# What is a generator of a finite cyclic group? (General)

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.

• That is meant To say G=<g> – Weyman Jan 6 '15 at 13:57
• Could you give some examples of the groups you are looking at? – Tobias Kildetoft Jan 6 '15 at 14:00
• @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? – Weyman Jan 6 '15 at 14:03
• Sure there is, it just takes some tries to find it. – Tobias Kildetoft Jan 6 '15 at 14:05
• But most of these elements are prime? @TobiasKildetoft – Weyman Jan 6 '15 at 14:07

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.