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By John Perry
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Extra info for Algebra: Monomials and Polynomials
Show that xy −1 ∈ G. 34. Suppose that H is an arbitrary group. Explain why we cannot assume that for every a, b ∈ H , (ab )−1 = a −1 b −1 , but we can assume that (ab )−1 = b −1 a −1 . 35. Let ◦ denote the ordinary composition of functions, and consider the following functions that map any point P = ( x, y ) ∈ R2 to another point in R2 : I (P ) = P , F ( P ) = ( y, x ) , X ( P ) = (−x, y ) , Y ( P ) = ( x, −y ) . (a) (b) (c) (d) (e) Let P = (2, 3). Label the points P , I ( P ), F ( P ), X ( P ), Y ( P ), ( F ◦ X ) ( P ), (X ◦ Y ) ( P ), and ( F ◦ F ) ( P ) on an x-y axis.
Since 〈g 〉 < ∞, there must exist a, b ∈ N+ such that a < b and g a = g b . Using the inverse property and substitution, g 0 = e = g b ( g a )−1 . 54, g 0 = g b −a . By definition, b − a ∈ N+ . Hence S is nonempty. By the well-ordering property of N, there exists a smallest element of S; call it d . Since 〈g 〉 contains n elements, 1 < d ≤ n. If d < n, that would contradict assertion (A) of this theorem (with a = 0 and b = d ). Hence d = n, and g n = e, and we have shown (A). (C) (D) In (B), S is the set of all positive integers m such that g m = e; we let the smallest element be d , and we found that d = n.
Thus, D3 is the smallest non-abelian group. To describe D3 , we start with an equilateral triangle in R2 , with its center at the origin. We want to look at its group of symmetries, where a symmetry of the triangle is a distance-preserving function on R2 that maps points on the triangle back onto itself. 38. Two obvious symmetries of an equilateral triangle are a 120◦ rotation through the origin, and a flip through the y-axis. 1. What functions are symmetries of the triangle? To answer this question, we divide it into two parts.