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Math Help - Group - How to transform

  1. #1
    Junior Member
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    Sep 2008
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    Freiburg i. Brgs. Germany
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    Question Group - How to transform

    Hello

    I have problems forming a term.

    The exercise is:

     \text{Let G be a group and } x,y,z,u \in G

    \mathrm{Z\kern-.3em\raise-0.5ex\hbox{Z}}: ~ \left(x \left( \left( \left( y^{-1} \left( x^{-1} \cdot z \right) \right) ^{-1} \cdot u \right) \cdot \left( y \cdot u \right)^{-1} \right) ^{-1} \right) = z

    I kwon that I have to show that:

    i.) Associative Law

    ii.) Identity Element

    iii.) Inverse Element

    If I look the term it is clear that I have to form it such that I only have z=z and the other elements x,y,u should be transformed into the identity because of their inverse elements.

    I don't know how to form this concretely

    Thanks for help
    Greetings
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  2. #2
    Senior Member
    Joined
    Nov 2008
    From
    Paris
    Posts
    354
    That's how you use your law's properties:

    \left(x \left(\left( \left( y^{-1} \left( x^{-1} \cdot z \right) \right) ^{-1} \cdot u \right) \cdot \left( y \cdot u \right)^{-1} \right) ^{-1} \right)
    = \left(x \left(\left( \left( y^{-1} \cdot x^{-1} \cdot z \right) ^{-1} \cdot u \right) \cdot \left( y \cdot u \right)^{-1} \right) ^{-1} \right) associativity
    = \left(x \left(\left( \left( z^{-1} \cdot x \cdot y \right) \cdot u \right) \cdot \left( y \cdot u \right)^{-1} \right) ^{-1} \right) inverse formula
    = \left(x \left(\left( z^{-1} \cdot x \cdot y \cdot u \right) \cdot \left( y \cdot u \right)^{-1} \right) ^{-1} \right) associativity
    = \left(x \left(\left( z^{-1} \cdot x \cdot y \cdot u \right) \cdot \left( u^{-1} \cdot y^{-1} \right) \right) ^{-1} \right) inverse formula
    = \left(x \left( z^{-1} \cdot x \cdot y \cdot u \cdot u^{-1} \cdot y^{-1} \right) ^{-1} \right) associativity
    = \left(x \left( z^{-1} \cdot x \cdot y \cdot y^{-1} \right) ^{-1} \right) associativity and inverse
    = \left(x \left( z^{-1} \cdot x \right) ^{-1} \right) associativity and inverse
    = \left(x \left( x^{-1} \cdot z \right) \right) inverse formula
    = \left(x \cdot x^{-1} \cdot z \right) associativity
    = z associativity and inverse

    where "inverse formula" is (a_{1}...a_{n})^{-1}=a_{n}^{-1}...a_{1}^{-1}
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  3. #3
    Junior Member
    Joined
    Sep 2008
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    Freiburg i. Brgs. Germany
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    Thank you very much

    That helps really a lot. Now I know how to show it in the future.
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