1. ## Prove

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

2. Originally Posted by mandy123
Prove that:
a) In a group (ab)^-1=b^-1 * a^-1
$b^{-1}*a^{-1}$
Multiply it to the right by (ab)

$b^{-1}*a^{-1}*a*b$

Since it is a group, * is associative.
So we can write :

$b^{-1}*a^{-1}*a*b=b^{-1}*\underbrace{(a^{-1}*a)}_{1}*b$
$=b^{-1}*b=1$

So we have $(b^{-1}*a^{-1})*(ab)=1$
Therefore...

b) A group G is Abelian iff (ab)^-1=a^-1 * b^-1
Abelian means that * is commutative.

If it is Abelian, $(ab)^{-1}=(ba)^{-1}=a^{-1}*b^{-1}$ by the previous relation.

If $(ab)^{-1}=a^{-1}*b^{-1}$, then since $(ab)^{-1}=b^{-1}*a^{-1}$, we have $b^{-1}*a^{-1}=a^{-1}*b^{-1}$
Thus it is commutative --> Abelian

The equivalence is proved.

c) In a group (a^-1)^-1=a for all a
$(a^{-1})^{-1}=a \Longleftrightarrow \underbrace{(a^{-1})^{-1}*(a^{-1})}_{1}=\underbrace{a*(a^{-1})}_{1}$

Sorry for the wording & the method... just trying myself to explain

3. ## thanks

Thank you so much!!! This helps me a lot!!
I like your animated cow by the way!!!

4. Originally Posted by mandy123
c) In a group (a^-1)^-1=a for all a
$a\cdot a^{-1} = e$ therefore inverse of $a^{-1}$ is $a$.
Thus, $(a^{-1})^{-1} = a$.