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Let $G = \mathbb{R} / \mathbb{Q}$. Is this an interesting group to study?

Is it isomorphic to any more natural mathematical objects?

Daniel Pietrobon
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2 Answers2

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Its structure is completely determined, being it a divisible torsion free abelian group; it's torsion free because, for $n\in\mathbb{N}$, $n>0$, and $x\in\mathbb{R}$,

$$ n(x + \mathbb{Q}) = \mathbb{Q} $$

is equivalent to $nx\in\mathbb{Q}$, so to $x\in\mathbb{Q}$.

Thus $\mathbb{R}/\mathbb{Q}$ is a vector space of dimension $\mathfrak{c}=\lvert\mathbb{R}\rvert$ over $\mathbb{Q}$, by a cardinality argument. This implies, as observed in comments, that, as $\mathbb{Q}$-vector spaces and so as abelian groups,

$$ \mathbb{R}/\mathbb{Q}\cong\mathbb{R}. $$

Finding a basis for it would imply finding a Hamel basis of $\mathbb{R}$ over $\mathbb{Q}$, because of the canonical $\mathbb{Q}$-linear mapping

$$ \pi\colon \mathbb{R}\to\mathbb{R}/\mathbb{Q} $$

Just choose elements $x_\alpha\in\mathbb{R}$ that are mapped to a basis of $\mathbb{R}/\mathbb{Q}$; adding $1$ to this family would produce a Hamel basis for $\mathbb{R}$.

egreg
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    Doesn't that imply that $\mathbb R/\mathbb Q\cong \mathbb R$? Basically, they are both vector spaces over $\mathbb Q$ with bases of the same cardinality. – Thomas Andrews Apr 09 '13 at 12:56
  • @ThomasAndrews Yes, of course. But you can't name such an isomorphism without naming a Hamel basis. – egreg Apr 09 '13 at 12:58
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    Sure, my point isn't that there is any constructable isomorphism, but to let the OP realize that the purely "abelian group" structure of $\mathbb R$ is not what he thinks it is. The fact that $\mathbb R/\mathbb Q\equiv \mathbb R$ is the most bold-faced way of making that strangeness apparent. – Thomas Andrews Apr 09 '13 at 13:01
  • "But you can't name such an isomorphism without naming a Hamel basis. " Why is that so? Of course the existence of a Hamel basis implies the existence of an isomorphism. But why the converse? I'm not really convinced. – Martin Brandenburg Nov 17 '13 at 13:56
  • @MartinBrandenburg I'm not convinced too. ;-) – egreg Nov 17 '13 at 14:07
  • I thought it was possible that $\mathbb{R}/\mathbb{Q} > \mathbb{R}$ if you didn't have the axiom of choice, and specifically if every subset of $\mathbb{R}$ is measurable. – Zemyla Dec 21 '19 at 07:15
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Well, it is an abelian torsion-free with an uncountable ammount (cardinal) of elements and, thus, it is also a non finitely-generated group.

You may also be interested in the fact that

$$\Bbb R/\Bbb Q\cong\left(\Bbb R/\Bbb Z\right)/\left(\Bbb Q/\Bbb Z\right)$$

so that $\,\Bbb R/\Bbb Q\,$ is a quotient of the circle group

DonAntonio
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