PlanetMath (more info)
 Math for the people, by the people.
Encyclopedia | Requests | Forums | Docs | Wiki | Random | RSS  
Login
create new user
name:
pass:
forget your password?
Main Menu
sections
Encyclopædia
Papers
Books
Expositions

meta
Requests (205)
Orphanage
Unclass'd
Unproven (490)
Corrections (8)
Classification

talkback
Polls
Forums
Feedback
Bug Reports

downloads
Snapshots
PM Book

information
News
Docs
Wiki
ChangeLog
TODO List
Legalese
About
Owner confidence rating: High Entry average rating: No information on entry rating
[parent] generator (Definition)

If $ G$ is a cyclic group and $ g \in G$, then $ g$ is a generator of $ G$ if $ \langle g \rangle =G$.

All infinite cyclic groups have exactly $ 2$ generators. To see this, let $ G$ be an infinite cyclic group and $ g$ be a generator of $ G$. Let $ z \in \mathbb{Z}$ such that $ g^z$ is a generator of $ G$. Then $ \langle g^z \rangle =G$. Then $ g \in G= \langle g^z \rangle$. Thus, there exists $ n \in {\mathbb{Z}}$ with $ g=(g^z)^n=g^{nz}$. Therefore, $ g^{nz-1}=e_G$. Since $ G$ is infinite and $ \vert g\vert=\vert\langle g \rangle \vert=\vert G\vert$ must be infinity, $ nz-1=0$. Since $ nz=1$ and $ n$ and $ z$ are integers, either $ n=z=1$ or $ n=z=-1$. It follows that the only generators of $ G$ are $ g$ and $ g^{-1}$.

A finite cyclic group of order $ n$ has exactly $ \varphi(n)$ generators, where $ \varphi$ is the Euler totient function. To see this, let $ G$ be a finite cyclic group of order $ n$ and $ g$ be a generator of $ G$. Then $ \vert g\vert=\vert\langle g \rangle \vert=\vert G\vert=n$. Let $ z \in \mathbb{Z}$ such that $ g^z$ is a generator of $ G$. By the division algorithm, there exist $ q,r \in \mathbb{Z}$ with $ 0 \le r<n$ such that $ z=qn+r$. Thus, $ g^z=g^{qn+r}=g^{qn}g^r=(g^n)^qg^r=(e_G)^qg^r=e_Gg^r=g^r$. Since $ g^r$ is a generator of $ G$, it must be the case that $ \langle g^r \rangle =G$. Thus, $ \displaystyle n=\vert G\vert=\vert\langle g^r \rangle\vert=\vert g^r\vert=\frac{\vert g\vert}{\gcd(r,\vert g\vert)}=\frac {n}{\gcd(r,n)}$. Therefore, $ \gcd(r,n)=1$, and the result follows.



"generator" is owned by Wkbj79.
(view preamble)

View style:

See Also: generating set of a group


This object's parent.
Log in to rate this entry.
(view current ratings)

Cross-references: division algorithm, Euler totient function, order, finite, integers, infinity, infinite, infinite cyclic groups, cyclic group
There are 61 references to this entry.

This is version 7 of generator, born on 2003-03-11, modified 2007-11-10.
Object id is 4094, canonical name is Generator.
Accessed 7184 times total.

Classification:
AMS MSC: 20A05 (Group theory and generalizations :: Foundations :: Axiomatics and elementary properties)
Pending Errata and Addenda
None.
[ View all 1 ]
Discussion
Style: Expand: Order:
forum policy

No messages.

Interact
post | correct | update request | add derivation | add example | add (any)