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We know that $\mathbb{Q}\cong\mathbb{Z}\times\mathbb{Z}/\sim$, where the isomorphism is a ring isomorphism and the equivalence relation is defined as

$$(a,b)\sim(c,d)\Longleftrightarrow ad=bc$$

Then the relation is stated in terms of the integers' multiplication. So, if I want to give a presentation for the additive group of $\mathbb{Q}$, how should I modify the presentation

$$\mathbb{Z}\times\mathbb{Z}\cong\, \langle a,b|aba^{1}b^{-1} \rangle$$

as I can't use the multiplication? The equivalence relation that defines $\mathbb{Q}$ can be stated in additive terms as

$$(a,b)\sim(c,d)\Longleftrightarrow \begin{cases}a=nc\\b=nd\end{cases}$$

where of course $nx$ stands for $\sum_{i=1}^{n}x$, but I have no idea as how to insert it in the presentation. I'm in doubt that the additive group of $\mathbb{Q}$ is different from that of $\mathbb{Z}\times\mathbb{Z}$, but it seems to me that fractions should be identified on the (additive) group structure independently of the subsequent definition of the multiplication.

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    $\begingroup$ Note that the rationals are not finitely generated as a group, so the presentation of $\Bbb{Z}\times\Bbb{Z}$ you give cannot (easily) be turned into a presentation of $\Bbb{Q}$. $\endgroup$
    – Servaes
    Commented Jun 4, 2015 at 13:30
  • $\begingroup$ I had heard of that, but I can't understand why, as $\Bbb{Z}\times\Bbb{Z}$ is finitely generated. $\endgroup$
    – Giorgio Comitini
    Commented Jun 4, 2015 at 13:58
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    $\begingroup$ Your first sentence is false. This is not a ring isomorphism, it is just a bijection between two sets. $\endgroup$
    – Derek Holt
    Commented Jun 4, 2015 at 15:22
  • $\begingroup$ As @DerekHolt mentions, your first sentence is false. Indeed, the group of nonzero rational numbers under multiplication is not finitely generated. (It is isomorphic to $\mathbb{Z}_2 \oplus \bigoplus_{n=1}^\infty \mathbb{Z}$, where the $\mathbb{Z}_2$ is the group $\{-1,1\}$, and each $\mathbb{Z}$ summand is the cyclic group generated by a prime number.) $\endgroup$
    – Jim Belk
    Commented Jun 4, 2015 at 17:02
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    $\begingroup$ Well, now that I think about it, it is obvious. $\Bbb{Z}\times(\Bbb{Z}\setminus\{0\}\to\Bbb{Q}$ is only a bijection. Thank you for the answers. $\endgroup$
    – Giorgio Comitini
    Commented Jun 4, 2015 at 17:41

2 Answers 2

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By the way, a presentation of $(\mathbb{Q},+)$ is

$$\langle x_1,x_2,\ldots \mid x_n^n=x_{n-1}, \ n \geq 2 \rangle.$$

To understand why, it is possible to consider the morphism induced by $x_n \mapsto \frac{1}{n!}$.

For more information, see Johnson's book, Presentations of groups, Chapter 5.7.

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    $\begingroup$ Or in additive notation, $\langle x_1, x_2, \ldots \mid n x_n = x_{n-1}, n \ge 2\rangle$. $\endgroup$
    – Dustan Levenstein
    Commented Jun 4, 2015 at 15:22
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    $\begingroup$ Note that the proof in Johnson's book is not super-trivial. (I mean, fine, it's not that hard, but it's a couple of pages long.) Just in case a reader was wondering, as I did, if they should just be "seeing" the answer in a totally trivial way. $\endgroup$
    – Hugh Thomas
    Commented Nov 9, 2020 at 13:43
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$\mathbb{Q}$ is not a quotient of $\mathbb{Z}\times \mathbb{Z}$, or indeed any finite product of copies of $\mathbb{Z}$. By the classification of finitely generated abelian groups, a quotient of $\mathbb{Z}^N$ has a free component (which is a product of finitely many $\mathbb{Z}$'s) and a torsion component.

If $\mathbb{Q}$ were isomorphic to such a quotient, it couldn't have a torsion component because $\mathbb{Q}$ doesn't have any torsion (there is no integer $n$ and nonzero element $x$ such that $nx=0$). However, $\mathbb Q$ is not free because for any element $x$ and any integer $n$ there is an element $y$ such that $ny=x$; this is not true for free abelian groups. We can see a simple example where this fails in a free abelian group in the element $x=(1,0,0,\ldots,0)$; there is no element $y$ such that $ny=x$ for any $n>1$.

Any abelian group is a quotient of a free abelian group, but for $\mathbb{Q}$ we would need a free abelian group with infinitely many generators. In particular, $\mathbb{Z}\times\mathbb{Z}$ won't work.

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  • $\begingroup$ Or alternatively, it cannot be a quotient of such a product since it is not finitely generated. $\endgroup$
    – Tobias Kildetoft
    Commented Jun 4, 2015 at 15:03
  • $\begingroup$ @TobiasKildetoft Essentially this is a proof that $\mathbb{Q}$ is not finitely generated using the classification. $\endgroup$
    – Matt Samuel
    Commented Jun 4, 2015 at 15:06
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    $\begingroup$ You don't need the classification theorem to prove that $\mathbb Q$ is not finitely generated; it simply follows from the fact that every finitely generated subgroup of $\mathbb Q$ is cyclic, and $\mathbb Q$ is not a cyclic group. $\endgroup$
    – Dustan Levenstein
    Commented Jun 4, 2015 at 15:25

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