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

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132 Problems of Chapter 1 and are two symmetries of the tetrahedron‚ then is a rotation about the axis through the vertex A. Since the commutant is a normal subgroup (see 116 and 121) the commutant in the group of symmetries of the tetrahedron coincides with the subgroup of rotations. 123. Answer. 24. For the cube: 1) the identity; 2) the rotations (they are 9) by 90°‚ 180°‚ and 270° about the axes through the centres of opposite faces; 3) the rotations (6 in total) by 180° about the axes through the middle pints of opposite edges; 4) the rotations (8 in total) by 120° and 240° about of the axes through opposite vertices. 124. If we join the centres of the adjacent faces of the cube we obtain an octahedron. Thus to every rotation of the cube there corresponds a rotation of the octahedron and vice versa. Moreover‚ to every composition of two rotations of the cube there corresponds a composition of two rotations of the octahedron and we obtain an isomorphism of the group of rotations of the cube in the group of rotations of the octahedron. 125. If we fix the position of the cube and we consider as different two colourings for which at least one face takes a different colour‚ then there are in all 6 · 5 · 4 · 3 · 2 = 720 colourings: indeed‚ by the first colour one can colour any one of the 6 faces‚ by the second‚ any one of the remaining 5‚ and so on. Since for every colouring one obtains 24 distinct colourings by means of rotations of the cube (see 123)‚ we have in all 720/24 = 30 ways of colouring the cube. There exist only 4 rotations transforming a box of matches into itself: the identity and the three rotations by 180° about the axes through the centres of opposite faces. Hence we have 720/4 = 180 ways of colouring a box of matches with 6 colours. 126. Answer. The group of symmetries of the rhombus and the group 127. Hint. a) See 57. b) Use the result that both and either exchange the tetrahedra or fix them. 128. The rotations and of the cube either both exchange the tetrahedra and (see Figure 8) or fix them. Thus each commutator transforms every tetrahedron into itself. Hence to every el-
Solutions 133 ement of the commutant of the group of rotations of the cube there corresponds a rotation of the tetrahedron Let be a rotation of the cube by 90° about the axis through the centres of the faces ABCD and and such that the vertex B is sent onto A. Let be the rotation of the cube by 120° about the axis through the vertices and and such that the vertex A is sent onto Thus the rotation sends the vertex A onto itself (verify) and the vertex onto D‚ i.e.‚ it is a non-trivial rotation of the cube about the axis through the vertices A and This rotation is also a rotation of the tetrahedron about the axis through the vertex A. Now it is easy to show (see 121) that the commutant in the group of rotations of the cube contains all the rotations fixing the tetrahedron. But since it contains only these rotations‚ the commutant in the group of rotations of the cube is isomorphic to the group of rotations of the tetrahedron. 129. Let A and B be two arbitrary cosets and their representant. Since the element belongs to the commutant‚ and therefore AB = BA. 130. Let be two arbitrary elements of the group and let A and B be the cosets to which they belong. Since AB = BA then E. The commutator therefore belongs to a normal subgroup N. In this way all the commutators belong to N and hence N is the commutant. 131. Let be two arbitrary elements of N and an arbitrary element of the group G. Since N is a normal subgroup the elements and belong to N. Thus is a commutator in the normal subgroup N‚ i.e.‚ it belongs to K(N). Hence K(N) is a normal subgroup of the group G. 132. Let and be two arbitrary elements of the group F. Since is a surjective homomorphism of the group G onto the group F there exist two elements and of the group G such that and Thus one has This means that the group F is commutative. The converse proposition is not true: see Example 12 in §1.13. 133. Let Thus Therefore and
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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)
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The complex numbers 69 REMARK. If a
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The complex numbers 71 2.8 Images o
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The complex numbers 73 where all th
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The complex numbers 75 in such func
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The complex numbers 77 FIGURE 26 no
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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 and 136: Solutions 119 hypothesis. The group
Page 137 and 138: Solutions 121 The given group is th
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: Solutions 131 sends every into itse
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
Page 187 and 188: Solutions 171 265. 266. 267. If the
Page 189 and 190: Solutions 173 following way. If the
Page 191 and 192: Solutions 175 for all such that and
Page 193 and 194: Solutions 177 b) In order for to va
Page 195 and 196: Solutions 179 to the condition coin
Page 197 and 198: 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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