When is $A$ isomorphic to $A^3$?

Question

This is totally elementary, but I have no idea how to solve it: let $A$ be an abelian group such that $A$ is isomorphic to $A^3$. is then $A$ isomorphic to $A^2$? probably no, but how construct a counterexample? you can also ask this in other categories as well, for example rings. if you restrict to Boolean rings, the question becomes a topological one which makes you think about fractals: let $X$ be Stone space such that $X \cong X + X + X$, does it follow that $X \cong X + X$ (here $+$ means disjoint union)?

Edit: In the answers there are already counterexamples. but you may add others in other categories (with products/coproducts), especially if they are easy to understand :).

Answer 1

The answer to the first question is no. That is, there exists an abelian group $A$ isomorphic to $A^3$ but not $A^2$. This result is due to A.L.S. (Tony) Corner, and is the case $r = 2$ of the theorem described in the following Mathematical Review.

MR0169905 Corner, A.L.S., On a conjecture of Pierce concerning direct decomposition of Abelian groups. 1964 Proc. Colloq. Abelian Groups (Tihany, 1963) pp.43--48 Akademiai Kiado, Budapest.

It is shown that for any positive integer $r$ there exists a countable torsion-free abelian group $G$ such that the direct sum of $m$ copies of $G$ is isomorphic to the direct sum of $n$ copies of $G$ if and only if $m \equiv n \pmod r$. This remarkable result is obtained from the author's theorem on the existence of torsion-free groups having a prescribed countable, reduced, torsion-free endomorphism ring by constructing a ring with suitable properties. It should be mentioned that the question of the existence of algebraic systems with the property stated above has been considered by several mathematicians. The author has been too generous in crediting this "conjecture" to the reviewer.

Reviewed by R.S. Pierce

Answer 2

Given a class of structures equipped with a product $(K, \times)$, the question of whether $X^3 \cong X \implies X^2 \cong X$ holds for every $X \in K$ is sometimes called the cube problem for $K$, and if it has a positive answer then $K$ is said to have the cube property. For the question to be nontrivial there need to be infinite structures $X \in K$ that are isomorphic to $X^3$. If there are such structures, it is usually possible to find one that witnesses the failure of the cube property for $K$, that is, an $X \in K$ such that $X \cong X^3$ but $X \not\cong X^2$. On the other hand, in rare cases the cube property does hold nontrivially.

I worked on the cube problem for the class of linear orders under the lexicographical product, and while doing so had a chance to look into the history of the problem for other classes of structures. The following list contains most of the results that I am aware of.

When the cube property fails

-- As far as I know, the first result concerning the failure of the cube property is due to Hanf, who showed in [1] that there exists a Boolean algebra $B$ isomorphic to $B^3$ but not $B^2$. Hanf's example is of size $2^{\aleph_0}$.

-- Tarski [2] and Jónsson [3] adapted Hanf's result to get examples showing the failure of the cube property for the class of semigroups, the class of groups, the class of rings, as well as other classes of algebraic structures. Most of their examples are also of size continuum.

It was unknown for some time after these results were published whether there exist countable examples witnessing the failure of the cube property for these various classes. Especially famous was the so-called "Tarski cube problem," which asked whether there exists a countable Boolean algebra isomorphic to its cube but not its square.

-- As Tom Leinster answered, Corner [4] showed, by a very different route, that indeed there exists a countable abelian group isomorphic to its cube but not its square. Later, Jones [5] constructed a finitely generated (necessarily non-abelian) group isomorphic to its cube but not its square.

-- Around the same time as Corner's result, several authors [6, 7] showed that there exist modules over certain rings isomorphic to their cubes but not their squares.

-- As Asher Kach answered, Tarski's cube problem was eventually solved by Ketonen, who showed in [8] that there does exist a countable Boolean algebra isomorphic to its cube but not its square.

Ketonen's result is actually far more general. Let $(BA, \times)$ denote the class of countable Boolean algebras under the direct product. If $(S, \cdot)$ is a semigroup, then $S$ is said to be represented in $(BA, \times)$ if there exists a map $i: S \rightarrow BA$ such that $i(a \cdot b) \cong i(a) \times i(b)$ and $a \neq b$ implies $i(a) \not\cong i(b)$. The statement that there exists a countable Boolean algebra isomorphic to its cube but not its square is equivalent to the statement that $\mathbb{Z}_2$ can be represented in $(BA, \times)$. Ketonen showed that every countable commutative semigroup can be represented in $(BA, \times)$.

