# Show that o(ab) = o(ba)

• Jul 31st 2013, 07:01 PM
phys251
Show that o(ab) = o(ba)
a, b are in group G. Show o(ab) = o(ba).

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 (**).
• Aug 1st 2013, 09:23 AM
emakarov
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.

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

If you need a precise proof, you can prove that $a(ba)^m = (ab)^ma$ by induction on m.
• Aug 1st 2013, 11:26 AM
phys251
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.