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Group isomorphism (C,+) to (R,+)

  1. Feb 20, 2013 #1
    1. The problem statement, all variables and given/known data
    Prove [itex](\mathbb{R},+)[/itex] and [itex](\mathbb{C},+)[/itex] are isomorphic as groups.


    2. Relevant equations
    An isomorphism is a bijection from one group to another that preserves the group operation, that is [itex]\phi(ab)=\phi(a)\phi(b)[/itex]


    3. The attempt at a solution
    I'm trying to find a bijection, but I can only find a bijection from [itex]\mathbb{C}[/itex] to [itex]R\times R[/itex].
     
  2. jcsd
  3. Feb 20, 2013 #2
    Well all real numbers are lie within the complex set of numbers so you could define a function as σ:ℝ→ℂ such that σ(a)= a+0i where a [itex]\in[/itex]ℝ. This is a bijective map. Now you just need to show that this function preserves the group operation....
     
  4. Feb 20, 2013 #3

    jbunniii

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    No, it's not surjective.
     
  5. Feb 20, 2013 #4

    jbunniii

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    I don't think you can write down an explicit isomorphism. Can you argue indirectly by treating them both as vector spaces over the rationals, and comparing the cardinalities of their dimensions?
     
  6. Feb 21, 2013 #5
    I have seen a proof that does that, however they omit how to construct a basis for the vector space, to show what its dimension is.
     
  7. Feb 21, 2013 #6

    micromass

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    You cannot explicitely find a basis of [itex]\mathbb{R}[/itex] over [itex]\mathbb{Q}[/itex].

    However, you can show that a basis exists. This uses Zorn's lemma and is nonconstructive. In general, you can show that any vector space has a basis.

    Finding the dimension of [itex]\mathbb{R}[/itex] over [itex]\mathbb{Q}[/itex] is not so difficult though. Basically, you need to use that every vector space over [itex]\mathbb{Q}[/itex] is isomorphic to [itex]\mathbb{Q}^X[/itex], where X is a set. It turns out that if X and Y have the same cardinality, then [itex]\mathbb{Q}^X\cong \mathbb{Q}^Y[/itex].
    So we can write [itex]\mathbb{R}\cong \mathbb{Q}^X[/itex]. Then do you see a way of writing [itex]\mathbb{R}^2[itex]??
     
  8. Feb 21, 2013 #7
    So I would have to prove [itex]|\mathbb{R}|=|\mathbb{R}^{2}|[/itex], because [itex]|\mathbb{R^{2}}|=|\mathbb{C}|[/itex].
     
  9. Feb 21, 2013 #8

    MathematicalPhysicist

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    If you know already that C and R have the same cardinality then all you need to look for is a monomorphism to prove that they are isomorphic. Inclusion will do.
     
  10. Feb 21, 2013 #9

    micromass

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    I don't think this is true. The [itex]\mathbb{Q}[/itex]-vector spaces [itex]\mathbb{Q}[/itex] and [itex]\mathbb{Q}^2[/itex] have the same cardinality and there is a monomorphism between [itex]\mathbb{Q}\rightarrow \mathbb{Q}^2[/itex], but the space are not isomorphic.

    It might be that the argument in fact works in this case, but then that still needs to be proven.
     
  11. Feb 21, 2013 #10

    MathematicalPhysicist

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    Well, I am not sure, but (R,+) is a subgroup of (C,+), and they have the same cardinality.

    Well, I am not sure there's a bijection between them.

    I mean look at some number a in R, what are our options to send it to C, either a+ia (which is not surjective), or exp(ia), but exp(ia) isn't single valued.

    I don't think there's such a bijection, at least I don't see it.

    For RxR I am not either that there's a way (now I remember that the proof that aleph x aleph =aleph goes through Zorn's Lemma, which I am never really fond of).
     
  12. Feb 21, 2013 #11

    jbunniii

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    There is a bijection which you can construct explicitly. It won't be a homomorphism, of course. Start with a number ##z = a + bi \in \mathbb{C}##, and interleave the decimal (or binary, or whatever) expansions of ##a## and ##b## to get a new number ##r \in \mathbb{R}##. This isn't well defined as it stands, because the decimal expansion is not unique (##0.9999\ldots = 1.0000\ldots##, for example), but that it can be made well defined if you fix a rule for which expansion to use. (I think it suffices to always use the non-terminating one.) I'm glossing over some details, but this is the basic idea and it can be made to work.
     
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