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Abstract Algebra - Cyclic groups

  1. Oct 26, 2011 #1
    1. Problem: Suppose a is a group element such that |a^28| = 10 and |a^22| = 20. Determine |a|.

    I was doing some practice problems for my exam next week and I could not figure this out. (This is my first post on PF btw)

    2. Relevant Equations: Let a be element of order n in group and let k be a positive integer. Then <a^k> = <a^gcd(n,k)> and |a^k| = n/gcd(n,k).

    3. Attempt at solution:

    10 = n/gcd(n, 28); 20 = n/gcd(n, 22)

    Setting n equal to each other, 10gcd(n, 28) = 20gcd(n,22)

    gcd(n, 28) = 2gcd(n, 22)

    The possible values for n are 4, 8, 12, 16, 20, 24, ... , so on.

    Not sure where to go from here.
     
  2. jcsd
  3. Oct 26, 2011 #2

    Deveno

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    gcd(440,280) = 40.

    this means we can find integers a,b with

    440a + 280b = 40, that is

    11a + 7b = 1. a = 2, b = -3 will work.

    thus a^40 = (a^(880))(a^-(840))

    = (a^(440))^2(a^(280))^(-3)

    = ((a^22)^20)^2((a^28)^10)^(-3)

    = e^2(e^-3) = ee = e.

    that means |a| divides 40, which gives us just 1,2,4,5,8,10,20 and 40 as possibilities.

    since a^28 ≠ e, we can rule out 1,2, and 4 as possibilities.

    suppose a^5 = e. then a^28 would have order 5,

    since (a^28)^5 = (a^5)^28. similarly, we can rule out

    every divisor of 10 and 20, leaving just 8 and 40.

    so suppose |a| = 8.

    then (a^28)^2 = a^56 = (a^8)^7 = e, but |a^28| = 10.

    what's left?
     
  4. Oct 26, 2011 #3
    Thanks for the response. But I don't understand why you are starting out with gcd(440,280) = 40.
     
  5. Oct 26, 2011 #4

    micromass

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    We know that the order of [itex]a^{28}[/itex] is 10. So [itex]a^{280}=e[/itex]. Similarly, we know that [itex]a^{440}=e[/itex].

    For each integers x and y, it follows that [itex]a^{280x+440y}=e[/itex]. We wish to minimize 280x+440y (since this will be a small number such that the order divides this number). The minimal number is exactly gcd(280,440). This is why he started by this.
     
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