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.
If $gh = hg$ in a group and $|g|$ and $|h|$ are finite, is $|gh|$ finite too?
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abstract-algebra
group-theory
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2HINT: If $gh=hg$, then can you rewrite $(gh)^n$ as a product of a power of $g$ and a power of $h$? – 2012-02-13
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0Can you see what's the value of a power $(gh)^n$? – 2012-02-13
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2**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. – 2012-02-13
1 Answers
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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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0This 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.$ – 2012-02-13
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4This 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. – 2012-02-13
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0Ahh that is true, I didn't read the question carefully enough and assumed it was what it usually is. Sorry. – 2012-02-13
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0@RagibZaman Do I have to show that (gh)^k \neq 1 since the question only askes to show |gh| is finite? – 2012-02-13
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0@Shannon No you don't have to. – 2012-02-13
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0Also, unless you assume some extra conditions, the order of $gh$ need not be $|g|\cdot |h|$ – 2012-02-13
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0@Ragib: If $g$ and $h$ commute, then $\text{ord}(gh) = \text{lcm}(\text{ord}(g), \text{ord}(h))$, and can be in general, less than $mn$. – 2012-02-17