Here's a couple of analogous questions, one in terms of finite-dimensional complex Lie algebras and one in terms of finite $p$-groups; I'd be interested in an answer to either:

1) Let $\mathcal{L}$ be an isomorphism-closed class of finite-dimensional nilpotent complex Lie algebras. Assume $\mathcal{L}$ is closed under taking finite direct products, subgroups, and quotients. Assume that it does not satisfy any common identity (i.e., any free Lie algebra is residually-$\mathcal{L}$). Does $\mathcal{L}$ necessarily consist of all finite-dimensional nilpotent complex Lie algebras?

2) Let $p$ be prime. Let $\mathcal{C}$ be an isomorphism-closed class of finite $p$-groups. Assume $\mathcal{C}$ is closed under taking finite direct products, subgroups, and quotients. Assume that it does not satisfy any common identity (i.e., any free group is residually-$\mathcal{C}$, or still equivalently the free group on 2 generators is residually-$\mathcal{C}$). Does $\mathcal{C}$ necessarily consist of all finite $p$-groups?

(Note: I ask both questions in the positive but I don't particularly expect a positive answer!)

  • $\begingroup$ Is the condition "does not satisfy any common identity" clearly enough formulated? (By the way, perhaps you mean to write e.g. rather than i.e.: e.g. means "for example", while i.e. means "that is".) $\endgroup$ Sep 21, 2014 at 19:23
  • $\begingroup$ I really mean "that is". A class $\mathcal{C}$ of Lie algebras satisfies a (nontrivial) common identity if there exists a free Lie algebra $\mathfrak{f}$ and $m\in\mathfrak{f}-\{0\}$ such that $m$ is an identity for every $\mathfrak{g}\in\mathcal{C}$ (i.e., every Lie algebra homomorphism $\mathfrak{f}\to\mathfrak{g}$ maps $m$ to 0). $\endgroup$
    – YCor
    Sep 21, 2014 at 19:47
  • $\begingroup$ Thanks. The language isn't familiar to me, but I'm aware of similarities between the two theories. It might be surprising if your two questions had different answers, but I have no idea how to approach either of them. (By the way, I guess the first question has the same answer over any algebraically closed field of characteristic 0.) $\endgroup$ Sep 21, 2014 at 21:03
  • $\begingroup$ Yes, the answer to the first question does not even depend on the ground field of characteristic zero. $\endgroup$
    – YCor
    Sep 21, 2014 at 21:35
  • $\begingroup$ For question 2), it appears that there is a literature on "pseudo-varieties of finite algebras". In particular, Baldwin and Berman proved that a class of finite algebras closed under subalgebras, homomorphisms and finite products is the set of finite algebras in an ascending union of varieties. yadda.icm.edu.pl/yadda/element/… $\endgroup$
    – Ian Agol
    Sep 21, 2014 at 22:34

1 Answer 1


Here is some idea, it is not very precise. Let $F$ be the free group on two generators and let $F_p$ be its pro-$p$ completion. Let $w$ be an infinite word in $F_p$, i.e., an element of $F_p$ which is not in $F$. Let $W$ be the closed verbal subgroup of $F_p$ generated by $w$. I think that it is possible to choose $w$ so that $W \cap F$ is trivial. For example, the first congruence subgroup of $SL_2(\mathbb{Z}_p)$ ($p>2)$ satisfies a pro-$p$ identity due to Zubkov, but I think it has a dense free group.

Let $D_n$ be the $n$-dimension subgroups of $F_p$. Then if I recall correctly there is a canonical way to write $w=w_nu_n$, where $u_n \in D_n$. Take your variety to be all the groups that satisfy $w_n$ for some $n$. I would guess they will satisfy your requirement.

I am not sure this idea will work, but if you like it and need more reference, then please contact me.

  • $\begingroup$ Well, there is still some stuff to check and some stuff to prove before we can conclude this. $\endgroup$ Sep 21, 2014 at 22:56
  • 4
    $\begingroup$ Let $P$ be an open pro-$p$-subgroup in $\mathrm{GL}_2(\mathbf{Z}_p)$. Clearly it has a free subgroup (e.g., in $\mathrm{SL}_2(\mathbf{Z}))$. On the other hand according to Zubkov (at least for $p>2$), $P$ satisfies a nontrivial pro-$p$-identity $I$. Let $\mathcal{C}$ be the class of finite $p$-groups satisfying the identity $I$ (or alternatively the smallest class of groups containing finite quotients of $P$ and stable under taking subgroups, quotients, and direct products). Then $\mathcal{C}$ satisfies no common group identity, but does not contain all $p$-groups, answering (2). $\endgroup$
    – YCor
    Sep 21, 2014 at 23:07
  • $\begingroup$ Yes, this formulation is indeed more convincing. $\endgroup$ Sep 22, 2014 at 7:18
  • $\begingroup$ I think one can try for the Lie algebra case to look at the first congruence subalgebra of $sl_2(\mathbb{C}[[X]])$ and prove similar results. I am not sure how difficult this might be. $\endgroup$ Sep 22, 2014 at 11:56
  • $\begingroup$ About case $p=2$: arxiv.org/abs/1910.05805 (On Pro-2 Identities of 2×2 Linear Groups, David El-Chai Ben-Ezra, Efim Zelmanov) $\endgroup$
    – YCor
    Oct 15, 2019 at 5:19

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.