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Partition of Symmetry Group.

  1. Apr 8, 2004 #1
    I reading a great book called Symmetry by Roy McWeeny. For those that love Dover Books this one's a gem.

    Anyway, I have a question.

    How do you partiton a particular group into distinct classes?


    The author was discussing the symmetry group C3v the rotation, and reflection of a triangle.

    The author was able to partition this group by using its multiplication table. The classes are {E}, {C3,-C3} and {r1,r2,r3}.

    Where E is identity, C3,-C3 are positive and negative rotations, r1,r2,r3 are the reflections.

    How did he get this answer?
    I hope my ? is clearly stated.
     
  2. jcsd
  3. Apr 9, 2004 #2

    matt grime

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    That would depend upon which classes you mean. I presume you mean conjugacy classes. In general you need to take some element x and work out what all the product zxz^-1 ,which we presume the author did by looking at the multiplication table, are, that set of products is then the conjugacy class.

    Here it's simple to do by hand. For larger cases one must be cleverer, but in the case of the symmetric group the conjugacy class is uniquely determined by the cycle type. C3v, also called D_3 or D_6 is the same as S_3 the permutation group on three letters. The letters here being the corners of the triangle. A reflection swaps two of them and fixes one, so it can be written as (12)(3) for some labelling of the corners. It has cycle type 2.1, or just 2. A rotation permutes three of them and can be written as (123) say which has cycle type 3. Read the bit on symmetric/permutation groups to see that in more detail.
     
  4. Apr 12, 2004 #3
    Thanks for the help. I cut out a triangle, labeled the points and worked the multiplication table manually. Learned a lot.

    Now onward to vector spaces and lattices with the rest stop at representaion theory. :biggrin:
     
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