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Cyclic Group Generators <z10, +> Mod 10 group of additive integers

  1. Jul 11, 2014 #1
    So I take <z10, +> this to be the group


    Z10 = {0,1,2,3,4,5,6,7,8,9} Mod 10 group of additive integers and I worked out the group generators, I won't do all of them but here's an example :

    <3> gives {3,6,9,2,5,8,1,4,7,0}
    on the other hand

    <2> gives {2,4,6,8,0} and that's it! but my book says prove that 2 and 5 are generators....these two I thought we not generators...what am I doing wrong?
     
  2. jcsd
  3. Jul 11, 2014 #2

    jbunniii

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    Are you sure that you are reading the question correctly? As you have noted, neither 2 nor 5 is a generator of ##Z_{10}##. However, the set ##\{2,5\}## does generate ##Z_{10}##.
     
  4. Jul 11, 2014 #3
    Can you give me more info about {2,5} generating Z10? feel like I must be misunderstanding how to Generate Z10. Thanks for the help :)
     
  5. Jul 11, 2014 #4

    jbunniii

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    We say that ##\{2,5\}## is a generating set for ##Z_{10}##. Here is the definition:

    http://en.wikipedia.org/wiki/Generating_set_of_a_group

    Briefly, if ##S## is any subset of a group ##G##, then we define ##\langle S\rangle## to be the smallest subgroup of ##G## which contains ##S##. To be more precise, ##\langle S\rangle## is the intersection of all subgroups of ##G## which contain ##S##. We call ##\langle S\rangle## the subgroup generated by ##S##, and we say that ##S## is a generating set for ##\langle S\rangle##.

    In the special case where ##S## contains one element, say ##S = \{x\}##, we usually write ##\langle x \rangle## instead of ##\langle \{x\}\rangle##.

    Applying these notions to your question, we want to prove that if ##S = \{2,5\}##, then the smallest subgroup of ##Z_{10}## which contains ##S## is ##Z_{10}## itself. In other words, we want to prove that ##\langle S\rangle = Z_{10}##.

    To prove this, note that since ##S## contains ##2##, it must be true that ##\langle S\rangle## contains ##\langle 2\rangle##. (Why?) Likewise, since ##S## contains ##5##, it must hold that ##\langle S \rangle## contains ##\langle 5\rangle##. What can you conclude?
     
    Last edited: Jul 11, 2014
  6. Jul 12, 2014 #5

    HallsofIvy

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    A group, G, is generated by a set, S, of members of G if and only if by combining the members of S using the group operation, we get every member of G. The set {3} gives, as you say all members of Z10 because 3= 3, 3+ 3= 6, 3+ 3+ 3= 9, 3+ 3+ 3+ 3= 12= 2 (mod 10), 3+ 3+ 3+ 3+ 3= 15= 5 (mod 10), 3+ 3+ 3+ 3+ 3+ 3= 18= 8 (mod 10), 3+ 3+ 3+ 3+ 3+ 3+ 3= 21= 1 (mod 10), 3+ 3+ 3+ 3+ 3+ 3+ 3+ 3= 24= 4 (mod 10), 3+ 3+ 3+ 3+ 3+ 3+ 3+ 3+ 3= 27= 7 (mod 10), and 3+ 3+ 3+ 3+ 3+ 3+ 3+ 3+ 3+ 3= 30= 0 (mod 10).

    {2} does not generate Z10 because 2= 2, 2+ 2= 4, 2+ 2+ 2= 6, 2+ 2+ 2+ 2= 8, 2+ 2+ 2+2+ 2= 10= 0 (mod 10), 2+ 2+ 2+ 2+ 2+ 2= 12= 2 (mod 10) and now it starts over. We do NOT get 1, 3, 5, 7, or 9.

    {5} does not generate Z10 because 5= 5, 5+ 5= 10= 0 (mod10), 5+ 5+5 = 15= 5 (mod 10) and it just repeats. We do not get 1, 2, 3 4, 6, 7, 8, or 9.

    But {2, 5} contains all of those and 2+ 5= 7, 2+ 2+ 5= 9, 2+ 2+ 2+ 5= 11= 1 (mod 10), 2+ 2+ 2+ 2+ 5= 13= 3 (mod 10) so that we have all of 1, 2, 3, 4, 5, 6, 7, 8, 9, and 0.
     
  7. Jul 12, 2014 #6
    HallsofIvy EGGGZELLENT Thx :) u2
    jbunniii
     
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