This Exercise 4.2.7(ii) of Robinson's "A Course in the Theory of Groups (Second Edition)". According to Approach0, it is new to MSE.
Earlier, I asked about Exercise 4.2.7(i):
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.
The Question:
If $H$ is a subgroup of an abelian group $G$, prove that [. . .] $r_p(H)+r_p(G/H)\ge r_p(G)$, with inequality in general.
(I have included the examples-counterexamples tag for the "inequality in general" part.)
Thoughts:
Like last time, each subgroup of an abelian group is itself abelian and is normal, and the quotient of an abelian group is abelian, so the question makes sense.
Unlike last time, however, I don't think the axiom of choice is required; I think this because the fact that each finite dimensional vector space has a basis, does not seem to require it. The analogy breaks down.
Therefore, I'm not sure how to build on @ArturoMagidin's answer to the first part. There's a fair amount I can transfer over though. For example, I could start with some $\{h_i\}_{i\in I}$, a maximal independent subset of elements of $p$-power order. I'm not sure what to do after that.
If $K$ is a torsionfree abelian group, then it has no elements of order $p$ for any prime $p$, so if $G$ is torsionfree, implying $H$ and $G/H$ are torsionfree,${}^{\dagger}$ then the exercise is trivial.
My guess is that an infinite group is needed to satisfy the inequality part. I don't know how to justify this hunch.
This seems like a question I could answer myself with a little more patience. I have spent too long on it already and I don't want to lose my momentum in the book.
The type of answer I'm hoping for is either a full solution or a few good hints.
Please help :)
$\dagger$: Right? Because $G/H$ cannot gain any torsion elements from $G$ by quotienting by $H$; it can only lose them. Strange things happen with infinite groups, however, so I'm tentative here . . .
The argument is essentially the same as that used in the previous part; for variety's sake, I will cast it as an ordinal induction this time, in the manner I hinted at in that post.
Let $\{x_i\}_{i\in I}$ be a set of elements of $G$ of $p$-power order that is independent in $G$. Note that $x_iH$ has $p$-power order in $G/H$.
Once again, let us take $I$ to be an ordinal (I'm invoking choice here; again, don't know if I can do it without Choice). As before, I proceed by ordinal induction to define two disjoint, $\{y_i\}_{i\in I_1}$ and $\{x_i\}_{i\in I_2}$, with $I$ the disjoint union of $I_1$ and $I_2$, each consisting of elements of $p$-power order, $y_i\in H$ for each $I$, with $\{y_i\}_{i\in I_1}$ independent in $H$, and $\{x_iH\}_{i\in I_2}$ independent in $G/H$, and each $y_i$ is a nontrivial multiple of the corresponding $x_i$ (that is, $y_i=kx_i$ for some $k$; note that this guarantees that $\{y_i\}_{i\in I_1}\cup\{z_i\}_{i\in I_2}$ is independent in $G$ and consists of elements of $p$-power order). If we succeeed, we will obtain that $|I|=|I_1|+|I_2|\leq r_p(H)+r_p(G/H)$, thus proving that $r_p(G)\leq r_p(H)+r_p(G/H)$.
If $x_1\in H$, then let $1\in I_1$, and let $y_1=x_1$. Otherwise, let $1\in I_2$.
Assume we have proceeded up to $x_{\alpha}$, $\alpha\lt I$, so that $\{y_i\}_{i\leq \alpha, i\in I_1}$ is an independent subset of $H$, with each $y_i$ a nontrivial power of $x_i$, and $\{x_iH\}_{i\leq \alpha, i\in I_2}$ is independent in $G/H$. If $\{x_iH\}_{i\leq \alpha,i\in I_2}\cup\{x_{\alpha+1}H\}$ is independent in $G/H$, then add $\alpha+1$ to $I_2$. Otherwise, there must exist $i_1,\ldots,i_k\in I_2$, integers $a_1,\ldots,a_k$ and $a_{\alpha+1}$ such that $a_{\alpha+1}x_{\alpha+1}\neq 0$ and $$\left(\sum_{j=1}^ka_jx_{i_j} \right) + a_{\alpha+1}x_{\alpha+1} = y_{\alpha+1} \in H.$$ Note that $h_{\alpha+1}\neq 0$, since the $x_i$ are independent in $G$. Note $y_{\alpha+1}$ is independent from $\{y_i\}_{i\leq \alpha, i\in I_1}$, as otherwise we would have a nontrivial linear dependence in our original independent set (after replacing $y_i$ with $a_ix_i$ for suitable nonzero integer $a_i$). Put $\alpha+1\in I_2$, and let $y_{\alpha+1}$ be the element above. Note also that $\{y_i\}_{i\leq\alpha+1,i\in I_1}\cup\{x_i\}_{i\leq \alpha, i\in I_2}$ remains linearly independent in $G$.
Now say $\gamma\leq I$ is a limit ordinal and for every $\alpha\lt \gamma$ we have succeeded in this "divide and where necessary replace" process up to $\alpha$. If $\{x_iH\}_{i\leq\gamma,i\in I_2}$ is linearly independent in $G/H$, then add $\gamma$ to $I_2$. Otherwise, there is a finite set of indices and integers, as above with $a_{\gamma}x_{\gamma}\neq 0$, with $$\left(\sum_{j=1}^ka_{i_j}x_{i_j} \right) + a_{\gamma}x_{\gamma} = y_{\gamma}\in H.$$ Then place $\gamma\in I_2$ with $y_{\gamma}$ defined as above. By the same argument as above, the resulting (potentially partial) list also satisfies the required conditions.
Inductively, we have taken the linearly independent set of $p$-power order elements $\{x_i\}_{i\in I}$ of $G$ and obtained a linearly independent subset of $p$-power order elements $\{h_i\}_{i\in I_1}$ of $H$, and a liunearly independent set $\{x_iH\}_{i\in I_2}$ of $p$-power order elements of $G/H$, with $I=I_1\cup I_2$ and $I_1\cap I_2=\varnothing$. this gives $|I|\leq r_p(H) + r_p(G/H)$. Thus the supremum of the cardinalities of all such indices is bounded above by $r_p(H)+r_p(G/H)$, so $$r_p(G)\leq r_p(H)+r_p(G/H).$$ That we cannot hope to obtain an equality (as we did in the case of $r_0$) is witnessed by $G=\mathbb{Z}$, $H=p\mathbb{Z}$, where the left hand sidee is $0$, but the right hand side is $1$.