# When is $G$ isomorphic to $G \times G$?

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 $G/G'$)
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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

## 2 Answers

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

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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

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.

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