Cyclic Group U(14): Generating Elements with <3> to Show Cyclic Property

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Discussion Overview

The discussion centers on the cyclic nature of the group U(14) and the demonstration of its generators, specifically focusing on the elements <3> and <5>. Participants explore the definitions and implications of cyclic groups within the context of group theory.

Discussion Character

  • Technical explanation
  • Conceptual clarification
  • Debate/contested

Main Points Raised

  • One participant asserts that U(14) can be generated by the element <3>, listing its elements as 1, 3, 5, 9, 11, and 13.
  • Another participant provides a definition of a cyclic group, stating that a finite group is cyclic if there exists an element whose order matches the order of the group.
  • A third participant mentions that if the mapping from n to r^n is surjective, then the group is cyclic with generator r.
  • One participant expresses uncertainty about how to demonstrate that U(14) is cyclic and questions whether generating the same elements repeatedly is sufficient.
  • Another participant confirms that U(14) equals <3> and states that it can similarly be shown that <3> equals <5> in U(14).

Areas of Agreement / Disagreement

Participants generally agree on the definitions and properties of cyclic groups, but there is some uncertainty regarding the demonstration of U(14) being cyclic and the implications of generating elements.

Contextual Notes

There are unresolved aspects regarding the specific steps needed to demonstrate the cyclic property of U(14) and the implications of the mappings discussed.

Who May Find This Useful

Readers interested in group theory, particularly those studying cyclic groups and their properties, may find this discussion relevant.

semidevil
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to show that U(14) = <3> = < 5 >.

i.e, U(14) is cyclic.

ok, so U(14) = 1, 3, 5, 9, 11, 13

and <3> = 3^1 = 3, 3^2 = 9, 3 ^3 = 13, 3^4 = 11, 3^5 = 5, and 3^6 = 1.

So what am I doing to show that it is Cyclic? Am I showing that with <3>, I can generate the same elements over and over? is that what it means that it is cyclic?
 
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Surely your notes define cyclic?

A finite group is cyclic if there is an element whose order is the same as the order of the group.

Or a group is cyclic if there is a single element x such that {x^r | 1<=r<=ord(x)}=G. Ie all elements are a power of x for some x.
 
i.e. if the map taking n to r^n is surjective, then your group is cyclic with generator r.
 
semidevil said:
to show that U(14) = <3> = < 5 >.

i.e, U(14) is cyclic.

ok, so U(14) = 1, 3, 5, 9, 11, 13

and <3> = 3^1 = 3, 3^2 = 9, 3 ^3 = 13, 3^4 = 11, 3^5 = 5, and 3^6 = 1.

So what am I doing to show that it is Cyclic? Am I showing that with <3>, I can generate the same elements over and over? is that what it means that it is cyclic?
I'm sorry, but I can't see the problem. You've shown that U(14)=<3>. You can show similarly that <3>=<5> in U(14).
 

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