Let $G$ be an abelian group and suppose that $G$ has elements of order $m$ and $n$ respectively. Prove that $G$ has an element of order $\mathrm{lcm}[m,n]$
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5See http://math.stackexchange.com/questions/10616/order-of-elements-in-abelian-groups?rq=1 – Kelenner Jul 30 '14 at 15:18
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If $m\mid n$ or $n\mid m$, there is nothing to be shown. Let $a,b$ have orders $m,n$ and let $d=\gcd(m,n)$. Then consider $x=a+b$. Then $mx=mb\ne 0$, $nx=na\ne0$ and $\frac{nm}dx = \frac nd mx =\frac nd mb=\frac md nb=0$ we conclude that the order of $x$ divides $\operatorname{lcm}(m,n)$, but neither $m$ nor $n$.
user26857
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Hagen von Eitzen
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1This solution is not complete, even incorrect. Consider $G=\Bbb Z/30$, $a=5$, $b=-3$, then $x=2$. – Quang Hoang Jul 30 '14 at 13:22
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1@QuangHoang I can agree that this solution is not complete. (What's next after finding such $x$?) But your example doesn't contradict anything from the answer: the order of $x$ is 15 which divides the lcm of 6 (order of $a$) and 10 (order of $b$). – user26857 May 06 '16 at 08:08