Difference between revisions of "ApCoCoA-1:GroupsToCheck"
From ApCoCoAWiki
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read the word that describes the property #generators = #relations ("Deficing zero?")? Thanks in advance! :-) | read the word that describes the property #generators = #relations ("Deficing zero?")? Thanks in advance! :-) | ||
Comment: I changed a little bit in the function. I also have problem to read the words in that page. :-( | Comment: I changed a little bit in the function. I also have problem to read the words in that page. :-( | ||
| − | + | P.S.: groups 1) to 7) are generalized triangle groups. | |
Other group 2/3 | Other group 2/3 | ||
Checked: Done | Checked: Done | ||
Notes: I didn't found the original paper of Prof. Rosenberger so I referred to another paper. It seems that the Groebner basis | Notes: I didn't found the original paper of Prof. Rosenberger so I referred to another paper. It seems that the Groebner basis | ||
is infinite or at least not feasible to determine. | is infinite or at least not feasible to determine. | ||
| + | Comment: Check literature about generalized triangle groups. | ||
Other group 4 | Other group 4 | ||
Checked: Done | Checked: Done | ||
Notes: -- | Notes: -- | ||
Other group 11 | Other group 11 | ||
| − | Checked: | + | Checked: Done |
Notes: In the implementation we need 2 generators (+ 1 additional invers) because one invers is given implicit with t^n = 1. | Notes: In the implementation we need 2 generators (+ 1 additional invers) because one invers is given implicit with t^n = 1. | ||
Perhaps I'm wrong, but I think that group 11 is isomorphic to group 12 for r=1,n=2 and a=b=1, am I right? Thank you very much! | Perhaps I'm wrong, but I think that group 11 is isomorphic to group 12 for r=1,n=2 and a=b=1, am I right? Thank you very much! | ||
| + | Comment: Acutally, you are right. We need three generators x,z and t, where x and z are inverse to each other. | ||
Other group 12 | Other group 12 | ||
| − | Checked: | + | Checked: Done |
Notes: -- | Notes: -- | ||
Other group 13 | Other group 13 | ||
| Line 110: | Line 112: | ||
Notes: Are there any restrictions for the parameters a,b,c and d? I assumed greater/equal one, but I don't know. Thank you very | Notes: Are there any restrictions for the parameters a,b,c and d? I assumed greater/equal one, but I don't know. Thank you very | ||
much. | much. | ||
| + | Comment: Your assumption on the parameters are proper. | ||
Revision as of 02:43, 24 September 2013
Inserted Groups
Baumslag-Gersten Group
Checked: Done Notes: --
Braid Group
Checked: Done Notes: --
Cyclic Group
Checked: Done Notes: --
Dicyclic Group
Checked: Done Notes: I added two different implementations, one with explicit invers elements and one without. I think the second one is the right one. The computation of the first implementation results in a GB with size 2812, the second one with size 901. Comment: The implementation in the page is correct.
Dihedral Group
Checked: Done
Notes: It follows, that a^{-1} = a^{2n-1} and that b^{4} = 1 (second equation) --> b^{-1} = b^{3}
My question is, do I have to implement the last equation with b^{3} instead of b^{-1} or should
I use 4 generators (a invers to c, b invers to d)?
Comment: The implementation in the page is already enough for this group. For your question, I would like to
suggest that we should try to add as few extra relations as possible.
von Dyck Group
Checked: Done Notes: A useful reference is still missing
Free abelian Group
Checked: Done Notes: --
Free Group
Checked: Done Notes: --
Fibonacci Group
Checked: Done Notes: --
Heisenberg Group
Checked: Done Notes: The matrix in the description will be added as a picture, then it will look much better. At the moment we cannot upload pictures to the server, but I contacted Stefan, there will be a solution soon.
Higman Group
Checked: Done Notes: --
Ordinary Tetrahedron Groups
Checked: Done
Notes: I used the implicit inverse elements: We know that x^{e_1} = 1, it follows that x^{e_1 - 1} is the inverse, and so on..
Please check, if I'm right.
Comment: You are correct.
Lamplighter Group
Checked: Done Notes: Since I cannot implement "for all n in Z" the user has to define a maximum n (= MEMORY.N). Until this boundary the group will be created.
Tetraeder group
Checked: Done Notes: --
Oktaeder group
Checked: Done Notes: --
Ikosaeder group
Checked: Done Notes: --
Symmetric groups
Checked: Done Notes: --
Quaternion group
Checked: Done Notes: Prof. Kreuzer gave me a list of groups and on this list the representation differs a lot with the one I used. Please check if I'm right with this representation. Comment: It is right.
Tits group
Checked: Done Notes: --
Special linear group
Checked: Done Notes: --
Modular group
Checked: No Notes: I didn't find an efficient representation in the Internet, I used the one Prof. Kreuzer gave me. I only found an article about the projective linear special group PSL. Please check my results, thank you very much!
Alternating group
Checked: Done Notes: --
Hecke group
Checked: Done Notes: I referred to the preprinted paper of Prof. Dr. Kreuzer and Prof. Dr. Rosenberger, is that okay? Comment: It is ok. I will try to find other resource. Or you can ask Prof. Dr. Kreuzer or Prof. Dr. Rosenberger for help.
Other group 1
Checked: Done
Notes: I'm not sure whether I get it right that k is congruent to 3 mod 6. It is very hard to read in the copy you gave me. (In
your paper it is the number 3). I couldn't find a paper or any reference since I didn't know the name of this group. I can't
read the word that describes the property #generators = #relations ("Deficing zero?")? Thanks in advance! :-)
Comment: I changed a little bit in the function. I also have problem to read the words in that page. :-(
P.S.: groups 1) to 7) are generalized triangle groups.
Other group 2/3
Checked: Done Notes: I didn't found the original paper of Prof. Rosenberger so I referred to another paper. It seems that the Groebner basis is infinite or at least not feasible to determine. Comment: Check literature about generalized triangle groups.
Other group 4
Checked: Done Notes: --
Other group 11
Checked: Done Notes: In the implementation we need 2 generators (+ 1 additional invers) because one invers is given implicit with t^n = 1. Perhaps I'm wrong, but I think that group 11 is isomorphic to group 12 for r=1,n=2 and a=b=1, am I right? Thank you very much! Comment: Acutally, you are right. We need three generators x,z and t, where x and z are inverse to each other.
Other group 12
Checked: Done Notes: --
Other group 13
Checked: No Notes: Are there any restrictions for the parameters a,b,c and d? I assumed greater/equal one, but I don't know. Thank you very much. Comment: Your assumption on the parameters are proper.
