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Thread: find all generators of a cyclic group

  1. #1
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    find all generators of a cyclic group

    Question;
    Given that 2 is a generator of cyclic group U(25), find all generators.

    how do i find other generator asides 2 please.

    I am only conversant with the finding the mod which is very long with this question. Pls can someone enlighten me on how to get it done faster.
    I am new to the cyclic group.
    solution

    U(25) = {1,2,3,4,6,7,8,9,11,12,13,14,16,17,18,19,21,22,23, 24}

    2^20 = 1 (mod 25)


    Thanks.
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  2. #2
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    Re: find all generators of a cyclic group

    Can I presume you know the definition of "generator"? A "generator", g, of a group is a member of the group such that for every x in the group there exist an integer n such that g^n= x. That is we can get all members of the group by taking powers of g.

    2 is a generator of the group of integers, mod 25, because 2^0= 1, 2^2= 4, 2^3= 8, 2^4= 16, 2^5= 32= 25+ 7= 7, 2^6= 64= 2(25)+ 14= 14, 2^7= 128= 5(25)+ 3= 3, 2^8= 256= 10(25)+ 6= 6, 2^9= 512= 20(25)+ 12= 12, 2^10= 1024= 40(25)+ 24= 24, 2^11= 2048= 81(25)+ 23, 2^12= 4096= 163(25)+ 21, 2^13= 8192= 327(25)+ 17= 17, 2^14= 16384= 655(25)+ 9= 9, 2^15= 32768= 1310(25)+ 18= 18, 2^16= 65536= 2621(25)+ 11, 2^17= 131072= 5242(25)+ 22, 2^18= 262144= 10485(25)+ 19= 19, 2^19= 524288= 20971(25)+ 13= 13, 2^20= 1048576= 41943(25)+ 1= 1, 2^21= 2097152= 83886(25)+ 2= 2, 2^22= 4194304= 167772(25)+ 4= 4, 2^23= 8388608= 335544(25)+ 8= 8, 2^24= 16777216= 671088(25)+ 16= 16 (all "mod 25"). All of those are different and, since they are all less than 25, they include all numbers from 1 to 25.

    On the other hand if we did the same with, say 5, we would get 5^0= 1, 5^1= 5, 5^2= 25= 0, 5^3= 125= 0, etc. The only numbers we get are "0", "1" and "5" precisely because "5" is a divisor of "25". The numbers that generate all of Z25 are simply those numbers that do not divide 25.
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  3. #3
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    Re: find all generators of a cyclic group

    if k is a generator of the integers mod 20
    then 2^k is a generator of U(25)

    so you are looking at 2^k where k={1, 3, 7, 9, 11, 13, 17, 19}
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  4. #4
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    Re: find all generators of a cyclic group

    The following attachment shows that the computation of the powers of 2 mod 25 requires only simple arithmetic. Then the generators of U(25) are easy:
    find all generators of a cyclic group-mhfgroups47.png
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