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Math Help - 4th symmetric group, order

  1. #1
    Member Maccaman's Avatar
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    4th symmetric group, order

    I have the following revision question:

    Recall that  S_4 (the fourth symmetric group) consists of the set of bijections  f: \{ 1,2,3,4 \} \rightarrow \{ 1,2,3,4 \}. The group operation is composition of functions. Let
     p = \begin{pmatrix} 1&2&3&4 \\ 4&2&1&3 \end{pmatrix}

    Find the order of  p and write down all the elements of the cyclic subgroup  <p> .

    I think that to find the order of  p I need to calculate  p^{something} = 1. i.e.  p^{something} = \begin{pmatrix} 1&2&3&4 \\ 1&2&3&4 \end{pmatrix} and then the order is the  something

    Then I think that if, for example, the order was 2 then  <p> = \{ p,p^1 \}.... but Im not sure.

    Anyway my main problem is calculating  p^2, p^3, p^4, ect . Can anyone please show me how this is done? Thanks
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  2. #2
    Super Member flyingsquirrel's Avatar
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    Quote Originally Posted by Maccaman View Post
    Then I think that if, for example, the order was 2 then  <p> = \{ p,p^1 \}.... but Im not sure.
    Don't forget the identity element: \langle p \rangle =\{1,p,p^2,\ldots,p^{\text{something}-1}\} (and "something" is the least positive integer such that p^\text{something}=1).
    Quote Originally Posted by Maccaman View Post
    Anyway my main problem is calculating  p^2, p^3, p^4, ect . Can anyone please show me how this is done? Thanks
    We have \begin{pmatrix} {\color{blue}1}&{\color{green}2}&3&{\color{red}4} \\ {\color{blue}4}&{\color{green}2}&1&{\color{red}3} \end{pmatrix}. To compute p^2 we need to compute p^2(1),\ p^2(2),\ p^2(3) and p^2(4).

    • p^2(1)=p( p(1)) = p(4)=3 because {\color{blue}p(1)=4} and {\color{red}p(4)=3}.
    • p^2(2)=p( p(2)) = p(2)=2 because {\color{green}p(2)=2}.
    • p^2(3)=p( p(3)) = p(1)=4 because {p(3)=1} and {\color{blue}p(1)=4}.
    • p^2(4)=p( p(4)) = p(3)=1 because {\color{red}p(4)=4} and {p(3)=1}.

    hence
    p^2=\begin{pmatrix} 1&2&3&4 \\ 3&2&4&1 \end{pmatrix}

    Can you take it from here ?
    Last edited by flyingsquirrel; June 12th 2009 at 03:36 AM.
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