MATH 111 m3 - MODEL ANSWERS TO THE THIRD HOMEWORK 1(b...

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MODEL ANSWERS TO THE THIRD HOMEWORK 1. (b) Circles centre the origin. (c) The real line union , where the number m R { } represents the slope. 2. Chapter 2, Section 4: 9. [0] = 0 + H = { [0] , [4] , [8] , [12] } [1] = 1 + H = { [1] , [5] , [9] , [13] } [2] = 2 + H = { [2] , [6] , [10] , [14] } [3] = 3 + H = { [3] , [7] , [11] , [15] } 2. Chapter 2, Section 4: 10. Four. 2. Chapter 2, Section 4: 13. First we write down the elements of U 18 . These will be the left cosets, generated by integers coprime to 18. Of the integers between 1 and 17, those that are coprime are 1, 3, 5, 7, 11, 13 and 17. Thus the elements of U 18 are [1], [5], [7], [11], [13] and [17]. We calculate the order of these elements. [1] is the identity, it has order one. Consider [5]. [5] 2 = [5 2 ] = [25] = [7] , as 25 = 7 mod 18. In this case [5 3 ] = [5][5 2 ] = [5][7] = [35] = [17] , as 35 = 17 mod 18. We could keep computing. But at this point, we can be a little more sly. By Lagrange the order of g = [5] divides the order of G . As G has order 6, the order of [5] is one of 1, 2, 3, or 6. As we have already seen that the order is not 1, 2 or 3, by a process of elimination, we know that [5] has order 6. As [17] = [5] 3 , [17] 2 = [5] 6 = [1]. So [17] has order 2. Similarly, as [7] = [5] 2 , [7] 3 = [5] 6 = [1]. So the order of [7] divides 3. But then the order of [7] is three. It remains to compute the order of [11] and [13]. Now one of these is the inverse of [5]. It must then have order six. The other would then be [5] 4
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