Question

3) Which one of the following groups cyclic? a) Z, ZA b) Z, Z c) Z, Z d) Z, Z6 e) Z.Z f) None

Please explain why each answer is wrong or correct

Thank you

Answer 1

3 a) Zzx 24 0,3 - 62 (0, 1), (0, 2) (1,0), (1, 1), (1, 2), (1,3 order (10,0)) = 0 order (0, 1) = 4 01) +(0,1)+(0,1) H0,1 = (0,0 order (0, 2) = (16, 2) +(0,2)=100) order (0, 3) = 4 order (61; 0)) = 4 order (C1,1)) = 4 order ((1, 2)) = 2 order ((1,3)) = 4 | Zzx 74=2 there doesn't exist an element of ordero .. 22 X 24 is not cyclic 2 But

b - 12 Z₂ X ZG f(0, 1), 60,2), (0,3), (0,4), (0,5) (0, 0, 1, 1),(1,2),(13),(4) (1,2), (40) order (10,0)) = 1 ordes ((enl)= order {(1,0)) - order(0,5) = = order in sy = orderony: Order (012) Order (14) = 6 order ((0, 2)) = order (2014)) = 3 Order (0.2)) = order (C1, 3)) = 4 - No element of arder exist in Z 2X 22 2 X Z1G is not cyclic Zyx 24 = f(0, 0), cond), (0, 2), (03) (1,0), (1, 1), (1,2),(1,3) (2,0), (2,1),(2;2), (2,3) (WDE (111)+(41) = (2/2) CD= (ul) +(2,2) = (0,3) (11)' = (1,1)+(03) = (1, 0) (1,1)= (1,1)+(,0) = (2,1) - (111)+(2, 1) = (0,2) Herce

Date (113) CWD (11) + (0,2) = 0 (1,99 = (1,1)+(1.3) = (2,0) (1, 1) = (1,1)+(2,0) (0, 1) (wwe= (1,) + (0,0) = (1/2) (LU" = (1,1)+ (1,2) = (2,3) (111) = (11) +(2,3) = (0,0) e order (6,1)) =12 f Cul) is generator of ZzxZ4 Z 3 x 24 is cyclic d) Z ₂ x Z6 There is one result ZnX Zim is cyclic iff gcd com = 1 ged (36) = 3 = Zzx 76 is not cyclic Here

24 X 76 grd (4,6)=2 #1 . 24 X 26 is not cyclic 6 .