-- Beginning in the 1970s, examples began to appear showing the failure of the cube property for various classes of relational structures. For example, Koubek, Nešetril, and Rödl showed that the cube property fails for the class of partial orders, as well as many other classes of relational structures in their paper [9].

-- Throughout the 70s and 80s, Trnková and her collaborators showed the failure of the cube property for a vast array of topological and relational classes of structures. Like Ketonen's result, her results are typically much more general.

Her topological results are summarized in [10], and references are given there. Some highlights:

• There exists a compact metric space $X$ homeomorphic to $X^3$ but not $X^2$. More generally, every finite abelian group can be represented in the class of compact metric spaces.
• Every finite abelian group can be represented in the class of separable, compact, Hausdorff, 0-dimensional spaces.
• Every countable commutative semigroup can be represented in the class of countable paracompact spaces.
• Every countable commutative semigroup can be represented in the class of countable Hausdorff spaces.

Her relational results mostly concern showing the failure of the cube property for the class of graphs. For example:

• Every commutative semigroup can be represented in $(K, \times)$, where $K$ is the class of graphs and $\times$ can be taken to be the tensor (categorical) product, the cartesian product, or the strong product [11].
• There exists a connected graph $G$ isomorphic to $G \times G \times G$ but not $G \times G$, where $\times$ can be taken to be the tensor product, or strong product. As of 1984, it was unknown whether $\times$ could be the cartesian product [12].

--Answering a question of Trnková, Orsatti and Rodino showed that there is even a connected topological space homeomorphic to its cube but not its square [13].

--More recently, as Bill Johnson answered, Gowers showed that there exists a Banach space linearly homeomorphic to its cube but not its square [14].

--Eklof and Shelah constructed in [15] an $\aleph_1$-separable group $G$ isomorphic to $G^3$ but not $G^2$, answering a question in ZFC that had previously only been answered under extra set theoretic hypotheses.

--Eklof revisited the cube problem for modules in [16].

When the cube property holds

There are rare instances when the cube property holds nontrivially.

-- It holds for the class of sets under the cartesian product: any set in bijection with its cube is either infinite, empty, or a singleton, and hence in bijection with its square. This can be proved easily using the Schroeder-Bernstein theorem, and thus holds even in the absence of choice.

-- Also easily, it also holds for the class of vector spaces over a given field.

-- Less trivially, it holds for the class of $\sigma$-complete Boolean algebras, since there is a Schroeder-Bernstein theorem for such algebras.

-- Trnková showed in [17] that the cube property holds for the class of countable metrizable spaces (where isomorphism means homeomorphism), and in [18] that it holds for the class of closed subspaces of Cantor space. The cube property fails for the class of $F_{\sigma \delta}$ subspaces of Cantor space. It is unknown if it holds or fails for $F_{\sigma}$ subspaces of Cantor space. See [10].

-- Koubek, Nešetril, and Rödl showed in [9] that the cube property holds for the class of equivalence relations.

-- I recently showed that the cube property holds for the class of linear orders under the lexicographical product. (My paper is here. See also this MO answer.)

A theme that comes out of the proofs of these results is that when the cube property holds nontrivially, usually some version of the Schroeder-Bernstein theorem is in play.

References:

