a, b are in group G. Show o(ab) = o(ba).

I'm stumped. This would be trivial if G were abelian, but it's not. Please help.

EDIT: The back of the book has:

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 (**).

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.

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