If $G$ is abelian, show the Prüfer rank $r(G)$ is finite iff $\max\{ d(H)\}$ is finite, for $d(H)$ the min number of generators of f.g. $H\le G$

Question

This is part of Exercise 4.2.2 of Robinson's "A Course in the Theory of Groups (Second Edition)". According to Approach0 and this search, it is new to MSE.

The Details:

On page 98 to 99, ibid., we have

Let $G$ be an abelian group and let $S$ be a nonempty subset of $G$. Then $S$ is called linearly independent, or simply independent, if $0\notin S$ and, given distinct elements $s_1,\dots, s_r$ of $S$ and integers $m_1,\dots, m_r$, the relation $m_1s_1+\dots+m_rs_r=0$ implies $m_is_i=0$ for all $i$.

. . . and . . .

If $p$ is prime and $G$ is 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 $G$,

$$r(G)=r_0(G)+\max_{p} r_p(G).$$

(See here for how the $\sup$ instead of the $\max$ is more appropriate. I don't think it effects the exercise is question though.)

The Question:

If $G$ is an abelian group, show that $r(G)$ is finite if and only if $\max\{ d(H)\}$ is finite where $H$ ranges over the finitely generated subgroups of $G$. [. . .] [Note: $d(H)$ is the minimum number of generators of $H$.]

Thoughts:

I doubt there's an easy way to tackle this using iff-statements, so I proceed as follows . . .

Suppose $G$ is an abelian group such that $r(G)=n$ for some $n\in\Bbb N$. Then

$$r_0(G)+\max_{p} r_p(G)=n.$$

This implies $r_0(G)=s$ and $\max_{p}r_p(G)=t$ for some $s,t\in\Bbb N\cup\{0\}$ such that $n=s+t$.

Where do I go from here?

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Of course, if $G$ is finite, then $r_0(G)=0$. In fact, upon reflection, the question is trivial for finite $G$. (Right? Because each (finitely-generated) $H\le G$ must be finite . . . )

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Next, I should look at the sufficiency.

Suppose $G$ is an (infinite) abelian group such that

$$\max\{ d(H)\mid H\le G, H \text{ is finitely-generated}\}$$

is finite, where $d(H)$ is the minimum number of generators of $H$.

Where do I go from here?

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I can see, sort of, that $r_0(G)=0$, but I don't know how to justify it and it wouldn't shock me if it's not necessary. I feel the same way about the fact that $r_0(G)$ is finite, except that it must be true in order for $r(G)$ to be finite.

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Nowhere have I used that $G$ is abelian.

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This seems like a problem I ought to be able to answer myself, given enough time. The type of answer I'm hoping for is a full solution but I'd be happy with a good hint or two.

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Please help :)

Answer 1

Here is a quick sketch proof - more detail on request.

First suppose that $\max\{ d(H) : H \le G, H\ \mbox{f.g.} \} = n$ is finite. We claim that $r(G) \le 2n$. If not then, either $r_0(G) > n$ or $r_p(G) > n$ for some prime $p$.

In the first case, there is a torsion-free subgroup $H$ with $r(H) > n$, so $H$ contains a set $s_1,\ldots,s_{n+1}$ independent elements. Then, putting $H = \langle s_1,\ldots,s_{n+1} \rangle$. we have $d(H) = n+1$, contradiction. The argument when $r_p(G) > n$ is similar.

Conversely, suppose that $r(G) = n$ is finite. We claim that $\max\{ d(H) : H \le G, H\ \mbox{f.g.} \} \le 2n+1$. If not, then there exists $H \le G$ with $d(H) = 2n+1$.

By the fundamental theorem of abelian groups, we have $$H \cong {\mathbb Z}^r \oplus \frac{\mathbb Z}{d_1{\mathbb Z}} \oplus \cdots \oplus \frac{\mathbb Z}{d_s{\mathbb Z}}$$ for integers $r,s \ge 0, d_i>1$ with each $d_i|d_{i+1}$.

Then $d(H) =r+s$, so either $r>n$ or $s>n$. If $r>n$ then the generators of the subgroup ${\mathbb Z}^r$ are independent, contradicting $d(G) = n$, and similarly if $s>n$.