# Example of a torsion-free abelian group of rank zero

Discussion in 'Mathematics' started by StefanH, Oct 8, 2018.

1. ### StefanHGuest

In Derek Robinsons A course in the theory of groups, on page 98 the rank of an abelian group is defined:

Let $G$ be an abelian group and let $S$ be a nonempty subset of $G$. Then $S$ is called lineary independent [...] if $0 \notin S$ and, given distinct elements $s_1, \ldots, s_r$ of $S$ and integers $m_1, \ldots, m_r$, the relation $m_1 s_1 + \ldots + m_r s_r = 0$ implies that $m_i s_i = 0$.

Note that this is slightly different than the usual definition in the sense that he just asserts the implication $m_i s_i = 0$, and not that the coefficients $m_i$ should vanish. Then he goes on:

If $p$ is a prime and $G$ an abelian group, the $p$-rank of $G$ $$r_p(G)$$ is defined as the cardinality of a maximal independent subset of elements of $p$-power order. Similarly the $0$-rank or torsion-free rank $$r_0(G)$$ is the cardinality of a maximal independent subset of elements of infinite order. Also important is the Prüfer rank, often just called the rank of $G$ $$r(G) = r_0(G) + \max_{p} r_p(G).$$

Then he proves that these notions are well-defined. Later on in a section on torsion-free groups of rank $1$, he introduces the notion of the type of an abelian group, and proves that two torsion-free abelian groups of rank $\le 1$ are isomorphic if and only if the y have the same type.

If we assume $G$ is torsion-free, I think the $p$-rank vanishes for every $p$ as there is no element of order $p$ by definition. Hence $r(G) = r_0(G)$. But in this case, how should a torsion-free group of rank zero look like? Let $G$ be any abelian torsion-free subgroup and $g \in G$, then $g$ itself is lineary indepedent, hence $r(G) \ge 1$. Or do I miss anything? Could anybody give me an example of an torsion-free abelian group of rank $0$? I am afraid I have misunderstood something because I cannot come up with one...