1
$\begingroup$

I have asked a few questions about this but I am still confused. So, in general, what is a generator of a finite cyclic group and how is it found? I have seen in books and my notes a lot of descriptions using the term all elements of $G$ are given by $g^n$, for some integer $n$, and where $G = \langle g\rangle$. But I don't see how this can be the case for many of the groups I'm looking at.

$\endgroup$
  • $\begingroup$ That is meant To say G=<g> $\endgroup$ – Weyman Jan 6 '15 at 13:57
  • $\begingroup$ Could you give some examples of the groups you are looking at? $\endgroup$ – Tobias Kildetoft Jan 6 '15 at 14:00
  • $\begingroup$ @TobiasKildetoft for example U(Z54) = {1,5,7,11,13,17,19,23,25,29,31,35,37,41,43,47,49,53} but there is no one element that multiplies into all other elements? $\endgroup$ – Weyman Jan 6 '15 at 14:03
  • $\begingroup$ Sure there is, it just takes some tries to find it. $\endgroup$ – Tobias Kildetoft Jan 6 '15 at 14:05
  • $\begingroup$ But most of these elements are prime? @TobiasKildetoft $\endgroup$ – Weyman Jan 6 '15 at 14:07
1
$\begingroup$

An element $g\in G$ generates the group $G$ if $\{g^k\>|\>k\in{\mathbb Z}\}=G$. A group containing such a $g$ is called cyclic. Such a group is automatically abelian, so that we may write the group operation additively.

Let $G$ be a finite cyclic group of order $n\geq1$. If $g$ is a generator of $G$ then $G=\{ k\,g\>|\, 0\leq k<n\}$, and $n\, g=0$. This $g$ is not uniquely determined: Let $r$ be any number relatively prime to $n$. Then $h:=r\, g$ is again a generator of $G$. Conversely: Any generator of $G$ can be obtained from $g$ in this way.

The proof of these facts involves no group theory per se, but the elements of divisibility theory of integers.

$\endgroup$

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.