I'm trying to construct a counterexample for my student. Does anyone know if there exists (or doesn't exist) a non-trivial group homomorphism:

$$g: \mathbb R/\mathbb Q \to S^1$$

where $S^1$ denotes the unit circle in $\mathbb C$ or equivalently ${[0,2\pi]}/_{0\,\sim\,\pi}$.



1 Answer 1


Yes, one exists (using, though, the infamous Axiom of Choice): let $e_i$ be a $\mathbb{Q}$-basis of $\mathbb{R}$ with $e_{i_0}=1$.

Define arbitrary real numbers $a_i$ with $a_{i_0}=0$. The homomorphism is defined by taking the quotient of $x=\sum_i{q_ie_i} \longmapsto \exp{2i\pi \sum_i{q_ia_i}}$.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .