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Let $a,b\in G$, with finite $G$. Assume the order of element $|a|$, $|b|$ is finite, then what i want to know is $|ab|$ finite?

First what i know is if $ab=ba$, $i.e$, $G$ is abelian, $|ab|$ is finite.

For $(|a|,|b|)=1$, $|ab| = |a| |b|$, and general case i notice that $|ab| = \textrm{lcm}(|a|,|b|) = \frac{|a||b|}{gcd(a,b)}$

I want to relax this by neglecting abelian condition.

Then is $|ab|$ finite?, If so how one can prove this?

phy_math
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  • What do you know about $(ab)^n$? (It should be equal to something strictly in terms of a,b, and n). – Theo C. Jun 08 '17 at 03:34
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    In general, $|ab|$ does not have anything to do with $|a|$ and $|b|$. Consider that $|(1,2)| = |(2,3)| = 2$ and $|(1,2)(2,3)| = 3$. – Kenny Lau Jun 08 '17 at 03:37
  • Ah, i made some stupid things. Of course if $G$ is finite, then every element of $G$ has finite order. Since Kenny Lau already made some answer, i post another question https://math.stackexchange.com/questions/2314299/let-a-b-in-g-if-a-b-finite-then-ab-finite-for-arbitrary-group-g – phy_math Jun 08 '17 at 03:51

1 Answers1

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$G$ is finite, so $|ab|$ must be finite, because $ab \in G$.

(If you remove this condition that $G$ is finite, an easy counter-example is to consider the free product $\Bbb Z_2 * \Bbb Z_3$.)

Kenny Lau
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