Abstract Algebra: help session, nth roots of unity, order of inverse, 8-31-18

[triciad]

@43:59 If you don't use the fact that ab = ba so by uniqueness their inverses must be equal, you can just apply a^{-1} and b^{-1} to both sides:

ab = ba
(a^{-1} b^{-1}) ab (b^{-1} a^{-1}) = (a^{-1} b^{-1}) ba (b^{-1} a^{-1})
(a^{-1} b^{-1}) a a^{-1} = a^{-1} a (b^{-1} a^{-1})
a^{-1} b^{-1} = b^{-1} a^{-1}

[AndrewLi]

isn't it easier to prove by asserting (ab)x=(ba)x=e, and so x=(ab)^{-1}=(ba)^{-1} ? (for 43:59)