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This question is not directly related to, but was inspired by, this question. We know that a finitely generated residually nilpotent group is residually of prime-power order. However, we may need to use different primes for different elements. Classes of groups for which residual nilpotence forces there to be a single prime that will do for all elements (i.e., for which the group in question must be residually $p$-finite, for some $p$) seem to be interesting, and include, for instance, free products of cyclic groups.

Is there a (non-cyclic) one-relator group that is residually nilpotent, but is not residually a finite $p$-group, for any prime number $p$?

Such a group must be torsion-free, with trivial centre.

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  • $\begingroup$ My answer was wrong, thanks for pointing to a mistake. I have deleted the answer. $\endgroup$
    – user6976
    Commented Oct 11, 2011 at 23:23

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The answer is yes, such a group exists: this is the main result of [1]. In fact, one can take the Baumslag-Solitar group $\operatorname{BS}(p^r, -p^r) = \langle a, b \mid ba^{p^r}b^{-1} = a^{-p^r} \rangle$ where $p$ is an odd prime and $r \geq 1$. These are residually nilpotent, but are not residually a finite $q$-group for any prime $q$.

In particular, the smallest example of this form is $\operatorname{BS}(3, -3) = \langle a, b \mid ba^3b^{-1}a^3 = 1 \rangle$.

(Note that OP's question was posed by McCarron [2] in 1996).

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[1] Moldavanskii, D. I., Residual nilpotence of groups with one defining relation, translation from Mat. Zametki 107, No. 5, 752-759 (2020). ZBL07215590.

[2] McCarron, James, Residually nilpotent one-relator groups with non-trivial centre, Proc. Amer. Math. Soc. 124, No. 1 (1996).

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    $\begingroup$ 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. $\endgroup$
    – YCor
    Commented Mar 6, 2023 at 17:10

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