4
$\begingroup$

Does there exist a variety of groups in which all finitely generated groups are residually nilpotent, and which contains some finitely generated group that is not nilpotent? That is, can a variety be locally residually nilpotent but not locally nilpotent?

Note that a consequence of Theorem 4.5 of [Traustason, Gunnar. Milnor groups and (virtual) nilpotence. J. Group Theory 8 (2005), no. 2, 203–221. MR2126730] is that no such variety can be metabelian (since the varieties $\mathcal{A}_p \mathcal{A}$ and $\mathcal{A} \mathcal{A}_p$ contain finite non-nilpotent groups). It appears the difficulty is that residual nilpotence does not pass to quotients in general.

$\endgroup$
10
  • $\begingroup$ A classical result of Kropholler is that every f.g. solvable group is either minimax, or has a subquotient isomorphic to a lamplighter $C_p\wr\mathbf{Z}$ for some prime $p$. It's probably not hard to deduce that there is no solvable example (i.e., every solvable group in a variety consisting of residually nilpotent groups is locally nilpotent). $\endgroup$
    – YCor
    Dec 6, 2017 at 16:46
  • $\begingroup$ Any variety where all f.g. groups are residually nilpotent has all f.g. groups residually finite. Now apply Theorem 3.24 of the paper you cite and note that a variety $V$ where all f.g. groups are residually nilpotent cannot contain $C_p\wr C$. Hence by part 1 of the theorem all f.g. groups in the variety are virtually nilpotent, hence nilpotent. $\endgroup$
    – user6976
    Dec 6, 2017 at 16:56
  • $\begingroup$ Thank you both. @MarkSapir how do you promote virtually nilpotent to nilpotent? $\endgroup$ Dec 8, 2017 at 14:05
  • $\begingroup$ @GilesGardam: See groupprops.subwiki.org/wiki/Nilpotent_residual $\endgroup$
    – user6976
    Dec 10, 2017 at 19:54
  • $\begingroup$ @MarkSapir that page claims that the quotient of a virtually nilpotent group by its nilpotent radical is nilpotent, but this is not true for $\mathbb{Z} \rtimes \mathbb{Z}/2$, for example. $\endgroup$ Dec 11, 2017 at 16:07

1 Answer 1

2
$\begingroup$

Since the page I referred to contains a wrong statement. Here is a complete answer.

Any variety where all finitely generated groups are residually nilpotent has all finitely generated groups residually finite. Now apply Theorem 3.24 of the paper you cite and note that a variety $V$ where all finitely generated groups are residually nilpotent cannot contain $C_p\wr C$. Hence by part 1 of the theorem all finitely generated groups in the variety are virtually nilpotent, hence nilpotent

Here is a more or less detailed (but far from optimal) proof of the last "hence". Suppose that the variety $V$ contains a non-nilpotent finitely generated group $G$. By the assumption, $G$ has a normal nilpotent subgroup $N$ with $G/N$ finite and nilpotent. Hence $G$ is solvable. By Gruenberg's theorem (Th. 2.2 here) $G$ is not Engel. Hence there exists $n\in N$ and $g\in G$ such that $[n,g,g,g,...]$ is never 1. We can assume that the nilpotency class of $N$ is the smallest possible. Then if $N$ is not Abelian, both groups $N$ and $G/[N,N]$ are nilpotent, and by the well known by Theorem 7 of P. Hall , $G$ is nilpotent. Hence we can assume that $N$ is Abelian.

Hence $G$ is Abelian-by-cyclic (generated by the image of $g$). Let $N_0$ be the maximal finite subgroup of $N$. If we can find our element $n$ in $N_0$, we can assume that $N=N_0$, and the group $G$ is finite, hence $G$ is nilpotent. So we can assume that we cannot find such an $n$ in $N_0$. Therefore we can assume that $N_0=\{1\}$ (take $G/N_0$). So $N$ is torsion-free. If some power $g^m$ of $g$ is in $N\setminus \{1\}$, we have that $g^m$ is central, and we can take the quotient by $\langle g^m\rangle$. So we can assume that $C=\langle g\rangle$ intersects $N$ trivially. But that means $g$ is of finite order (since $N$ is of finite index in $G$) and $G$ is a semidirect product of a finitely generated torsion-free Abelian group $N$ by $C$.

