Order of an element in a group

[missavvy]

Hey guys!

I'm having some trouble trying to solve this question.. Any advice/help is appreciated!

Homework Statement

Suppose that a and b belong to a group such that:
|a| = 4, |b| = 2, and suppose a³b=ba
Find the order of ab.

Homework Equations

The Attempt at a Solution

So I am unsure of which theorems I should be looking at... but anyways... my horrible attempt:
so a,b are in G.
|ab| = (ab)ⁿ = 1

Let n = order(ab), aⁿbⁿ = 1 --> aⁿ=b^-n

I don't know if I should continue from this point because I don't know how exactly I would go about finding the order.

I think |ab| = |ba|, yes or no?
If it does then I'm guessing I can use that to find the order of a³b?
Any hints or suggestions on how?

Thank you.

 

[Dick]

(ab)^n and a^n*b^n are not the same thing. Clearly your group isn't abelian. Hint: (ab)(ab)=a(ba)b.

 

[missavvy]

Hey, thanks for the reply.
So is this a matter of manipulating a and b ?

For your hint, I'm just wondering why is it (ab)(ab)?
I see that a(ba)b = a^4b^2, which we have as the original orders, but why (ab)^2 and not some other number n?

 

[Dick]

Hey, thanks for the reply.
So is this a matter of manipulating a and b ?

For your hint, I'm just wondering why is it (ab)(ab)?
I see that a(ba)b = a^4b^2, which we have as the original orders, but why (ab)^2 and not some other number n?

Just because I noticed (ab)^2 worked. You can probably figure the order can't be too high with a relation like ba=a^3*b. You could try it for other n. You know n=4 will also work. The given relation tell you how to commute a and b. I.e. how to turn a 'ba' type expression into an 'ab' type expression.