A NEW PROOF OF A THEOREM CONCERNING DECOMPOSABLE GROUPS

WOJCIECH CHOJNACKI

Abstract. We give an elementary proof of the following result: If G is a compact non-zero Abelian group with dual isomorphic to a subgroup of ℚ, such that U ∪(−U) = G\G_{(2)} and U ∩(−U) = ∅ for some open subset U ⊂ G, where G_{(2)} = {a ∈ G: 2a = 0}, then G is topologically isomorphic with 𝕋.

1. Introduction

Let G be a locally compact Abelian group with dual Ĝ. Denote by G^{(2)} and G_{(2)} the image and kernel of the homomorphism G ∋ a ↦ 2a ∈ G, respectively. Given a subset X ⊂ G, let

−X = {a ∈ G: − a ∈ X}.

In agreement with the terminology introduced in [1], G will be said to be decomposable if there exists an open subset U ⊂ G such that U ∪ (−U) = G \ G_{(2)} and U ∩ (−U) = ∅.

Let 𝕋 be the multiplicative group of complex numbers with unit modulus, endowed with the usual topology. Let ℚ be the additive group of rational numbers, equipped with the discrete topology. For each n ∈ ℕ, let ℤ(n) be the cyclic group with n elements. Assume that the ℤ(n) are endowed with the discrete topology. Given Abelian groups G_{i} (i = 1, …, n), denote by G_{1} × ⋯ × G_{n} the direct product of the G_{i}. For a cardinal number 𝔪 and a compact Abelian group H, designate by H^{𝔪} the direct product of 𝔪 copies of H, enriched with the product topology (under which H^{𝔪} is compact).

In [1] the following characterisation of decomposable compact Abelian groups is given:

Theorem 1. Let G be a compact Abelian group. Then G is decomposable if and only if either (Ĝ)^{(2)} is a countable torsion group or G is topologically isomorphic with 𝕋 × ℤ(2)^{𝔪} × F, where 𝔪 is a cardinal number and F is a finite Abelian group.

The above theorem is a consequence of a number of results describing certain subclasses of the class of all decomposable compact Abelian groups. One of these results reads as follows:

Theorem 2. Any decomposable compact connected Abelian group different from a singleton is topologically isomorphic with 𝕋.

The main part of the proof to Theorem 2 is embodied by the following result:

1991 Mathematics Subject Classification. Primary 22C05; Secondary 22B05.

Key words and phrases. Compact, connected, decomposable, Abelian group.

Theorem 3. Suppose that G is a decomposable compact Abelian group different from a singleton. Suppose, moreover, that Ĝ is isomorphic with a subgroup of ℚ. Then G is topologically isomorphic with 𝕋.

The proof of Theorem 3 given in [1] (as part of the proof to Proposition 4.1) is short but quite involved. This note offers a longer but more elementary proof. While the first of these proofs utilises a rather special result concerning compact cancellative semigroups, the second uses only standard tools from general topology. This notwithstanding, both proofs invoke freely a basic lore on locally compact Abelian groups.

2. Proof of the main result

This section gives the proof of Theorem 3 alluded to above.

Proof of Theorem 3. We commence by showing that, for every a ∈ G, G \ {a} is connected. Since Ĝ is isomorphic with a subgroup of ℚ, it is torsion free. Hence, being compact, G is connected (cf. [2, Thm. 24.25]). We see that G is a continuum with more than one element. By a theorem of Moore-Wallace [4, 6] (see also [3, §47, Sec. IV, Thm. 5]), any continuum different from a singleton contains at least two elements, each of which has a connected complement. Therefore there exists b ∈ G for which G \ {b} is connected. Now, to conclude that G \ {a} is connected for each a ∈ G, it suffices to observe that G \ {a} is the image of G \ {b} via the translation by a − b (defined as G ∋ h ↦ a − b + h ∈ G), which is a homeomorphism.

Denote by 0 the neutral element of G. Let U be an open subset of G such that U ∪ (−U) = G \ G_{(2)} and U ∩ (−U) = ∅. It is clear that G \ G_{(2)} is disconnected. In view of the assertion established in the preceding paragraph, G \ {0} is connected. Therefore G_{(2)} \ {0} is non-empty.

Let ρ be a monomorphism mapping Ĝ into ℚ. For each n ∈ ℕ, let K_{n} be the cyclic subgroup of ℚ given by

K_{n} = {k/n! | k ∈ ℤ}

and let Γ_{n} be the subgroup of Ĝ given by

Γ_{n} = ρ^{−1}(K_{n} ∩ ρ(Ĝ)).

It is clear that, for each n ∈ ℕ, Γ_{n} is cyclic and Γ_{n} ⊂ Γ_{n+1}. Furthermore, Ĝ = ⋃_{n=1}^{∞} Γ_{n}.

