20100720, 16:18  #1 
Noodles
"Mr. Tuch"
Dec 2007
Chennai, India
10011101001_{2} Posts 
Number of groups for given order
For the past two weeks, I have been researching about all groups of small order. Different types of groups that are not isomorphic for any given order. I collected information from all over internet and built all groups upto order 30, which I will post within a few days. Number of groups for some given order can be found out at GAP's database.
For groups of order that are product of two distinct primes p×q, I noticed that number of groups is 1 (cyclic only) if p does not divide q1. 2 if p divides q1. Why this is the case? Why is this condition that pq1? What will be that structure of this noncyclic (nonabelian) group will be like? For p = 2, then it is the obvious dihedral group... As a result, 15 and 33 have only that cyclic group for each, while numbers such as 21 and 39 have two groups. The structure of dihedral group is D_{2N} = <a, b  a^{N} = b^{2} = e, ba = a^{1}b> For N = 21, the second group structure is given to be <a, b  a^{3} = b^{7} = e, ba = ab^{2}> Such type of groups does not exist at all, or is rather isomorphic to that cyclic group itself, when p does not divide q1? For N = 21, this works out because ba^{3} = a^{3}b^{8} = b. So, for my sequence: 21, 155, 889, 106483, 2228207, 9961453, 66571993057... the nonabelian group is similar to that for N = 21 155: <a, b  a^{5} = b^{31} = e, ba = ab^{2}> 889: <a, b  a^{7} = b^{127} = e, ba = ab^{2}> For N = 22517, there exists 13 different structures of groups! Though for N = 2001, there is only that cyclic group, though 3 divides 666, though 3 does not divide 22 or 28. Why? 1001, that is the product of three primes also has only one group, but here 7 does not divide either 10 or 12, 11 does not divide 12 as well, that's why it's so the case over here... According to the assertion, 203, 253, 301, 497, 689, 737, 791, 979, 1027 all have two groups each. What will be the second group structure will be like for (say) 689, or 1027  the recently factored Mersenne number, but it is too elusive to discern over here. I tried out for 39, with <a, b  a^{3} = b^{13} = e, ba = ab^{2}>, but over here ba^{3} = a^{3}b^{8} = b^{8} ≠ b. Then, what will be the nontrivial group structure be like? I tried with <a, b  a^{3} = b^{13} = e, ba = ab^{3}> It seems alright that ba^{3} = a^{3}b^{27} = b. Ok, then if pq1, will there always exist a value of r < q, such that r^{p} ≡ 1 (mod q)? For every such case, will all such groups be isomorphic to each other? For N = 33, where p = 3, q = 11. For numbers such as these, for values of N like this case, any values for r does not exist at all, in order to satisfy with that above equation, or that such groups so formed are being isomorphic to that cyclic group itself? 
20100720, 18:05  #2 
Sep 2002
Vienna, Austria
3×73 Posts 
The Sylow qsubgroup have to be normal in this case, and thus we can prove that the group must be a semidirect product.

20100904, 20:32  #4  
Noodles
"Mr. Tuch"
Dec 2007
Chennai, India
2351_{8} Posts 
Alas, that I have created that list (Cayley tables rather) for all that finite groups upto order of 30 elements. It can be viewed over here.
Quote:
For N = 39, the second group is this? <a, b  a^{3} = b^{13} = e, ba = ab^{3}> There may be other powers for b that satisfy this similar condition, that r^{p} ≡ 1 (mod q) For example, for where N = 689, p = 13, q = 53. 10^{13} ≡ 1 (mod 53). Thus, <a, b  a^{13} = b^{53} = e, ba = ab^{10}> is a valid group for that nonabelian group for that order of 689? Also, that similarly 16^{13} ≡ 1 (mod 53). So, is <a, b  a^{13} = b^{53} = e, ba = ab^{16}> again a valid group? Is that isomorphic to that above group? 24 is not an inverse for that element of 10 (mod 53), with the property that 24^{13} ≡ 1 (mod 53). Then, what about that for <a, b  a^{13} = b^{53} = e, ba = ab^{24}>? Quote:
The link that I mentioned within my first post contains that upto order 2000 exhaustively. http://wwwpublic.tubs.de:8080/~hubesche/number.html Ok. Looking up at that number of groups in depth, consider the following: Orders 1216, 1472, 2752, 3008, 4288, 4544, 5056, 5312, 6592, 6848, 8128, 8384, 8896, 9664, 10432, 704 all have 1387 groups, but rather that order 1984 has 1388 groups. Why? What is that extra group? As a result, while orders of 1408, 2944, 2432, 5504, 7552, 8576, 9088, 10112, 10624, 6016 have exactly 19324 groups only, 3968 has 19326 groups. 896 has 19349. Similarly, multiplying by 2 again, 2816, 4864, 5888, 11008 each have 1083472 groups in common, 7936 has 1083477, whereas that order of 1792 has 1083553 groups. On the other hand, in some cases I have noticed that a special case has got some number of fewer groups lesser than from a general pattern, as well. For example, 11072, 10048, 6976, 2368, 3904, 6464, 9536, 1856, 832 each has got only 1630 groups actually, but that 4672, 2624, 5696, 8768 each have 1672 groups, while 1088, 7232 have 1681; 6208 has 1683. Rather that way, what about that for 1664, 3712, 4736, 6784, 7808 each has got exactly 21507 groups; 11388, 9344, 5248 have got 21766 groups each; 21808 groups for that order of 2176 elements; while only 20169 groups for that order of 384 elements. 3328, 7424, 9472 each has got only with 1116308 groups; order of 1280 has got 1116461; that for 4342 has got up with 1118911. Finally, that 480, 1632 each has got only 1213 groups; for that order of 1248 elements, has got upto exactly 1460 groups... Last fiddled with by Raman on 20100904 at 20:37 

20100905, 01:54  #5  
Aug 2006
5,987 Posts 
Quote:
But I posted it more for the references and links than the list itself. Last fiddled with by CRGreathouse on 20100905 at 01:55 

20100905, 04:01  #6  
Noodles
"Mr. Tuch"
Dec 2007
Chennai, India
10011101001_{2} Posts 
Quote:
Ok, that is being rather available up right at this place, over here. Are they all being correct, or any chances of mistakes? What is that best algorithm, that is being available so far in order to compute up with that number of groups for any randomly given order, for example, as hard enough as 2016, 2176, 2187 rather? within that way, rather Last fiddled with by Raman on 20100905 at 04:23 

20100905, 17:43  #7 
Aug 2006
5,987 Posts 
They're all correct.
I don't know of any good algorithms. The problem is very hard. 
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