If $H\le G$ for an abelian group $G$, then $r_p(H)+r_p(G/H)\ge r_p(G)$, where $r_p(K)$ is the $p$-rank of $K$.

Question

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):

For $H\le G$, abelian $G$, show that $r_0(H)+r_0(G/H)=r_0(G)$, where $r_0(K)$ is the torsionfree rank of $K$

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.

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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.

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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.

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My guess is that an infinite group is needed to satisfy the inequality part. I don't know how to justify this hunch.

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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.

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The type of answer I'm hoping for is either a full solution or a few good hints.

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Please help :)

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$\dagger$: 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 . . .

Answer 1

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 $1$.