I'm struggling with this problem:
Let $\mathcal{G}=(G,+,-,0)$ be an abelian group. Work in the language $L=\{+,-,c_0\}$ and expand it with a constant symbol $c_g$, for each $g\in G\setminus \{0\}$. Let $\Sigma$ be the set of divisble group axioms and let:
$$\Sigma' = \Sigma \cup\{c_{g_1}+c_{g_2}=c_h: g_1,g_2,h\in G \text{ and } g_1+g_2=h\} \cup \{c_g\neq c_h: g,h\in G \text{ and } g\neq h\}$$
(a) Prove that $\Sigma'$ is satisfiable;
(b) Conclude that, up to isomorphism, every abelian group $\mathcal{G}$ embeds into some divisible abelian group $\mathcal{H}$.
(c) Show that, if $\mathcal{G}$ is infinite, there exists $\mathcal{H}$ as in (b) such that $|G|=|H|$.
Now I have problems with point (a), I can't manage to prove the satisfiability of $\Sigma'$.
I posted also points (b), (c) because I get the idea but I'm struggling to formalize it. Any help is very much appreciated :)
First show that every finitely generated abelian group can be embedded in a divisible one. This uses the structure theorem for finitely generated abelian groups plus the divisibility of $\mathbb Q$, $\mathbb Q/\mathbb Z$, and direct sums of copies of these. This result for finitely generated abelian groups gives you that every finite subset of $\Sigma'$ is satisfiable. Apply the compactness theorem.