Re: Show that o(ab) = o(ba)

Quote:

Originally Posted by

**phys251** o(ab) = m

=> (ba)^m = (a^-1)(a)(ba)^m

= (a^-1)(ab)^m(a) (**)

I have NO clue how they got the step marked by (**).

Consider some examples.

$\displaystyle a^{-1}a(ba)^3 =a^{-1}a(ba)(ba)(ba) =a^{-1}(ab)(ab)(ab)a =a^{-1}(ab)^3a$

If you need a precise proof, you can prove that $\displaystyle a(ba)^m = (ab)^ma$ by induction on m.

Re: Show that o(ab) = o(ba)

Ohhhh, you just associate everything once to the left, so now instead of ba's inside parentheses, we have ab's. Thanks.