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Oct 12, 2020 at 22:33 vote accept lab
Oct 9, 2020 at 19:29 comment added YCor A variant in which $N$ is finitely generated: write the Baumslag-Solitar group as $G=\langle t,x\mid txt^{-1}=x^n\rangle$, with $n\ge 2$. Then in $G^2$, the subgroup $N$ generated by $\{(t,t^{-1}),(x,1),(1,x)\}$ is finitely generated, and normal. Then $G^2$ admits a 4-dimensional finite classifying space. But $N$ is not finitely presented so doesn't admit any finite classifying space. (Nevertheless I'm not claiming that $N$ has no 3-dimensional classifying space.)
Oct 9, 2020 at 18:42 history answered Denis T CC BY-SA 4.0