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College Algebra Exam Review 206

# College Algebra Exam Review 206 - The telling observation...

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216 4. SYMMETRIES OF POLYHEDRA There are also three 4–fold axes through the centroids of opposite faces, giving nine nonidentity group elements, three of order 2 and six of order 4. See Figure 4.1.6 on the previous page . Finally, the cube has six 2–fold axes through centers of opposite edges, giving six order 2 elements. See Figure 4.1.7 . With the identity, we have 24 rotations. Figure 4.1.7. Two– fold axis of the cube. Figure 4.1.8. Diago- nals of cube. As for the tetrahedron, we can show that we have accounted for all of the rotational symmetries of the cube. See Exercise 4.1.8 . Now we need to find an injective homomorphism into some permuta- tion group induced by the action of the rotation group on some set of geo- metric objects associated with the cube. If we choose vertices, or edges, or faces, we get homomorphisms into S 8 , S 12 , or S 6 respectively. None of these choices look very promising, since our group has only 24 elements.
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Unformatted text preview: The telling observation now is that vertices (as well as edges and faces) are really permuted in pairs. So we consider the action of the rotation group on pairs of opposite vertices, or, what amounts to the same thing, on the four diagonals of the cube. See Figure 4.1.8 . This gives a homomor-phism of the rotation group of the cube into S 4 . Since both the rotation group and S 4 have 24 elements, to show that this is an isomorphism, it sufﬁces to show that it is injective, that is, that no rotation leaves all four diagonals ﬁxed. This is easy to check. Proposition 4.1.4. The rotation group of the cube is isomorphic to the permutation group S 4 . The close relationship between the rotation groups of the tetrahedron and the cube suggests that there should be tetrahedra related geometrically...
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