Classification of finite rank Abelian groups An Abelian group $A$ is said to be finite rank if there is a natural number $n$ such that any finitely generated subgroup of $A$ can be generated by no more than $n$ elements. It is well-known that a torsion-free finite rank group can always be embedded into a finite dimensional $\mathbb Q$-vector space. However, the general case seems to be more complicated. But I still want to know if there is any convenient characterization for such groups. More specifically, I'd like to know:

1) Is a finite rank Abelian group always a direct sum of its torsion subgroup and a torsion free group?
2) Is a finite rank Abelian group always countable?

 A: 1) No. 2) Yes.
(2) Let's start with the easier 2. Suppose that $A$ has finite rank, and $T_A$ denotes its torsion subgroup. Then $A/T_A$ and $T_A$ both have finite rank. So it's enough to prove the result in the torsion-free and the torsion case. The torsion-free case is fine, as you already mentioned ($A$ embeds into $\mathbf{Q}^n$ for some $n$, so is countable). Suppose that $A$ is torsion, and of finite rank. Then $A=\bigoplus A_p$, where $A_p$ is its $p$-component (the largest $p$-subgroup of $A$); $\bigoplus$ denotes restricted direct product, $p$ ranges over primes. This reduces to $A_p$. It is not hard and standard to prove that an abelian $p$-group whose kernel of multiplication by $p$ is finite, is isomorphic to a subgroup of $(\mathbf{Z}[1/p]/\mathbf{Z})^n$ for some $n$ (take an injective hull). Whence the result.
(1) Consider the uncountable group $G=\prod_p\mathbf{Z}/p\mathbf{Z}$, where $p$ ranges over primes. Its torsion-free quotient by its torsion subgroup $\bigoplus\mathbf{Z}/p\mathbf{Z}$ is easily seen to be divisible, and hence is isomorphic to a vector space over $\mathbf{Q}$. Consider a copy of $\mathbf{Q}$ in this quotient, and let $A$ be its preimage in $G$. So $T_A$ is isomorphic to $\bigoplus\mathbf{Z}/p\mathbf{Z}$, $A/T_A$ is isomorphic to $\mathbf{Q}$; both have rank (in your sense) 1 and hence $A$ has rank $\le 2$ (actually $=2$ since $A$ admits a subgroup isomorphic to $\mathbf{Z}\oplus \mathbf{Z}/2\mathbf{Z}$). Then $T_A$ is not a direct summand of $A$ since $A$, being residually finite, does not include any subgroup isomorphic to $\mathbf{Q}$.
