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Groups and Subgroups

  1. Mar 5, 2009 #1
    1. The problem statement, all variables and given/known data

    Recall that Vm is the set of all invertibles in Z/m
    a) List the elements in V15
    b) Find all the subgroups of V15
    c) Is V15 cyclic? why?

    2. Relevant equations

    3. The attempt at a solution

    From my notes:

    a) V15 = {1, 2, 4, 7, 8, 11, 13, 14}
    b) {2,4,8,1}{4,1}{7,4,13,1}{11,1}{14,13}{1}
    c) No, no generator for whole group.

    To be honest, I don't understand how any of these answers were found. I'm lost
     
  2. jcsd
  3. Mar 5, 2009 #2
    If I'm listing the invertibles, are they the relatively prime numbers to 15?
     
  4. Mar 5, 2009 #3
    there should be a theorem that states that: in Z_m a is said to be an invertibile if and only if gcd(a,m)=1.

    proof:

    let gcd(a,m)=1. we want to show that a is invertible. First let's see what does it mean for a to be invertible?

    a is said to be invertible if there exist a b such that [a]=1.

    Now, since gcd(a,m)=1=> that ax+my=1, where x,y are integers.

    =>ax-1=(-y)m=> ax=1mod(m) =>[a][x]=[1], so a is invertible.

    Now, if a is invertible, it means that there exists some b such that [a]=[1]=> ab=1mod(m)=> ab-1=km=> ab+(-k)m=1=> gcd(a,m)=1, so a and m are relatively prime.

    Note:[a] means a mod m

    Then a group G, of ord m, is said to be cyclic if there is an element g of order m. In other words, G is cyclic if [g]=G, that is if there is an element such that [g]={g^m:m integer} is the group G itself.


    But, in Z_m, there is also a theorem i believe, that says sth like this: a is a generator of V_m if and only if gcd(a,m)=1
     
    Last edited: Mar 5, 2009
  5. Mar 5, 2009 #4
    Oohhhh ok thanks!
    Can you provide any help on B) or C)?
     
  6. Mar 5, 2009 #5
    Well, if G is a group and g is any element of that group, then the group generated by g, i.e [g] is also a subgroup of G. so this is one way of finding some subgroups of G. For part C) i already gave u a hint in my previous post. What is the order of V_15?

    Is there any a in V_15 such that (a,ord{V_15})=1 ?
     
  7. Mar 5, 2009 #6
    I'm sorry, I don't understand. Do you think you could use {2,4,8,1} as an example?
     
  8. Mar 5, 2009 #7
    How does this differ from an invertible?
     
  9. Mar 5, 2009 #8
    It doesn't. Every invertible element is a generator. Why? if a is an invertible element of V_m,what it means is gcd(a,m)=1. So, this means that m is not an integral multiple of a. hence, a^m=1 or ma=1(when dealing with additive groups). i.e. order of a will be m, which would mean that a is a generator as well.
     
  10. Mar 5, 2009 #9
    Well, assuming that this is a group, then it would be cyclic as well. WHy, the order of this group is simply the nr. of its elements. so ord =5. But, since 5 is a prime nr. it means that every nr. smaller than 5, is relatively prime to it, and also every non integral multiple of it.
     
  11. Mar 5, 2009 #10
    Why is the order 5 instead of 4?
     
  12. Mar 5, 2009 #11
    Well, because obviously,i have forgotten to count!

    In that case it would mean that it is not cyclic, since none of the elements is relatively prime with 4. 1 is not a generator, since it is the identity(iif the operator is multiplication) and thus 1^m =1 for any m.
     
  13. Mar 5, 2009 #12
    lol.
    I still don't understand how this subgroup was found (part b). Maybe that part of my brain just doesn't work. Can you try one more time to explain it?
     
  14. Mar 5, 2009 #13
    how do you generate a subgroup from a group? specifically {2,4,8,1} from {0,1,2,4,7,8,10,13,14}
     
  15. Mar 5, 2009 #14
    [2]={2^k:k in Z}={2^1=2,2^2=4,2^3=8,2^4=16=1}={2,4,8,1}

    notice that 16=1mod(15)

    now take 4


    [4]={4^k:k in Z}={4^1=4,4^2=16=1}={4,1}

    As soon as we reach the multiplicative identity, 1, the elements start to repeat.

    Now you do the same with others.
     
  16. Mar 5, 2009 #15
    great! you have no clue how much you've helped... i've been so upset the past couple of hours. I have a test in the morning and was going to freak if i didn't figure these out.

    Could you maybe explain how to find a generator one more time?
     
  17. Mar 9, 2009 #16
    I made a 89 on my abstract algebra test (should have been a 90, but w/e >.>) !!
    Thank you!!
     
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