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Group tables of Z

  1. Feb 13, 2008 #1
    [SOLVED] Group tables of Z

    1. The problem statement, all variables and given/known data
    Write down the group tables of Z[tex]^{}4[/tex] and Z[tex]^{}2[/tex] x Z[tex]^{}2[/tex] and for every element a in Z[tex]^{}4[/tex] and Z[tex]^{}2[/tex] x Z[tex]^{}2[/tex] determine the smallest positive integer m such that ma equals the identity element.


    2. Relevant equations



    3. The attempt at a solution

    Z[tex]^{}4[/tex]:

    + 0 1 2 3
    0 0 1 2 3
    1 1 2 3 0
    2 2 3 0 1
    3 3 0 1 2

    I know Z[tex]^{}2[/tex] is:

    + 0 1
    0 0 1
    1 1 0

    but i dont know how to perform the binary operation the question asks for, can anyone explain how to do it please?
     
  2. jcsd
  3. Feb 13, 2008 #2

    Dick

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    An element of Z^2xZ^2 is a pair (a,b) where a and b are in Z^2. The binary operation is (a,b)+(c,d)=(a+c,b+d).
     
  4. Feb 13, 2008 #3
    Ok, so in this case assuming a=c and b=d, the group table looks like:

    + 0 1
    0 0 2
    1 2 0

    im not sure whether the + sign is correct and whether the 0 and 1 are correct??
     
  5. Feb 13, 2008 #4

    Dick

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    Noooooo. There are 4 elements in the group. (0,0), (0,1), (1,0) and (1,1). Those should be your four horizontal and vertical labels for the table. Eg. (0,1)+(0,1)=(0,0), (1,0)+(0,1)=(1,1) etc.
     
  6. Feb 13, 2008 #5
    Ok, great, so i get:

    + 0,1 0,1 1,0 1,1
    0,0 0,0 0,1 1,0 1,1
    0,1 0,1 0,0 1,1 1,0
    1,0 1,0 1,1 0,0 0,1
    1,1 1,1 1,0 0,1 0,0

    I don't think i understand the second part of the question. I need to find an integer value m for each element in each group that when multiplied by the element will produce the identity element. So is m the inverse element? But if this is true then how do i make for example, an element in Z^4 equal to 2 become an identity element which i assume is equal to 1 by multiplying by m, which must be an integer?
     
  7. Feb 13, 2008 #6
    For Z^4 is the identity element = 0 and the inverses for 0,1,2,3 are 0,3,2,1 respectively?
     
  8. Feb 13, 2008 #7

    Dick

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    There's a typo or two in your Z^2xZ^2 table, but that's ok, I think they are just typos. On the second part, m is not an element of the group. It's an integer. Take a=3 in Z^4, 1a=3, 2a=a+a=2, 3a=a+a+a=1, 4a=a+a+a+a=0. So the m for a=3 is 4. I know this is easy to confuse with the group operation since you are using the same symbols to represent the group elements and m. Now in Z^2xZ^2 take a=(0,1). 1a=(0,1). 2a=(0,1)+(0,1)=(0,0). So for a=(0,1), m=2. Work this value of m out for each member of Z^4 and Z^2xZ^2.
     
  9. Feb 14, 2008 #8
    Oh yes, there is a typo, it should be:

    + 0,0 0,1 1,0 1,1
    0,0 0,0 0,1 1,0 1,1
    0,1 0,1 0,0 1,1 1,0
    1,0 1,0 1,1 0,0 0,1
    1,1 1,1 1,0 0,1 0,0

    So for Z^4, the integer m for each element are as follows:
    a=0, m =0
    a=1, 2a = a+a=2, 3a= 2+1=3, 4a= 3+1 = 0 so m =4
    a=2, 2a=2+2=0 so m =2
    a=3, 2a=3+2=2, 3a = 2+3 = 1, 4a=1+3=0 so m =4

    For Z^2 x Z^2:
    a=0,0 m = 0,0
    a=0,1 2a = 0,1+0,1 = 0,0 so m =2
    a=1,0 2a = 1,0+1,0 = 0,0 so m=2
    a=1,1 2a=1,1+1,1 = 0,0 so m =2
     
  10. Feb 14, 2008 #9

    Dick

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    Looks pretty good to me. But I would say m=1 for the zero elements.
     
  11. Feb 14, 2008 #10
    Great, i also thought that but was unsure, thanks for the help. Question solved
     
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