Is $\Bbb C^* \times \Bbb Z$ isomorphic to a subgroup of $\Bbb C^*$?

Question

I would like to know if $\Bbb C^* \times \Bbb Z$ is isomorphic (as an abelian group) to a subgroup of $\Bbb C^*$.

Of course, they are not isomorphic: one is divisible, while the other is not. We can't apply some version of Cantor-Schröder-Bernstein, because it is wrong for the category of abelian groups (see ex. 3.1. here). I know that $\Bbb C^* \cong S^1 \times \Bbb R_+^* \cong \Bbb R/\Bbb Z \times \Bbb R$, and that a morphism $f : \Bbb C^* \times \Bbb Z \to \Bbb C^*$ is determined by $f(a,1)$ and $f(a,0)$ for $a \in \Bbb C^*$, since

$$f(re^{i\theta};n) = f(\sqrt[n]{r}e^{i\theta/n}\,;1)^n \qquad (n \geq 1)$$

But I don't know how to go further, even if my problem may be very easy.

Thank you very much for your help.

Answer 1

View $\mathbb{R}$ as a vector space over $\mathbb{Q}$. Since $\dim_{\mathbb{Q}} \mathbb{R} = \mathfrak{c}$, it follows that there is a $\mathbb{Q}$-vector space isomorphism $\mathbb{R}\times \mathbb{Q} \to \mathbb{R}$. Thus we have an injective group homomorphism $\mathbb{R}\times \mathbb{Z} \hookrightarrow \mathbb{R}$. This induces an injective group homomorphism

$$S^1 \times \mathbb{R} \times \mathbb{Z} \hookrightarrow S^1 \times \mathbb{R}.$$

Answer 2

Here is a proof of a variant of your question. I feel this would be beneficial. In essence this is same as Daniel Fischer's answer. But more explicit for this case.

I will prove this: Denoting by $\mathbf{Q}_{>0}$ the group of positive rational numbers (for multiplication) we will show $\mathbf{Q}_{>0} $ is isomorphic to $\mathbf{Q}_{>0}\times \mathbf{Z}$.

Fundamental theorem of arithmetic can be interpreted as saying that positive rational numbers form a free abelian group with the set of prime numbers as basis.

So a bijection of this basis $\{2,3,5,7,11,\ldots\}$ to an enlarged basis $\{t ,2,3,5,7,11,\ldots\}$ for $\mathbf{Q}_{>0}\times \mathbf{Z}$ gives the isomorphism we want.

(I believe it would be possible to adapt this proof to give a proof for what you ask).