It is known that torsion-free compact abelian groups are exactly the product of the maximal solenoid group $\Sigma_{(2,3,\cdots)}$ (which is the Pontryagin dual of the additive group $\mathbb{Q}$ of rational numbers equipped with the discrete topology) and the additive $p$-adic integers $\Delta_p$ (cf. 25.4 and 25.8 of Abstract Harmonic Analysis by Hewitt & Ross). This makes me wonder if a similar classification result exists for connected (instead of torsion free) compact abelian groups.

As in the torsion-free case, this reduces to the problem of classifying all discrete abelian groups for which the Pontryagin dual is connected, which seems a little intractable for me. Perhaps a simpler, more tractable problem is, does there exist any compact connected abelian group which is not a product of $\mathbf{a}$-adic solenoids $\Sigma_\mathbf{a}$ for various sequence $\mathbf{a} = (a_1, a_2, \cdots)$ of integers greater than $1$ (cf. 10.12 and 10.13 ibid) and the circle group $\mathbb{T}$ (with products of $\mathbb{T}$ accounts for the torsion part and solenoids for the rest, if such an idea can be made precise)? In particular, is it true that any compact connected abelian group with a dense torsion subgroup a product of $\mathbb{T}$? More particularly, is it true that any compact connected abelian Lie group a finite product of $\mathbb{T}$? I think the last question has an affirmative answer (it is related to this question), but I don't know a rigorous proof.

  • 2
    $\begingroup$ For a discrete abelian group, having connected dual is equivalent to be torsion-free. The classification of torsion-free abelian groups is not easy but there's a large literature on it. $\endgroup$
    – YCor
    Nov 13, 2018 at 10:23
  • 2
    $\begingroup$ Yes any compact connected Lie group is a torus. By Pontryagin duality this is a restatement of the fact that any finitely generated torsion-free abelian group is free abelian. $\endgroup$
    – YCor
    Nov 13, 2018 at 10:24
  • $\begingroup$ @YCor Could you please point out a reference for the proof of the fact that a discrete abelian group has connected dual if and only if it is torsion-free? $\endgroup$ Nov 13, 2018 at 10:29
  • 3
    $\begingroup$ I'm fairly sure you'd find this in: Sidney A. Morris, Pontryagin Duality and the Structure of Locally Compact Abelian Groups, London Math. Soc. Lecture Notes 29, Cambridge U. Press, 1977. $\endgroup$
    – Todd Trimble
    Nov 13, 2018 at 12:05
  • 2
    $\begingroup$ @ToddTrimble Thanks, it is indeed the corollary 4 on page 99 of this nice little book. $\endgroup$ Nov 13, 2018 at 12:26

1 Answer 1


Perhaps a simpler, more tractable problem is, does there exist any compact connected abelian group which is not a product of $\mathbf{a}$-adic solenoids $\Sigma_\mathbf{a}$ for various sequence $\mathbf{a} = (a_1, a_2, \cdots)$ of integers greater than $1$ (cf. 10.12 and 10.13 ibid) and the circle group $\mathbb{T}$

Following YCor's comments, the Pontryagin dual question is whether there exists a torsion-free abelian group which is not a direct sum of localizations of $\mathbb{Z}$. And the answer is yes: for example, you can consider the $p$-adic integers $\mathbb{Z}_p$ as a discrete group. Cf. the accepted answer at this MO question about simply presented abelian groups; YCor's answer is also educational and suggests simpler examples such as suitable subgroups of $\mathbb{Z}[1/p]^2$.

Edit: Let me see if I can flesh out YCor's claim about subgroups of $\mathbb{Z}[1/p]^2$. The interesting ones are the ones of rank $2$, and for simplicity I'll restrict my attention to subgroups containing $\mathbb{Z}^2$ (I thought I had an argument that every subgroup of rank $2$ is isomorphic to such a subgroup but now I'm not so sure). These correspond to subgroups of the quotient $\mathbb{Z}[1/p]^2/\mathbb{Z}^2 \cong \mu_{p^{\infty}}^2$, the product of two copies of the Prüfer $p$-group.

