Let G be the symmetric group S5. Can you find a cyclic subgroup of G of order(i) 1 (ii) 2 (iii) 3 (iv) 4 (v) 5 (vi) 6 (vii) 7?

Sure, let's find the cyclic subgroups of the symmetric group S5 of the given orders.

(i) The cyclic subgroup of order 1 is the trivial group {e}, where e is the identity element.

(ii) The cyclic subgroup of order 2 can be generated by a 2-cycle. For example, the subgroup generated by (12) is {(12), e}.

(iii) The cyclic subgroup of order 3 can be generated by a 3-cycle. For example, the subgroup generated by (123) is {(123), (132), e}.

(iv) The cyclic subgroup of order 4 can be generated by a 4-cycle. For example, the subgroup generated by (1234) is {(1234), (1243), (1342), e}.

(v) The cyclic subgroup of order 5 can be generated by a 5-cycle. For example, the subgroup generated by (12345) is {(12345), (12453), (12534), (13524), e}.

(vi) The cyclic subgroup of order 6 does not exist in S5. This is because the order of a cyclic subgroup is equal to the order of its generator, and there is no element in S5 of order 6.

(vii) The cyclic subgroup of order 7 does not exist in S5. This is because the order of a cyclic subgroup is equal to the order of its generator, and there is no element in S5 of order 7.