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Consider a group homomorphism $f:\Bbb{C}^*\to\Bbb{R}^*$. In a recent question we easily established that $\ker(f)$ is necessarily infinite. The homomorphisms that can be easily described are of the form $$ f(z)=|z|^a $$ for some real constant $a$. All those homomorphisms contain the unit circle in their kernel. This raises the suspicion:

Is $\ker (f)$ necessarily uncountable?

Not all the homomorphisms are of the above form. The group of positive real numbers is divisible, i.e. all the positive real numbers have positive roots of a given integer order. This implies that in the category of Abelian groups the group $\Bbb{R}_{>0}$ is an injective object (Zorn's lemma is needed to prove this). This implies the existence of other homomorphisms as follows. Let $\omega\in\Bbb{C}$ be a number such that i) $|\omega|=1$, and ii) $\omega^n\neq1$ for all $n\in\Bbb{Z}$, IOW $\omega=e^{2\pi i r}$ for some irrational real number $r$. Let us select $a\in\Bbb{R}_{>0}$, $a\neq1$. Consider the subgroup $H=\langle \omega\rangle\times\Bbb{R}_{>0}\le\Bbb{C}^*$. The rule $$ f_a(\omega^nx)=a^nx, $$ for all $x\in\Bbb{R}_{>0}$, then defines a homomorphism $f_a:H\to\Bbb{R}_{>0}$. By injectivity of the target group we can extend this to a homomorphism $f'_a$ from all of $\Bbb{C}^*$ to $\Bbb{R}_{>0}$ such that $f'_a(\omega)=a\neq1$. But does that really help answer the question? Neither $\Bbb{C}^*$ nor $\Bbb{R}^*$ is finitely generated, so we don't know whether they can be written as a direct product of a torsion group and a free abelian group, or do we?

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Jyrki Lahtonen
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1 Answers1

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Both groups $\mathbb{R}^* / \mu(\mathbb{R})$ and $\mathbb{C}^* / \mu(\mathbb{C})$ (where $\mu$ is the group of roots of unity) are uniquely divisible, and thus are vector spaces over $\mathbb{Q}$.

Since the positive reals are closed under multiplication, it's easy to see that

$$ \mathbb{R}^* \cong \mu(\mathbb{R}) \oplus \mathbb{R}^* / \mu(\mathbb{R}) $$

I doubt that $\mu(\mathbb{C})$ is a direct summand of $\mathbb{C}^*$. However, this is still enough to give a positive answer to the question: we can, using the axiom of choice, construct a group homomorphism

$$ \mathbb{C}^* \to \mathbb{C}^* / \mu(\mathbb{C}) \cong \mathbb{Q}^{\mathfrak{c}} \cong \mathbb{R}^* / \mu(\mathbb{R}) \to \mathbb{R}^*$$

Interestingly, we can also use this analysis to show there is no surjective group homomorphism $\mathbb{C}^* \to \mathbb{R}^*$, by using the fact that there are no nontrivial homomorphisms $\mathbb{Q} \to \mathbb{Z}/2\mathbb{Z}$ nor from $\mu(\mathbb{C}) \cong \mathbb{Q}/\mathbb{Z} \to \mathbb{Z}/2\mathbb{Z}$.

EDIT: My intuition is miscalibrated: $\mu(\mathbb{C})$ is a direct summand of $\mathbb{C}^*$. This must be true because

$$ \operatorname{Ext}(\mathbb{C}^*/\mu(\mathbb{C}), \mu(\mathbb{C})) \cong \operatorname{Ext}(\mathbb{Q}^{\mathfrak{c}}, \mathbb{Q}/\mathbb{Z}) \cong 0$$

(because $\mathbb{Q} / \mathbb{Z}$ is injective), and so the exact sequence

$$ 0 \to \mu(\mathbb{C}) \to \mathbb{C}^* \to \mathbb{C}^* / \mu(\mathbb{C}) \to 0$$

must be split.

  • This still leaves open the question of a surjective homomorphism. Or if we can arrange for the kernel to be $2 \mu(\mathbb{C})$ rather than $\mu(\mathbb{C})$. –  Feb 14 '14 at 09:38
  • Very nice! I guess by $2\mu$ you mean $\mu^2$? – Alex B. Feb 14 '14 at 09:40
  • @Alex: Yes; the subgroup of squares. –  Feb 14 '14 at 09:55
  • @Alex: Oh duh, the subgroup of squares in $\mu(\mathbb{C})$ is just $\mu(\mathbb{C})$ itself! The only homomorphism $\mu(\mathbb{C}) \to \mu(\mathbb{R})$ (i.e. from $\mathbb{Q} /\mathbb{Z} \to \mathbb{Z} / 2\mathbb{Z}$) is the trivial homomorphism. Silly oversight on my part. –  Feb 14 '14 at 10:07
  • Nice! The structure as vector spaces over the rationals is the key. – Jyrki Lahtonen Feb 14 '14 at 10:09