# Countable reduced abelian group containing all countable reduced abelian groups

Is there a countable abelian group for which its subgroups are exactly all of the countable "reduced" abelian groups? (Reduced means that its divisible subgroup is zero)

Is the following group a possible candidate: $\bigoplus_{p}\bigoplus_{r=1}^{\infty}(\mathbb{Z}_{(p)}\oplus\mathbb{Z}/p^r\mathbb{Z})$? Here $p$ runs through all primes. $\mathbb{Z}_{(p)}$ stands for integers localized at $p$ or integers adjoin the inverses of all the primes except for $p$.

Also does anyone know a good survey for the modern theory of infinite abelian groups?

• Note that the corresponding question for countable abelian groups has a positive answer, namely $\bigoplus_{\aleph_0}(\mathbf{Q}\times\mathbf{Q}/\mathbf{Z})$ works.
– YCor
May 30, 2014 at 18:45
• I should add that the given "possible candidate" is hopeless because it is residually finite and hence cannot contain any reduced but not residually finite abelian group (which do exist among countable groups), although I showed it also fails among residually finite abelian groups (follows from my answer).
– YCor
May 20, 2017 at 9:19

Edit (May 2017):

I initially claimed to answer the question, but I precisely only obtain the following:

(a) There is no countable torsion-free reduced abelian group $A$ containing copies of all $\mathbf{Q}$-rank 2 torsion-free countable reduced abelian groups.

Fix a prime $p$. In a group $G$, let us define a $p$-maximal subgroup as the kernel of a nontrivial homomorphism onto the quasi-cyclic group $C_{p^\infty}=\mathbf{Z}[1/p]/\mathbf{Z}$.

Claim 1: in $\mathbf{Q}^2$, there are uncountably (continuum) many $p$-maximal subgroup, only countably many of which are non-reduced.

Proof of claim 1: first we count the non-reduced ones. Each non-reduced one is in the $\mathrm{GL}_2(\mathbf{Q})$-orbit of one containing $\mathbf{Q}\times\{0\}$. Then we conclude by arguing that $\mathrm{Hom}(\mathbf{Q},C_{p^\infty})$ is countable (exercise). It remains to check there are uncountably many $p$-maximal subgroups, and actually uncountably containing $\mathbf{Z}_{(p)}^2$ ($\mathbf{Z}_{(p)}$ is the set of rationals with denominator coprime to $p$). We have $\mathbf{Q}/\mathbf{Z}_{(p)}\simeq C_{p^\infty}$, and $C_{p^\infty}^2$ has uncountably (continuum) many subgroups (with quotient isomorphic to $C_{p^\infty}$ as well). This finishes the proof of the first claim. (Reference for these standard counting facts: B. Boyer, Enumeration theorems in infinite Abelian groups, Proc. AMS 7:565–570, 1956, who first characterized abelian groups with countably many subgroups.

Claim 2: in $\mathbf{Q}^2$, there are uncountably (continuum) many non-isomorphic reduced $p$-maximal subgroups.

Proof: by Claim 1 there are uncountably many. Since any isomorphism between any two extends (uniquely) to a automorphism of $\mathbf{Q}^2$, and since the automorphism group of $\mathbf{Q}^2$ is countable (this is $\mathrm{GL}_2(\mathbf{Q})$), we deduce Claim 2.

Let us finish the proof of (a). Assume by contradiction that $A$ exists. For a subgroup $H$ of $A$; denote by $\widehat{H}$ the inverse image in $A$ of the torsion subgroup of $A/H$. Note that $\widehat{H}$ and $H$ have the same $\mathbf{Q}$-rank, and more precisely that if $\overline{H}$ is the divisible hull of $H$, there is a canonical embedding of $\widehat{H}$ into $\overline{H}$ that is the identity on $H$; it cannot be surjective because $A$ is reduced.

Now assume that $H$ is $p$-maximal in its divisible hull. Then $\overline{H}/H\simeq C_{p^\infty}$, so by nonsurjectivity, we deduce that $\widehat{H}/H$ is finite.

Furthermore, if $H$ has finite $\mathbf{Q}$-rank $d$, $\widehat{H}$ is equal to $\widehat{H}$ for any free abelian subgroup of rank $d$; in particular, since $A$ is countable, there are only countably possibilities for $\widehat{H}$.

If we assume that $H$ is both $p$-maximal in its divisible hull and has finite $\mathbf{Q}$-rank, we combine the previous two paragraphs; since $\widehat{H}$ has only countably many subgroups of finite index (finitely many for each given index), we deduce that there are countably many possibilities for $H$.

We thus conclude that $A$ contains copies of only countably many of the $p$-maximal subgroups of $\mathbf{Q}^2$. Given Claim 2, this means that $A$ fails to contain some reduced $p$-maximal subgroup of $\mathbf{Q}^2$. This concludes the proof of (a).

Remark: the above argument also shows, regardless of the continuum hypothesis, the following:

The least cardinal of a reduced torsion-free abelian group containing isomorphic copies of all reduced subgroups of $\mathbf{Q}^2$ is $2^{\aleph_0}$.

Edit (May 2017): I've removed a confusing paragraph I initially wrote, which does not address the intended question anyway.

I think I made a confusion between being reduced and residually finite for abelian groups. For abelian groups, clearly residually finite implies reduced. The converse is true for torsion-free abelian groups. But it is easy to find counterexamples among countable torsion abelian $p$-groups.

If I could remove the first occurrence of "torsion-free" in (a), it would answer the question. But the quotient of even a residually finite abelian group by its torsion subgroup can fail to be reduced (a classical instance being $\prod_p\mathbf{Z}/p\mathbf{Z}$, or the preimage of a copy of $\mathbf{Q}$ modulo the torsion if we want a countable example). So we cannot naively reduce to the torsion-free case.

• Thank you for your answer. I was aware of all the things you said except for the claim about any reduced group being residually finite. Could you elaborate more on this part? May 30, 2014 at 13:26
• I don't know what you call a "reduced group".
– YCor
May 30, 2014 at 14:01
• Oh I think I see it. You consider quotients of the form G/nG and the kernels intersect inside the divisible subgroup. May 30, 2014 at 14:03
• So my question becomes: is there a characterization of reduced groups as all the subgroups of a fixed group? And also the reference for a survey of recent developments in abelian groups. I am a little unsure of the math overflow etiquette when the question changes like this. May 30, 2014 at 14:17
• I forgot to say countable, which is an assumption I actually want to emphasize. May 30, 2014 at 14:30