We can describe such subgroups using Goursat's lemma, which says in this case that if $H \subseteq \mu_{p^{\infty}}^2$ is a subgroup such that the two projections $\pi_1, \pi_2 : H \to \mu_{p^{\infty}}$ are surjective (this is the interesting case), then there exist subgroups $N_1, N_2 \subseteq \mu_{p^{\infty}}$ and an isomorphism $\varphi : \mu_{p^{\infty}}/N_1 \cong \mu_{p^{\infty}}/N_2$ such that

$$H = \{ (q_1, q_2) \in \mu_{p^{\infty}}^2 : \varphi(q_1) \equiv q_2 \bmod N_2 \}.$$

The proper subgroups of $\mu_{p^{\infty}}$ take the form $\mu_{p^n}$ for $n \in \mathbb{Z}_{\ge 0}$ (if we consider all of $\mu_{p^{\infty}}$ then $H$ is all of $\mu_{p^{\infty}}^2$). The quotient $\mu_{p^{\infty}}/\mu_{p^n}$ is isomorphic to $\mu_{p^{\infty}}$ again, and its automorphism group is the group of $p$-adic units $\mathbb{Z}_p^{\times}$, of which there are uncountably many.

So there are uncountably many choices for $H$, and hence uncountably many subgroups of $\mathbb{Z}[1/p]^2$ containing $\mathbb{Z}^2$. Of these, the subgroups isomorphic to $\mathbb{Z}^2, \mathbb{Z} \times \mathbb{Z}[1/p]$, or $\mathbb{Z}[1/p]^2$ are determined by the image of the copy of $\mathbb{Z}^2$ in each of them, and there are countably many choices for this image. So as promised, at most countably many subgroups can be isomorphic to a direct sum of localizations of $\mathbb{Z}$. Unfortunately I don't quite see how to explicitly exhibit a particular choice of $H$ which is not such a direct sum.

Edit #2: Let's take $N_1 = N_2 = 0$ to be trivial above, and $\varphi : \mu_{p^{\infty}} \cong \mu_{p^{\infty}}$ to be multiplication by a $p$-adic unit $u \in \mathbb{Z}_p^{\times}$. Then

$$H = \{ (q_1, q_2) \in \mu_{p^{\infty}}^2 : u q_1 = q_2 \}$$

is a subgroup of $\mu_{p^{\infty}}^2$ isomorphic to $\mu_{p^{\infty}}$, and it lifts to a subgroup

$$\widetilde{H} = \{ (q_1, q_2) \in \mathbb{Z}[1/p]^2 : u (q_1 \bmod 1) \equiv q_2 \bmod 1 \}.$$

Thinking of $H$ as a "line with slope $u$" suggests that $\widetilde{H}$ is isomorphic to a direct sum of localizations of $\mathbb{Z}$ (in fact to $\mathbb{Z} \times \mathbb{Z}[1/p]$) iff $u$ is rational. Indeed, if $u = \frac{a}{b}$ is rational, so that $a, b$ are integers relatively prime to $p$ and each other, then the condition that $u (q_1 \bmod 1) \equiv q_2 \bmod 1$ is equivalent to the condition that $aq_1 - bq_2 \in \mathbb{Z}$, and consequently the map

$$\widetilde{H} \ni (q_1, q_2) \mapsto (q_1, aq_1 - bq_2) \in \mathbb{Z}[1/p] \times \mathbb{Z}$$

is an isomorphism.

