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Q+ not isomorphic to Q

  1. Oct 10, 2011 #1
    I'm taking a first course in algebra, and in my textbook, there is the following problem:

    a) Show that, for every natural # n, there is a subgroup A of Q+ such that |Q+/H| = n.

    b) Suppose that B is a proper subgroup of Q. Show that |Q / B| = ∞.

    c) Conclude that Q+ ≠ Q.

    I did parts a and b just fine, but I'm not sure if I see how to (properly) conclude from them that Q+ ≠ Q.

    I'm thinking that if there is to be a bijection from Q+ → Q, then there should be a bijection from Q+ / A → Q / B, but that's not possible since Q+ / A and Q / B should at least have the same order.

    So basically if P : Q+ → Q is a bijection, the following are also bijections:
    For g in Q+, h in Q,
    R taking g to gA
    S taking h to hB

    So the composition S o P o R-1 should be a bijection from Q+ / A → Q / B, which is (I think) a contradiction, so there should be no such P.

    Am I doing this properly or am I missing something more obvious?
     
  2. jcsd
  3. Oct 10, 2011 #2

    micromass

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    Let [itex]f:\mathbb{Q}^+\rightarrow \mathbb{Q}[/itex] be an isomorphism. What can you say about f(A)? (where A is the subgroup with [itex]|\mathbb{Q}^+/A|=3[/itex] or another number).
     
  4. Oct 10, 2011 #3
    I cannot think of anything helpful. f(A) will be a subgroup of Q? I feel like we should have | Q / f(A) | = n but I don't know how to show this.
     
  5. Oct 10, 2011 #4

    micromass

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    Show that

    [tex]\mathbb{Q}^+/A \rightarrow \mathbb{Q}/f(A): x+A\rightarrow f(x)+f(A)[/tex]

    is a bijection. So show

    1) it is well-defined
    2) it is injective
    3) it is surjective
     
  6. Oct 10, 2011 #5
    Thank you very much for your help :)
     
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