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Background: Example 10.3. An introduction to the theory of groups, J.Rotman, chapter 10. Abelian groups.

$\mathbb{G} = \mathbb{Q}$ has generators $\left\{ x_1, \cdots, x_n, \cdots \right\}$ and relations $\left\{ x_1-2x_2, x_2-3x_3, \cdots, x_{n-1}-nx_n, \cdots \right\}$.

Surely, $\mathbb{Q}^+$ can be generated using a finite number of generators, or... am I mistaken? To be frank I don't understand this example at all.

Question: Please explain the presentation of $\mathbb{Q}^+$.

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$\mathbb Q^+$ certainly can't be finitely generated, as if it has generators $a_1/b_1,\ldots,a_n/b_n$ then $1/(b_1\cdots b_n+1)\notin \mathbb Q^+$, as this cannot be written as a sum of the generators. –  Alex Becker Mar 6 '12 at 14:05

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The group $\mathbb Q^+$ consists of the set of rational numbers and has the operation of addition. A set $S$ generates $\mathbb Q^+$ if any rational number can be written as a finite sum of the elements of $S$. This cannot happen if $S$ is finite, since if $$S=\left\{\frac{a_1}{b_1},\frac{a_2}{b_2},\ldots,\frac{a_n}{b_n}\right\}$$ then any finite sum of elements in $S$ is of the form $$\frac{k_1a_1(b_2b_3\cdots b_n)+k_2a_2(b_1b_3\cdots b_n)+\cdots+k_na_n(b_1b_2\cdots b_{n-1})}{b_1b_2\cdots b_n}$$ which rules out $\frac{1}{b_1b_2\cdots b_n+1}$. However, the infinite set $$S=\left\{\frac{1}{1},\frac{1}{1\cdot 2},\ldots,\frac{1}{n!},\ldots\right\}$$ does generate $\mathbb Q^+$, as any rational number $\frac{a}{b}$ can be made by adding $\frac{1}{b!}$ to itself $a\cdot(b-1)!$ times. But this is precisely the generating set you were given (with $x_n=\frac{1}{n!}$), as $\frac{1}{n!}=\frac{1}{n}\frac{1}{(n-1)!}$ or in other words $\frac{1}{(n-1)!}-n\frac{1}{n!}=0$, hence the relation you were given.

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