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Apparently $ord (xy)$ divides the least common multiple of $ord(x)$ and $ord(y)$. It can be easily proved that if $ord(x)$ and $ord(y)$ are coprime, then $$ord (xy) = ord(x).ord(y) = lcm (ord(x), ord(y)).$$ Must always be $$ord (xy) = lcm (ord(x), ord(y))?$$ If yes, can you prove it? If not, can you give a counterexample, i.e. find positive integers $ord(x), ord(y)$ with $gcd (ord(x), ord(y))\gt1$ such, that $ord (xy)\lt lcm(ord(x),ord(y))$?

Ivos
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    Welcome to stackexchange. If you [edit] the question to show us what you tried and where you are stuck we may be able to help. Or you can try to find the answer elsewhere on this site by searching with a few good key words. – Ethan Bolker Jun 13 '22 at 21:11
  • What properties of the LCM are you allowed to assume (or do you know)? Two are obvious - it is a common multiple, and it is least - have you deployed both aspects in your thinking? – Mark Bennet Jun 13 '22 at 21:13
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    There's a very simple family of counterexamples: take any finite-order, non-trivial $x$ and its inverse. – Theo Bendit Jun 13 '22 at 21:23
  • If $x$ and $y$ have finite order there is not even assurance that $xy$ will have finite order. – CyclotomicField Jun 13 '22 at 21:37
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    @CyclotomicField The $x$ and $y$ in the question are assumed to be commuting, so $xy$ will have finite order. – Theo Bendit Jun 13 '22 at 23:48

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