MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 :).

share|cite|improve this question
I saw a similar question called Zariski cancellation problem. Perhaps you find some interesting answers when you google for that. – user2146 Dec 30 '09 at 12:40
interesting. but why is zariski cancellation related to this? – Martin Brandenburg Dec 30 '09 at 13:12
I don't see any direct relation to Zariski cancellation. Moreover, I no longer feel confident that the result is true. But, Martin, why do you think it is probably false? – Pete L. Clark Dec 30 '09 at 13:44
a confirmative proof would knock me for a loop ;-). the naive approaches just don't work and rather (especially in geometric terms) suggest a counterexample. that's just intuition, nothing which I can make precise. – Martin Brandenburg Dec 30 '09 at 19:59
What is the situation when abelian groups are replaced with posets? – Richard Stanley Feb 5 '11 at 3:37
up vote 66 down vote accepted

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

share|cite|improve this answer

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.

share|cite|improve this answer
at the risk of labouring the response, thought I'd just elaborate on Bill Johnson's answer: Gowers' construction provides a negative answer (i.e. X is isomorphic as a B. space to X \oplus X \oplus X but not to X \oplus X ) – Yemon Choi Dec 31 '09 at 0:30

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.

share|cite|improve this answer
Doesn't the first paragraph contradict Tom's proof? – Martin Brandenburg Feb 2 '12 at 8:48
The second one is very interesting, 1+. – Martin Brandenburg Feb 2 '12 at 8:48

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.

share|cite|improve this answer
A metrizable Stone space can have infinitely many isolated points: take e.g. $\{0\}\cup\{1/(n+1):n\in\omega\}\subseteq\mathbb R$. In fact, there are countable Stone spaces of arbitrary large countable Cantor–Bendixson rank. – Emil Jeřábek Feb 2 '12 at 11:41
Thanks, Emil. – Tom Leinster Feb 2 '12 at 13:15
Thank you Emil for this clarification. Somehow I took the finiteness of $I(X)$ for granted. – Martin Brandenburg Feb 3 '12 at 7:59

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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