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This is Exercise 4.3.9(a) of Robinson's "A Course in the Theory of Groups (Second Edition)". According to Approach0, it is new to MSE.

It is an exercise noted as being referred to later on in the text.

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.

The Question:

Let $G$ be an abelian such that $r_p(G)<\infty$ if $p=0$ or a prime.

(a) If $H\le G$, show that $r_p(G/H)\le r_0(G)+r_p(G)$ for all $p>0$.

Thoughts:

I have asked questions here on related concepts:

I'm completely stuck.

However, to be honest, I haven't given it much time: I'm approaching the end of my postgraduate research degree so I've been too busy. I'm asking so that I can maintain a good pace in the book without forgetting the preceding material.

The kind of answer I'm looking for is a full solution, please.

The exercise is in the section on pure subgroups and $p$-groups. This suggests that I could use these concepts. I don't see the connection. There is the following theorem on page 110:

4.3.13. An abelian $p$-group $G$ has finite $p$-rank if and only if it is a direct sum of finitely many cyclic and quasicyclic groups.

That's the only direct mention of $p$-rank in the section.

I know, since $G$ is abelian, that $H$ is normal and $G/H$ is abelian.

I hope this is enough context.

Please help :)

Shaun
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    Haven't thought this through, but my first impulse would be the following: if you take a maximal independent $p$-subset of $G/H$, and lift it, some of the elements will have finite order, some will have infinite order; that's why you get the $r_0$ and the $r_p$ in the inequality. So try to construct an independent set of elements in $G$; the torsion ones will have order a multiple of a power of $p$ so you can replace them with a multiple to get order power of $p$. As I said, just a very rough idea of what I might try as a first approach. – Arturo Magidin Sep 19 '21 at 00:27

1 Answers1

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The two things you need to realize is that

  1. The image of a linearly dependent set is linearly dependent, and
  2. If $\{x_1^{a_1},\dots,x_n^{a_n}\}$ is a linearly dependent then so is $\{x_1,\dots,x_n\}$.

It's the contrapositives of these that you need, but this way round is much more obvious for some reason!

Suppose that $r_p(G/H)=n$ (including the possibility that $n=\infty$), and let $Hx_1,\dots,Hx_n$ be a linearly independent set of $p$-elements of $G/H$. (OK, if $n=\infty$, replace it by some incredibly large finite number.) Taking preimages yields a linearly independent set of $G$ (by 1.) Some of these are of infinite order and some of finite order. Let $x_1,\dots,x_m$ be infinite order and $y_1,\dots,y_r$ be finite order. Since $Hy_i$ is a non-trivial $p$-element, $o(y_i)$ is divisible by $p$. Thus some power of $y_i$ is a non-trivial $p$-element, $y_i'$. The $y_i'$ are linearly independent by 2. Thus $m\leq r_0(G)$ and $r\leq r_p(G)$, as needed for the equation $$ r_p(G/H)=n=m+r\leq r_0(G)+r_p(G).$$

David A. Craven
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