Is it true that the order of $ab$ is always equal to the order of $ba$? 
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?
 A: By associativity, $(ab)^p=a(ba)^{p-1}b$ for $p\geqslant 1$. If $(ab)^p=e$ then $a(ba)^{p-1}b=e$, so $a(ba)^p=a$ and $(ba)^p=e$. We conclude that for $p\geqslant 1$, 
$$(ab)^p=e\Leftrightarrow (ba)^p=e.$$
A: Here's an approach that allows you to do some hand-waving and not do any calculations at all. $ab$ and $ba$ are conjugate: indeed, $ba=a^{-1}(ab)a$. It is obvious (and probably already known at this point) that conjugation is an automorphism of the group, and it is obvious that automorphisms preserve orders of elements.
A: Hint: Suppose $ab$ has order $n$, and consider $(ba)^{n+1}$.
Another hint is greyed out below (hover over with a mouse to display it):

 Notice that $(ba)^{n+1} = b(ab)^na$.

A: If $(ab)^n=e$ then $(ab)^na=a$. Since $(ab)^na=a(ba)^n$, $(ba)^n=e$. This proves that the order of $ba$ divides the order of $ab$. By symmetry, the order of $ab$ divides the order of $ba$. Hence the order of $ab$ and the order of $ba$ coincide.
A: (1)  
$(ab)^n = e$ 
$\Rightarrow$ 
$(ba)^n = (ba)^nbb^{-1} = b(ab)^nb^{-1} = beb^{-1} = e$.  
(2)  
$(ba)^n = e$ 
$\Rightarrow$ 
$(ab)^n = (ab)^naa^{-1} = a(ba)^na^{-1} = aea^{-1} = e$.  
A: Another elementary way
On contrary suppose $|ab|,|ba|$ are different 

With out loss of generality assume $|ab|=n>|ba|=k$

$(ab)^n=
abababab........ab=e$
$a(ba)^{n-1}b=e$ as form assumption k
$a(ba)^{n-1-k}b=e=(ab)^{n-k}$ that implies order of ab is n-k which contradition to assumption.
n-k
