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1. IZ ,l = 12; IU(10)1 = 4; IU(12)1 = 4; IU(20)1 = 8; ID,1 = 8 In Z,,, 101 = 1; 111 = 151 = 171 = (11/ = 12; 121 = 1101 = 6; 131 = 191 = 4; 141 = 181 = 3; 161 = 2. In U(10), 111 = 1; 131 = 171 = 4; 191 = 2. In U(12), |11 = 1; 151 = 2; 171 = 2; 1111 = 2. In U(20), 111 = 1; 131 = 171 = 1131 = 1171 = 4; 191 = 111/ = 1191 = 2. In DA, Rol = 1; IRgo! = IR2701 = 4; IR 1801 = IHI = IVI = IDI = ID'1 = 2. In each case, notice that the order of the element divides the order of the group. 3. In Q, 101 = 1 and all other elements have infinite order. In Q*, |1| = 1, 1-11 = 2, and all other elements have infinite order. 5. Each is the inverse of the other. 7. (atc-2b4) -1 = b-4cza-4 =b3cza- 9. DA; D ; it contains { Ro, R180, H, V} 11. If n is a positive integer, the real solutions of x" = 1 are I when n is odd and + 1 when n is even. So, the only elements of finite order in R* are +1. 13. H is a subgroup. To prove this we need only show that if a E H then a ' E H. But if a ! & H, then the given property says that a = (a -])-! & H. 15. Since lal = 7 we have a = alta = all = (as). 17. If a and b are distinct elements of order 2, then ab has order 2 and is distinct from a and b. If c is a fourth element of order 2, then ac, be, and abc make at least 7 elements of order 2. D, has exactly five elements of order 2. 19. Suppose that m 0, use induction. For n