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 $m \equiv n (\mod r)$. 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

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 3

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.

Answer 4

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 uncondtional basis.

Answer 5

It follows that $A \cong A^{2 k + 1}$ for any $k$ because we can always increase the exponent by two: $A \cong A^{n} \cong A \times A^{n-1} \cong A^3 \times A^{n-1} \cong A^{n + 2}$. But I can't get $A$ to be isomorphic to an even power. That could be a hint as to how to find a counter-example in groups.

I think the topological version has better chances of being true, but I have to think a bit about it.