Timeline for Is there a residually nilpotent one-relator group that is not residually a finite p-group for any prime p?

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Mar 6, 2023 at 17:10	comment	added	YCor		One half is easy to explain in a comment. Let $q$ be any odd number and $G=BS(q,-q)=\langle t,x:x^qtx^q=t\rangle$. Then $G$ is not residually a finite $p$-group for any prime $p$; more generally $G$ is not residually a finite $2$-group, and is not residually finite of odd order. Indeed, in any quotient of $G$ of odd order, the image of $x^q$ is conjugate to its inverse, so has to be trivial. In any 2-group that is a quotient of $G$, $x$ is a power of $x^q$ and hence $t^2$ commutes with $x$. That is, $[t^2,x]$ is killed in any 2-group quotient.
Mar 1, 2023 at 3:34	vote	accept	James
Feb 28, 2023 at 19:06	history	edited	Carl-Fredrik Nyberg Brodda	CC BY-SA 4.0	deleted 3 characters in body
Feb 28, 2023 at 14:56	history	edited	Carl-Fredrik Nyberg Brodda	CC BY-SA 4.0	added 278 characters in body
Feb 28, 2023 at 12:10	history	answered	Carl-Fredrik Nyberg Brodda	CC BY-SA 4.0