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Thread: Permutation Group

  1. #1
    Sep 2009

    Permutation Group

    Let $\displaystyle A=\{1.2.3\}$ and $\displaystyle (S_A,\circ)$ be the symmetric group. Then $\displaystyle (S_A,\circ)$ has 3!=6 elements.

    Let $\displaystyle S_3=\{\alpha_1,\alpha_2,\alpha_3,\alpha_4,\alpha_5 ,\alpha_6\}$ and

    \alpha_1=\left(\begin{array}{ccc}1&2&3\\1&2&3\end{ array}\right)\:\\\alpha_2=\left(\begin{array}{ccc} 1&2&3\\1&3&2\end{array}\right)\;\alpha_3=\left(\be gin{array}{ccc}1&2&3\\3&2&1\end{array}\right)

    \alpha_4=\left(\begin{array}{ccc}1&2&3\\2&1&3\end{ array}\right)\:\alpha_5=\left(\begin{array}{ccc}1& 2&3\\2&3&1\end{array}\right)\;\alpha_6=\left(\begi n{array}{ccc}1&2&3\\3&1&2\end{array}\right)

    The following are the inverses:
    $\displaystyle \alpha_1^{-1}=\alpha_1,$
    $\displaystyle \alpha_2^{-1}=\alpha_2, $
    $\displaystyle \alpha_3^{-1}=\alpha_3, $
    $\displaystyle \alpha_4^{-1}=\alpha_4, $
    $\displaystyle \alpha_5^{-1}=\alpha_6,$ and $\displaystyle \alpha_6^{-1}=\alpha_5$

    I have noticed that $\displaystyle \{\alpha_1,\alpha_2,\alpha_3,\alpha_4,\alpha_5,\al pha_6\}=\{\alpha_1^{-1} ,\alpha_2^{-1} ,\alpha_3^{-1},\alpha_4^{-1},\alpha_5^{-1},\alpha_6^{-1}\}$, although not in the same order.

    Question 1: Is it always true that $\displaystyle \{\alpha_1,\alpha_2,...,\alpha_n\}=\{\alpha_1^{-1} ,\alpha_2^{-1} ,...,\alpha_n^{-1}\}$?

    Question 2: While $\displaystyle \{\alpha_1,\alpha_2,...,\alpha_n\} $ is labeled as S_n, how do we label $\displaystyle \{\alpha_1^{-1} ,\alpha_2^{-1} ,...,\alpha_n^{-1}\}$?

    Can we write them like this?

    $ and

    $\displaystyle S_{n^{-1}}=\{\alpha_1^{-1} ,\alpha_2^{-1} ,...,\alpha_n^{-1}\}$

    or can they be written like this?

    $\displaystyle S_{\alpha}=\{\alpha_1,\alpha_2,...,\alpha_n\}
    $ and

    $\displaystyle S_{\alpha^{-1}}=\{\alpha_1^{-1} ,\alpha_2^{-1} ,...,\alpha_n^{-1}\}$
    Last edited by novice; Aug 16th 2010 at 12:21 PM.
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  2. #2
    Senior Member
    Jul 2010
    Answer 1. For every finite group $\displaystyle G$, every element has its own inverse, so yes, that equality is true. I think this should also hold for infinite groups.

    Answer 2. The $\displaystyle n$ in $\displaystyle S_n$ signifies the number of elements that we are permuting, so $\displaystyle S_n$ has $\displaystyle n!$ elements, so all these notations are incorrect. In fact, $\displaystyle S_n = \{a_1,a_2,\ldots,a_{n!}\} = \{a_1^{-1},a_2^{-1},\ldots,a_{n!}^{-1}\}$.
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