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Proof involving group of permutations of {1,2,3,4}.

  1. Feb 10, 2016 #1
    1. The problem statement, all variables and given/known data
    Let ##\sigma_4## denote the group of permutations of ##\{1,2,3,4\}## and consider the following elements in ##\sigma_4##:

    ##x=\bigg(\begin{matrix}1&&2&&3&&4\\2&&1&&4&&3\end{matrix}\bigg);~~~~~~~~~y=\bigg(\begin{matrix}1&&2&&3&&4\\3&&4&&1&&2\end{matrix}\bigg)##
    ##\sigma=\bigg(\begin{matrix}1&&2&&3&&4\\2&&3&&1&&4\end{matrix}\bigg);~~~~~~~~~\tau=\bigg(\begin{matrix}1&&2&&3&&4\\2&&1&&3&&4\end{matrix}\bigg)##

    and put ##K=\{1,x,y,xy\},~~~~~~~ Q=\{1,\sigma,\sigma^2,\tau,\sigma \tau,\sigma^2\tau\}##

    Show that if ##q\in Q## and ##k\in K## then ##qkq^{-1}\in K##.

    2. Relevant equations

    I have shown ##K## and ##Q## are subgroups of ##\sigma_4## that ##\sigma_4=KQ=\{kq~~;~~k\in K,~q\in Q\}##.

    And I have found the following relations: ##x^2=1,y^2=1,yx=xy;~~\sigma^3=1,\tau^2=1,\tau\sigma=\sigma^2\tau##


    3. The attempt at a solution

    I would go through and calculate ##qkq^{-1}## for each ##k## and each ##q## but I know there must be a shorter way. Do you know of it?
     
  2. jcsd
  3. Feb 10, 2016 #2

    Orodruin

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    Just check the relation for the (sub)group generators. You do not need to check every element.
     
  4. Feb 10, 2016 #3
    Ok thanks. So I only need to calculate ##\sigma x (\sigma)^{-1},~\tau x (\tau)^{-1}, \sigma y (\sigma)^{-1} \text{ and } \tau y (\tau)^{-1}##?
     
  5. Feb 10, 2016 #4

    Orodruin

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    Right. You should however think about and convince yourself why this is sufficient as well.
     
  6. Feb 10, 2016 #5

    Orodruin

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    Also a personal preference: I would suggest you learn and start using the more short-hand notation for the symmetric groups in terms of cycles. In this notation, your elements would be
    $$
    x = (12)(34), \ y = (13)(24), \ \sigma = (123), \ \tau = (12).
    $$
    It really simplifies performing the group multiplications as well.
     
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