# Prove

• Sep 17th 2008, 06:09 AM
mandy123
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
• Sep 17th 2008, 06:17 AM
Moo
Quote:

Originally Posted by mandy123
Prove that:
a) In a group (ab)^-1=b^-1 * a^-1

$\displaystyle b^{-1}*a^{-1}$
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...

Quote:

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

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.

Quote:

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

Sorry for the wording & the method... just trying myself to explain (Worried)
• Sep 17th 2008, 06:22 AM
mandy123
thanks
Thank you so much!!! This helps me a lot!!
I like your animated cow by the way!!!
• Sep 17th 2008, 12:57 PM
ThePerfectHacker
Quote:

Originally Posted by mandy123
c) In a group (a^-1)^-1=a for all a

$\displaystyle a\cdot a^{-1} = e$ therefore inverse of $\displaystyle a^{-1}$ is $\displaystyle a$.
Thus, $\displaystyle (a^{-1})^{-1} = a$.