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Proving Basic Exponent Properties for a Group

  1. May 3, 2013 #1
    When proving that [itex]x^m x^n = x^{m+n} [/itex] and that [itex] (x^m)^n = x^{mn} [/itex] for all elements [itex]x[/itex] in a group, it's easy enough to show that they hold for all [itex]m \in \mathbb{Z} [/itex] and for all [itex] n \in \mathbb{N} [/itex] using induction on [itex]n[/itex]. The case [itex] n = 0 [/itex] is also very easy. But how does one prove this for [itex] n \in \mathbb{Z}^{-} [/itex]?

    I tried to do it by using the fact that [itex] n = - \nu [/itex] for some [itex] \nu \in \mathbb{N} [/itex], but this didn't get me anywhere. Do you have to do induction on the negative integers separately? I'm sure there's a simple answer to this question that I'm just not seeing.

    Note that I'm working with the standard recursive definition of exponents, and the definition [itex] x^{-n} = (x^{-1})^n [/itex].
     
    Last edited: May 3, 2013
  2. jcsd
  3. May 3, 2013 #2
    Update: I think I figured it out. I'd still appreciate a simpler way, if anyone has one.

    For a negative integer [itex] n [/itex], set [itex] n = - \nu [/itex] as before.
    Then
    [tex]
    \begin{align}
    x^m x^n
    &= x^m x^{- \nu} \\
    &= x^m (x^{-1})^{\nu} \\
    &= ((x^{-1})^{-1})^m (x^{-1})^{\nu} \tag*{since $x = (x^{-1})^{-1}$} \\
    &= (x^{-1})^{-m} (x^{-1})^{\nu} \tag*{by the definition of negative exponents} \\
    &= (x^{-1})^{-m+\nu} \tag*{by the rule for positive exponents}\\
    &= (x^{-1})^{-(m+ (-\nu))}\\
    &= (x^{-1})^{-(m+n)} \\
    &= ((x^{-1})^{-1})^{m+n} \\
    &= x^{m+n} \tag*{since $x = (x^{-1})^{-1}$}
    \end{align}
    [/tex]
     
    Last edited: May 3, 2013
  4. May 3, 2013 #3

    Office_Shredder

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    If you liked induction for positive n you can do induction for negative n also. Show if it holds for n, then it holds for n-1 as well
     
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