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Exercise 4: Task 1

Verfasst: 3. Feb 2009 21:36
von Xelord
b) How can the group \(\mathbb{F}_{256}\) have a group generator ? Groups only have a group generator, if p is prime ?
Can anybody explain for what I need the generator ?

Re: Exercise 4: Task 1

Verfasst: 3. Feb 2009 22:32
von Christoph-D
Xelord hat geschrieben:b) How can the group \(\mathbb{F}_{256}\) have a group generator ? Groups only have a group generator, if p is prime ?
The order of a group with a generator (AKA "cyclic group") doesn't need to be prime. Take for example \((\mathbb{Z}/n\mathbb{Z}, +)\). The element \(\overline{1} \in \mathbb{Z}/n\mathbb{Z}\) is always a generator, no matter what n is.

Since \(\mathbb{F}_{256}\) is a finite field, the multiplicative group \((\mathbb{F}_{256} \setminus \{0\}, {\cdot})\) must be cyclic, so it certainly has a generator.


(I'm not actually taking this class, so consider my posting "uninformed".)

Re: Exercise 4: Task 1

Verfasst: 3. Feb 2009 22:57
von mädchen
Das sehe ich nicht so. Betrachten wir \(\mathbb{F}_8\) multiplikativ. Dann gilt doch
\(1^1 = 1, 1^2 = 1\) => Kein Generator
\(2^1 = 2, 2^2 = 4, 2^3 = 0\) => Kein Generator
\(3^1 = 3, 3^2 = 1\) => Kein Generator
\(4^1 = 4, 4^2 = 0\) => Kein Generator
\(5^1 = 5, 5^2 = 1\) => Kein Generator
\(6^1 = 6, 6^2 = 4, 6^3 = 0\) => Kein Generator
\(7^1 = 7, 7^2 = 1\) => Kein Generator

Wo ist mein Fehler?

Re: Exercise 4: Task 1

Verfasst: 3. Feb 2009 23:02
von Christoph-D
mädchen hat geschrieben:Das sehe ich nicht so. Betrachten wir \(\mathbb{F}_8\) multiplikativ. Dann gilt doch
\(1^1 = 1, 1^2 = 1\) => Kein Generator
\(2^1 = 2, 2^2 = 4, 2^3 = 0\) => Kein Generator
\(3^1 = 3, 3^2 = 1\) => Kein Generator
\(4^1 = 4, 4^2 = 0\) => Kein Generator
\(5^1 = 5, 5^2 = 1\) => Kein Generator
\(6^1 = 6, 6^2 = 4, 6^3 = 0\) => Kein Generator
\(7^1 = 7, 7^2 = 1\) => Kein Generator

Wo ist mein Fehler?
\(\mathbb{Z}/8\mathbb{Z}\) ist nicht \(\mathbb{F}_8\).

Re: Exercise 4: Task 1

Verfasst: 4. Feb 2009 12:58
von Xelord
Fine. What is the generator of \(\mathbb{F}_8\) ?