Question

Hello, please kindly show all your steps without skipping any part. Please also cite any defintions if necessary generously to help me understand this question. Thanks in advance.

4) [3 marks) Recall that in Question 2a from Assignment 6, you proved that H = {{0), (4), (8]} is a subgroup of Zi2. Write out all of the distinct cosets of H in Z12. Determine (Z12: H).

Answer 1

We Z 12 know that co [0],[1].[2].[3].[4],[5],[6],[7], [8], [9],[10],[11] } a is group with respect to addition modulo 12. Now, H = [0], [4],[8]} S Z22 H is a have to show that we sub group First of all, of Zez. additive identity the [0] E H. Then, [4] tu [o] [4] to, [o] = [4 [8] tu [o] - [O] ts [8] = [8] and with respect to Also, [4] too [8] = [12] = [o] (with addition modulo 12) So, (H , too) is closed addition modulo have that every Also, element additive inverse has 12. seen we So. is (H, +42) Therefore, t group itself subgroup of a is

in Z12 H is Distinct Cosets of h i [o] t22 H itself coset of it in Z12. i) [1] + 2 H = = {1+], [5). [9]} in) [2] + H = {[2]. [6], [207] iv) [3] tazh [3], [7], [11] } +22H = And then [a] tiz H= [4], [12] . [8] -{[0], [4], [8]} [o +12H H ( أن in Z 12 So, in and i are the distinet Cosets of H o(242) (12) Now [Z12:H] - - 4 where o (2/12) = number of elements in 2412 = number of elements in H. & 3