Finding the order of all the elements in Group $\mathbb{Z}_{12}$

Question

I know that the order of an element $a$ in a group $G$ is the smallest positive integer $m$ such that $a^m=e$ and so for $(\mathbb{Z}_{12},+)$ we have

$[0]$ is the identity of order 1.

$[1]$ is order 12 because $[1]+[1]+[1]+[1]+[1]+[1]+[1]+[1]+[1]+[1]+[1]+[1] = [0]$ and so $[1]^{12} = [0]$.

$[2]$ is order 6.

$[3]$ is order 4.

$[4]$ is order 3 because $[4]+[4]+[4] = [0]$ and so $[4]^3 = 0$.

$[5]$ No order!?

$[6]$ is order 2.

$[7]$ No order.

$[8]$ No order.

$[9]$ No order.

$[10]$ No order.

$[11]$ No order.

Did I get that right? And how would I prove that there is no order for $[5]$, for example?

Answer 1

In $\mathbb Z_{12}$, the order of an element $g$ is the smallest positive number $m$ for which $$\underbrace{g +g + \ldots +g}_{m \ \textrm{times}} \equiv 0 \pmod {12}.$$

It's easy to find the order for divisors of $12$, as you have done. But all the other elements have orders too; for example, to find the order of $5$:

$1 \cdot 5 = 5$

$2 \cdot 5 = 10$

$3 \cdot 5 = 15 \equiv 3 \pmod {12}$

$\cdots$

$12 \cdot 5 = 60 \equiv 0 \pmod {12}$

The order of $5$ is $12$, since this is the first time that we get back to the identity element (i.e., $0$).

Answer 2

This is not correct. All elements have an order. For example, let $[a,b]$ be the least common multiple of $a$ and $b$.

$$[5,12]=60=5\cdot 12$$

So, $[5]$ has order $12$.

$$[9,12]=36=9\cdot 4$$

So, $[9]$ has order $4$.