Timeline for Some questions on a paper of Baumslag and Solitar
Current License: CC BY-SA 4.0
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May 20, 2021 at 16:40 | comment | added | HJRW | @user101010: I think the fact you want is true; it's very similar to Magnus' famous result that $\langle\langle u\rangle\rangle=\langle\langle v\rangle\rangle$ in a free group implies that $u$ is conjugate to $v^{\pm1}$. But you still then have to apply the solution to the word problem in multiple different one-relator groups, so I don't think it's going to be easy! | |
May 20, 2021 at 16:01 | comment | added | Carl-Fredrik Nyberg Brodda | I wonder how Baumslag and Solitar proved the result - perhaps there is an easier argument. An electronic copy of the proceedings from that conference can be found on whichever pseudo-legal book source you want :-) (there is an excellent paper in there by Baumslag on the history of one-relator groups). | |
May 20, 2021 at 15:27 | comment | added | user101010 | Thanks for the reference - I'll check it out! I thought a bit more about proving this by brute force. The elements $y^{-3}x^{-1}y^2x$ and $y^{-6}x^{-1}y^4x$ are in the kernel. I'd guess that given a cyclically reduced word in a free group, that word is the shortest element in the normal group generated by that element. Assuming that, it would remain to look at words $w$ shorter than both of these and see that they are not in the kernel or if they are that both these elements are not in the normal closure of $w$. But I'm not sure if that works. | |
May 20, 2021 at 15:11 | history | answered | HJRW | CC BY-SA 4.0 |