Let $G$ be an abelian group. Let $a, b \in G $ and let their order be $m$ and $n$ be respectively. Is it always true that order of $ab$ is $lcm(m,n)$?
What if $m$ and $n$ are coprime to each other?
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29Certainly the order of $ab$ need not be the lcm. As a simple example, $a$ and $a^{-1}$ have the same order, but $a^{-1}a$ has very small order. – André Nicolas Sep 24 '11 at 13:56
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When the group is finite and Abelian, you can show that if $d=\text{ord}(ab)$ where $m=\text{ord}(a)$ and $n=\text{ord}(b)$ then $$d \mid \frac{mn}{\gcd(m,n)}=\text{lcm}(m,n)$$ and $$\frac{mn}{\gcd(m,n)^2} \mid d.$$ ${}{}{}{}{}$ In particular this means that if $m$ and $n$ are coprime the order is multiplicative.
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Eric Naslund
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So, if $m$ and $n$ are coprime to each other, the relation is true. right? – ABC Sep 24 '11 at 14:18
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4$m,n$ should be $mn$ but edits have to have at least 6 characters changes, so I cannot edit – geo909 Jun 27 '14 at 16:25
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@karhas For the proof of the first formula see MathSE 1051177 and Yashinski's notes, and for proof of the second MathSE 2445367. Both and more are proved in Jungnickel, On the Order of a Product in a Finite Abelian Group. – Conifold Mar 06 '24 at 12:54
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You can prove that "anything" could happen for the order of a product. In more precise words:
For any integers $m;n;r > 1$, there exists a finite group $G$ with elements $a$ and $b$ such that $a$ has order $m$, $b$ has order $n$, and $ab$ has order $r$.
For a beautiful proof, see Theorem 1.64 in http://www.jmilne.org/math/CourseNotes/GT.pdf
Nathan Portland
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Best bounds for Abelian groups can be found in Jungnickel, On the Order of a Product in a Finite Abelian Group. Anything can only happen within them, unlike in the general case. – Conifold Mar 06 '24 at 13:04