Two questions about cycles (algebra)

  • Thread starter Thread starter Artusartos
  • Start date Start date
  • Tags Tags
    Algebra Cycles
Artusartos
Messages
236
Reaction score
0
I have two questions:

1) For the example on the second page, I don't understand why they say \alpha\gamma\alpha^{-1} = (\alpha1 \alpha3)(\alpha2 \alpha4 \alpha7)(\alpha5)(\alpha6) instead of \alpha\gamma\alpha^{-1} = (\alpha1\alpha^{-1} \alpha3\alpha^{-1})(\alpha2\alpha^{-1} \alpha4\alpha^{-1} \alpha7\alpha^{-1})(\alpha5\alpha^{-1})(\alpha6\alpha^{-1}).

2) For the tables at the top of the 2nd page, I don't know how they computed those numbers...

Thanks in advance
 

Attachments

  • 20121225_114029.jpg
    20121225_114029.jpg
    13.8 KB · Views: 405
  • 20121225_114339.jpg
    20121225_114339.jpg
    32.2 KB · Views: 446
Last edited:
Physics news on Phys.org
Artusartos said:
I have two questions:

1) For the example on the second page, I don't understand why they say \alpha\gamma\alpha^{-1} = (\alpha1 \alpha3)(\alpha2 \alpha4 \alpha7)(\alpha5)(\alpha6) instead of \alpha\gamma\alpha^{-1} = (\alpha1\alpha^{-1} \alpha3\alpha^{-1})(\alpha2\alpha^{-1} \alpha4\alpha^{-1} \alpha7\alpha^{-1})(\alpha5\alpha^{-1})(\alpha6\alpha^{-1}).

They say what is consistent with what the theorem says. The theorem says to "apply \alpha" to the symbols in the cycles.

If \alpha,\ p,\ q are cycles, It is true that \alpha (\ p \ q) \ \alpha^{-1} =( \alpha \ p \ \alpha^{-1})(\alpha \ q \ \alpha^{-1}) but this is not the content of the theorem. A cycle is not the same as the product of the individual symbols in the cycle. The cycle (1,2,3) is not equal to (1)(2)(3).

2) For the tables at the top of the 2nd page, I don't know how they computed those numbers...

For example, In the permutation group S_4, there are 8 different elements of the group that are cycles of length 3. The example (1,2,3) in the table illustrates one of them.
(There are 24 = (4)(3)(2) different permutations that can be formed by taking 3 distinct numbers from the set of numbers {1,2,3,4}. However, each permutation such as (1,2,3) is one of 3 representations of the same cycle. (1,2,3) = (2,3,1) = (3,1,2) So there are 8 = 24/3 distinct cycles of length 3 )
 
In a nutshell, notice that alpha gamma alpha inverse takes alpha of 1 to alpha of 3. ;)
 
Stephen Tashi said:
They say what is consistent with what the theorem says. The theorem says to "apply \alpha" to the symbols in the cycles.

If \alpha,\ p,\ q are cycles, It is true that \alpha (\ p \ q) \ \alpha^{-1} =( \alpha \ p \ \alpha^{-1})(\alpha \ q \ \alpha^{-1}) but this is not the content of the theorem. A cycle is not the same as the product of the individual symbols in the cycle. The cycle (1,2,3) is not equal to (1)(2)(3).



For example, In the permutation group S_4, there are 8 different elements of the group that are cycles of length 3. The example (1,2,3) in the table illustrates one of them.
(There are 24 = (4)(3)(2) different permutations that can be formed by taking 3 distinct numbers from the set of numbers {1,2,3,4}. However, each permutation such as (1,2,3) is one of 3 representations of the same cycle. (1,2,3) = (2,3,1) = (3,1,2) So there are 8 = 24/3 distinct cycles of length 3 )

Thank you
 

Similar threads

Back
Top