# Abelian Group

• Mar 29th 2009, 05:11 AM
Jimmy_W
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}$

• Mar 29th 2009, 05:38 AM
PaulRS
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)