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.
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$\begingroup$ See this post. $\endgroup$– Dietrich BurdeJan 21, 2021 at 14:57
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$\begingroup$ 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? $\endgroup$– Leonardo LoveraJan 21, 2021 at 15:02
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$\begingroup$ Because $\Bbb Q/\Bbb Z$ is the torsion subgroup of $\Bbb R/\Bbb Z$, so different from it. Where does it say that $\Bbb Q/\Bbb Z\cong S^1$? $\endgroup$– Dietrich BurdeJan 21, 2021 at 15:03
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$\begingroup$ And what does that mean? I'm just a beginner in the subject, sorry. $\endgroup$– Leonardo LoveraJan 21, 2021 at 15:04
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$\begingroup$ 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 $\mathbb{R/Z}$? $\endgroup$– Leonardo LoveraJan 21, 2021 at 15:45
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3 Answers
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$\Bbb Q/\Bbb Z$ has no infinite cyclic subgroup, whereas $\Bbb R/\Bbb Z$ does.
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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. $\sqrt{2}+\mathbb Z$).
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$\begingroup$ @downvoter, I would like to improve my answer, if I could knew the reason for the downvote, please. $\endgroup$– user700480Jan 21, 2021 at 14:50
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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.
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$\begingroup$ What about the homomorphism $f: \mathbb{Q/Z}→S^1 : q + \mathbb{Z} ↦ exp(2\pi q)$? Isn't it an isomorphism? $\endgroup$ Jan 21, 2021 at 14:55
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$\begingroup$ @LeonardoLovera An isomorphism must be surjective. The map you write down is not surjective. $\endgroup$– J. De RoJan 21, 2021 at 15:01
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$\begingroup$ Give me a counterexample, please; I don't see it. $\endgroup$ Jan 21, 2021 at 15:02
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$\begingroup$ 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. $\endgroup$– J. De RoJan 21, 2021 at 15:04
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$\begingroup$ 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 $\mathbb{R/Z}$? $\endgroup$ Jan 21, 2021 at 15:33