# Is $\mathbb{R/Z}$ isomorphic to $\mathbb{Q/Z}$? [closed]

I’ve seen a proof that those quotient groups are isomorphic to the circle group, but I don’t know if they’re isomorphic to each other. By transitivity, they should be, but I cannot prove it directly.

• See this post. Jan 21, 2021 at 14:57
• So the post says that $\mathbb{Q/Z}$ is isomorphic to $S^1$ but if $\mathbb{R/Z}$ is isomorphic to $S^1$ as well, why aren't $\mathbb{Q/Z}$ and $\mathbb{R/Z}$ isomorphic? Jan 21, 2021 at 15:02
• Because $\Bbb Q/\Bbb Z$ is the torsion subgroup of $\Bbb R/\Bbb Z$, so different from it. Where does it say that $\Bbb Q/\Bbb Z\cong S^1$? Jan 21, 2021 at 15:03
• And what does that mean? I'm just a beginner in the subject, sorry. Jan 21, 2021 at 15:04
• Wait, so $\mathbb{Q/Z}$ is isomorphic to all of the $n$ th roots of unity, which is a subgroup of all the complex numbers $z$ where $|z|=1$ i.e. $S^1$ which is isomorphic to $\mathbb{R/Z}$? Jan 21, 2021 at 15:45

$$\Bbb Q/\Bbb Z$$ has no infinite cyclic subgroup, whereas $$\Bbb R/\Bbb Z$$ does.

A group-theoretic answer why they are not isomorphic is that $$\mathbb Q/\mathbb Z$$ is a torsion group, i.e. each element has a finite order. (The order of $$p/q+\mathbb Z$$ divides $$q$$.)

On the other side, some elements in $$\mathbb R/\mathbb Z$$ are of infinite order (e.g. $$\sqrt{2}+\mathbb Z$$).

• @downvoter, I would like to improve my answer, if I could knew the reason for the downvote, please.
– user700480
Jan 21, 2021 at 14:50

They can't be isomorphic because they have different cardinalities. Indeed, note that $$\mathbb{R} \to S^1: t \mapsto \exp(2 \pi it)$$ induces an isomorphism $$\mathbb{R}/\mathbb{Z}\cong S^1$$, so $$\mathbb{R}/\mathbb{Z}$$ is uncountable. On the other hand, $$\mathbb{Q}/\mathbb{Z}$$ is countable because $$\mathbb{Q}$$ is countable.

• What about the homomorphism $f: \mathbb{Q/Z}→S^1 : q + \mathbb{Z} ↦ exp(2\pi q)$? Isn't it an isomorphism? Jan 21, 2021 at 14:55
• @LeonardoLovera An isomorphism must be surjective. The map you write down is not surjective. Jan 21, 2021 at 15:01
• Give me a counterexample, please; I don't see it. Jan 21, 2021 at 15:02
• The image of your map is $\{\exp(2\pi i q)\mid q \in \mathbb{Q}\} \subsetneq S^1$. The problem is basically the same as asking if $\mathbb{Q}= \mathbb{R}$. You end up with a circle that has "points missing" in very much the same way as $\mathbb{Q}$ has holes missing. Jan 21, 2021 at 15:04
• So $\mathbb{Q/Z}$ is isomorphic to all of the $n$ th roots of unity, which is a subgroup of all the complex numbers $z$ where $|z|=1$ i.e. $S^1$ which is isomorphic to $\mathbb{R/Z}$? Jan 21, 2021 at 15:33