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Unable to find generators of {G, Xn}

  1. Apr 18, 2009 #1
    1. The problem statement, all variables and given/known data
    Consider the group {G, Xn} where
    G is the set containing the n - 1 residue
    classes modulo n excluding 0.

    Which members are generators of {G, Xn} when
    n = 7?

    3. The attempt at a solution

    The right answers are 3 and 5.

    I get 2 only.

    My table of results

    [tex]1: 1^1 = 1[/tex]
    [tex]2: 2^1 = 2, 2^2 = 4, 2^3 = 4, 2^4 = 2[/tex]
    [tex]3: 3^1 =3, 3^2 = 2, 3^3 = 4[/tex]
    [tex]4: 4^1 =4, 4^2 = 2[/tex]
    [tex]5: 5^1 = 5, 5^2 =4[/tex]
    [tex]6: 6^1 = 6, 6^2 = 1[/tex]
    [tex]7: 7^1 = 7, 7^2 = 7[/tex]
     
  2. jcsd
  3. Apr 18, 2009 #2
    [tex]2^3\neq4[/tex]

    [tex]3^3\neq4[/tex]

    Find [tex]3^4, 3^5,\dots[/tex] and [tex]5^3,5^4, 5^5,\dots[/tex]

    (And 7=0.)
     
  4. Apr 18, 2009 #3
    I do not mean with 3^3 3 to the power of 3 only.
    I mean with it 3^3 3 to the power of 3 modulo 7.
     
  5. Apr 18, 2009 #4
    I cannot find [tex]3^4[/tex].
    There is no such number which is 3 times 3 times 3 times 3.
    The only way to get that is when n = 9, that is for 9 times 9.
     
  6. Apr 18, 2009 #5

    matt grime

    User Avatar
    Science Advisor
    Homework Helper

    You do understand the operation is multiplication and reduction modulo 7. It doesn't make sense to say that 'there is no such number in the set that is 3^4 mod 7'. There is. 3^2 is 2 mod 7. 3^3 is 6 mod 7, and 3^4 is 4 mod 7.
     
  7. Apr 18, 2009 #6
    I think I got it.

    The order of a group is the number of order of elements.
    For example, in this case,

    2^1 = 1, 2^2 = 4, 2^3 = 1
    However, the group already has the order 1 so we stop.

    I get that 3 and 5 have 7 different results from modulo 7.
    This means that the Cayley table is really useful here too, since I can use it repeatedly in the calculations.

    Thank you for your answers!
     
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