Describe the group generated by K the following cases: i. K = {tau_v}. ii. K = o

Question

Describe the group generated by K the following cases: i. K = {tau_v}. ii. K = ohm_eta. iii. K = {ohm_epsilon, ohm_m}, l perpendicular m. iv. K = {tau_v, tau_w}. v. K = {tau_v middot ohm_l}, l = P + [v]. If l, m, and n are lines of a pencil, prove that ohm_l ohm_m ohm_n = ohm_n ohm_m ohm_l. Let p be a nontrivial rotation with center P. Let v be any vector. Show that ta_v rho is a rotation. Find its center in terms of the given information. Let P, Q, R, and S be four points, no three of which are collinear. Let A, B, C, and D be the respective midpoints of the segments PQ, QR, RS, and SP. Prove that bidirectional AB || bidirectional CD and bidirectional AD || bidirectional BC or they coincide. i. Prove the remark following the definition of perpendicular bisector. ii. Find the perpendicular bisector of the segment joining (-2, 6) and (4, 8).

Explanation / Answer

the group of rotations of a dodecahedron is isomorphic to the alternating groupA5.

Each of the 12 pentagonal faces of the dodecahedron has 5 diagonals, all of the same length. The key observation is that these 60 diagonals are edges of 5 cubes. (Each cube has 12 edges — one in each face of the dodecahedron.) The 5 cubes are permuted by the symmetry group of the dodecahedron.

This defines a homomorphism G S5 of the group G of rotations of the dodecahedron to the group of permutations. It is easy to see geometrically that rotations of order 5 (about centers of faces) 3 (about vertices), and 2 (midpoints of edges) generate permutations of the 5 cubes which are: a 5-cycle, a 3-cycle, and the product of two commuting 2-cycles respectively.

All such permutations are even and non-trivial. Thus the range of the homomorphism lies in A5, the kernel is trivial, and since both groups G and A5 are known to have the same order 60, the monomorphism must be bijective.

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