Proving a group of rotaions is cyclic

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Homework Help Overview

The problem involves proving that a finite group of rotations of the plane about the origin is cyclic. The original poster attempts to understand the implications of cyclic groups and how to demonstrate that all elements can be expressed as powers of a single element.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the concept of a smallest nonzero rotation and its implications for the group structure. They explore the relationship between elements of the group and question how to demonstrate the properties of cyclicity through the division algorithm.

Discussion Status

The discussion is ongoing, with participants exploring various lines of reasoning and attempting to clarify their understanding of the problem. Some guidance has been offered regarding the implications of the smallest rotation and the potential contradiction arising from the assumptions made about the group.

Contextual Notes

There is a focus on the properties of finite groups and the implications of having a smallest element in the context of rotations. Participants express uncertainty about how to resolve contradictions and the implications of their findings on the cyclic nature of the group.

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Homework Statement


Let G be a finite group of rotation of the plane about the origin. Prove that G is cyclic.


The Attempt at a Solution



What it means to be cyclic is that every element of the group can be written as a^n for some integer n.

I can see this is true if i take some examples. i.e. {0,pi/2, pi, 3pi/2} is cleraly cyclic. but i can't for the life of me figure out how to prove this.
 
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There are lots of ways to prove this depending on what tools you happen to have around. The group must have a smallest nonzero rotation r. All multiples of r must also be in the group. That makes a nice cyclic subgroup, right? If it's noncyclic there must be an element s which is not a multiple of r. Now what?
 
Dick said:
There are lots of ways to prove this depending on what tools you happen to have around. The group must have a smallest nonzero rotation r. All multiples of r must also be in the group. That makes a nice cyclic subgroup, right? If it's noncyclic there must be an element s which is not a multiple of r. Now what?

By division algorithm s= qr+ m . I am guessing we have to somehow show that m=0? Or show that if s is in a group then r is not.

dont know how to do that.
 
s=qr+m. If r is in the group and s is in the group, s-qr=m is in the group too.The notation is a little bit non-rigorous here, but I think it gets the idea across
 
SNOOTCHIEBOOCHEE said:
By division algorithm s= qr+ m . I am guessing we have to somehow show that m=0? Or show that if s is in a group then r is not.

dont know how to do that.

The idea is that the rotation s is between nr and (n+1)r for some n. Then, yes, s-nr must be in the group. Which is bigger s-nr or r? Then remember we picked r to be the smallest nonzero rotation.
 
s-nr is smaller than r.

because we defined s to be between nr and nr+r. so s is at most (by most i mean slightly less than) nr+r. so s<nr+r ==> s-nr<r.

since we chose r to be the smallest non zero rotation we have a contradiction. But i have no clue what this contradiction tells us.
 
It contradicts that the group is finite. An infinite group of group of rotations doesn't need to have a smallest member. And doesn't need to be cyclic.
 

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