Question

Problem 6. The set (Z19 − {0}, ·19) is a group with the indicated operation; see the attached table. a.) Show that H = {1, 7, 8, 11, 12, 18} is a subgroup. b.) List all the right cosets of H. c.) Show that if Hy = Hx then xy−1 ∈ H. [Make sure to give a reason for each step.] d.) Show that φ : H → Hx defined by φ(h) = hx is one-to-one and onto. [Use the fact that H is a group and the property of inverses to do this. Part (c) may also come in handy.] e.) Show that the operation ⊗ defined by (Ha) ⊗ (Hb) ≡ Hab is well-defined.

2. 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 1 1 1 4 5 6 7 8 10 11 12. 13. 14 15 17. 18. 214 2] 2 8 10 12. 14 18. 1 3 5 7 9 11 15 17 16. 3 3 6 12 15 18. 2 8 11 14 17 4 7 13 1 4] 8 16 1 5 9 17. 2. 6-5-31 210 6 101 3. 11 4 15) 6-310-71411-7 5 5 10 1 6 11 16 7 12 17 3 18. 9 14 6 111 3-619125-8125-81114 6 12. 5 11. 17 4 16 3 9 418 2519 15 8-31 8/317 14 13. 기 7 14 9 16 4 18 6 13 1 8. 715 10 12 111 8] 8 13 2 10 7 15 4 12 1 14 3 11 9 9 7 16 6 5 14 4 13 3 12. 2. 11 1 10 1 5 17 2 100 10 12 3 13 4 14 5 15 6 16 17 21712 18. 9. 11 11 61811-3-5-719 6 17 9 1 12. 4. 15. 7 18) 100 13 5 16 8. 122 12 10 3 8 13 6 18. 11 4 16, 9 14 7 1 5 13. 21 1917 14 2 15 3 16, 10 4 17 11 5 18 12. 6 1 1 1 1 1 13 14. | 11417 14 8-3 18 8. 3 12. 7 2. 11 6 1 15 101 5. 15. 11 3 00 14 10 6 2 17.. 13 5 1 16. 12. 8. 4 516 619-24 CO 13 10 7 4 1 17 14 11 8. 5 18. 15 12. 9 CC 3. 17 17. 15 13 11 9 7 5 1 18, 16 12 10 8. 6 4 22 311 18. 18 17 16 15 14 12. 10 9 00 7 6 5 4 3 2. 1 w \cdot_{19}

Answer 1

OR SU Ha {17 7,8, 11, 12, 18} Here 1 is the identity element also 7.8= 18 7.11=1, 7.12=8 7.18= 12 11.18=8 8.11 = 12 8.12=1, 8.18 = 11 11.12-18 12.18= 7, 7.7 = 11, 8.8=7, 11.11-7 12.12 = 11, 18.18=1 (H..) is closed 18 = 18 Now 7 = 11, 8-1 = 12, 11 = 7, 12 = inverse of each element belongs in H H is a subgroup of (3-103 , is) (b) Hi= H H2 = = {1,7,8, 11, 12, 18} {2, 14, 16, 3, 5, 17 } {4,9,13,6, 10, 15 } H4= we know that weight Cosets are Rither disjoint of equal Hence all the right Cosets of H H, H2 H4 we have Hy= Hn Then for some hi, h₂EH hiy= hqx xy! hilhi EH [ H is a subgroup nyl EH each has for any a E (7/19 - 203, is), HR forms a Coset Again all the Cosets of t are H, H2, Hu same number of elements, Then Hx is one of these cosets 91 H HR is one-to one and onto and hence

e for cosets (Ha) ® H b) = Hab Now for a,bt (249-{0}, in) abe (2,9 -403, ig) Then Ha Hb and Hab all, are cosets of H Again we know that Cosets are either disjoint or equal Therefore the lunction (Ha) (46)= Hab is well defined