# How do $1$ and $5$ generate the cyclic group $\Bbb Z_6$?

How do $1$ and $5$ generate the cyclic group $\Bbb Z_6$?

I was reading through my text and this example was brought up. I'm pretty sure the binary operation they are using is simply addition. I understand $1$: $1+1$, $1+1+1$,... But how does $5$ generate the $\Bbb Z_6$?

• Modulo $6$, what are $5, 5+5, 5+5+5, \dots$? Does that form all of $\Bbb Z_6$? Does that also hold for $2$ or $3$? – TMM Mar 12 '14 at 1:40

$5 = -1$ who generates the same way 1 does.

• Could you write out how $5$ generates the group, like how I did with $1$? – atherton Mar 12 '14 at 1:40
• @RoyM. $5,5+5=10=4,5+4=9=3,3+5=8=2,2+5=7=1,1+5=6=0$, listing we have $5,4,3,2,1,0$; which is all of $\Bbb Z/(6)$. – Pedro Tamaroff Mar 12 '14 at 1:44

Let's consider $\mathbb{Z}_n$, and $\bar{x} \in \mathbb{Z}_n$, so that $(x, n) = 1$. Since the order of $\bar{1}$ is clearly $n$, we can find the order of $\bar{x}$ as follows:
In a finite cyclic group $G=<g>$, we know that $o(g^i) = \dfrac{n}{(i, n)}$.
In our case, the group is $\mathbb{Z}_n = < \bar{1} >$, and therefore $o(\bar{x}) = o(x \bar{1}) = \dfrac{n}{(x, n)}=n$, since $(x, n) = 1$. Thus, $\mathbb{Z}_n = < \bar{x} >$.
In your example, since $(1, 6) = (5, 6) = 1$, we've got that $\mathbb{Z}_6 = <\bar{1}> = <\bar{5}>$.
Numbers that are relatively prime to 6 generate $\mathbb{Z_6}$. It's a theorem.