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...


1 Answer 1


First, note that any 1-element set other than $\{0\}$ is linearly independent: the condition is "$ms = 0$ implies $ms = 0$" which is clearly true! (I think you've realised this, but it's worth pointing out explicitly).

You are right in saying that if $G$ is torsion-free then $r(G) = r_0(G)$, and if we then take $g \neq 0$ then $\{g\}$ is indeed linearly independent and so $r(G) \geq 1$ - and so if we know that $r(G) = 0$ then no such $g$ exists, and $G = \{0\}$.

It may help to realise that the rank of an abelian group is analogous to the dimension of a vector space - and a $0$-dimensional vector space is also just $\{0\}$.


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