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A student came to me showing a question from his exam in basic group theory, in which they are asked to prove that $\mathbb{C}^*$ modulo the subgroup of roots of unity is isomorphic to $\mathbb{R}^+$ (in both cases we mean the multiplicative groups).

Now this seems to me to be a simple error in the question. I believe they meant to ask to prove that $\mathbb{C}^*$ modulo all the elements of absolute value 1 is isomorphic to $\mathbb{R}^+$ which is very easy to prove (take the homomorphism mapping $z$ to $|z|$ and use the first homomorphism theorem). However, I couldn't prove the claim about the roots of unity is wrong; is there an easy way to show this?

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I think it means: allowing discontinuous, in fact non-measurable, isomorphisms, there is a (set theoretic, uncountably Axiom Of Choice dependent) proof that such an isomorphism exists. Both groups are uncountable, same cardinality, torsion-free, divisible, abelian, so are isomorphic according to the general characterization. It is like the set-theoretic existence of isomorphism between algebraic closures of real and p-adic fields. – T.. Dec 1 '10 at 17:56
I agree that they are isomorphic, but that the exam question probably was not the one that the examiners intended. The proof that they are isomorphic is above the level of "basic group theory." – Grumpy Parsnip Dec 1 '10 at 18:18
Gadi, it is equivalent to the statement that infinite vector spaces (over the rational numbers) are classified by their cardinality. Because the groups are torsion-free and divisible, they are uniquely divisible, so it is necessary and sufficient to get an isomorphism of vector spaces. And this is where the set theory is used. – T.. Dec 1 '10 at 21:20
Yes, I understand this now, thanks everyone (why has no one formally "answered"? I think your answers here were good enough). – Gadi A Dec 1 '10 at 21:31
@T..: Uncountably infinite vector spaces over $\mathbb{Q}$ are classified by their cardinality. $\mathbb{Q}$ and $\mathbb{Q}^2$ are nonisomorphic $\mathbb{Q}$-vector spaces of the same infinite cardinality. – Pete L. Clark Dec 9 '10 at 0:49
up vote 11 down vote accepted

For the sake of completeness: $\mathbb{C}^{\ast}$ is isomorphic to $\mathbb{R}^{+} \oplus \mathbb{R}/\mathbb{Z}$ in the obvious way, and $\mathbb{C}^{\ast}$ modulo the roots of unity is then isomorphic to $\mathbb{R}^{+} \oplus \mathbb{R}/\mathbb{Q}$. As a $\mathbb{Q}$-vector space this is abstractly isomorphic to $\mathbb{R}^{+}$, but the construction of such an isomorphism is likely to require the axiom of choice.

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Right. Perhaps it is worth pointing out that the multiplicative group $\mathbb{R}^+$ is isomorphic to the additive group $\mathbb{R}$ via the logarithm. Thus the quotient is a $\mathbb{Q}$-vector space of continuum dimension, as is $\mathbb{R}$. – Pete L. Clark Dec 9 '10 at 0:48

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