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Homework Help: Proving a group of rotaions is cyclic

  1. Nov 13, 2008 #1
    1. The problem statement, all variables and given/known data
    Let G be a finite group of rotation of the plane about the origin. Prove that G is cyclic.


    3. 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 cant for the life of me figure out how to prove this.
     
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  3. Nov 13, 2008 #2

    Dick

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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?
     
  4. Nov 13, 2008 #3
    By division algorithm s= qr+ m . Im 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.
     
  5. Nov 13, 2008 #4

    Office_Shredder

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    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
     
  6. Nov 13, 2008 #5

    Dick

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    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.
     
  7. Nov 13, 2008 #6
    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.
     
  8. Nov 13, 2008 #7

    Dick

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    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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