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

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

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

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

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

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

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

    Can you take it from here ?
    Last edited by flyingsquirrel; Jun 12th 2009 at 02:36 AM.
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