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 $(ab)^{mn} = a^{mn} b^{mn} = \ldots$