College Algebra Exam Review 236

College Algebra Exam Review 236 - 246 5 ACTIONS OF GROUPS...

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Unformatted text preview: 246 5. ACTIONS OF GROUPS Let r be the rotation of 2 =9 of the nonagon. For any k (1 Ä k Ä 8), r k either has order 9, and hr k i acts transitively on vertices, or r k has order 3, and hr k i has three orbits, each with three vertices. In either case, there are no fixed arrangements, since it is not possible to place beads of one color at all vertices of each orbit. Now consider any rotation j of about an axis through one vertex v of the nonagon and the center of the opposite edge. The subgroup fe; j g has one orbit containing the one vertex v and four orbits containing two vertices. In any fixed arrangement, the vertex v must have a white bead. Of the remaining four orbits, two must be colored red, one white and one yellow; there are 4Š D 12 2Š1Š1Š ways to do this. Thus, j has 12 fixed points in X . Since there are 9 such elements, there are 1 1X jFix.g/j D .1260 C 9.12// D 76 jG j 18 g 2G possible necklaces. Example 5.2.4. How many different necklaces can be made with nine beads of three different colors, if any number of beads of each color can be used? Now the set X of arrangements of beads has 39 elements; namely, each of the nine vertices of the nonagon can be occupied by a bead of any of the three colors. Likewise, the number of arrangements fixed by any g 2 D9 is 3N.g/ , where N.g/ is the number of orbits of hg i acting on vertices; each orbit of hg i must have beads of only one color, but any of the three colors can be used. We compute the following data: n-fold rotation axis, nD 9 9 2 order of rotation 1 9 3 2 Thus, the number of necklaces is 1X 1 jFix.g/j D .39 C 2 jG j 18 N.g/ number of such group elements 9 1 1 6 3 2 5 9 33 C 6 3C9 35 / D 1219: g 2G Example 5.2.5. How many different ways are there to color the faces of a cube with three colors? Regard two colorings to be the same if they are related by a rotation of the cube. ...
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