# Thread:

1. ## Prove that in a group

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

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

2. 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?

3. 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.

4. 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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