I was wondering how to prove that $\mathbb{R}/\mathbb{Q}$ contains a subgroup isomorphic to $\mathbb{Q}$. I know that $\mathbb{R}/\mathbb{Q}$ is torsion-free, but I don't know if using that would help me prove it. In another problem I also saw that $\mathbb{Q}/\mathbb{Z}$ is isomorphic to a subgroup of $S^1$. Do you think I could use that too?
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See also: assuming the axiom of choice, $\mathbb R/\mathbb Q$ is isomorphic to $\mathbb R$, and so every group theoretic property of $\mathbb R$ is shared by $\mathbb R/\mathbb Q$. However, we don't need to use the axiom of choice to prove the specific fact asked above, as my answer shows. – Joe Dec 27 '23 at 16:19
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Consider the cosets of the form $r\sqrt 2 +\mathbb Q$, where $r\in\mathbb Q$. One can show, using the irrationality of $\sqrt 2$, that if $r\sqrt2+\mathbb Q=s\sqrt 2+\mathbb Q$ with $r,s\in\mathbb Q$, then $r=s$. Thus, there are a countably infinite number of these cosets, one for each rational.
The map $\varphi:\mathbb Q\to\mathbb R/\mathbb Q$ given by $\varphi(r)=r\sqrt2+\mathbb Q$ is an injective homomorphism. Injectivity follows from the first paragraph. To see that it is a homomorphism, note that $\varphi$ is the composition of the homomorphism $\mathbb Q\to\mathbb R$, $r\mapsto r\sqrt 2$ with the canonical quotient homomorphism $\mathbb R\to\mathbb R/\mathbb Q,r\mapsto r+\mathbb Q$.
Hence, $\varphi(\mathbb Q)$ is isomorphic to $\mathbb Q$.
Joe
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