We now prove that G_{(2)} \ {0} has precisely one element. Select g ∈ G_{(2)} \ {0} arbitrarily. Given n ∈ ℕ, let χ_{n} be a generator of Γ_{n}. Since 2g = 0, we have (g, χ_{n}) = ±1 for each n ∈ ℕ. Here (·, ·) represents the pairing between elements of G and Ĝ. Now either there is a sequence {n_{k}}_{k∈ℕ} in ℕ diverging to infinity such that (g, χ_{n_{k}}) = 1 for each k ∈ ℕ, or (g, χ_{n}) = −1 for all but finitely many n ∈ ℕ. Suppose that the first possibility holds. Since {Γ_{n}}_{n∈ℕ} is an increasing sequence of subgroups eventually exhausting all of Ĝ, any given γ ∈ Ĝ can be written as γ = lχ_{n_{k}} for some k, l ∈ ℕ. It then follows that (g, γ) = 1, which, in view of the arbitrariness of γ, implies that g = 0, a contradiction. The first possibility being excluded, let n_{0} ∈ ℕ be such that (g, χ_{n}) = −1 for each integer n greater than n_{0}. Given γ ∈ Ĝ, choose l ∈ ℕ and n ∈ ℕ with n > n_{0} such that γ = lχ_{n}. Then, clearly, (g, γ) = (−1)^{l}, which shows that (g, γ) does not depend on the particular choice of g. Consequently, g is uniquely determined, and so G_{(2)} \ {0} is a singleton.

Denote by g the unique element of G_{(2)} \ {0}. We clearly have G_{(2)} = {0, g}. For each subset X ⊂ G, denote by ∂X the boundary of X relative to G. We now show that

(1)  ∂U = {0, g}.

It is evident that ∂U ⊂ {0, g}. Since the inversion G ∋ a ↦ −a ∈ G is a homeomorphism, we have ∂(−U) = −∂U. Taking into account that g = −g, we see that ∂(−U) = ∂U. Now ∂U ⊂ ∂(U ∪ (−U)), since U is open. Moreover,

∂(U ∪ (−U)) ⊂ ∂U ∪ ∂(−U) = ∂U.

It follows that ∂U = ∂(U ∪ (−U)). In particular, the set U ∪ (−U) ∪ ∂U is closed. Suppose that {0, g} \ ∂U ≠ ∅. Being a finite set, {0, g} \ ∂U is closed. Since G is the union of {0, g} \ ∂U and U ∪ (−U) ∪ ∂U, we arrive at a contradiction with G being connected. Thus {0, g} \ ∂U = ∅, establishing (1).

We contend that U ∪ {0} is connected. Suppose, on the contrary, that U ∪ {0} = A ∪ B, where A and B are non-empty disjoint closed subsets of U ∪ {0}. In view of (1), U ∪ {0} is closed in G \ {g}. Correspondingly, A and B are closed in G \ {g}. It is now clear that A ∪ (−A) and B ∪ (−B) are non-empty disjoint closed subsets of G\{g}, whose union is the whole of G\{g}. But this contradicts the connectedness of G\{g} (which follows from the assertion from the first paragraph) and establishes the contention.

Let

V_{1} = (U + g) ∩ (−U) and V_{2} = (U + g) ∩ U.

We claim that V_{1} is not empty. Since G is compact and connected, it is also divisible (cf. [2, Thm. 24.25]). In particular, g = 2h for some h ∈ G. Since g is non-zero, we see that h is a member of G \ G_{(2)}, and so either h or −h falls into U. If h ∈ U, then, taking into account that 2g = 0, we have −h = h + g, whence h ∈ V_{1}. If −h ∈ U, then, in view of h = −h + g, we have −h ∈ V_{1}. In either case, V_{1} is non-empty, as claimed.

We shall now focus our attention on the set V_{1} ∪ {0, g}. We first show that it is closed in G.

Given a subset X ⊂ G and a ∈ G, let

X + a = {b ∈ G: b − a ∈ X}.

Clearly, since G is connected, ∂V_{1} is non-void. We have

∂V_{1} ⊂ ∂(U + g) ∪ ∂(−U) = {0, g}.

Since V_{1} is invariant under the composition of the inversion and the translation by g, so too is ∂V_{1}. It is easily seen that any non-empty subset of ∂V_{1} invariant under the same composition coincides with ∂V_{1}. Therefore ∂V_{1} = {0, g}, implying that V_{1} ∪ {0, g} is closed.

