The Prüfer rank of an abelian group.

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 $$G$$, $$r(G)=r_0(G)+\max_p r_p(G).$$

Hi: Let's call the second term in the right side of the equality $$L$$. What guaranties the existence of $$L$$? I can have an infinite sequence of primes $$p_1, p_2, ...$$ with $$r_{p_i}(G) \lt r_{p_{i+1}}(G)$$. And there will clearly be no maximum. Is there a mistake in this text?

• Which book is this from? Commented Mar 18, 2020 at 17:39
• Note that they are defined as cardinals. If $G$ is a set, then each $r_p$ is a cardinal, and you have a countable collection of cardinals. As such, it has a maximum, which is a cardinal. Commented Mar 18, 2020 at 17:47
• @Shaun: Derek Robinson, A Course in the Theory of Groups, 2nd ed, New York, 1996. Commented Mar 18, 2020 at 20:21
• @Magidin: are you saying every countable set (or collection) has a maximum element? Commented Mar 18, 2020 at 20:27
• @stf91: As cardinals, it does have an upper bound: no element there can be larger than $|G|$, which is a cardinal. Commented Mar 19, 2020 at 19:12

You are correct that $$\max$$ is not technically correct, it should be $$\sup$$.

Now, Robinson states that the purpose of that section is to describe the structure of finite abelian groups, abelian groups with the maximal condition, and abelian groups with the minimal condition.

For finite abelian groups, $$r_0=0$$, each $$r_p$$ is finite, and $$r_p(G)=0$$ for almost all $$p$$; they are all bounded by $$|G|$$, and so $$\max$$ actually makes sense.

The "maximal condition", as given by Robinson, is that every nonempty collection of subgroups has maximal elements; equivalently ACC on subgroups. For abelian groups, this is equivalent to being finitely generated (Proposition 4.2.8 in Robinson); for such a group, the torsion subgroup is finite, so again each $$r_p$$ is finite and it equals $$0$$ for almost all $$p$$, and $$r_0$$ is finite, so again $$\max$$ makes sense.

The "minimal condition" is that every nonempty collection of subgroups has minimal elements, or equivalently that the group has DCC on subgroups. In Proposition 4.2.11 Robinson gives the Theorem of Kuros that says that an abelian group satisfies the minimal condition if and only if it is a direct sum of finitely many quasicyclic and cyclic groups of prime power order. For such a group, $$r_0=0$$, $$r_p=0$$ for almost all $$p$$, and all $$r_p$$ are finite, so again we get that the right hand side makes sense and $$\max$$ is adequate.

So for the context that Robinson is interested in, the expression for the Prüfer rank is sensible with $$\max$$.

That said, if we go to a more general setting, we can still make the definitions given if we use cardinals rather than finite quantities, and we replace $$\max$$ with $$\sup$$. In that case, since both $$r_0$$ and each $$r_p$$ is bounded above by $$|G|$$, the collection of cardinals is a set that is bounded above and so have a supremum, and the Prüfer rank is certainly at most $$|G|$$ (and could be equal to $$|G|$$; for example, the direct sum of $$\aleph_0$$ copies of $$\mathbb{Z}$$ has torsionfree rank $$\aleph_0=|G|$$.

• I see. |Z|= aleph 0 and aleph 0 times aleph 0 = aleph 0. Thanks Arturo. A question: G is an abelian group and for each prime p, G has an element of order some power of p. Can such a group G exist? And if it does not, why? Of course the torsion subgroup of G would then be infinite. Commented Mar 20, 2020 at 19:02
• @stf91: The direct sum, over all primes $p$, of $C_p$ is an abelian group in which every element has finite order and there are elements of order $p$ for every $p$. More generally, if you take the direct sum $\oplus_{n=1}^{\infty}C_n$ you get an abelian group in which every element is of finite order, and for every $n$ you have an element of order $n$. Commented Mar 20, 2020 at 19:08
• I understand, thanks. Commented Mar 20, 2020 at 20:20