Question

In Exercises 27 through 35, find the order of the cyclic subgroup of the given group generated by the indicated element. 27. The subgroup of Z4 generated by 3 28. The subgroup of V generated by c (see Table 5.9) 29. The subgroup of U6 generated by cos 3 + i sin 30. The subgroup of Z10 generated by 8 31. The subgroup of Z16 generated by 12 32. The subgroup of the symmetric group S8 generated by (2,4,6,9)(3,5,7) 33. The subgroup of the symmetric group S1o generated by (1, 10)(2,9)(3.8)(4, 7)(5,6) 34. The subgroup of the multiplicative group G of invertible 4 x 4 matrices generated by

Answer 1

The order of a cyclic subgroup generated by an element in a group is determined by the smallest power of the element that results in the identity element. In the examples given, the order is found by examining the powers of given elements in their respective modular arithmetic or by finding the LCM of the cycle lengths in permutations.

Step-by-step explanation:

The order of a cyclic subgroup generated by an element in a group is the smallest positive integer n such that the nth power of the element is the identity element of the group. For instance, in Z4, the cyclic subgroup generated by 3 consists of {3, 2, 1, 0} since 31 = 3, 32 = 1 (mod 4), 33 = 3 (mod 4), and 34 = 1 (mod 4), which is the identity element. Thus, the order is 4. For the cyclic subgroup of Z10 generated by 8, we get {8, 4, 2, 6, 0} since 81 = 8, 82 = 4 (mod 10), 83 = 2 (mod 10), 84 = 6 (mod 10), and 85 = 0 (mod 10), which is the identity, so the order is 5.

For the symmetric group S8 generated by the permutation (2,4,6,9)(3,5,7), one should note that the order of a permutation is the least common multiple (LCM) of the lengths of its disjoint cycles. Here, the cycles (2,4,6,9) and (3,5,7) have lengths 4 and 3, respectively. Therefore, the order of the subgroup generated by this element is the LCM of 4 and 3, which is 12.

In S10, the permutation (1, 10)(2,9)(3,8)(4, 7)(5,6) is the product of 5 transpositions, each of order 2. Because these transpositions are disjoint, the order of this permutation is the LCM of 2, 2, 2, 2, and 2, which is simply 2.