alexander! :) – Peter Jan 02 '16 at 21:07
  • I hoped that the cubefree-case would be easy, but it is not. For numbers beyond $50,000$, GAP takes some time to calculate the gnu's. – Peter Jan 02 '16 at 21:08
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    Hi, Peter! :) I hope you will find the new tag [groups-enumeration](http://math.stackexchange.com/questions/tagged/groups-enumeration) useful. – Olexandr Konovalov Jan 04 '16 at 13:35
  • Yes, thank you for the additional tag. – Peter Jan 05 '16 at 16:54

3 Answers3

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I believe the answer is 91.

This is calculated by looking at where the construction process hangs (conjugacy test of small cyclic subgroups with many regular orbits) and reducing this test for subgroups (which does not fit well with the reductions for subgroup conjugacy, but so far never had been critical in applications) by conjugacy tests of elements (i.e. for subgroups $<g>$, $<h>$, test whether $g$ is conjugate to $h^e$ for $e$ coprime to the element order) -- I'll see to put this into a future version of GAP.

ahulpke
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  • Is $91$ a sure lower bound ? And can you say anything about the upper bound (for the case the calculation is wrong) ? – Peter Jan 02 '16 at 22:16
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    Unless I have made a stupid mistake the answer is exactly 91, and if I made a mistake then 91 is an upper bound. – ahulpke Jan 02 '16 at 22:18
  • With GAP 4.7.9 and GrpConst 2.3 it took almost 4 hours to compute ConstructAllGroups(2058) and the result is 91. It may be worth to double-check with the new version of GrpConst which fixes the bug described here: http://mail.gap-system.org/pipermail/forum/2015/005127.html. It is available from http://www.icm.tu-bs.de/~beick/so.html and will be included in the next GAP release. – Olexandr Konovalov Jan 03 '16 at 11:29
  • With the improvement in the conjugacy test for cyclic subgroups this will take under 2 minutes. – ahulpke Jan 03 '16 at 15:38
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$\mathtt{ConstructAllGroups(2058)}$ completed for me in a little over two hours (8219 seconds on a 2.6GHz machine) and returned a list of $91$ groups, which confirms Alexander Hulpke's results.

Many serious computations in group theory take a long time - in some cases I have left programs running for months and got useful answers at the end! So this does not rate for me as a difficult calculation.

Derek Holt
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This is not full answer; but I would progress for construction of groups of this order as follows: if $|G|$ is $2.3.7^3$ then Sylow-$7$ is normal (and solvable); then quotient by Sylow-7 is also solvable. Thus $G$ is solvable. Then $|G|=6.7^3$ where factors are relatively prime, so $G$ has subgroup of order $6$. Thus, $G$ will look like $$G=\mbox{(group of order $7^3$)$\rtimes S_3$ or $G$= (group of order $7^3$)$\rtimes Z_6$}.$$ For non-abelian groups of order $7^3$, the automorphism group is quite complicated than that in abelian groups of order $7^3$. Certainly some other ideas have to be used for constructing non-isomorphic semi-direct products.

p Groups
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  • Why must $G$ have a subgroup of order $6$? It sounds like you are saying this because $6$ is relatively prime to $7^3$, or maybe because $2$ is relatively prime to $3$? Is this a consequence of $G$ being solvable? Because $A_5$ occurs to me as an example where $\left\lvert A_5\right\rvert=15\cdot2^2$, yet there is no subgroup of order $15$. Of course $A_5$ is not solvable. – 2'5 9'2 Jan 03 '16 at 05:40
  • If G is solvable and has order $m.n$ where $(m,n)=1$ then (P. Hall)- $G$ has a subgroup of order $m$ and $n$. – p Groups Jan 03 '16 at 05:58
  • [This paper](https://people.kth.se/~boij/kandexjobbVT11/Material/pgroups.pdf) describes all of the automorphism groups of the five groups of order $p^3$. – 2'5 9'2 Jan 03 '16 at 07:06