Question

IV. Orders of Elements. (a). List the elements of the group 2, XZ, and give the orders of each. (b). Is Z4 Z2 cyclic? If so, find a generator. If not, explain why not.

Answer 1

q @ ze = 50, 1, 2, 3) Z2 = {0,13 24 x Z2 = {(0,0), (010), (1.0), (6), (2.0), (2,1) (3,0), (3,1)) List of all elements of the group Z4 X Zg is {roio (ov). COCW, (2,0,0,1),(310,(3,1); 0 (Z¢ X Zg) = 0 (29) 0 (22) = 9.2=8 ® Order of each element of zq X Zg is. (010) E ZqXz2. Then of(0,0)) = 1 (0) EZ4X22Then o (CO) = 2 B (CO) E Zq XZ2. Then (og) = 4 (111) € 2q X Z₂. Then o((1)) = 4 (2,0) E 2q X 22. Then 0 (1210)) = 2
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(21) € 2qx 23. Then o (C20) = 2 (310) E Z q X 22. Then of (3,0)) = 4 (3,1) EZq XZg. Then O( (3.1)) = 4 6 Za X Z2 = {(010), (0.), Clio), Cri), (2,0), (21) (3,0), (311) } 0 (24X22) = 0(24.0(29) = 4.258 O ( 29 X 22)=8 24 X 22 is hat cyclic group since, O (26X22)=8 and ho elements of zq X 2g has order 8. Hence, no elements of Zq X Zo generates all elements of Zq X Za Therefore, Zq x Zg is hot cyclic
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