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The relation $gh = hg$ means this group $G$ is commutative. $\langle g\rangle$ and $\langle h \rangle$ are cyclic subgroups of G. Still have no idea how to conclude $|gh|$ is finite.

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HINT: If $gh=hg$, then can you rewrite $(gh)^n$ as a product of a power of $g$ and a power of $h$? –  Arturo Magidin Feb 13 '12 at 4:27
    
Can you see what's the value of a power $(gh)^n$? –  Mariano Suárez-Alvarez Feb 13 '12 at 4:27
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Note: The condition $gh=hg$ does not mean that the group $G$ is commutative. It only means that $g$ and $h$ commute. It is perfectly possible for two elements in a noncommutative group to commute. –  Arturo Magidin Feb 13 '12 at 4:30

1 Answer 1

Take powers of $gh$. The condition $gh = hg$ will tell you that $(gh)^n = g^n h^n$. If $g^m = e$ the identity, and if $h^n = e$, then we see that $gh^{mn} = g^{mn}h^{mn} = e$.

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This alone is not enough. The more difficult part of the problem is to show that $ (gh)^k \neq 1 $ for $ k=1,2,3\cdots ,mn-1.$ –  Ragib Zaman Feb 13 '12 at 5:04
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This alone is not enough to determine the order of $gh$, I agree. But the OP is only asking about showing that $gh$ has finite order, for which this is sufficient. –  NKS Feb 13 '12 at 5:05
    
Ahh that is true, I didn't read the question carefully enough and assumed it was what it usually is. Sorry. –  Ragib Zaman Feb 13 '12 at 5:11
    
@RagibZaman Do I have to show that (gh)^k \neq 1 since the question only askes to show |gh| is finite? –  Shannon Feb 13 '12 at 6:26
    
@Shannon No you don't have to. –  Ragib Zaman Feb 13 '12 at 7:02

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