Abelian group is the additive group $\mathbb{Q}$ Prove that the following abelian group is the additive group $\mathbb{Q}$:
$$G=\langle x_1, x_2,\ldots,x_n,\ldots\mid x_1 - 2x_2, x_2 - 3x_3,\ldots,x_{n-1} - nx_n,\ldots\rangle.$$
Thanks for your help!
 A: Try writing down a map!  Here is one idea:
$$x_n\longmapsto \frac{1}{n!}$$
and extend additively.  This means for an element $\sum a_nx_n$ where almost all $a_n$ are zero, define the map as follows:
$$\sum a_nx_n\longmapsto\sum\frac{a_n}{n!}$$
To check if this map is well-defined, show that it vanishes on all elements of the form $x_{n-1}-nx_n$, as it should.  Having done this, the map is automatically a homomorphism because we've defined it to split across sums.
We need to show this is bijective.  One way to do this is to write down an inverse.
$$\frac{a}{b}\longmapsto a(b-1)!x_b$$
Using the relations, you should be able to show this inverse is well-defined, that is, it is independent of the choice of representative of the fraction $a/b$.  To see this, suppose $a/b=c/d$.  Then we have:
$$\begin{align}a(b-1)!x_b&=\frac{bc}{d}(b-1)!(b+1)x_{b+1}\\[.1in]&=\frac{c}{d}(b-1)!b(b+1)(b+2)x_{b+2}\\[.1in]&\vdots\\[.1in]&=\frac{c}{d}d!x_d=c(d-1)!x_d\end{align}$$
The following equalities show that the inverse is additive.
$$\begin{align}a(b-1)!x_b+c(d-1)!x_d&=a(b-1)!(b+1)\ldots(bd)x_{bd}+c(d-1)!(d+1)\ldots(bd)x_{bd}\\[.1in]&=\frac{a}{b}(bd)!x_{bd}+\frac{c}{d}(bd)!x_{bd}\\[.1in]&=(ad+bc)(bd-1)!x_{bd}\end{align}$$
Everything works out!
A: Hint: Identify $x_n$ with $\cfrac1{n!}$.
