Group Theory, cyclic group proof

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Homework Help Overview

The discussion revolves around proving that the group of integers modulo n, denoted as Z_n, is cyclic. The original poster is exploring the properties of cyclic groups and the nature of their generators.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to define the group operation and considers whether to use additive or multiplicative operations. They express confusion about how to proceed after rewriting the group definition. Other participants suggest using a specific generator for the group.

Discussion Status

Participants are engaging with the original poster's reasoning and providing suggestions for potential generators. There is an ongoing exploration of the concept of generators in cyclic groups, but no consensus has been reached on the proof itself.

Contextual Notes

The original poster notes difficulty in finding the appropriate notation for the group and expresses uncertainty about the choice of operation. There is also a mention of the multiplicative inverse not being an integer, which influences their approach.

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Homework Statement



Prove that Z sub n is cyclic. (I can't find the subscript, but it should be the set of all integers, subscript n.)

Homework Equations

Let (G,*) be a group. A group G is cyclic if there exists an element x in G such that G = {(x^n); n exists in Z.}

(Z is the set of all integers)

The Attempt at a Solution



* is a binary operation, and for my purposes, is either additive (+) or multiplicative (x).

Multiplicative does not work because the multiplicative inverse of, say, 2 is not an integer. So the operation must be additive. So I can rewrite the equation for (G,+) as:

G = {nx; n exists in Z}

but that's where I get stuck. Thanks for the help!
 
Last edited:
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How about taking x = 1 as your generator?
 
Every cyclic group has a generator.

What is your generator in this case?

edit: nm already beaten too it
 
thanks to both!
 

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