TITLE: Is it true that the order of $ab$ is always equal to the order of $ba$?
QUESTION [18 upvotes]: How do I prove that if $a$, $b$ are elements of group, then $o(ab) = o(ba)$?

For some reason I end up doing the proof for abelian(ness?), i.e., I assume that the order of $ab$ is $2$ and do the steps that lead me to conclude that $ab=ba$, so the orders must be the same. Is that the right way to do it?

REPLY [2 votes]: (1)  
$(ab)^n = e$ 
$\Rightarrow$ 
$(ba)^n = (ba)^nbb^{-1} = b(ab)^nb^{-1} = beb^{-1} = e$.  
(2)  
$(ba)^n = e$ 
$\Rightarrow$ 
$(ab)^n = (ab)^naa^{-1} = a(ba)^na^{-1} = aea^{-1} = e$.