Please i, ii, iii, iv, v, vi.

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Let T be the group of the 12 rotational symmetries of a tetrahedron. (i) T acts on the set of 4 vertices of the tetrahedron. Is this action faithful? (ii) Dcfinc an OEP (opposite edge pair) of the tetrahedron as follows: Two edges form an OEP if they do not share a vertex. The tetrahedron has 6 edges, which are partitioned into 3 OEPs. T acts faithfully on the set of 6 edges. However, show that the action of T on the set of OEPs is not faithful. (iii) By making use of part (ii) and the permutation representation of the action, show that T has a normal subgroup other than {1} and T. In other words, show T is not a simple group. (iv) How many elements of order 3 are in T? How many elements of order 2? How many elements of order 4? How many elements of order 6? (v) Find the class equation of T. (Note that the elements of order 3 are not all conjugate to each other; explain why.) (vi) Use the class equation to prove that T does not have a subgroup of order 6. TT 1 mo A normal subgroup is closed ut under Conjugation. Therefore in T, it a permutation of a a cycle is exists, then all the permutations of this cycle type are in norunal Subqoroup (1234) Now, we will calculate the number of types of each Cycle, 1 only one cycle of the typec). d) 4.3% 6 cycles of the type (18). 3) 4. 3. 2/2 = g cycles of the type (125). +) 4:3.211668.812) - 6 cycles of the type E27 (39) 5) 4.3.2. // (2.2.2) = 3 cycles of the type (12) (34) up of all I 6, 8, 6 and 3-24 to get 2 in two different way. But, there is only one way of doing this Hence, there Can only be a subgroup of order Tsy sum . 12 is SA Hence, This action is tarthbul. Let T be the group of the 12 rotational symmetries of a tetrahedron. (i) T acts on the set of 4 vertices of the tetrahedron. Is this action faithful? (ii) Dcfinc an OEP (opposite edge pair) of the tetrahedron as follows: Two edges form an OEP if they do not share a vertex. The tetrahedron has 6 edges, which are partitioned into 3 OEPs. T acts faithfully on the set of 6 edges. However, show that the action of T on the set of OEPs is not faithful. (iii) By making use of part (ii) and the permutation representation of the action, show that T has a normal subgroup other than {1} and T. In other words, show T is not a simple group. (iv) How many elements of order 3 are in T? How many elements of order 2? How many elements of order 4? How many elements of order 6? (v) Find the class equation of T. (Note that the elements of order 3 are not all conjugate to each other; explain why.) (vi) Use the class equation to prove that T does not have a subgroup of order 6. TT 1 mo A normal subgroup is closed ut under Conjugation. Therefore in T, it a permutation of a a cycle is exists, then all the permutations of this cycle type are in norunal Subqoroup (1234) Now, we will calculate the number of types of each Cycle, 1 only one cycle of the typec). d) 4.3% 6 cycles of the type (18). 3) 4. 3. 2/2 = g cycles of the type (125). +) 4:3.211668.812) - 6 cycles of the type E27 (39) 5) 4.3.2. // (2.2.2) = 3 cycles of the type (12) (34) up of all I 6, 8, 6 and 3-24 to get 2 in two different way. But, there is only one way of doing this Hence, there Can only be a subgroup of order Tsy sum . 12 is SA Hence, This action is tarthbul