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Group inverses

  1. Feb 1, 2009 #1
    1. The problem statement, all variables and given/known data

    x is in the arbitrary multiplicative group, and a,b are positive integers.
    given that
    [tex] x^{a+b} = x^ax^b[/tex] and [tex] (x^a)^b =x ^{ab} [/tex]
    [tex] (x^a)^-1 = x^{-a} [/tex]

    2. Relevant equations

    3. The attempt at a solution

    I) (x)^{-1} = x ^{-1}
    II) Assume [tex] (x^n)^{-1} = x^{-n} [/tex], to prove that [tex] (x^{n+1})^{-1} = x^{-(n+1)} [/tex].

    [tex] (x^{n+1})^{-1}) = (x^nx)^{-1} = x^{-1}x^{-n} = x^{-n-1} [/tex]

    Is the last step justified?
  2. jcsd
  3. Feb 1, 2009 #2


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    Staff Emeritus
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    Gold Member

    Seeing how you didn't provide a justification, no. :wink:

    More seriously, if you cannot see a rigorous reason why that last step should be true, then you definitely haven't written a valid proof.
  4. Feb 1, 2009 #3
    Actually, this is what gets me!

    The text explicitly states the following
    "[tex] x^{-1}x^{-1}x^{-1} \ldots x^{-1} \text{ n terms }[/tex]"

    It should follow from this that
    [tex] x^{n} = x^{-1}x^{-(n-1)} [/tex]

    I can't see a rigorous road from this description, per se.
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