Proving (ab)^n=e: A Group Theory Question | Homework Statement

[eddyski3]

Homework Statement

If a and b are in a group, show that if (ab)^n=e then (ba)^n=e.

Homework Equations

The Attempt at a Solution

I'm not sure how one would prove this. The question is obviously for non-abelian groups.

 

[Dick]

If (ab)^2=e then (ab)(ab)=e. So b(ab)(ab)a=bea=ba. Now use associativity and 'cancellation'. Do you see how to do the same trick for (ab)^n?

 

[eddyski3]

I'm not sure how this helps us show that (ba)^2=e? When generalizing to (ab)^n I see we'll get a similar result but I'm not sure how this shows that (ba)^n=e.

 

[Dick]

b(ab)(ab)a=ba, yes? That's the same as (ba)(ba)(ba)=(ba). Do you see it now?

 

[Bacle]

abab=e , so ab=b^-1a^-1

ababab=e , so baba=a^-1b^-1

Play around with these until you can figure out one, then , if you don't have a general
argument for all n, maybe induction on n will help.

 

[eddyski3]

Oh, ok. Now I understand the argument. Thank you.