Abel's theorem in problems and solutions - School of Mathematics

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120 Problems of Chapter 1 2) The orders of elements and are equal to 2, i.e., We are now looking for the element corresponding to product It is possible neither that (otherwise and nor that (otherwise nor that (otherwise There remains therefore only Similarly one obtains and The multiplication table is complete and we obtain (see 6) the group of symmetries of the rhombus, isomorphic to (see 76). b) The elements of the group can be of order 1, 2, 3 or 6 (see 83). We will consider some cases. 1) Suppose that there exists an element of order 6. The given group is thus the cyclic group 2) Suppose that all elements are of order 2. Thus the group is commutative (see 25), and if and are elements of the initial group the elements form a subgroup of it. This is not possible (see Lagrange’s theorem) and this case must be excluded. 3) All elements are of order 2 or 3, and there exists an element of order 3. Let be the element of order 3 and c an element which is not a power of Thus is a left cosets of the subgroup and the six elements are thus all distinct (see 82). We will prove that in this set of 6 elements there exists only one way to define a multiplication table. At first we prove that In fact, it is not possible that (otherwise If (or ), then we should have (or but we have supposed that all elements have order 2 or 3. Thus Since is an arbitrary element which does not belong to the subgroup we also have that and In this way the product of all pairs of the 6 elements of the group has been defined. Indeed, Consequently there is only one possibility of constructing a multiplication table in order to obtain a group. Hence there exists only one group of 6 elements, with orders equal to 1, 2, and 3. We know this group: it is the group of symmetries of the equilateral triangle. c) the elements of the group can have order 1, 2, 4, or 8 (see 83). Consider some cases. 1) Suppose that there exists an element of order 8. The given group is thus the cyclic group 2) Suppose that all the elements different from the unit have order 2.
Solutions 121 The given group is thus commutative (see 25). In this case let and be two distinct elements of order 2 of the initial group. Thus form a subgroup. If the element does not belong to this subgroup then the elements form a right coset of the subgroup and the 8 elements are thus all distinct. The products of these elements are well defined, because the group is commutative and (for example, Therefore if all elements have order 2 then there is only one possible group. This group is, in fact, 3) Suppose the element have order 4 and that amongst the elements, different from the powers of there is an element of order 2, i.e., In this case and are the two left cosets of the subgroup and the 8 elements that they contain are thus all distinct. We look for which of these elements may be equal to product It is possible neither that (otherwise nor (otherwise ). If then and (since Thus one obtains the contradiction This means that either or Let us consider the two cases: The table of multiplication is in this case uniquely defined. Indeed, We can therefore have only one group: this group, in fact, does exist: it is the group If and are the unit and the generator of and the unit and the generator of then it suffices to write for all the properties listed above being satisfied. Also in this case the table of multiplication is uniquely defined. Indeed, In this case we therefore have only one group. In fact, this group does exist: it is the group of symmetries of the square. It suffices to put = rotation by 90°, = reflection with respect to a diagonal, for all the properties listed above being satisfied. 4) Suppose that there exists an element of order 4 and that all elements different from are of order 4 as well. Let be an arbitrary element different from Thus the elements are all distinct. We look for which of these elements is equal to product It is possible neither that (otherwise nor (because the order of is 4). If (or ) one has the
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ABEL’S THEOREM IN PROBLEMS AND SO
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Abel’s Theorem in Problems and So
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Contents Preface for the English ed
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A.10.3 The integrable case A.11 Top
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Preface to the English edition by V
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of the differential equations. Unfo
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Preface In high school algebraic eq
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Introduction We begin this book by
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Introduction 3 where by is indicate
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Introduction 5 By Viète’s theore
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Introduction 7 We will be able to p
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Chapter 1 Groups 1.1 Examples In ar
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Groups 11 the triangle ABC into its
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Groups 13 7. Find all symmetries of
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Groups 15 notations (see the soluti
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Groups 17 element, which we will de
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Groups 19 35. Suppose that is the m
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Groups 21 If a group is isomorphic
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Groups 23 tude); (rotation by 180°
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Groups 25 Hence left cosets generat
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Groups 27 as a permutation of the t
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Groups 29 102. Let be the order of
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Groups 31 111. Find all normal subg
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Groups 33 FIGURE 8 127. Prove that
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Groups 35 137. Prove that ker is a
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Groups 37 We now observe what happe
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Groups 39 FIGURE 12 vertices; 4) ro
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Groups 41 Every permutation of degr
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Groups 43 180. Prove that by multip
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Chapter 2 The complex numbers When
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The complex numbers 47 195. Prove t
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The complex numbers 49 where is the
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The complex numbers 51 Consequently
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The complex numbers 53 For complex
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The complex numbers 55 think that n
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The complex numbers 57 214. Let us
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The complex numbers 59 219. Prove t
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The complex numbers 61 functions (s
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The complex numbers 63 of continuit
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The complex numbers 65 erations of
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The complex numbers 67 a) b) c) d)
Page 85 and 86: The complex numbers 69 REMARK. If a
Page 87 and 88: The complex numbers 71 2.8 Images o
Page 89 and 90: The complex numbers 73 where all th
Page 91 and 92: The complex numbers 75 in such func
Page 93 and 94: The complex numbers 77 FIGURE 26 no
Page 95 and 96: The complex numbers 79 C from to th
Page 97 and 98: The complex numbers 81 DEFINITION.
