Let n E N. n 23. In Exercise 2.67 and Definition 2.65, the dihedral groups are...

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Let n E N. n 23. In Exercise 2.67 and Definition 2.65, the dihedral groups are defined as the group of rigid motions of a regular n-gon. The following figures represent such a polygon form odd and even: n-10 n odd 84 71-10 Therefore i +2 We may give a different presentation for D, using permutations. First we label the vertices of a regu- 2x lar n-gon as in the figure above. Then a rotation by radians can be represented as a function sending 1+2+3+1. This can be represented by the n-cycle r= (1 2 3 ..n). Hence, all the rotations of the n-gon correspond to the elements in (r). n Since we know that ID,|- 2n, and ((r)]=n, all the non-rotation elements (i.c. reflections) of D, are given by a fixed reflection followed by a rotation, for some k with 0 5k<n. Now we need to know what r looks like, as a permutation. n even . If n is odd, then r must fix a vertex. Without loss of generality we will assumer fixes 1. In this case, 1=(2n) (3 n-1)...(n+¹+3) n If n is even, we may choose t such that r fixes 1 and +1. Then 1=(2n) (3 n-1)--(+2). D={¹; j=0,1 and k=1,2,---,n}. With this presentation, D, may be thought of as a subset of S. The presentation of D, as a subset of S, is very useful, as it allows us to avoid the geometry associated with the first definition of D. However, we know D, is a group, so the product of two elements of the form t should also be of that form, a fact that is not so obvious. The following exercise will give us the tools we need to perform multiplications in D, with ease. Letr and t be as defined above. (a) Describe the product trt geometrically to conclude that tr7 is a rotation. (b) Prove that trt=r¹. (c) Prove that tr] =rt, for all j=1,2,...,n. Let n E N. n 23. In Exercise 2.67 and Definition 2.65, the dihedral groups are defined as the group of rigid motions of a regular n-gon. The following figures represent such a polygon form odd and even: n-10 n odd 84 71-10 Therefore i +2 We may give a different presentation for D, using permutations. First we label the vertices of a regu- 2x lar n-gon as in the figure above. Then a rotation by radians can be represented as a function sending 1+2+3+1. This can be represented by the n-cycle r= (1 2 3 ..n). Hence, all the rotations of the n-gon correspond to the elements in (r). n Since we know that ID,|- 2n, and ((r)]=n, all the non-rotation elements (i.c. reflections) of D, are given by a fixed reflection followed by a rotation, for some k with 0 5k<n. Now we need to know what r looks like, as a permutation. n even . If n is odd, then r must fix a vertex. Without loss of generality we will assumer fixes 1. In this case, 1=(2n) (3 n-1)...(n+¹+3) n If n is even, we may choose t such that r fixes 1 and +1. Then 1=(2n) (3 n-1)--(+2). D={¹; j=0,1 and k=1,2,---,n}. With this presentation, D, may be thought of as a subset of S. The presentation of D, as a subset of S, is very useful, as it allows us to avoid the geometry associated with the first definition of D. However, we know D, is a group, so the product of two elements of the form t should also be of that form, a fact that is not so obvious. The following exercise will give us the tools we need to perform multiplications in D, with ease. Letr and t be as defined above. (a) Describe the product trt geometrically to conclude that tr7 is a rotation. (b) Prove that trt=r¹. (c) Prove that tr] =rt, for all j=1,2,...,n.

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