Show Rationals and Integers are not Isomorphic

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Homework Statement


Show that the groups [tex]\textbf{Q}[/tex] and [tex]\textbf{R}[/tex] are not isomorphic (both under addition).

This was already answered before https://www.physicsforums.com/showthread.php?t=294687", but using different theory (generators and cyclic groups). We haven't covered that stuff in class yet. See below.

Homework Equations


Two groups, G and H, are NOT isomorphic if:
1) |G| [tex]\neq[/tex] |H|
2) Let [tex]\varphi[/tex] be a map from G to H. Then G, H are not isomorphic if for some [tex]x \in \textbf{G}, |x| \neq |\varphi(x)|[/tex]
3) G is abelian, H is not abelian or vice versa.

Two groups are isomorphic iff there exists [tex]\varphi:\textbf{G} \rightarrow \textbf{H} [/tex] s.t:
1) [tex]\varphi[/tex] is a homomorphism.
2) [tex]\varphi[/tex] is a bijection.

Homomorphism:
[tex]\varphi(xy) = \varphi(x)\varphi(y)[/tex].

The Attempt at a Solution


I know it is possible to set up a bijection from Z to Q because Q is countable (isomorphic to N) and so is Z. Then the solution must be that no homomorphism exists from Z to Q.

Proof
p,q are rational. p < q. a,b are integers. [tex]\varphi:{\bftext{Q} \rightarrow \bftext{R}[/tex], assume it is a homomorphism.
If [tex]\varphi(p)=a, \varphi(q) = b[/tex], then [tex]\varphi(p + q)=\varphi(p)+\varphi(q) = a + b[/tex]. But consider p + (q - p)/2. This must map to an integer in between a and b. Why? (I'm not sure, i'll have to investigate this rigorously). Since there are finitely many integers between a and b, but infinitely many rationals between p and q, [tex]\varphi[/tex] cannot be bijective. Which means that there is no homomorphic map from the rationals to the integers that is also bijective. So Q cannot be isomorphic to R.

Is this sufficient?
 
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Answers and Replies

  • #2
Dick
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It's a little sketchy. Let f be an isomorphism Q->Z. Let p=f^(-1)(1). Then what is f(p/2)? It must be an integer that when added to itself gives 1.
 
  • #3
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Yeah got it.

If [tex]f^{-1}(1) = p[/tex], then [tex]\varphi(\frac{p}{2}+\frac{p}{2})=1[/tex] but that implies [tex]0<\varphi(\frac{p}{2})<1[/tex] which is impossible since no integer exists between 0 and 1.

Thanks man.
 

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