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The finite subgroups of SO(3).Math 126 lecture 3Morphisms.More examples of morphisms.Counting fix points.isCounting fixed points, 2.We have the mapCounting the fix points 3.soUsing the formula for finite subgroups of SO(3).Using the formula for finite subgroups of SO(3), continued.We havewherer = 2, the cyclic groups.The possibilities for r = 3.TetrahedronCubeOctahedronThe icosahedral group is denoted by I . It is isomorphic to the alternating group on five letters, A .5To visualize this isomorphism it is best to pass to the buckyball which is obtained by truncating each vertex of the icosahedron at each vertex along a plane perpendicular to the radius at each vertex. Since each vertex is formed by the intersection of five triangles, the new faces created in this way are pentagons. The remaining portions of the original triangular faces are converted into hexagons.Icosahedron DodecahedronThe buckyball.Proof of the isomorphism of I with A .There are 15 elements of degree 2. There are 10 three fold axes and so 20 elements of degree 3. There are 6 five fold axes so 24 elements of degree 5. Together with the identity element we get 1+15+20+24 =60 accounting for all the elements in I. The group I acts transitively on the set of 15 axes of degree 2. So the isotropy group of one such axis has 4 elements. Each of these four elements other than the identity must therefore be of degree 2 and so they all commute. This implies that their axes must be mutually perpendicular. So the set of 15 axes of degree 2 breaks up into 5 sets of configurations each consiting of 3 mutually perpendicular axes. So the group I acts transitively on a set with 5 elements and no element of I other than the identity acts trivially. So we have an isomorphism of I with a subgroup of order 60 of the permutation group on 5 objects, and the only such subgroup is the subgroup consisting of even permutations, i.e the group A


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HARVARD MATH 126 - Lecture 3

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