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Math Help - Permutation Group

  1. #1
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    Permutation Group

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

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

     <br />
\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)<br />

     <br />
\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)<br />

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

    I have noticed that \{\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 \{\alpha_1,\alpha_2,...,\alpha_n\}=\{\alpha_1^{-1} ,\alpha_2^{-1} ,...,\alpha_n^{-1}\}?

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

    Can we write them like this?

     <br />
S_n=\{\alpha_1,\alpha_2,...,\alpha_n\}<br />
and

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

    or can they be written like this?

    S_{\alpha}=\{\alpha_1,\alpha_2,...,\alpha_n\}<br />
and

    S_{\alpha^{-1}}=\{\alpha_1^{-1} ,\alpha_2^{-1} ,...,\alpha_n^{-1}\}
    Last edited by novice; August 16th 2010 at 01:21 PM.
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  2. #2
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    Answer 1. For every finite group 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 n in S_n signifies the number of elements that we are permuting, so S_n has n! elements, so all these notations are incorrect. In fact, S_n = \{a_1,a_2,\ldots,a_{n!}\} = \{a_1^{-1},a_2^{-1},\ldots,a_{n!}^{-1}\}.
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