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Let $G$ be a group, $X$ a set of generators of G and assume that $f: X \rightarrow G$ is such that $G=\left<f(X)\right>$ (for example $f(x)=x$, for $x\in X$). Denote by $F(X)$ the free group with the free generating set $X$.

Then by the universal property there exists exactly one homomorphism $f^*:F(X)\rightarrow G$ such that $f=f^*\circ \mu$ (clear $f^*$ is an epimorpism), where $\mu:X\rightarrow F(X)$, $\mu(x)=[x],$ for $x\in X$.

If a set of relations $\mathcal{R}\subset \mbox{Ker}f^*$ is such that $Ker f^*$ is the smallest normal group in $F(X)$ containing $\mathcal{R}$ then the couple $(X, \mathcal{R})$ is called a genetic code of the group $G$ (of course it is not unique) and we write $G=\mbox{gr}(X\|\mathcal{R})$.

We have a very useful theorem:

Let $G=\mbox{gr}(X\|\mathcal{R})$ and $f: X \rightarrow G$ be a function satisfied conditions above. Assume that $K$ is a group and $h: f(X) \rightarrow K$ is an arbitrary mapping. Then the following conditions are equivalent:

(i) there exists exactly one homomorphism $h':G\rightarrow K$ such that $h'|_{f(X)}=h$,

(ii) for any $[{x_1}^{\epsilon^1}\cdot...\cdot{x_n}^{\epsilon^n}]\in\mathcal{R}$, where $x_i\in X, \epsilon_i\in\{-1,1\}$ we have $(h\circ f)(x_1)^{\epsilon_1}\cdot...\cdot (h\circ f)(x_n)^{\epsilon_n}=1_K$

By the theorem above it is easy to see for example that $\mathbb{Z}_2\times\mathbb{Z}_2=gr(\{a,b\}\|a^2,b^2,aba^{-1}b^{-1})$ or more intuitively $\mathbb{Z}_2\times\mathbb{Z}_2=gr({a,b}\|a^2,b^2,ab=ba$).

Could anyone help me to prove that a genetic code of the additive group of $\mathbb{Q}$ is as follows $\mathbb{Q}=\mbox{gr}(x_1,x_2,...\|x_1=x_2^2,x_2=x_3^3,x_3=x_4^4...)$? Can one use for this purpose the theorem above? I will be grateful for your help.

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The $n$th relation is $x_{2n-1}=(x_{2n})^{2n}$, is this correct? – Did Oct 23 '11 at 17:19
No, it was a mistake, thank you very much for the remark. Now is correctly. – dawid Oct 23 '11 at 17:27
Have you tried mapping $x_n$ to $1/n!$ ? – user641 Oct 23 '11 at 17:49
Right, and surjectivity is somewhat easy, and injectivity can be checked locally (if some element was mapped to $0$, it would be a word in only finitely many of the $x_i$). – user641 Oct 23 '11 at 18:10
$\frac{1}{q}=\frac{(q-1)!}{q!}=(q-1)!h'(x_q)=h'({x_q}^{(q-1)!})$, is it correct? – dawid Oct 23 '11 at 18:47
up vote 2 down vote accepted

We need a bijective homomorphism from your group $G$ to the rationals $\mathbb{Q}$.

First, we define our potential homomorphism by specifying that it maps $x_n$ to $\frac{1}{n!}$. Since $\frac{n}{n!}=\frac{1}{(n-1)!}$, your quoted theorem above implies this is a homomorphism.

To see it is surjective, we need to show any fraction $\frac{p}{q}$ is in the image. But if $k=p(q-1)!$, then $x_q^k$ maps to $\frac{p}{q}$.

Before we check injectivity, note that $G$ is abelian. Indeed if $i<j$, then $x_i\in\langle x_j\rangle$, so certainly $x_i$ and $x_j$ commute.

To see it's injective, suppose some element of $G$ mapped to $0$ in $\mathbb{Q}$. Then this element is a finite product of some of the $x_i$, let's say we can write it as $x_{i_1}^{n_1}\cdots x_{i_m}^{n_m}$, with $i_1<i_2<\cdots <i_m$ [we can do this because $G$ is abelian]. But the relations of $G$ allows us to write, for example, $x_{i_1}$ as some power of $x_{i_m}$. Similarly this works for $x_{i_2}$, etc., so that we can actually write this word as $x_{i_m}^N$, for some integer $N$. Thus saying this element maps to zero is the same as saying $\frac{N}{i_m!}=0$, which of course means $N=0$, so that our initial element was really $x_{i_m}^0$ - the identity. So our map is injective.

I will also mention what I mentioned in the comments above: the idea here generalizes quite a bit. You have some countable group $H$, and you want to embed it in the rationals. The necessary and sufficient condition here is that $H$ is abelian, and $H$ is locally infinite cyclic. The reason this works is because $\mathbb{Q}$ is divisible, so you can lift a "local" map to a bigger map, then rinse, repeat. Again, this is really like lifting local data to global data: you can check $H$ is abelian locally, you can check the cyclic condition locally (by definition), and so you can build a "local" homomorphism, where you check injectivity of the map, again, locally. Then the divisibility kicks in and recursive definition gives you a global homomorphism.

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Thank you very much for the comprehensive response – dawid Oct 25 '11 at 11:59

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