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Group elements order

  1. Aug 15, 2012 #1
    Hi
    i need a little help
    i was given group (Z3 x Z3,+) and i should find order of every elements
    so the elements are {(0,0),(0,1),(0,2),(1,0),(1,1),(1,2),(2,0),(2,1),( 2,2)} and the order of every element is
    (0,0) has order 1
    (0,1)*3=(0(mod 3),3(mod 3)) = (0,0) order 3
    (0,2)*3=(0(mod 3),6(mod 3)) = (0,0) order 3
    (1,0)*3=(3(mod 3),0(mod 3)) = (0,0) order 3
    (1,1)*3=(3(mod 3),3(mod 3)) = (0,0) order 3
    (1,2)*3=(3(mod 3),6(mod 3)) = (0,0) order 3
    (2,0)*3=(6(mod 3),0(mod 3)) = (0,0) order 3
    (2,1)*3=(6(mod 3),3(mod 3)) = (0,0) order 3
    (2,2)*3=(6(mod 3),6(mod 3)) = (0,0) order 3

    (a,b) + (a,b) + (a,b) = (3a(mod3), 3b(mod3))=(0,0) so max order is 3

    next is group (Z2 x Z4, *)
    the elements are {(0,0),(0,1),(0,2),(0,3),(1,0),(1,1),(1,2),(1,3)}
    so here i should multiply every element n times till i get (an(mod 2),bn(mod 4)) = (1,1) so the order is n (i'm not sure about this correct me if i'm wrong)
    the element (0,0) always have order one
    and what about the other elements???

    example the element (0,2)
    there isn't ANY n with (0,2)n (mod 2, mod 4) = (1,1)
    please help :)
    thanks for your time
     
  2. jcsd
  3. Aug 15, 2012 #2

    micromass

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    But [itex]\mathbb{Z}_2\times\mathbb{Z}_4[/itex] isn't even a group under multiplication. Are you sure they didn't say that the operation is addition?
     
  4. Aug 15, 2012 #3
    On top of that, [itex]\mathbb{Z}_4[/itex] itself isn't a group under multiplication either. At best, it could be a field (if 4 were prime), so talking about a product like [itex](\mathbb{Z}_2\times\mathbb{Z}_4,*)[/itex] doesn't even make sense. This has to be a typo.
     
    Last edited: Aug 16, 2012
  5. Aug 16, 2012 #4
    they say to try with multiplication to see what is gonna happen
     
  6. Aug 16, 2012 #5
    thanks for the answer and explanation :)
     
  7. Aug 16, 2012 #6
    ok and what about (Z3xZ5,*)
    (0,0),(0,1),(0,2),(0,3),(0,4) will have order 1 or what??
    how can (an(mod 3),bn(mod 5)) = (1,1) when the element have (0,b)?? :D
    what should i do here??
     
    Last edited: Aug 16, 2012
  8. Aug 16, 2012 #7
    These will have order equal to the order of the "right" element. This is not a group under multiplication (see my earlier edit - I was mistaken. Zero never has a multiplicative inverse.); only addition.
     
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