Abelian Group

**Jimmy_W** · March 29th 2009, 05:11 AM

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

**PaulRS** · March 29th 2009, 05:38 AM

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

If it is abelian then we can write: $ 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 $ aba^{ - 1} b^{ - 1} = e $ for all $a,b\in G$ then $ G $ is abelian

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

And: $ 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: $ ab = ba;\forall a,b \in G $ and so it follows that it is commutative (Abelian)

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