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Z and Q are not isomorphic

  1. Feb 23, 2009 #1
    1. The problem statement, all variables and given/known data

    Prove that the integers (under addition) are not isomorphic to the rationals (under addition).

    2. Relevant equations

    Two groups are isomorphic if there is an isomorphism between them.

    If there is an isomorphism from G to H, f : G --> H, then G is cyclic iff H is cyclic.

    A group G is cyclic if [tex] \{ x^n | n \in \mathbb{Z} \} = G, for some x \in G [/tex] .

    3. The attempt at a solution

    The integers are generated by [tex] <1> [/tex]. We can show that Z and Q are not isomorphic if we show that the rationals cannot be generated. Thus assume they are. Then there is an a such that

    [tex] <a> = Q [\tex].
    [tex] 0a = 0, 1a = \frac{l}{m} , 2a = \fract{2l}{m}[\tex].

    Because the rationals are dense there is a [tex] b \in Q s.t. \frac{l}{m} < b < \frac{2l}{m} [\tex]

    We must show that [tex] b = ka = \frac{kl}{m}, thus \frac{l}{m} < \frac{kl}{m} < \frac{2l}{m} [\tex].

    Now I don't know what to do. The above is not a contradiction. Any ideas?
    Last edited: Feb 23, 2009
  2. jcsd
  3. Feb 23, 2009 #2
    Suppose Q is cyclic.

    Let p/q be a generator.

    Can you find a rational number which is not an integer multiple of p/q?
    Think of a rational involving p and q somehow.
  4. Feb 23, 2009 #3
    sure, continuing

    Observe p/2q


    p/2q = p/q where p,q are not equal to zero. (if p is zero then the set is finite).

    From the above conclude that
    pq=2pq, therefore 1=2.
  5. Feb 23, 2009 #4


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    No, no. <p/q> is the set of number k*p/q where k is an integer. Set kp/q=p/(2q) and derive a different contradiction.
  6. Feb 23, 2009 #5
    Thanks! Yeah, I don't know what I was thinking there. I went and got some food, came back and hit myself in the head on that one.
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