BS(1,-1) is a right-ordered and locally indicable group. Additionally, BS(1,n) where n is a positive integer are ordered groups, therefore they are also locally indicable groups. What can be said about BS(1,n) where n<-1? They are right-ordered groups, but are they also locally indicable groups? Furthermore, what about BS(m,n) where m is not equal to 1? Are they locally indicable groups?
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1$\begingroup$ Nomenclature note: the groups are called "Baumslag–Solitar", not "Baum-Slag Solitar". I have edited accordingly. $\endgroup$– LSpiceCommented Sep 4 at 15:40
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Even more is true: All torsion-free 1-relator groups are locally indicable:
Brodskij, S. D., Equations over groups, and groups with one defining relation, Sib. Math. J. 25, 235-251 (1984); translation from Sib. Mat. Zh. 25, No. 2(144), 84-103 (1984). ZBL0579.20020.
(And a 1-relator group is torsion-free if the defining relator is not a proper power.)
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$\begingroup$ I am wondering: $BS(n,m)$ has a natural surjection onto $\mathbb{Z}$, but is there an explicit description of the kernel? Naively, I would say it is locally free. If so, it would provide a rather elementary argument in the specific case of Baumslag-Solitar groups. $\endgroup$ Commented Sep 5 at 7:16
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3$\begingroup$ @AGenevois the kernel is not locally free (unless $m$ or $n$ is $\pm 1$). It's a line of $\mathbb{Z}$'s and contains the "torus knot group" $\langle x, y | x^m = y^n \rangle$. $\endgroup$ Commented Sep 5 at 8:14