I'm trying to prove the below equation, where $ a,b \in G $ and $(G, *) $ is a group.
$(a*b)^{-1} = (a^{-1}) * (b^{-1}) $
I'm not really sure how to do it though. I tried doing something like
$(a*b)^{-1} * (a*b) = e = a * a^{-1} * b^{-1} * b $
$(a*b)^{-1} * (a*b) = e = a * (a^{-1} * b^{-1}) * b $
However, I can't get the RHS to $ (a * b)* (a^{-1} * b^{-1}) $ without assuming communitivity.
How would I go about doing this? Any help is appreciated.
Thanks
Edit: Given that correction, would this be correct?
$ (a*b)^{-1} * (a*b) = e = a*(b*b^{-1})*a^{-1}$
Then
$ a*(b*b^{-1})*a^{-1} = a*b* (b^{-1}*a^{-1}) = (a*b)*(b^{-1}*a^{-1})$
Thus
$ (a*b)^{-1} = (b^{-1}*a^{-1})$