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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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Certainly 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

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

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$G$ is assumed to be abelian. – Alex B. Oct 10 '11 at 1:07

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 $$d|\frac{mn}{\gcd(m,n)}=\text{lcm}(m,n)$$ and $$\frac{mn}{\gcd(m,n)^2}|d.$$ ${}{}{}{}{}$ In particular this means that if $m$ and $n$ are coprime the order is multiplicative.

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So, if $m$ and $n$ are coprime to each other, the relation is true. right? – Norman Sep 24 '11 at 14:18
$m,n$ 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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