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Thread: Abelian Group

  1. #1
    Junior Member
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    Abelian Group

    Show that $\displaystyle \mathbf{G} \ $ is Abelian iff $\displaystyle aba^{-1} b^{-1} = 1 $ for every $\displaystyle a,b \ \in \ \ $$\displaystyle \ \mathbf{G} $

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  2. #2
    Super Member PaulRS's Avatar
    Joined
    Oct 2007
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    Direct: If $\displaystyle
    G
    $ is abelian then $\displaystyle
    aba^{ - 1} b^{ - 1} = e
    $ for all $\displaystyle a,b\in G$

    If it is abelian then we can write: $\displaystyle
    aba^{ - 1} b^{ - 1} = a\left( {ba^{ - 1} } \right)b^{ - 1} = a\left( {a^{ - 1} b} \right)b^{ - 1} = \left( {aa^{ - 1} } \right)\left( {bb^{ - 1} } \right) = e
    $ since commutativity holds

    Reciprocal: If $\displaystyle
    aba^{ - 1} b^{ - 1} = e
    $ for all $\displaystyle a,b\in G$ then $\displaystyle
    G
    $ is abelian

    $\displaystyle
    aba^{ - 1} b^{ - 1} = e \Leftrightarrow aba^{ - 1} b^{ - 1} ba = eba = ba
    $

    And: $\displaystyle
    aba^{ - 1} b^{ - 1} ba = aba^{ - 1} \underbrace {\left( {b^{ - 1} b} \right)}_ea = aba^{ - 1} a = ab\underbrace {\left( {a^{ - 1} a} \right)}_e = ab
    $ thus: $\displaystyle
    ab = ba;\forall a,b \in G
    $ and so it follows that it is commutative (Abelian)
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