1. William Hanf, MR 108451 On some fundamental problems concerning isomorphism of Boolean algebras, Math. Scand. 5 (1957), 205--217.
2. Alfred Tarski, MR 108452 Remarks on direct products of commutative semigroups, Math. Scand. 5 (1957), 218--223.
3. Bjarni Jónsson, MR 108453 On isomorphism types of groups and other algebraic systems, Math. Scand. 5 (1957), 224--229.
4. Corner, A. L. S., "On a conjecture of Pierce concerning direct decompositions of Abelian groups." Proc. Colloq. Abelian Groups. 1964.
5. Jones, JM Tyrer, "On isomorphisms of direct powers." Studies in Logic and the Foundations of Mathematics 95 (1980): 215-245.
6. P. M. Cohn, MR 197511 Some remarks on the invariant basis property, Topology 5 (1966), 215--228.
7. W. G. Leavitt, MR 132764 The module type of a ring, Trans. Amer. Math. Soc. 103 (1962), 113--130.
8. Jussi Ketonen, MR 491391 The structure of countable Boolean algebras, Ann. of Math. (2) 108 (1978), no. 1, 41--89.
9. V. Koubek, J. Nešetril, and V. Rödl, MR 357669 Representing of groups and semigroups by products in categories of relations, Algebra Universalis 4 (1974), 336--341.
10. Vera Trnková, MR 2380275 Categorical aspects are useful for topology—after 30 years, Topology Appl. 155 (2008), no. 4, 362--373.
11. Trnková, Věra, and Václav Koubek, "Isomorphisms of products of infinite graphs." Commentationes Mathematicae Universitatis Carolinae 19.4 (1978): 639-652.
12. Trnková, Věra, "Isomorphisms of products of infinite connected graphs." Commentationes Mathematicae Universitatis Carolinae 25.2 (1984): 303-317.
13. A. Orsatti and N. Rodinò, MR 858335 Homeomorphisms between finite powers of topological spaces, Topology Appl. 23 (1986), no. 3, 271--277.
14. W. T. Gowers, MR 1374409 A solution to the Schroeder-Bernstein problem for Banach spaces, Bull. London Math. Soc. 28 (1996), no. 3, 297--304.
15. Paul C. Eklof and Saharon Shelah, MR 1485469 The Kaplansky test problems for $\aleph_1$-separable groups, Proc. Amer. Math. Soc. 126 (1998), no. 7, 1901--1907.
16. Eklof, Paul C., "Modules with strange decomposition properties." Infinite Length Modules. Birkhäuser Basel, 2000. 75-87.
17. Trnková, Věra, "Homeomorphisms of powers of metric spaces." Commentationes Mathematicae Universitatis Carolinae 21.1 (1980): 41-53.
18. Vera Trnková, MR 580990 Isomorphisms of sums of countable Boolean algebras, Proc. Amer. Math. Soc. 80 (1980), no. 3, 389--392.

Answer 3

The Banach space version of this, where A is a Banach space and "isomorphism" means "linear homeomorphism", was a famous problem solved by Tim Gowers (Bull. London Math. Soc. 28 (1996), 297-304), using the space he and Bernard Maurey constructed that had no subspace with an unconditional basis.

Answer 4

The answer is negative for the class of countable Boolean algebras. The reference is Jussi Ketonen's "The Structure of Countable Boolean Algebras" (Annals of Mathematics [Second Series], Vol. 108, 1978, No. 1, pp. 41-89). There, Ketonen shows any countable commutative semigroup can be embedded into the monoid of countable Boolean algebras. The proof of this is rather involved.

The answer is positive for the class of linear orders (replacing product with concatenation). Lindenbaum showed for any linear orders $y$ and $z$, if $y$ is an initial segment of $z$ and $z$ is an end segment of $y$, then $y \cong z$. Taking $x+x$ for $y$ and $x = x+x+x$ for $z$ suffices. A reference is Joseph Rosenstein's "Linear Orderings" (Academic Press Inc., New York, 1982, p.22). The proof of this is rather straightforward.

Answer 5

Edit: In the comment below, Emil Jeřábek points out that my proof is wrong. But I'll leave this answer here for posterity.

Here's a partial answer to the Stone space question. The answer is yes for metrizable Stone spaces: if $X \cong X + X + X$ then $X \cong X + X$. I assume you're using $+$ to denote coproduct of topological spaces.

Proof: Write $I(X)$ for the set of isolated points of a topological set $X$. (A point is isolated if, as a singleton subset, it is open.) Then $I(X + Y) \cong I(X) + I(Y)$ for all $X$ and $Y$. So, supposing that $X \cong 3X$, we have $I(X) \cong 3I(X)$. But $X$ is compact, so $I(X)$ is finite, so $I(X)$ is empty. Hence $X$ is a compact, metrizable, totally disconnected space with no isolated points. A classical theorem then implies that $X$ is either empty or homeomorphic to the Cantor set. In either case, $X \cong X + X$.

I guess metrizability of the Stone space corresponds to countability of the corresponding Boolean ring.

The topological theory of Stone spaces is more subtle in the non-metrizable case, if I remember correctly.