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The orders of the elements in a cyclic group of order 12

Find the orders of each element of the additive group $\mathbb{Z}/(12)$.

Solution: For an element $n$ in $\mathbb{Z}/(12)$, the order of $\bar n$ is the smallest positive number $m$ such that $mn$ is a multiple of 12. Therefore, we have the following table.

element $\bar n$order
$\bar 0$1
$\bar 6$2
$\bar 4$, $\bar 8$3
$\bar 3$, $\bar 6$, $\bar 9$4
$\bar 2$, $\bar 10$6
$\bar 1$, $\bar 5$, $\bar 7$, $\bar 11$12


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This Post Has One Comment

  1. Dans Z/12Z, 6 n'est-il pas d'ordre 2, car 6 + 6 =12 = 0? (No 1.1.11 Dummit)

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