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Math Help - Abelian Group

  1. #1
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    Abelian Group

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

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  2. #2
    Super Member PaulRS's Avatar
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    Direct: If <br />
G<br />
is abelian then <br />
aba^{ - 1} b^{ - 1}  = e<br />
for all a,b\in G

    If it is abelian then we can write: <br />
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<br />
since commutativity holds

    Reciprocal: If <br />
 aba^{ - 1} b^{ - 1}  = e<br />
 for all a,b\in G then <br />
 G<br />
 is abelian

    <br />
aba^{ - 1} b^{ - 1}  = e \Leftrightarrow aba^{ - 1} b^{ - 1} ba = eba = ba<br />

    And: <br />
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<br />
thus: <br />
ab = ba;\forall a,b \in G<br />
and so it follows that it is commutative (Abelian)
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