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HW9sol.F09 - Math 150A Fall 09 Homework 9 – Selected...

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Unformatted text preview: Math 150A - Fall 09 - Homework 9 – Selected Solutions (Based on solutions prepared by Jeff Ferreira ) 5.4.1 Prove that a discrete group G consisting of rotations about the origin is cyclic and is generated by ρ θ where θ is the smallest angle of rotation in G . Proof. This is very similar to a homework problem from last week. In fact, we can use almost the same proof. To convince yourself that a smallest angle of rotation θ exists, notice that the fact that G is discrete assures us the if ρ α and ρ β are in G , then | α- β | ≥ . So this means rotations “aren’t too close together.” It follows that there exists a smallest angle θ . Now to do the proof: Pick an arbitrary ρ α ∈ G . We may write α = kθ + r where 0 ≤ r < θ . Write ρ α = ρ kθ + r = ρ kθ ρ r and since ρ α ∈ G and ρ kθ = ( ρ θ ) k ∈ G , we must have ρ r ∈ G . But this contradicts that θ was minimal, so we must have r = 0. So we showed ρ α = ρ kθ = ( ρ θ ) k , so ρ θ generates G , and G is cyclic. 5.5.2 Prove that (5.4) is an equivalence relation. Proof. (5.4) says: Let S be a G-set. Then we have the relation s ∼ s if s = gs for some g ∈ G . For reflexivity, s ∼ s for all s ∈ S since s = (1) s where 1 is the identity in G . For symmetry, suppose s ∼ s , where s = gs . Then s ∼ s by noticing s = g- 1 s . For transitivity, suppose....
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HW9sol.F09 - Math 150A Fall 09 Homework 9 – Selected...

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