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Nov 12, 2021 at 17:55 history edited HJRW CC BY-SA 4.0
Minor edits
Nov 11, 2021 at 16:43 vote accept Ross Griebenow
Nov 11, 2021 at 15:00 comment added YCor Yes, this is the simple observation I had in mind. For some reason it seemed to be unnoticed by several people.
Nov 11, 2021 at 14:46 comment added HJRW @YCor: it's easy to see: $BS(m,n)$ is an HNN extension of the torus-link group $\mathbb{Z}*_{m=n}\mathbb{Z}$, and one can see this has a $\mathbb{Z}^2$ subgroup in various ways. For instance, the fact that it's the torus-link group proves it immediately geometrically, or one can construct the embedding explicitly. I don't know a reference, though, if that's what you're asking for.
Nov 11, 2021 at 10:47 comment added YCor I'm curious where the fact that $\mathrm{BS}(2,3)$ possesses a $\mathbf{Z}^2$ subgroup can be located. When I was PhD I asked whether it holds to a few people and at least two "famous" people proved me shaking hands that it doesn't. But it indeed does contain such a $\mathbf{Z}^2$ subgroup (as does $\mathrm{BS}(m,n)$ whenever $\min(|m|,|n|)\ge 2$), and this is not hard to check. I'm asking it here because it's implicit in passing the question from $\mathrm{BS}(m,n)$ subgroups (as it appears in Bestvina's list) to $\mathrm{BS}(1,n\ge 1)$ subgroups that the other cases are actually redundant.
Nov 11, 2021 at 10:41 history edited YCor CC BY-SA 4.0
replaced ill-formated --, fixed BS
Nov 11, 2021 at 8:37 history answered HJRW CC BY-SA 4.0