Prove that:
a) In a group (ab)^-1=b^-1 * a^-1
b) A group G is Abelian iff (ab)^-1=a^-1 * b^-1
c) In a group (a^-1)^-1=a for all a
$\displaystyle b^{-1}*a^{-1}$
Multiply it to the right by (ab)
$\displaystyle b^{-1}*a^{-1}*a*b$
Since it is a group, * is associative.
So we can write :
$\displaystyle b^{-1}*a^{-1}*a*b=b^{-1}*\underbrace{(a^{-1}*a)}_{1}*b$
$\displaystyle =b^{-1}*b=1$
So we have $\displaystyle (b^{-1}*a^{-1})*(ab)=1$
Therefore...
Abelian means that * is commutative.b) A group G is Abelian iff (ab)^-1=a^-1 * b^-1
If it is Abelian, $\displaystyle (ab)^{-1}=(ba)^{-1}=a^{-1}*b^{-1}$ by the previous relation.
If $\displaystyle (ab)^{-1}=a^{-1}*b^{-1}$, then since $\displaystyle (ab)^{-1}=b^{-1}*a^{-1}$, we have $\displaystyle b^{-1}*a^{-1}=a^{-1}*b^{-1}$
Thus it is commutative --> Abelian
The equivalence is proved.
$\displaystyle (a^{-1})^{-1}=a \Longleftrightarrow \underbrace{(a^{-1})^{-1}*(a^{-1})}_{1}=\underbrace{a*(a^{-1})}_{1}$c) In a group (a^-1)^-1=a for all a
Sorry for the wording & the method... just trying myself to explain