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If G is abelian, then ab = b a. 4 has reduced the number of requirements for a subgroup from four to three. Amazingly, we can simplify this further, to only one criterion. 5 (The Subgroup Theorem). Let H ⊆ G be nonempty. The following are equivalent: (A) H < G; (B) for every x, y ∈ H , we have xy −1 ∈ H . 6. Observe that if G were an additive group, we would write x − y instead of xy −1 . Proof. 33 on page 31, (A) implies (B). Conversely, assume (B). 4, we need to show only that H satisfies the closure, identity, and inverse properties.

Compare ϕρ = −1 0 0 1 3 2 − 12 − 12 − 3 2 = 1 2 3 2 3 2 − 12 and ρ2 ϕ = = = − 12 − 3 2 3 − 12 2 − 12 23 − 23 − 12 1 3 2 2 3 − 12 2 3 2 − 12 − 12 − 3 2 −1 0 0 1 −1 0 0 1 . 41? It implies that multiplication in D3 is non-commutative! We have ϕρ = ρ2 ϕ, and a little logic (or an explicit computation) shows that ρ2 ϕ = ρϕ: thus ϕρ = ρϕ. 42. In D3 , ρ3 = ϕ 2 = ι. Proof. You do it! 43. Exercises. 43. Show explicitly (by matrix multiplication) that in D3 , ρ3 = ϕ 2 = ι. 44. The multiplication table for D3 has at least this structure: ◦ ι ϕ ρ ρ2 ρϕ ρ2 ϕ ι ι ϕ ρ ρ2 ρϕ ρ2 ϕ ϕ ϕ ρ2 ϕ ρ ρ ρϕ 2 ρ ρ2 ρϕ ρϕ ρ2 ϕ ρ2 ϕ Complete the multiplication table, writing every element in the form ρ m ϕ n , never with ϕ before ρ.

If • α does not move the origin; that is, α (0, 0) = (0, 0), and • the distance between α ( P ) and α ( R) is the same as the distance between P and R for every P , R ∈ R2 , then α has one of the following two forms: ρ= cos t − sin t sin t cos t ∃t ∈ R cos t sin t sin t − cos t ∃t ∈ R. or ϕ= The two values of t may be different. (You might wonder why we assume that the origin doesn’t move. Basically, this makes life easier. If it bothers you, try to see if you can prove that the origin must remain in the same place under the action of a function α that preserves both distance and a figure centered at the origin.

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Algebra: Monomials and Polynomials by John Perry


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