# 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?

• Do you have an example of a finite rank abelian group which is not finitely generated ? – GreginGre Mar 31 '18 at 9:54
• By the way, your definition fintie rank abelian group is weird...The rank of $A$ is usually defined as $dim_\mathbb{Q}(A\otimes\mathbb{Q})$. Clearly your notion of rank is not the usual one. With the usual definition, rank $0$ abelian groups are exactly abelian torsion groups. With yours, the only rank $0$ abelian group is the trivial group (indeed, let $A$ be an abelian group of rank $0$ in your sense, and let $x\in A$. Then $\mathbb{Z}x$ can be generated by $0$ elements so is the trivial group. In particular, $x=0$). – GreginGre Mar 31 '18 at 10:10
• @GreginGre Apparently $\mathbb Q$ is a finite rank and not f.g. group... – Censi LI Mar 31 '18 at 11:42
• @GreginGre I think your definition is the torsion-free rank of a group, which is of little use to study torsion group or mixed group. And like I said, in torsion free case, finite rank groups are fairly easy to characterize. – Censi LI Mar 31 '18 at 11:45
• @GreginGre By the way, if you are familiar with properties of finitely generated abelian group. You should notice that the definition of the rank of a group $A$ I used here is equivalent to the maximal cardinality of generalized independent subsets of $A$, where $\{a,b,c,\cdots\}\subset A$ is called generalized independent if the natural map $\langle a\rangle\oplus\langle b\rangle\oplus\langle c\rangle\cdots\to A$ is injective, which is rather natural and well accepted in my opinion... – Censi LI Mar 31 '18 at 12:00

(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 $\mathbf{Q}$.