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

[Max.Planck]

Homework Statement

Prove (\mathbb{R},+) and (\mathbb{C},+) are isomorphic as groups.

Homework Equations

An isomorphism is a bijection from one group to another that preserves the group operation, that is \phi(ab)=\phi(a)\phi(b)

The Attempt at a Solution

I'm trying to find a bijection, but I can only find a bijection from \mathbb{C} to R\times R.

 

[porroadventum]

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 \inℝ. This is a bijective map. Now you just need to show that this function preserves the group operation...

 

[jbunniii]

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 \inℝ. This is a bijective map.

No, it's not surjective.

 

[jbunniii]

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?

 

[Max.Planck]

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?

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.

 

[micromass]

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.

You cannot explicitely find a basis of \mathbb{R} over \mathbb{Q}.

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 \mathbb{R} over \mathbb{Q} is not so difficult though. Basically, you need to use that every vector space over \mathbb{Q} is isomorphic to \mathbb{Q}^X, where X is a set. It turns out that if X and Y have the same cardinality, then \mathbb{Q}^X\cong \mathbb{Q}^Y.
So we can write \mathbb{R}\cong \mathbb{Q}^X. Then do you see a way of writing \mathbb{R}^2??

 

[Max.Planck]

You cannot explicitely find a basis of \mathbb{R} over \mathbb{Q}.

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 \mathbb{R} over \mathbb{Q} is not so difficult though. Basically, you need to use that every vector space over \mathbb{Q} is isomorphic to \mathbb{Q}^X, where X is a set. It turns out that if X and Y have the same cardinality, then \mathbb{Q}^X\cong \mathbb{Q}^Y.
So we can write \mathbb{R}\cong \mathbb{Q}^X. Then do you see a way of writing \mathbb{R}^2??

So I would have to prove |\mathbb{R}|=|\mathbb{R}^{2}|, because |\mathbb{R^{2}}|=|\mathbb{C}|.

 

[MathematicalPhysicist]

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.

 

[micromass]

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.

I don't think this is true. The \mathbb{Q}-vector spaces \mathbb{Q} and \mathbb{Q}^2 have the same cardinality and there is a monomorphism between \mathbb{Q}\rightarrow \mathbb{Q}^2, 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.

 

[MathematicalPhysicist]

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).

 

[jbunniii]

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.

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.