Now suppose that $u$ is irrational. First let's show that $\widetilde{H}$ has no subgroup isomorphic to $\mathbb{Z}[1/p]$: equivalently, no nonzero element is $p$-divisible. If $(q_1, q_2) \in \widetilde{H}$ is a nonzero element, for it to be $p$-divisible would require that

$$u \left( \frac{q_1}{p^k} \bmod 1 \right) \equiv \frac{q_2}{p^k} \bmod 1$$

for all $k$, or equivalently that

$$u(q_1) - q_2 \in p^k \mathbb{Z}$$

for all $k$. Taking $k \to \infty$ gives $u = \frac{q_2}{q_1}$, but this contradicts $u$ irrational.

Hence in this case, if $\widetilde{H}$ is isomorphic to a direct sum of localizations of $\mathbb{Z}$ then it can only be isomorphic to $\mathbb{Z}^2$. But the projection to either coordinate gives a surjection onto $\mathbb{Z}[1/p]$, which $\mathbb{Z}^2$ does not possess.

Some additional comments. Projection onto the first coordinate shows that $\widetilde{H}$ is an extension

$$0 \to \mathbb{Z} \to \widetilde{H} \to \mathbb{Z}[1/p] \to 0.$$

The argument above shows that this sequence splits iff $u$ is rational. In general, this extension is classified by a class in $\text{Ext}^1(\mathbb{Z}[1/p], \mathbb{Z})$, which can be computed from $\text{Ext}^1(\mu_{p^{\infty}}, \mathbb{Z}) \cong \mathbb{Z}_p$ and the short exact sequence $0 \to \mathbb{Z} \to \mathbb{Z}[1/p] \to \mu_{p^{\infty}} \to 0$ to give

$$\text{Ext}^1(\mathbb{Z}[1/p], \mathbb{Z}) \cong \mathbb{Z}_p/\mathbb{Z}.$$

The point of this computation is to show that for general homological reasons there are interesting nontrivial extensions of torsion-free abelian groups by torsion-free abelian groups; taking Pontryagin duals, there are interesting nontrivial extensions of compact connected abelian groups by compact connected abelian groups.

It should be possible to match up this $\mathbb{Z}_p$ with the $\mathbb{Z}_p$ that $u$ lives in above but I'm not entirely sure how.

  • 1
    $\begingroup$ Indeed, $\mathbf{Z}[1/p]^2$ has uncountably many subgroups, only countably of which are isomorphic to a product of rank-1 groups (which in this case means isomorphic to one of $\{0\}$, $\mathbf{Z}$, $\mathbf{Z}[1/p]$, $\mathbf{Z}^2$, $\mathbf{Z}\times\mathbf{Z}[1/p]$, $\mathbf{Z}[1/p]^2$). $\endgroup$
    – YCor
    Nov 13, 2018 at 22:30
  • 2
    $\begingroup$ To exhibit an example, one way is to use that the inclusion $\mathbf{Z}[1/p]\to\mathbf{Q}_p$ induces an isomorphism $\mathbf{Z}[1/p]/\mathbf{Z}\to\mathbf{Q}_p/\mathbf{Z}_p$. If $D$ is a line in $\mathbf{Q}_p$, it corresponds to the subgroup $u_D=(D+\mathbf{Z}_p^2)\cap\mathbf{Z}[1/p]^2$. Then $u_D$ is isomorphic to a product of rank 1 groups if and only if $D$ is rational. Hence, choose $D$ irrational to have an explicit example. $\endgroup$
    – YCor
    Nov 14, 2018 at 22:43
  • 1
    $\begingroup$ Also algebraic number theory provides examples. Let $x$ be an algebraic number of degree 2 over $\mathbf{Q}$, with conjugate $x'$, such that $x$ has negative $p$-valuation while $x'$ has zero $p$-valuation (so the minimal polynomial is split over $\mathbf{Q}_p$). Then the subring generated by $x$ has an additive structure with the required properties. $\endgroup$
    – YCor
    Nov 14, 2018 at 23:01
  • $\begingroup$ The proofs above have one or two mistakes in them; hopefully I'll get around to fixing them soon. $\endgroup$ Nov 24, 2018 at 4:19

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.