Prove that in a group

**universalsandbox** · October 2nd 2008, 12:23 PM

(xy)^(-1) = b^(-1)a^(-1),

keeping in mind the group may not be commutative. Thanks.

**ThePerfectHacker** · October 2nd 2008, 12:27 PM

Originally Posted by universalsandbox

(xy)^(-1) = b^(-1)a^(-1),

keeping in mind the group may not be commutative. Thanks.

To show $(ab)^{-1} = b^{-1}a^{-1}$ you need to show $(ab)(b^{-1}a^{-1}) = e$ .
Can you do that now?

**universalsandbox** · October 2nd 2008, 12:29 PM

Originally Posted by ThePerfectHacker

To show $(ab)^{-1} = b^{-1}a^{-1}$ you need to show $(ab)(b^{-1}a^{-1}) = e$ .
Can you do that now?

Could you add a little more process to that please? This stuff is very new to me. Thanks.

**ThePerfectHacker** · October 2nd 2008, 12:31 PM

Originally Posted by universalsandbox

Could you add a little more process to that please? This stuff is very new to me. Thanks.

Say that $x^{-1} = y$ . What does that mean? It means that $xy = e$ and $yx=e$ . That is the definition of what it means to be a multiplicative inverse. In your problem you have $x=ab$ and $y=b^{-1}a^{-1}$ .

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