The element $g$ acts (by conjugation) on $N$ as a square matrix $A$ of finite size (the rank of $N$) with integer coefficients. Viewing $n$ as an integer vector we get that $[n,g]=An-n=(A-1)n$. Hence $[n,g,g,...] (t\ \mathrm{ times})= (A-1)^tn$. (In particular, the matrix $A-1$ is not nilpotent). Then (by standard linear algebra plus a little number theory) there exists a prime $p\gg 1$ such that $(A-1)^tn$ is never zero modulo $p$. Now let $N^p$ be the $p$-th power of $N$. Then $G/N^p$ is finite, belongs to $V$ and not nilpotent, a contradiction.

Update. Here is a little number theory needed in the proof (it corrects the not quite correct comment by @YCor below). First note that it is enough to prove that $A-1 \mod p$ is not nilpotent for some prime $p$ (indeed, then one can take one of the finite set of generators of $N$ for $u$).

Thus we need $p$ such that $trace(\lambda - (A-1)^s)$ modulo $p$ is not equal to $(\lambda-1)^k$ for every integer $s\ge 1$ where $k$ is the size of the matrix $A$ (the rank of $N$). Note that if $(A-1)^m=0 \mod p$ then $(A-1)^{m'}=0$ for every $m'>m$. Now take prime $p$ bigger than the absolute value of every coefficient of the characteristic polynomial of $A-1$. It is enough to show that $(A-1)^{p^m}$ is never $0$ mod $p$. Let $q_s(\lambda)$ be the trace polynomial $\det(\lambda-(A-1)^{p^s})$. Then modulo $p$, $q_s(\lambda^{p^k})=\det(\lambda^{p^s}-(A-1)^{p^s})=\det(\lambda - (A-1))^{p^s}= q_1(\lambda)^{p^s}=q_1(\lambda^{p^s})$ by the Newton binomial formula and the well known properties of binomial coefficients mod $p$. Since the coefficients of $q_1(\lambda^{p^s})$ are smaller than $p$, this polynomial modulo $p$ is not equal to $(\lambda^{p^{s}}-1)^k$. So $(A-1)^{p^s}$ is not equal to $0$ modulo $p$.

$\endgroup$
7
  • $\begingroup$ "A little number theory" is just that there are infinitely many primes :) one needs $p$ such that the characteristic polynomial of $A$ is not equal to $(X-1)^d$ modulo $p$. Also one needs a little argument to justify that the quotient by the central $\langle g^m\rangle$ preserve being non-Engel, but this is essentially clear once one has the linear algebra interpretation ($g^m$ lies in the characteristic subspace for the eigenvalue 1, which we can kill beforehand). $\endgroup$
    – YCor
    Dec 12, 2017 at 7:34
  • $\begingroup$ The statement proved in the last 3 paragraphs is: every non-nilpotent finitely generated virtually nilpotent group has a finite non-nilpotent subquotient. This is enough for the current purposes. Actually an easy adaptation shows that every non-nilpotent finitely generated virtually nilpotent group has a finite non-nilpotent quotient. $\endgroup$
    – YCor
    Dec 12, 2017 at 7:49
  • $\begingroup$ Quotient by any central subgroup preserves the property of being non-Engel, that is obvious. It is not enough that $p$ is bigger than the coefficients of the characteristic polynomial because there is also vector $n$. $\endgroup$
    – user6976
    Dec 12, 2017 at 12:04
  • $\begingroup$ You're right about killing a central element. Anyway, my comment is correct, maybe not quite understood :) Indeed, if modulo $p$ the characteristic polynomial is not $(X-1)^d$, then there exists $n'\in A$ such that all iterated brackets $[n,g,\dots,g]$ is never trivial. Namely any $n$ not in the kernel of $(g-1)^d$ works. $\endgroup$
    – YCor
    Dec 13, 2017 at 5:44
  • $\begingroup$ ""A little number theory" is just that there are infinitely many primes :) " is wrong. The phrase " Namely any $n$ not in the kernel of $(g−1)^d$ works." is wrong as written. $\endgroup$
    – user6976
    Dec 13, 2017 at 9:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.