Page 99 and 100: The complex numbers 83 FIGURE 32 cu
Page 101 and 102: The complex numbers 85 to infinity
Page 103 and 104: The complex numbers 87 property is
Page 105 and 106: The complex numbers 89 means will b
Page 107 and 108: The complex numbers 91 For example,
Page 109 and 110: The complex numbers 93 of the Riema
Page 111 and 112: The complex numbers 95 in Figure 38
Page 113 and 114: The complex numbers 97 permutation
Page 115 and 116: The complex numbers 99 2.13 Monodro
Page 117 and 118: The complex numbers 101 First we pr
Page 119 and 120: The complex numbers 103 and prove t
Page 121 and 122: Chapter 3 Hints, Solutions, and Ans
Page 123 and 124: Solutions 107 positive integer unde
Page 125 and 126: Solutions 109 on the left and by on
Page 127 and 128: Solutions 111 39. a) See Table 8; b
Page 129 and 130: Solutions 113 of the real numbers i
Page 131 and 132: Solutions 115 to all and therefore
Page 133 and 134: Solutions 117 is a subgroup of the
Page 135: Solutions 119 hypothesis. The group
Page 139 and 140: Solutions 123 of symmetries of the
Page 141 and 142: Solutions 125 i.e., there exists an
Page 143 and 144: Solutions 127 Answer. The normal su
Page 145 and 146: Solutions 129 In this way the quoti
Page 147 and 148: Solutions 131 sends every into itse
Page 149 and 150: Solutions 133 ement of the commutan
Page 151 and 152: Solutions 135 143. Let be the three
Page 153 and 154: Solutions 137 150. See 57. Suppose
Page 155 and 156: Solutions 139 subgroup. Hence if a
Page 157 and 158: Solutions 141 is commutative. Since
Page 159 and 160: Solutions 143 180. Since the lower
Page 161 and 162: Solutions 145 FIGURE 43 190. The pe
Page 163 and 164: Solutions 147 permutations of type
Page 165 and 166: Solutions 149 is a field. 195. We h
Page 167 and 168: Solutions 151 necessarily that and
Page 169 and 170: Solutions 153 Let C be the field of
Page 171 and 172: Solutions 155 in as numerators and
Page 173 and 174: Solutions 157 c) d) e) where 226. W
Page 175 and 176: Solutions 159 function of real argu
Page 177 and 178: Solutions 161 i.e., Consider the re
Page 179 and 180: Solutions 163 Choose as the smalles
Page 181 and 182: Solutions 165 Since the function is
Page 183 and 184: Solutions 167 FIGURE 55 FIGURE 56 F
Page 185 and 186: Solutions 169 FIGURE 60 257. Suppos
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Solutions 171 265. 266. 267. If the
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Solutions 173 following way. If the
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Solutions 175 for all such that and
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Solutions 177 b) In order for to va
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Solutions 179 to the condition coin
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Solutions 181 are continuous functi
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Solutions 183 turn around the other
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Solutions 185 sought is shown in Fi
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Solutions 187 FIGURE 86 FIGURE 87 p
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Solutions 189 sum of multi-valued f
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Solutions 191 316. a) Let and be th
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Solutions 193 by the formal method,
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Solutions 195 The correct scheme is
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Solutions 197 branch of the functio
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Solutions FIGURE 110 FIGURE 111 329
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Solutions 201 It follows that and b
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Solutions 203 340. To every sheet o
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Solutions 205 Since by hypothesis t
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Solutions 207 moving along a curve
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Drawings of Riemann surfaces 209 Th
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FIGURE 119 211
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FIGURE 121 213
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FIGURE 123 215
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FIGURE 125 217
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FIGURE 127 219
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Appendix by A. Khovanskii: Solvabil
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Solvability of Equations 223 functi
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Solvability of Equations 225 Suppos
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Solvability of Equations 227 where
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Solvability of Equations 229 finds
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Solvability of Equations 231 quadra
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Solvability of Equations 233 More p
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Solvability of Equations 235 group.
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Solvability of Equations 237 Every
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Solvability of Equations 239 The gr
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Solvability of Equations 241 FIG. 1
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Solvability of Equations 243 are ob
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Solvability of Equations 245 corres
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Solvability of Equations 249 up to
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Solvability of Equations 251 fined
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Solvability of Equations 253 the qu
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Solvability of Equations 255 the no
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Solvability of Equations 257 THEORE
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Bibliography [1] [2] [3] [4] [5] [6
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[23] [24] [25] [26] [27] [28] 263 V
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Appendix (V.I. Arnold) The topologi
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Index Abel’s theorem, 6, 103 addi
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pack of sheets, 96 parametric equat
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Abel's theorem in problems and solutions - School of Mathematics

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