How many groups of order $2058$ are there?

Question

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 $2058$ ?

Answer 1

$\mathtt{ConstructAllGroups(2058)}$ 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.

Answer 2

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.

Answer 3

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.