11
$\begingroup$

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?

  • 1
    @Pete: yes, thanks, +1. Countable cardinality does not pin down the dimension but uncountable does, at least in a world with AC.2010-12-09

1 Answers 1

13

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.

  • 3
    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 $\mathbb{R}$.2010-12-09