A true or false question, is there a counter example or short proof for this statement?
Let G be an abelian group. For all a, b ∈ G the order of a + b is the lowest common multiple of the orders of a and b.
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False, consider $\mathbb{Z}/2\mathbb{Z}$, $1$ is of order 2 while $1+1=2$ is of order $1$. – Oolong Milktea May 07 '21 at 08:20
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See also this post. We need that $a+b\neq 0$. Note that $a\circ b=a+b$ for addition as composition. – Dietrich Burde May 07 '21 at 08:30
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Counterexample:
Take $\mathbb{Z}/4\mathbb{Z}$. Then $\operatorname{ord}(2)=2$ as $2+2=0$.
But $\operatorname{lcm}(2,2)=2$, and $\operatorname{ord}(0)\neq 2$
Cornman
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Just for information, this question comes straight from an online exam at the University of Warwick, which was taking place when the question was posted. – Derek Holt May 07 '21 at 16:55