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

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?

• Have you even tried any of the ones you say have no order? Is it possible for an element to have no order? Commented Jan 18, 2015 at 2:24
• If $g$ has order $k$, then $g^n$ will have order $k/(n,k)$. Prove this and profit. =)
– Pedro
Commented Jan 18, 2015 at 2:34

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., $$0$$).

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 $$4$$.