Question

Q3 Consider the group (Z3 x Z3, +), where again Z3 x Z3 = {(a,b) a, b e Z3}, and we define addition by (a, b) + (c,d) = (a + cb+d). So, for instance, (1, 2) + (0,2)=(1,1). (a) List all of the subgroups of (Z3 X Z3, +). You should explain why your list is complete (i.e., why there are no subgroups other than the ones you have written). (b) Which of these subgroups are cyclic? You do not need to justify your answers. (10 marks)

Answer 1

consider the group (Zig XZ8 , +) (a) As ordez of Z3 = 9 order of group Zgx Za = 3x3, Thus, by Lagranges theorem, Z3xZz can have subgroup of order 1,8,9 CAs order of subgroup divides order of group). for order H1 = 3 Co,oof is the only subgroup of ordet I [only subgroup of ordet I is subgroup containing only identity] for order a since every group is a subgroup of itseff, ZzxZz is itself a subgroup of its Z3xZz . Only subgroup of order g is H2 = 23x23 for order 3 Hg = {(1,0), (2,0), (0,0} = "Zx 203* H4={(0,09, 20,19,20,2)} = {0} x 23 H5= 2 0,0), (1,19,22,273 H6 = ?10,0), (1,2), (2,7)} There are 4 subgroups of order 3.

As every element other than pdentity hat order 3., there are 9-1 = 8 elements of order 3. ... If x has order of it also has orders This . x, xl & identity constitude 1 subgroup FG So there are such 8/2=4 pairs. - Exactly 4 subgroups of ordea 3 are these ? Thus. Hz ={e}, H2 = Zzx Z3, H3 = 303xZ, Hepa Z3x?o} H5= { 00,0), (1,2), (2,77}, H6 ={10,0), (1,1),(24293 are the only 6 subgroups of G=ZzXZz. Hi= 2e}= ke> Hiis cyclic Every group of order prime is cylic --- All 4 subgroups H3, H4, H5-4 Ho are cyclic. Thus Hi, H3, H5, H6 , 14 que cyclic subgroups. e=c0,0)