Abstract Algebra, order of ab is equal to the order of a times the order of b?

[zardiac]

Abstract Algebra, order of ab is equal to the order of a times the order of b??

Hi!
I am working on some problems in abstract algebra and I am stuck at the moment. I hope some of you guys could help me out a little.

Homework Statement

a and b are two elements in a group G.
Assume that ab=ba.
Show that if GCD(o(a),(ob))=1, then o(ab) = o(a)*o(b)

Where o(a) is the order of a. (i.e. a^(o(a))=1.)

Homework Equations

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The Attempt at a Solution

I call o(a)=n, o(b)=m, o(ab)=k, then show that k=mn.
Since SGD(m,n)=1, then m and n are coprime integers, and I have this relation: 1=sn+tm, where s and t are some integers.

However I am stuck now and I am not sure how to use this or where to start.
So any suggestions would be very appreciated.

Thanks in advance

 

[jbunniii]

Clearly (ab)^{mn} = 1, so o(ab) divides mn. Can you show that \langle ab\rangle, the subgroup generated by ab, contains subgroups of order m and n? If so, what does that imply?

 

[zardiac]

Clearly (ab)^{mn} = 1, so o(ab) divides mn. Can you show that \langle ab\rangle, the subgroup generated by ab, contains subgroups of order m and n? If so, what does that imply?

I am not sure why (ab)^{mn} = 1 is clear.
(a)^{n} = 1 and (b)^{m} = 1 but does that say anything about (ab)^{mn} ? Sorry maybe to early in the morning for this, Ill think about what you wrote during the day and see where I get!
Thanks for the help

 

[jbunniii]

I am not sure why (ab)^{mn} = 1 is clear.
(a)^{n} = 1 and (b)^{m} = 1 but does that say anything about (ab)^{mn} ? Sorry maybe to early in the morning for this, Ill think about what you wrote during the day and see where I get!
Thanks for the help

You are given that ab = ba, so (ab)^{mn} = a^{mn} b^{mn} = \ldots