Talk:(2,3,7) triangle group
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[edit] Typo
Hello, It appears that x=2*cos(2Pi/7), not eta=cos(2Pi/7), satisfies (2-x)^3=7(x-1)^2. ---- Kai-Cheong Chan
- Certainly, that was an error. Katzmik —Preceding comment was added at 08:58, 21 October 2007 (UTC)
[edit] Perfect?
Clearly the (2,3,7) triangle group is not simple because it has non-trivial finite quotient groups. Is it a perfect group? Scott Tillinghast, Houston TX (talk) 23:30, 15 March 2008 (UTC)
- Yes. In an abelian quotient, the order of a divides 2, the order of b divides 3, the order of ab should divide 6, but the relations say it also divides 7, so it divides 1, and ab=1, but then the order of a divides both 2 and 3, so its order is 1, etc. A group without nontrivial abelian invariants is perfect. JackSchmidt (talk) 02:31, 16 March 2008 (UTC)
Thank you. Scott Tillinghast, Houston TX (talk) 22:35, 16 March 2008 (UTC)
[edit] Finite?
Is this a finite group? Can we have the quaternion representation explicitly, which would answer the first question for some readers? Septentrionalis PMAnderson 16:06, 31 May 2008 (UTC)
- No it is not a finite group. Some infinite quotients are described at Talk:Hurwitz's automorphisms theorem#Commutator order. JackSchmidt (talk) 01:21, 1 June 2008 (UTC)
- I think quaternion representation given is fairly explicit, but it is a projective representation, not a true representation. The article seems to claim there is a true representation as well. If it is related to the one given, then it should be fairly easy to find, as the representation should split over the center. Note also that the representation is in *a* quaternion algebra, not *the* quaternions. JackSchmidt (talk) 01:27, 1 June 2008 (UTC)
- I would consider a fairly explicit representation would be one which would permit writing down elements of the group, at least the generators. Septentrionalis PMAnderson 16:52, 2 June 2008 (UTC)
- The generators are given explicitly? The generators are g2 and g3, and they are ij/n and (1 + (n^2-2)j+(3-n^2)ij)/2. These are elements of the quaternion algebra over Q[n] with i^2=j^2=n, ij=-ji, where n is z + 1/z where z is a primitive 7th root of unity. This defines a projective, (irreducible I think, assuming this infinite group is behaving like a finite group) representation of <a,b:a^2=b^3=(ab)^7=1> of dimension 4 over the complexes (or over Q[z] or Q[n]). JackSchmidt (talk) 17:18, 2 June 2008 (UTC)
- Ah, down there. It might be a good idea to move the existence statement The (2, 3, 7) triangle group admits a presentation... to the beginning of that section. Septentrionalis PMAnderson 17:43, 2 June 2008 (UTC)
- The generators are given explicitly? The generators are g2 and g3, and they are ij/n and (1 + (n^2-2)j+(3-n^2)ij)/2. These are elements of the quaternion algebra over Q[n] with i^2=j^2=n, ij=-ji, where n is z + 1/z where z is a primitive 7th root of unity. This defines a projective, (irreducible I think, assuming this infinite group is behaving like a finite group) representation of <a,b:a^2=b^3=(ab)^7=1> of dimension 4 over the complexes (or over Q[z] or Q[n]). JackSchmidt (talk) 17:18, 2 June 2008 (UTC)
- I would consider a fairly explicit representation would be one which would permit writing down elements of the group, at least the generators. Septentrionalis PMAnderson 16:52, 2 June 2008 (UTC)
