TITLE: Subgroup of $\mathbb{C}^*$ such that $\mathbb{C}^*/H \cong \mathbb{R}^*$
QUESTION [9 upvotes]: Define $\mathbb{C}^* = \mathbb{C}\setminus \{0\}$ and $\mathbb{R}^* = \mathbb{R}\setminus \{0\}$

Does there exist a subgroup $H$ of $\mathbb{C}^*$ such that $\mathbb{C}^*/H$ isomorphic to $\mathbb{R^*}$?

I knew that $\mathbb{C}^*/{U}\cong \mathbb{R}^+$, with $U = \{ z \in \mathbb{C}:|z| = 1\}$, but how about $\mathbb{R}^*$?
Thank a lot.

REPLY [14 votes]: Suppose there exists a group epimorphism $\phi \colon \mathbb{C}^\times \to \mathbb{R}^\times$. Then there exists some $z \in \mathbb{C}^\times$ with $\phi(z) = -1$. If $w$ is a root of $z$ then $\phi(w)^2 = -1$, which is not possible.
This show more generally that the image of every group homomorphism $\mathbb{C}^\times \to \mathbb{R}^\times$ already lies in $\mathbb{R}_{>0}$.