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

I wondered whether it is possible to find two finitely generated, virtually Abelian, torsionfree groups $G,H$ that are not isomorphic but that become isomorphic after crossing with $\mathbb{Z}$. I have the following candidates:

Consider $K:=(\mathbb{Z}[t]/(t^5+1))\rtimes_{\cdot t} \mathbb{Z}$. Let $\varphi$ be the automorphism of $K$ given by $\cdot t$ on $\mathbb{Z}[t]/(t^5+1)$ and $(0,s)\mapsto (1,s)$ , where $s$ denotes a generator of the other copy of $\mathbb{Z}$. My candidates are $G:=K\rtimes\mathbb{Z}$ and $H:=K\rtimes_{\varphi^3}\mathbb{Z}$.

Are these two groups isomorphic ?

Each one contains a finite index subgroup isomorphic to the other one. Crossing with $\mathbb{Z}$ gives $K\rtimes \mathbb{Z^2}$ where a basis of $\mathbb{Z}^2$ acts by $\varphi,\mbox{id}$ respectively $\varphi^3,\mbox{id}$. The isomorphism is given by a base change of $\mathbb{Z}^2$.

$G$ contains a finite index subgroup isomorphic to $H$ and vice versa. If it turns out that they are actually isomorphic, one might still hope to get an example by replacing $5$ with a bigger number (that is coprime to 3 to make the base change work).

Related: Hirshon, some cancellation theorems with applications to nilpotent groups (The example given there is torsionfree, nilpotent, but maybe not virtually Abelian).

share|improve this question
2  
I am not sure about your candidate but there certainly are non-homotopy equivalent closed flat manifolds that become affinely diffeomorphic after taking product with a circle, see Charlap's book on flat manifolds, section IV.8. If memory serves, this is also published in a journal (maybe annals?). –  Igor Belegradek Jan 30 '13 at 14:49
1  
I found the paper: Compact Flat Riemannian Manifolds, I by Charlap Annals of Mathematics, Vol. 81, No. 1 (Jan., 1965), pp. 15-30. –  Igor Belegradek Jan 30 '13 at 15:53
    
@Igor: Thank you. The precise place where this is mentioned is the remark after Theorem 3.10 on p. 29. In that chapter $\mathbb{Z}/p$-flat manifolds are studied. Flat manifolds are isometrically covered by $\mathbb{R}^n$. A $\mathbb{Z}/p$ manifold is a manifold where the quotient of its fundamental group by the subgroup of translations is $\mathbb{Z}/p$. Maybe its possible to cook up easier groups by unraveling the theory. I will have a look at it. –  HenrikRüping Jan 31 '13 at 14:49
    
Let me just mention, that the groups above seem to be $\mathbb{Z}/10\oplus \mathbb{Z}/10$-manifolds, so the theory from chapter 3 there does not apply directly. Still it would be nice to find an elementary argument that shows that these two groups are not isomorphic. –  HenrikRüping Jan 31 '13 at 14:50
add comment

Know someone who can answer? Share a link to this question via email, Google+, Twitter, or Facebook.

Your Answer

 
discard

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

Browse other questions tagged or ask your own question.