Show that order(ab) = order(ba) when order(ab) is finite

[Metahominid]

Homework Statement

G is a group. Let a,b be elements of G. If order(ab) is a finite number n, show order(ba) = n as well.

Homework Equations

order(a) = order(<a>) where <a> is the cyclic group generated by a.

The Attempt at a Solution

I do not know. I thought it may be related to how that if a finite cyclic group has order n it is isomorphic to (Zn,+n). Any hints would be good.

 

[jbunniii]

For an elementary proof, you could use this trick:

$$(ab)(ab)...(ab) = a(ba)(ba)...(ba)b$$

k parenthesized terms on the left side <--> k-1 on the right

 

[Dick]

If ab has order 3 then (ab)(ab)(ab)=e. Regroup that as a(ba)(ba)b=e. Think about that.

 

[Metahominid]

Hey, sorry it took me so long to reply. Thank you so much. I chose (ba)^n+1 = b((ab)^n)a = b(e)a =ba, so I took the inverse and got (ba)^n = e. I know I need to prove that n is the smallest integer s.t. that is true, so I assumed there was a k < n s.t. (ba)^k = e. Then I know this implies that (ab)^k = e as well, which is a contradiction so therefore k = n. Thanks again.