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Math Help - group and subgroup problem

  1. #1
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    group and subgroup problem

    let  \rho , \sigma , \tau be the element of  S_{6} given by

     \rho=(26) , \sigma = (123) , \tau = (45)

    (1). Find  \rho \sigma \tau and  \rho^{-1} \sigma^{-1} \tau^{-1}
    and verify  ( \rho \sigma \tau )^{-1}= \rho^{-1} \sigma^{-1} \tau^{-1}
    (2). find the order of  \sigma \tau and  \rho \sigma
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  2. #2
    MHF Contributor Swlabr's Avatar
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    Quote Originally Posted by ramiee2010 View Post
    let  \rho , \sigma , \tau be the element of  S_{6} given by

     \rho=(26) , \sigma = (123) , \tau = (45)

    (1). Find  \rho \sigma \tau and  \rho^{-1} \sigma^{-1} \tau^{-1}
    and verify  ( \rho \sigma \tau )^{-1}= \rho^{-1} \sigma^{-1} \tau^{-1}
    (2). find the order of  \sigma \tau and  \rho \sigma
    What is it that you do not understand about this question? It is a rather hard question to answer over the internet as it is about understanding, not about a simple `trick'. However, you should remember that what you are doing is the same as using a normal permutation. That is,

    (123) = \left(\begin{array}{cccccc} 1 & 2 & 3 & 4 & 5 & 6 \\ 2 & 3 & 1 & 4 & 5 & 6 \end{array}\right).

    To multiply to disjoint cycles, you look at where the preceeding permutation takes the element you are looking at.

    So, for example,

    2(253) = 5, 5(253)=3, 3(253)=2 and all the rest are kept as they are, and so 1(253)(15) = 1(15)=5, 2(253)(15) = 5(15) = 1 etcetera.

    Does that make sense?

    Now, to find the inverse of a (single) disjoint cycle, you should notice that if a_i \rightarrow a_{i+1} then you want a_{i+1}\rightarrow a_i in the inverse. This is equivalent to keeping the first element fixed and flipping all the others round,

    (a_1 a_2 a_3 a_4 \ldots a_n)^{-1} = (a_1 a_n a_{n-1} \ldots a_3 a_2).

    Now, does that make sense?

    It is now your task to apply all this to your question!
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