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Two questions about cycles (algebra)

  1. Dec 25, 2012 #1
    I have two questions:

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

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

    Thanks in advance

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    Last edited: Dec 25, 2012
  2. jcsd
  3. Dec 25, 2012 #2

    Stephen Tashi

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    Science Advisor

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

    If [itex] \alpha,\ p,\ q [/itex] are cycles, It is true that [itex] \alpha (\ p \ q) \ \alpha^{-1} =( \alpha \ p \ \alpha^{-1})(\alpha \ q \ \alpha^{-1}) [/itex] 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 [itex] S_4 [/itex], 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 )
  4. Dec 26, 2012 #3
    In a nutshell, notice that alpha gamma alpha inverse takes alpha of 1 to alpha of 3. ;)
  5. Dec 26, 2012 #4
    Thank you
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