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

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- Sep 17th 2008, 06:09 AMmandy123Prove
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 AMMoo
$\displaystyle b^{-1}*a^{-1}$

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

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

Sorry for the wording & the method... just trying myself to explain (Worried) - Sep 17th 2008, 06:22 AMmandy123thanks
Thank you so much!!! This helps me a lot!!

I like your animated cow by the way!!! - Sep 17th 2008, 12:57 PMThePerfectHacker