We now show that V_{1} ∪ {0, g} is connected. Suppose, on the contrary, that V_{1} ∪ {0, g} = A ∪ B, where A and B are non-empty disjoint closed subsets of V_{1} ∪ {0, g}. Since V_{1} ∪ {0, g} is closed in G, it follows that A and B are closed in G too. With no loss of generality, we may assume that 0 ∈ A. Then, necessarily, g ∈ B. For otherwise B would be an open subset of V_{1} and, since V_{1} is open in G, B would be open in G; as B is also closed in G, we would thus arrive at a contradiction with G being connected. Now A \ {0} is non-empty for otherwise {0} would be an open subset of V_{1} ∪ {0, g} contrary to the fact that 0 ∈ ∂V_{1} \ V_{1}. Since

A \ {0} ⊂ (U + g) ∪ {g} ⊂ G \ {0}

and since A is closed in G, it follows that A \ {0} is a closed subset of (U + g) ∪ {g}. On the other hand, since ∂V_{2} is contained in ∂(U + g) ∪ ∂U = {0, g}, we find that V_{2}∪{g} is a closed subset of (U +g)∪{g}. As B is closed in G, B∪V_{2} = B∪V_{2}∪{g} is closed in (U +g)∪{g}. We thus see that A\{0} and B ∪V_{2} are closed non-empty subsets of (U + g) ∪ {g}. Clearly, A \ {0} and B ∪ V_{2} are disjoint and their union is all of (U + g) ∪ {g}. This is, however, incompatible with the fact that, being the translate by g of the connected set U ∪ {0} (recall that the connectedness of U ∪ {0} was already shown earlier), (U + g) ∪ {g} is connected. The connectedness of V_{1} ∪ {0, g} is thus established.

In preparation for the next step, we show now that if V_{2} is non-void, then both V_{2} ∪ {0, g} and (−V_{2}) ∪ {0, g} are connected. Assume then that V_{2} ≠ ∅. Since G is connected, ∂V_{2} is not empty. V_{2} is invariant under the translation by g, and so too is ∂V_{2}. Since ∂V_{2} ⊂ {0, g} and since any non-empty subset of {0, g} invariant under the translation by g coincides with {0, g}, it follows that ∂V_{2} is all of {0, g}. Repeating the argument employed in the proof of the connectedness of V_{1} ∪ {0, g}, we conclude that V_{2} ∪ {0, g} is connected. Now (−V_{2}) ∪ {0, g} is connected too for it is the inverse of V_{2} ∪ {0, g}.

At this stage, we are in position to show that V_{1} ∪{0, g} is an arc with endpoints 0 and g. Note first that, since G is compact and Ĝ is countable, G is metrisable (cf. [2, Thm. 24.15]). In particular, V_{1} ∪ {0, g} is a metrisable continuum. By a theorem of Moore [4] (see also [3, §47, Sec. V, Thm. 1]), if every point in a metrisable continuum with the exception of two points a and b has a disconnected complement, then the continuum is an arc with endpoints a and b. Thus to prove that V_{1} ∪ {0, g} is an arc with endpoints 0 and g, it suffices to show that, for each a ∈ V_{1}, (V_{1} ∪ {0, g}) \ {a} is disconnected. Suppose that (V_{1} ∪ {0, g}) \ {a} is connected for some a ∈ V_{1}. Noting that (V_{1} ∪ {0, g}) \ {a} coincides with (V_{1} \ {a}) ∪ {0, g} and that the translate of (V_{1} \ {a}) ∪ {0, g} by g coincides with ((−V_{1}) \ {a + g}) ∪ {0, g}, we see that ((−V_{1}) \ {a + g}) ∪ {0, g} is connected. Now both (V_{1}\{a})∪{0, g} and ((−V_{1})\{a+g})∪{0, g} are connected and contain 0 and g, so their union C is connected. If V_{2} is empty, then, as is easily seen, C coincides with G \ {a, a + g}, and in particular G \ {a, a + g} is connected. If V_{2} is not empty, then both V_{2} ∪ {0, g} and (−V_{2}) ∪ {0, g} are connected and contain 0 and g, and so (V_{2} ∪ {0, g}) ∪ ((−V_{2}) ∪ {0, g}) ∪ C is connected. It is straightforwardly verified that the latter set coincides with G \ {a, a + g}. Thus, independently of whether or not V_{2} is empty, G \ {a, a + g} is connected. But G \ {a, a + g} is disconnected, since it is the translate by a of the disconnected set G \ {0, g} (= G \ G_{(2)}). This contradiction proves that V_{1} ∪ {0, g} is an arc with endpoints 0 and g.

Now that V_{2} is an open subset of G homeomorphic with the real line ℝ, G is locally connected. According to a theorem of Pontryagin [5, Thm. 42], any compact, metrisable, connected and locally connected group is the direct product of a finite or countable number of subgroups, each isomorphic with 𝕋. Applying this theorem and taking into account that G contains an open subset homeomorphic with ℝ (namely V_{2}), we find that G is topologically isomorphic with 𝕋. □

Acknowledgement. The author gratefully acknowledges the support of the Polish Research Council.

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Instytut Matematyki Stosowanej i Mechaniki, Uniwersytet Warszawski, ul. Banacha 2, 02-097 Warszawa, Poland

   E-mail address: wojtekch@appli.mimuw.edu.pl

Department of Computer Science, University of Adelaide, Adelaide, SA 5005, Australia

   E-mail address: wojtek@cs.adelaide.edu.au