1. Abelian Group

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

2. 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)