This is Exercise 4.2.7(i) of Robinson's "A Course in the Theory of Groups (Second Edition)". According to Approach0, it is new to MSE.
I've asked two previous questions on based on earlier exercises on a related concept:
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 . . .
[T]he $0$-rank or torsion-free rank
$$r_0(G)$$
is the cardinality of a maximal independent subset of elements of infinite order.
The Question:
If $H$ is a subgroup of an abelian group $G$, prove that [. . .] $$r_0(H)+r_0(G/H)=r_0(G).$$
Thoughts:
Since $G$ is abelian, so is $H$, which is then normal in $G$, and so $G/H$ is a group; the quotient of any abelian group by any subgroup is itself abelian. Hence the exercise makes sense.
If $K$ is a finite, abelian group, then $r_0(K)=0$, so the result is trivial if $G$ is finite.
I'm not sure how to handle when $H$ is finite while $G$ is infinite; of course, I would need to show that $r_0(G/H)=r_0(G)$.
If $G$ is finitely generated, then $H$ is not necessarily finitely generated. It might help to consider when $H$ is finitely generated. I'm stuck here though.
For some reason, I get the impression that the axiom of choice is required, since the axiom is equivalent to the statement that each vector space has a basis. I mean this in the sense that perhaps there is an analogy between the maximal independent subset of elements of infinite order and the basis of a vector space.
I am not well-versed in the axiom of choice.
This seems like a problem I ought to be able to answer myself. But I would like to move on: I've given it a few days and I doubt I could make fruitful progress any time soon.
Please help :)
I think this does it.
Let $\{h_i\}_{i\in I}$ be a linearly independent subset of elements of infinite order of $H$, and given a linearly independent subset $S$ of elements of infinite order of $G/H$, say indexed by $J$, for each $j\in J$ let $g_j$ be an element of $G$ such that $g_jH$ is an element of $S$, so $S=\{g_jH\}_{j\in J}$. Note that because $g_jH$ is of infinite order in $G/H$, then $g_j$ is of infinite order in $G$. Note also that $\langle g_j\mid j\in J\rangle\cap H$ must be trivial: otherwise, there exist $j_1,\ldots,j_k\in J$, and nonzero integers $a_1,\ldots,a_k$, such that $a_1g_{j_1}+\cdots+a_kg_{j_k}\in H$, which means this combination is trivial in $G/H$, which means that $a_tg_{j_t}$ is trivial in $G/H$, which means that $g_{j_t}$ has finite order in $G/H$, a contradiction.
Let us consider $\{h_i\}_{i\in I}\cup\{g_j\}_{j\in J}$. I claim that this is linearly independent in $G$. Indeed, say we have a linear combination of elements of $H$ and $g_j$ equal to $0$; but because $\langle g_j\mid j\in J\rangle$ intersects $H$ trivially, this combination leads to a linear combination of $h_i$ equal to $0$ and one of the $g_j$ equal to $0$. So these are themselves equal to $0$, and hence are the trivial linear combination.
This shows that $r_0(G)$ is at least $r_0(H)+r_0(G/H)$.
Now let $\{g_k\}_{k\in K}$ be a maximal linearly independent set of elements of infinite order in $G$. It is possible that some of the $g_k$ have finite order in $G/H$, so divide $K$ into a disjoint union, $K = K_1\cup K_2$, where for each $k\in K$, we have that $k\in K_1$ if and only if $g_k$ has finite order in $G/H$.
If $k\in K_1$, then there exists $r\gt 0$ such that $rg_k\in H$; but the element is nontrivial, since $g_k$ has infinite order in $G$. Fix such an $r$ and let $h_k=rg_k$. Note that because $\{g_k\}_{k\in K_1}$ is independent, it follows that $\{h_k\}_{k\in K_1}$ is an independent subset of $H$. Thus, $|K_1|\leq r_0(H)$.
We are now going to do a bit of a shuffle: it's possible for the $g_k$ with $k\in K_2$ to not be linearly independent in $G/H$ (here's the example I had in mind: take $G=\mathbb{Z}\oplus\mathbb{Z}$, $H=\mathbb{Z}\oplus\{0\}$, and $g_1=(1,1)$, $g_2=(0,1)$. Then both elements are of infinite order in $G/H$, but they are not independent).
Let's well-order $K$, since you are okay with the Axiom of Choice. I wonder if the following can be justified without this, but I don't know and this seems straightforward, so here goes.
If $\{g_kH\mid k\in K_2\}$ is linearly independent in $G/H$, we are done. Otherwise, there is a least index $r\in K_2$ such that $\{g_kH\mid k\in K_2,k\leq r\}$ is linearly dependent in $G/H$. Thus there is a nontrivial linear combination $0\neq h_r=\sum_{i\leq r}a_ig_i \in H$, with $a_r\neq 0$. Note that $h_r$ is linearly independent from $\{h_k\mid k\in K_1\}$, because the original family was linearly independent. Thus, if we replace $g_r$ with $h_r$, we obtain a set that remains linearly independent in $G$, and we've moved an index from $K_2$ to $K_1$. The least index in $K$ that witnesses that $\{g_kH\mid k\in K_2\}$ is linearly dependent in $G/H$ is now larger than $r$.
This process allows us to set up a transfinite induction on the well-ordered set $K$ that will replace our original collection with a new collection that is still linearly independent, still contains only elements of infinite order, is still partitioned into $K_1$ and $K_2$, but now $\{g_kH\mid k\in K_2\}$ is linearly independent in $G/H$. Because linear combinations are inherently finite, the limit step is exactly the same as the regular step.
At the end of the transfinite induction, we have a disjoint union $K=K_1\cup K_2$, with $\{h_k\mid k\in K_1\}$ linearly independent in $H$, and $\{g_kH\mid k\in K_2\}$ linearly independent in $G/H$. This shows that $|K_1|\leq r_0(H)$ and $|K_2|\leq r_0(G/H)$, hence $|K|\leq r_0(H)+r_0(G/H)$. Thus, $r_0(G)$ is bounded above by $r_0(H)+r_0(G/H)$, proving the other inequality.