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Unformatted text preview: MODEL ANSWERS TO THE SECOND HOMEWORK 1. Label the vertices of the square A , B , C , D , where we start at the top left hand corner and we go around the square clockwise. In particular A is opposite to C and B is opposite to D . There are three obvious types of symmetries. There are rotations. One obvious rotation R corresponds to rotation clockwise through π/ 2 radians. The others are R 2 , R 3 and the identity I . They represent rotation through π and 3 π/ 2. There are two sorts of flips. One set of flips are diagonal flips. The first D 1 fixes the diagonal AC and switches B and D . The other D 2 fixes the diagonal BD and A and C . The other possibility is to look at the flip F 1 which switches A and D and B and C and the flip F 2 which switches A and B and C and D . I claim that this exhausts all possible symmetries. In fact any symme try is determined by its action on the fours vertices A , B , C and D . Now there are 24 = 4! possible such permutations. On the other hand any symmetry of a square must fix opposite cor ners. Thus once we have decided where to send A , for which there are four possibilities, the position of C is determined, it is opposite to A . There are then two possible positions for B . So there are at most eight symmetries and we have listed all of them. We start looking for subgroups. Two trivial examples are D 4 and { I } . A nontrivial example is afforded by the set of all rotations { I, R, R 2 , R 3 } . Clearly closed under products and inverses. Note that rotation through π radians R 2 generates the subgroup { I, R 2 } . Simliarly, since any flip is its own inverse, the following are all sub groups, { I, F 1 } , { I, F 2 } , { I, D 1 } and { I, D 2 } . Now try combining side flips and diagonal flips. Now F 1 D 1 = R 3 . So any subgroup that con tains F 1 and D 1 must contain R 3 and hence all rotations. From there it is easy to see we will get the whole of G . So we cannot combine side flips with diagonal flips. Now consider combining rotations and flips. Note that F 1 F 2 = R 2 and D 1 D 2 = R 2 by direct computation. We then try to see if { I, F 1 , F 2 , R 2 } is a subgroup. As this is finite, it suffices to check that it is closed underis a subgroup....
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This note was uploaded on 07/15/2008 for the course MATH 111 taught by Professor Long during the Spring '08 term at UCSB.
 Spring '08
 LONG
 Math, Algebra

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