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Unformatted text preview: Homework #9 Solutions Due: November 18, 2011 1. For each of the following subsets of the plane, determine the symmetry group. Re member that the possible symmetry groups are C n and D n for some n . See De nition 5.2.4, the bottom of page 217 and the rst paragraph of page 218 for the description of C n and D n . (a) (b) (c) I Solution. (a) The only symmetries are rotations by multiples of 90 , so the sym metry group is C 4 . (b) Since the dots at the the midpoints of each edge of the square, the symmetries are all rotations by multiples of 90 and re ections through the horizontal and vertical lines through the center and re ections through the pairs of opposite diagonals. Thus, the symmetry group is the same as the symmetry group of a square, namely, D 4 . (c) The symmetries are the re ections through the horizontal and vertical lines through the center of the square, plus the rotation by 180 . Hence, the symmetry group is the group D 2 . J 2. If = 1 2 3 4 5 then G = ( ), the cyclic subgroup of S 5 generated by , is a subgroup of S 5 of order 6. As a subgroup of S 5 , G acts on the set X = f 1 ; 2 ; 3 ; 4 ; 5 g . (a) Determine the orbit, Orb( x ) for each x 2 X . I Solution. Orb(1) = k (1) : k 2 Z = f 1 ; 2 ; 3 g = Orb(2) = Orb(3), while Orb(4) = k (4) : k 2 Z = f 4 ; 5 g = Orb(5). J (b) Determine the stabilizer subgroup Stab( x ) for each x 2 X . I Solution. G = f e; ; 2 ; 3 ; 4 ; 5 g , so we can write each element of G in disjoint cycle form as e = (1)(2)(3)(4)(5) = 1 2 3 4 5 2 = 1 3 2 3 = 4 5 4 = 1 2 3 5 = 1 3 2 4 5 Math 4023 1 Homework #9 Solutions Due: November 18, 2011 Since Stab(1) = f g 2 G : g (1) = 1 g , we see from the above table that the only elements of G that do not move 1 are e and 3 . Thus, Stab(1) = f e; 3 g = Orb(2) = Orb(3). Similarly, the only elements of G that do not move 4 are e , 2 , and 4 . Thus, Stab(4) = f e; 2 ; 4 g = Stab(5). J (c) Use your results in part (a) to verify that...
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This document was uploaded on 12/28/2011.
 Fall '09

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