#9 please 1:31e s3.amazonaws.comChapter 4 Homework1. Find all the generators of 2»2. List the elements of the subgroups H = <3> and K = <5> in .3. Let a be an element of a group G with |a| = 15. Find the order ofthe following elements of G: - |. |). \|4. Let G = and |a| = 24. List all generators for the subgroup oforder 8.5. Determine the subgroup lattice for Z6. Consider the set G = {4, 8, 12, 16}. Show that this set is a groupunder multiplication modulo 20 (construct its Cayley table). What isthe identity element? Is this group cyclic? If so, find all of itsgenerators.7. Prove that a group of order 3 must be cyclic.8. For any element a in a group G, prove that is a subgroup ofC(a).9. Let G be a cyclic group with |G| = 24 and let a be an element of G.If a' - elidentity) and a" - e(identity) show that G = .10. Show that the group of positive rational numbers undermultiplication is not cyclic.H11. Prove that ""ilo 1f=<} is a cyclic subgroup of GL(2, R)

Let G be a group of order 24 and let a∈G. Given a8≠e and a12≠e. Prove G=<a>.To prove G=<a>, it is enough to show that a=24. Since the order of an element of a group divides the order of the group, the possible orders of a can be 1,2,3,4,6,8,12 and 24. If a=1, then a=e implies a8=e8=e, which is not possible. So, a≠1. If a=2 or 4, then a8=e8=e, which is not possible. So, a≠2 and a≠4. If a=3 or 6, then a12=e12=e, which is not possible. So, a can't be 3 or 6. Hence the only possible order of a is 24. Thus 'a' generates G, that is, G=<a>.