2

For which $(m,n,k,l) \in (\mathbb N\cup \{0\})^4$ , with $m\le n ; k\le l$ , does there exist a group $G$ with a finite subnormal series with torsion-free Abelian quotients such that $G \times \mathbb Z^n \cong G \times \mathbb Z^l$ but $G \times \mathbb Z^m \ncong G \times \mathbb Z^k$ ?

$A$ is isomorphic to $A \oplus \mathbb{Z}^2$, but not to $A \oplus \mathbb{Z}$ shows $(0,0,1,2)$ is such a pair .

NOTE : Also previously asked on Math SE https://math.stackexchange.com/questions/2340835/a-kind-of-cancellation-exchange-problem-for-groups

  • The construction in my answer to the linked question generalizes straightforwardly to give, for any $d$, a torsion free abelian group with $G\oplus\mathbb{Z}^r\cong G\oplus\mathbb{Z}^s$ if and only if $d$ divides $r-s$, so a sufficient condition is that $n-l$ does not divide $m-k$. For abelian groups, Walker's Cancellation Theorem shows that this is also a necessary condition. Is there any particular reason for the class of groups you're interested in? – Jeremy Rickard Jul 24 '17 at 10:20
  • @JeremyRickard : Well , for a field $k$ , and a group $G$ with a finite subnormal series with torsion-free Abelian quotients , Kaplansky's unit conjecture holds for $k[G]$ ; so if $G,H$ are two such groups then $k[G] \cong k[H]$ would imply $G \cong H$ if $k$ is finite ... I was trying to get some cancellation; exchange properties for Laurent polynomial rings from these observations ... your example is even better by the way ( though I haven't worked out it fully yet) .. because $G$ is abelian and then $k[G]$ is commutative ring ... –  Jul 24 '17 at 16:57

0 Answers0