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Prove (Q+, *) is isomorphic to a proper subgroup of itself

  1. Dec 9, 2016 #1
    1. The problem statement, all variables and given/known data

    Prove that Q+, the group of positive rational numbers under multiplication, is isomorphic to a proper subgroup of itself.

    2. Relevant equations
    None

    3. The attempt at a solution

    Not at all sure if this is legit.

    Let phi: Q+ --> G
    phi(x) = x2, x is in Q+
    We will demonstrate that G c Q+
    It is a subgroup: 1=e is in G, and ab-1 = x2 y-2 = (xy-1)2 is in G
    It is a proper subgroup: 2 is in Q+, but sqrt(2) is not in G and indeed not in Q+

    One-to-one:
    phi(x) = phi(y)
    x2 = y2
    x, y > 0
    x = y

    Onto:
    Take some g in G. We have that sqrt(g) satisfies phi(sqrt(g)) = sqrt(g)2 = g.
    Therefore, there is an element in Q+ such that phi(x)=g.

    Operation preservation:
    We have phi(x*y) = (xy)2 = x^2y2
    phi(x)phi(y) = x2y2
    So phi(x*y)=phi(x)*phi(y)

    Therefore, phi is an isomorphism between Q+ and a proper subgroup of itself.
     
  2. jcsd
  3. Dec 9, 2016 #2

    fresh_42

    Staff: Mentor

    Looks fine, beside some minor issues on the notation (the missing definition of G, sqrt cannot be defined, it should be ##\phi^{-1}## (preimage) instead, and the equation under "subgroup" is a bit short, i.e. doesn't introduce a,b, injectivity could be a little more explicitly, i.e. why does x=y follow, resp. what properties of ##\mathbb{Q}## do you use).
     
  4. Dec 9, 2016 #3
    Thank you so much again fresh_42!
     
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