0
$\begingroup$

I want to determine the number of groups of order $512$ with exponent $2,4,8,16,32,64,128,256$ and $512$

The first $500,000$ groups in GAP give the following result :

500000  4:90031  8:366214  16:41085  32:2425  64:212  128:27  256:5  512:1

But the calculation took several hours. It is a long way to go through the $10,494,213$ groups.

Is there a method to determine the numbers faster, or does someone know an internet link, where the numbers are given ?

$\endgroup$
6
  • 3
    $\begingroup$ OK, so the computation will take a few days. What's the problem - just do it! $\endgroup$
    – Derek Holt
    Apr 19, 2016 at 14:23
  • $\begingroup$ I do not understand the downvote. The question is clear and as Alexander Konovalov shows, it can be answered. Additionally, I showed what I did. $\endgroup$
    – Peter
    Apr 21, 2016 at 15:29
  • $\begingroup$ I was not the downvoter, and I don't agree with the downvote. But you seem to be implying that a few hours is a long time. I soemtimes run computations of this type for months at a time. $\endgroup$
    – Derek Holt
    Apr 22, 2016 at 16:05
  • $\begingroup$ I did not want to invest several days for such data ... $\endgroup$
    – Peter
    Apr 22, 2016 at 19:49
  • $\begingroup$ "Do not want to invest" or "do not have technical opportunities"? I'm afraid that the reader would not find the former formulation very motivating, if the question asks the reader to invest their (personal and CPU) time to answer this. $\endgroup$
    – Olexandr Konovalov
    May 2, 2016 at 12:46

1 Answer 1

Reset to default
3
$\begingroup$

I have these data, computed for the same project that I've mentioned in another question of yours:

gap> Length(exponent512);
10494213
gap> Collected(exponent512{[1..500000]});
[ [ 4, 90031 ], [ 8, 366214 ], [ 16, 41085 ], [ 32, 2425 ], [ 64, 212 ], 
  [ 128, 27 ], [ 256, 5 ], [ 512, 1 ] ]

so exactly as in your question. Now the numbers for the full list:

gap> Collected(exponent512);
[ [ 2, 1 ], [ 4, 8791062 ], [ 8, 1656695 ], [ 16, 43767 ], [ 32, 2443 ], 
  [ 64, 212 ], [ 128, 27 ], [ 256, 5 ], [ 512, 1 ] ]
$\endgroup$
3
  • 1
    $\begingroup$ I also have some more - e.g. the number of conjugacy classes, the order of the derived subgroup, etc. Should really make them available somewhere. $\endgroup$
    – Olexandr Konovalov
    Apr 19, 2016 at 14:36
  • $\begingroup$ How did you determine the data ? Did you precalculate it ? $\endgroup$
    – Peter
    Apr 19, 2016 at 14:38
  • 1
    $\begingroup$ Yes, somewhere in 2007-2008, using distributed parallel calculations with GAP. $\endgroup$
    – Olexandr Konovalov
    Apr 19, 2016 at 14:41

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .