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Suppose $G$ is an abelian group (not necessarily finite) and $a,b$ are two elements in $G$ such that $o(a) = m$ and $o(b) = n.$

I need to show that $\text{lcm}(m,n)/\gcd(m,n)$ divides $o(ab)$

Please note that $m$ and $n$ are not relatively prime.

I have tried looking at $m$ and $n$ as expansions of powers of primes and have recognized that $\text{lcm}(m,n) / \gcd (m,n) =A$ where $A$ is always an integer.

I have tried to show that $o(ab)/A$ is an integer but have gotten stuck and don't know what else to try.

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  • $\begingroup$ I'm pretty sure this has been asked several times in the past. $\endgroup$ – Xam Sep 26 '17 at 14:16
  • $\begingroup$ Please put a link to the repeat if it has. I could not find this question. On the versions of this question where m and n a relatively prime. $\endgroup$ – pictorexcrucia Sep 26 '17 at 14:17
  • $\begingroup$ I don't have to because I'm not asking the question. $\endgroup$ – Xam Sep 26 '17 at 14:20
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Since $G$ is abelian, $(ab)^{mn}=a^{mn} b^{mn}=(a^m)^n (b^n)^m=1$. Thus $o(ab)| mn$. So $mn=o(ab)s_1$, for some $s \in \Bbb{Z}$. On the other hand $A:=lcm(m,n)/gcd(m,n)=mn/gcd(m,n)^2$. Hence we have $o(ab)=\frac{mn}{s_1}=\frac{A (gcd(m,n))^2}{s_1} $. Put $s_2=\frac{gcd(m,n))^2}{s_1}$. So $o(ab)=As_2$, i.e. $A|o(ab)$, as desierd.

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  • $\begingroup$ You have to prove $s_2 \in \mathbb Z$. $\endgroup$ – Orat Sep 26 '17 at 7:10
  • $\begingroup$ You're right. And in fact, it's not an integrer. S2 is just 1/A $\endgroup$ – pictorexcrucia Sep 27 '17 at 2:33

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