0
$\begingroup$

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 $\Bbb Z_6$?

$\endgroup$
  • 1
    $\begingroup$ 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 $3$? $\endgroup$ – TMM Mar 12 '14 at 1:40
2
$\begingroup$

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

$\endgroup$
  • $\begingroup$ Could you write out how $5$ generates the group, like how I did with $1$? $\endgroup$ – atherton Mar 12 '14 at 1:40
  • $\begingroup$ @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 $\Bbb Z/(6)$. $\endgroup$ – Pedro Tamaroff Mar 12 '14 at 1:44
1
$\begingroup$

Expanding user133458's answer:

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 $\mathbb{Z}_6 = <\bar{1}> = <\bar{5}>$.

$\endgroup$
0
$\begingroup$

Numbers that are relatively prime to 6 generate $\mathbb{Z_6}$. It's a theorem.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.