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Algebra - Groups

  1. Jan 22, 2008 #1
    1. The problem statement, all variables and given/known data

    Prove that the following sets form infinite groups with respect to ordinary multiplication.

    a){2^k} where k E Z
    b){(1+2m)/(1+2n)} where m,n E Z

    2. Relevant equations

    3. The attempt at a solution

    I sort of know about


    ...and i know that multiplication is somewhere but what do i multiply? - apart form that im pretty stuck.
  2. jcsd
  3. Jan 22, 2008 #2
    You multiply the elements of the set. To show it's a group, show that the set is closed under the operation (multiplication), that there's an identity, and that arbitrary elements have an inverse. Technically you have to show the operation is associative too, but that's given by the fact that the operation is "ordinary" multiplication.
  4. Jan 22, 2008 #3


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    To prove closure for the easy one a), you just want to show that 2^(k1)*2^(k2) can be written as 2^(n) for some integer n. Can it? There's an identity if there is some integer n such that 2^n=1. Is there? There are inverses if for every integer k1, there is another integer k2 such that 2^k1*2^k2=1. Is there? You don't have to worry about associativity, it's a well known property of the real numbers.
  5. Jan 23, 2008 #4
    k1 + k2 = k3 which is an integer? therefore closure proven?


    inverses are just

  6. Jan 23, 2008 #5


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    That's pretty much it. Now try the second one.
  7. Jan 24, 2008 #6
    expand...? urghh...i dont know
  8. Jan 24, 2008 #7


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    Yes. Expand. Like this. (1+2m)(1+2n)=1+2m+2n+4mn=1+2(m+n)^2.
  9. Jan 24, 2008 #8
    cant say i understand

    why not (1+2m)(1+2m)/(1+2n)(1+2n)?
  10. Jan 24, 2008 #9


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    That was only a hint. I just did half the problem. You have to tell me if ((1+2m1)/(1+2n1))*((1+2m2)/(1+2n2)) can be written in the form (1+2m3)/(1+2n3) where m3 and n3 are integers, what's their relation to m1,n1,m2 and n2?
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