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Let $G$ and $H$ be two groups and let $g∈G$ and $h∈H$ be two elements of finite orders. Is the order of $(g,h)$ equals the least common multiple of $\operatorname{ord}(g)$ and $\operatorname{ord}(h)$? That is $$\operatorname{ord}(g,h)=\operatorname{lcm}(\operatorname{ord}(g),\operatorname{ord}(h))$$It is my conjecture, and I am not sure if it is correct or not. If it's true, how can I formally prove it?

Bernard
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JJW22
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    show that the least common multiple as an exponent indeed gives the neutral element. Then prove that this must be the smallest such positive integer (for example, by assuming that a smaller exponent achieving this would result in a contradiction). – Student Oct 07 '18 at 21:56
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    It is perfectly correct, and more or less obvious from the definition of the order of an element. – Bernard Oct 07 '18 at 21:58
  • @Student Wondering if it more easy to approach this problem using additive version instead of multiplicative version of group? I mean, using "+" instead of "*"? – JJW22 Oct 07 '18 at 22:01
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    That won't make it easier. It is the same. – David P Oct 07 '18 at 22:03

1 Answers1

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$(g,h)^n\! = (g^n,h^n) = (1,1)\!\iff\! g^n=1=h^n\! \iff o(g),o(h)\mid n\iff {\rm lcm}(o(g),o(h)\mid n,\, $ where the final step uses the lcm universal property: $\ a,b\mid n\iff {\rm lcm}(a,b)\mid n$

Bill Dubuque
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