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So the question of why an abelian torsion-free group can always be embedded in a divisible group has already been asked on this site; but I'm looking for a proof using the compactness theorem.

I have what one may call leads, but not a full solution.

Here's my attempt : let $G$ be an abelian torsion-free group, and let $L$ be the language of groups, to which we've added a constant symbol for each element in $G$. Let $T$ be the theory of abelian groups, with the tables of $G$ (i.e. if $g, h\in G$ and if $\underline{g}$ denotes the symbol added for $g$, then $T$ has the axiom $\underline{g}+\underline{h} = \underline{g+h}$). Then we consider, for each $n\in \mathbb{N}$ and $a\in G$, the following sentence $\Phi_{n,a} := \exists x, x+...+x = \underline{a}$ (where there are $n$ $x$'s).

Let $\Sigma := T\cup \{\Phi_{n,a}, n\in \mathbb{N}, a\in G\}$. Through compactness, it is easily seen that $\Sigma$ is consistent (indeed it suffices to show that for any abelian group $H$ and any $a\in H$, $H$ can be embedded in an abelian group with an "$n$th root of $a$", which is done by taking $H(x)$ a formal abelian group generated by $H$ and $x$, then taking the quotient of $H(x)$ by the subgroup $\langle n\cdot x - a\rangle$). But this is where I get stuck : what I now have is a group $H$ in which $G$ can be embedded, and in which every element of $G$ has an $n$th root. The first problem is that $H$ need not be torsion-free and so the root in question may not be unique. This is a problem since my next idea was to take the intersection of all subgroups of $H$ which have roots for all elements of $G$, and then showing that this would work, but I can't..

Any indications ?

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Instead, let $T$ be the theory of t.f. (torsion-free) abelian groups, enriched with the diagram of $G$ (for $g,h\in G$, add as axioms: $\underline{g+h}=\underline{g}+\underline{h}$ and, if $g\ne h$, also $\underline{g}\ne\underline{h}$), and let $\Sigma$ be the theory $T\cup \{\Phi_{n,a}\,| \, n\in \mathbb{N} \, \& \, a\in G\}$ as before.

Let $\Delta\subseteq\Sigma$ be finite. It contains finitely many sentences of the form $\Phi_{n,a} = \exists x, n.x = \underline{a}$. We treat them one by one. Given $n\in\mathbb{N}$ and $a\in G$, if $n=1$ we may skip $\Phi_{n,a}$. If not, factorize $n$ as $n=p_{1}p_{2}\cdots p_{r}$ with the $p_{i}$ prime (but not necessarily different, of course).

Start with $p:=p_{1}$, and let $H$ be our current t.f. abelian group (initially, we take $H:=G$), which we extend step by step. And let $b:=a$ be our current "working element" in $H$.

If $\exists x\in H, p.x = b$, do nothing; H is unchanged, and is still t.f.

If no such $x$ exists in $H$, let $K:=H\oplus\mathbb{Z}/<(b,-p)>$ (this is your $H(x)/<px-b>$). Then $H\subseteq K$ via $h\mapsto \overline{(h,0)}$, and we identify $H$ with it's image in $K$. The element $x:=\overline{(0,1)} \in K$ satisfies $p.x=\overline{(0,p)}=\overline{(b,0)}=b$. The group $K$ remains t.f.: suppose $s.\overline{(h,m)}=0$ in $K$ for some $s\in\mathbb{N},h\in H, m\in\mathbb{Z}$. Then $\overline{(s.h,s.m)}=0$, so $(s.h,s.m)\in <(b,-p)>\subseteq H\oplus\mathbb{Z}$, so $s.h=i.b$ and $s.m=-i.p$ for some $i\in\mathbb{Z}$. Then $(-ip)h=(sm)h=m.(sh)=i.m.b$; as $H$ is t.f, it follows that $-p.h=m.b$. If $p$ does not divide $m$, one has $1=u.p+v.m$ for suitable integers $u,v$, so $b=upb+vmb=upb-vph=p.(ub-vh)$ is a multiple of $p$ in $H$ after all - contradiction. Hence $p$ divides $m$, so $m=p.j$, say, and we find $-p.h=m.b= p.j.b$, so $-h=j.b$, and $\overline{(h,m)}=\overline{(-j.b,j.p)}=-j.\overline{(b,-p)}=-j.0=0$ in $K$. I.e., $K$ is indeed torsion-free. We now let our working group $H:=K$.

We take our new $b$ to be the unique $x\in H$ such that $p.x=b$ (our previous $b$), and we proceed to $p:=p_{2}$, the next prime factor of $n$, and argue in the same way. So if $\exists x\in H, p.x = b$, do nothing, and otherwise add a "$p$th root" of $b$ as above.

After finitely many steps we arrive at an $H\models T\cup \{\Phi_{n,a}\}$, and clearly we can continue in this way for the remaining $\Phi_{n',a'}$, to get a model $H\models \Delta$.

Using Compactness, we find a t.f. abelian group $H$ that contains $G$ and has the property that for every $a\in G$ and $n\in\mathbb{N}$, $H$ has an element $x$ such that $n.x=a$. The subgroup $H_{0}:=\{x\in H | \exists n\in\mathbb{N},n.x\in G \}$ clearly is t.f as well, contains $G$, and is divisible.

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  • $\begingroup$ Thanks ! I'll look at the details a bit later but this seems fine ! $\endgroup$ Commented May 30, 2017 at 18:27

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