Cyclic Groups and Subgroups

  • Thread starter FanofAFan
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  • #1
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Homework Statement


Find all of the subgroups of Z3 x Z3


Homework Equations


Z3 x Z3 is isomorphic to Z9


The Attempt at a Solution


x = (0,1,2,3,4,5,6,7,8)
<x0> or just <0> = {0}
<1> = {identity}
<2> = {0,2,4,6} also wasn't sure if I did this one correctly x o x for x2
<3> = {0,3,6}
and so on until I got
<8> = {0,8}
<9> = {0}

I feel like I might be completely wrong but is this even a cyclic group, and do I need to approach finding the subgroups differently? Thanks
 

Answers and Replies

  • #2
Dick
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Z3xZ3 is not cyclic. Any element of Z3xZ3 has order 3. And even if it were, your subgroups of Z9 have problems.
 
  • #3
Office_Shredder
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If you're going to compute the subgroup of Z9 generated by 2, you can't stop at 8. We start with 0, 2, 4, 6, 8, and then 2+8=10=1 mod 9, 1+2=3, then you get 5 and 7 so 2 generates the whole group.

Of course like Dick said those aren't the subgroups you're looking for anyway
 

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