Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Is there a finitely generated nontrivial group $G$ such that $G \cong G \times G$?

Here are some properties which such a group $G$ has to satisfy:

  • $G$ is not abelian (otherwise $G$ is a noetherian $\mathbb{Z}$-module, and the composition of the first projection $G \times G \to G$ with an isomorphism $G \cong G \times G$ will be bijective, i.e. $G$ is trivial).
  • $G$ is perfect (apply the first observation to $G/G'$)
share|improve this question
apart from the trivial group, I guess. –  wood Oct 27 '10 at 14:50
$G$ must also not be residually finite (as a finitely generated residually finite group is Hopfian, i.e. has no isomorphic proper quotients). –  Jonathan Kiehlmann Nov 1 '10 at 17:59
add comment

2 Answers

up vote 19 down vote accepted

Yes. Some Googling turns up J. M. Tyrer Jones, "Direct products and the Hopf property," J. Austral. Math. Soc. 17 (1974), 174-196.

share|improve this answer
Thanks! I used google, but didn't know to feed it properly ;). –  Martin Brandenburg Oct 27 '10 at 15:43
has anyone else the problem, that the link to the pdf doesn't work? –  HenrikRüping Oct 27 '10 at 15:44
@Henrik: hmm. The paper might not be freely available. I found it indirectly: it was referenced in a paper of Baumslag and Miller (rio.sci.ccny.cuny.edu/caissny.org/Publications/…) which I found on Google, and which may have a more general construction in it. –  Qiaochu Yuan Oct 27 '10 at 16:20
It's available for free: journals.cambridge.org/action/… –  Steve D Oct 27 '10 at 20:10
add comment

As a geometric group theorist, one would of course relax the question by allowing passage to finite index subgroup, i.e. one would ask for groups such that G and GxG have finite index subgroups, which are isomorphic. One then calls G an GxG commensurable, and for this weaker property there are a lot of interesting examples. My favourite one right now is the Grigorchuk group. But even commensurability to GxG is a very restrictive property: It implies, for instance, that if G is infinite, then it has infinite asymptotic dimension. I just stumbled over this result in the thesis of J. Smith. The proof is almost trivial: Since G is coarsely equivalent to GxG, it is coarsely equivalent to G^n for all n. Now Z embeds quasi-isometrically into G (since G is infinite), and hence Z^n embeds coarsely into G^n (hence G), so asdim G is at least asdim Z^n for all n, and we conclude. This is in particular the case if G and GxG are isomorphic. The upshot is, that for a group of finite asymptotic dimension one cannot have G=GxG, not even up to finite index.

share|improve this answer
add comment

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.