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Multiplication table for Z_n

  1. May 8, 2010 #1
    1. The problem statement, all variables and given/known data

    Create a multiplication table for the group of invertible elements in the ring [tex]Z_{10}[/tex]. Can you rename the elements and arrange them so that the multiplication table is transformed into a multiplication table for the group [tex]Z_n[/tex] for some n?



    2. Relevant equations



    3. The attempt at a solution

    If [tex]p \in Z_m[/tex], [tex]p[/tex] has an inverse iff [tex]GCD(p,m)=1[/tex], so the invertible elements in [tex]Z_{10}[/tex] are 1, 3, 7 and 9, and we end up with

    [tex]\begin{bmatrix}
    1 & 3 & 7 & 9\\
    3 & 9 & 1 & 7\\
    7 & 1 & 9 & 3\\
    9 & 7 & 3 & 1\\
    \end{bmatrix}[/tex]

    as the suspiciously matrix-looking multiplication table in [tex]Z_{10}[/tex].

    I don't know what the second sentence of the problem implies though. After attempting proof by asking IRC, I received the reply "Z/10Z =~ Z/2Z x Z/5Z -> (Z/10Z)* =~ (Z/2Z)* x (Z/5Z)* =~ Z/4Z", but I haven't seen any similar notation before. Where do I begin on this, or perhaps, what should I read to get a better understanding of similar problems?
     
    Last edited: May 8, 2010
  2. jcsd
  3. May 8, 2010 #2

    tiny-tim

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    Hi Combinatus! :smile:

    (nice LaTeX, btw! :biggrin:)
    Well, it's a 4x4 matrix, so it obviously can only be the table for Z5

    so re-name 3 7 and 9 as some permutation of 2 3 and 4 so that the table works. :wink:
     
  4. May 8, 2010 #3

    Thanks for your help! :)

    Okay, so in [tex]Z_5[/tex] we get

    [tex]\begin{bmatrix}
    1 & 2 & 3 & 4\\
    2 & 4 & 1 & 3\\
    3 & 1 & 4 & 2\\
    4 & 3 & 2 & 1\\
    \end{bmatrix}[/tex]

    The permutation that generates this table from the one of the inverses in [tex]Z_{10}[/tex] (i.e. the one in my previous post) is thus 1 -> 1, 3 -> 2, 7 -> 3, 9 -> 4.

    The key to the problem, however, says that "If the elements are arranged to 1, 3, 9, 7 and they are named 1 -> 0, 3 -> 1, 7 -> 3, 9 -> 2, you get the table for the group [tex]Z_4[/tex]." Even if you include 0 in the multiplication table for the group [tex]Z_4[/tex], they don't look similar at all, so I'm not sure what they're on about.
     
  5. May 8, 2010 #4

    tiny-tim

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    Hi Combinatus! :smile:

    hmm … some of Z4 is right, but if 1 -> 0, then there should be 1s (for 0s) all along the top and left of the table (and there aren't) :frown:

    the key is wrong … it must be Z5
     
  6. May 8, 2010 #5

    jbunniii

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    The notation in the key is a bit vague, but what it is saying is that [itex]Z_{10}^*[/itex] is isomorphic to [itex]Z_5^*[/itex] (the MULTIPLICATIVE group of units of [itex]Z_5[/itex]), which is isomorphic to [itex]Z_4[/itex] (the ADDITIVE group). This is indeed true.
     
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