Is a Discrete Group of Rotations Cyclic?

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Homework Help Overview

The discussion revolves around the properties of a discrete group of rotations about the origin, specifically whether such a group is cyclic and can be generated by the smallest angle of rotation.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants explore the implications of defining the smallest angle of rotation and whether this leads to the conclusion that the group is cyclic. Questions arise about the utility of the definition and the existence of other rotations in the group.

Discussion Status

Some participants suggest that if the smallest angle generates all rotations in the group, then the group is cyclic. Others raise questions about the existence of rotations that might contradict this, leading to a productive examination of the group's structure.

Contextual Notes

There is a focus on the definition of a discrete group and the implications of having a smallest angle of rotation, with participants considering the constraints of the problem and the definitions involved.

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Homework Statement



Prove that a discrete group G cosisting of rotations about the origin is cyclic and is generated by [tex]\rho_{\theta}[/tex] where [tex]\theta[/tex] is the smallest angle of rotation in G

The Attempt at a Solution



since G is by definition a discrete group we know that if [tex]\rho[/tex] is a rotation in G about some point through a non zero angle [tex]\theta[/tex] the the angle [tex]\theta[/tex] is at least [tex]\epsilon[/tex]:|[tex]\theta[/tex]|[tex]\geq\epsilon[/tex]

But i don't know how to apply this definition to show that G is cyclic. Is this definition even useful?
 
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Ok, if you pick theta to be the smallest angle (which you can do since the rotations are discrete), then all of the rotations n*theta for n an integer are in the group. If that's the whole group, then you are done since it's cyclic. If not there a rotation phi in the group that isn't equal to n*theta for any n. Can you take the next step?
 
with that i can show that there is a non zero positive rotation less than theta, a contradiction. Is that it?
 
SNOOTCHIEBOOCHEE said:
with that i can show that there is a non zero positive rotation less than theta, a contradiction. Is that it?

It sure is.
 

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