A pronilpotent variety whose locally nilpotent subvarieties are nilpotent

Problem

Must every nonnilpotent variety of pronilpotent groups contain a locally nilpotent subvariety that is not nilpotent?

\mathcal V\text{ nonnilpotent} \quad\stackrel{?}{\Longrightarrow}\quad \exists\mathcal W\subseteq\mathcal V:\quad \mathcal W\text{ locally nilpotent but nonnilpotent}.

All groups and subgroups in this formulation are profinite, and finite generation is topological. This article addresses part (a).

The counterexample variety

All profinite groups carry their profinite topology. Subgroups in a profinite variety are closed, quotient maps are continuous and surjective, and products have the product topology. A profinite group is topologically generated by $g_1,\ldots,g_d$ if $\overline{\langle g_1,\ldots,g_d\rangle}=G$ . A profinite variety is a nonempty class closed under these three operations. It is nilpotent if each of its members is nilpotent, and locally nilpotent if each of its topologically finitely generated members is nilpotent.

For abstract groups, use the conventions

\tag{1} [x,y]=xyx^{-1}y^{-1},\qquad \gamma_1(G)=G,\qquad \gamma_{j+1}(G)=[\gamma_j(G),G].

Thus $G$ has class at most $c$ if $\gamma_{c+1}(G)=1$ ; the trivial group has class zero. Write $\operatorname{cl}(G)$ for the class of a nilpotent group. Nilpotence of a profinite group refers to its underlying abstract group. A profinite group is pronilpotent if all its finite continuous quotients are nilpotent.

Problem 7.38(a) of the Kourovka Notebook, due to O. V. Mel’nikov, asks whether every nonnilpotent variety of pronilpotent groups contains a locally nilpotent subvariety that is not nilpotent (2). We construct a variety for which every locally nilpotent subvariety is nilpotent.

Let $\mathcal C$ consist of the finite groups $F$ satisfying

\tag{2} \begin{gathered} F\text{ is nilpotent},\qquad F''=1,\\ \gamma_p(P)=1 \quad(p\text{ prime},\ P\le F\text{ a }p\text{-group}). \end{gathered}

Since every $p$ -subgroup lies in a Sylow $p$ -subgroup, the last condition is equivalent to requiring class less than $p$ for the Sylow $p$ -subgroups. Put

\tag{3} \mathcal V=\operatorname{Pro}(\mathcal C) =\{G:\text{every finite continuous quotient of }G \text{ belongs to }\mathcal C\}.

Theorem 1. The class $\mathcal V$ is a variety of pronilpotent groups. It is not nilpotent, and every locally nilpotent subvariety of $\mathcal V$ is nilpotent.

The finite-group ingredients have classical antecedents. Gupta and Newman proved a factorial polarization lemma for metabelian commutators (1, Lemma 2.1, pp. 364–365). Weichsel characterized the critical finite metabelian $p$ -groups of class less than $p$ by two-generation and cyclicity of the centre (3, Theorem 1.14, p. 61), and described basic groups with unipotent cyclic actions (3, Definition 2.1 and Theorem 2.2, p. 62). Below we give the polarization argument, the three-generator consequence needed here, and the truncated-polynomial construction explicitly.

Conjugation differences and class detection

Let $G$ be metabelian. Write $A=G'$ additively and let $B=G/G'$ . Conjugation induces an action of $B$ on $A$ : an element of $G'$ acts trivially because $A$ is abelian. For $g\in G$ put

\tag{4} c_g(x)=gxg^{-1},\qquad \delta_g=c_g-1\in\operatorname{End}_{\mathbb Z}(A).

In the multiplicative notation of $G$ , $\delta_g(x)=[g,x]$ . The maps $c_g$ commute, factor through $B$ , and satisfy $c_{gh}=c_gc_h$ . Consequently

\tag{5} \delta_g\delta_h=\delta_h\delta_g,\qquad \delta_{gh}=\delta_g+\delta_h+\delta_g\delta_h.

Also,

\tag{6} x\in A\cap\gamma_j(G) \quad\Longrightarrow\quad \delta_{g_1}\cdots\delta_{g_r}(x)\in\gamma_{j+r}(G).

This follows by applying the definition of the lower central series once for each difference.

Lemma 2. For every $r\ge0$ ,

\tag{7} \gamma_{r+2}(G)= \big\langle \delta_{g_1}\cdots\delta_{g_r}([a,b]): a,b,g_1,\ldots,g_r\in G \big\rangle.

For $r=0$ , the expression inside the brackets is $[a,b]$ .

Proof. The case $r=0$ is the definition of $G'$ . Suppose the assertion holds for $r$ . All displayed generators belong to $A$ . Since $\delta_g$ is an endomorphism of the abelian group $A$ , applying it to any product of these generators and their inverses gives a product of their images and their inverses. Thus $[G,\gamma_{r+2}(G)]$ is generated by the expressions obtained by adding one further difference. This subgroup is $\gamma_{r+3}(G)$ . Conversely, each such expression belongs to $\gamma_{r+3}(G)$ by (6), completing the induction. ◻

Lemma 3. Suppose $\gamma_{r+3}(G)=1$ , where $r\ge1$ , and fix $x\in A$ . The map

\tag{8} T_x:B^r\longrightarrow A,\qquad T_x(g_1G',\ldots,g_rG') =\delta_{g_1}\cdots\delta_{g_r}(x)

is symmetric and $\mathbb Z$ -multilinear.

Proof. The action on $A$ factors through $B$ , so the map is well-defined. Symmetry follows from the first identity in (5). In any argument, the difference between evaluation at $gh$ and the sum of evaluations at $g,h$ is an expression with $r+1$ differences applied to $x\in\gamma_2(G)$ . It lies in $\gamma_{r+3}(G)$ by (6), and is therefore zero. The map is additive in each argument; additivity also gives compatibility with every integer multiple. ◻

Lemma 4. Let $M,A$ be abelian groups and let $T:M^r\to A$ be symmetric and $\mathbb Z$ -multilinear, with $r\ge1$ . For $b_1,\ldots,b_r\in M$ , set $b_S=\sum_{j\in S}b_j$ . Then

\tag{9} r!\,T(b_1,\ldots,b_r) =\sum_{S\subseteq\{1,\ldots,r\}} (-1)^{r-|S|}T(b_S,\ldots,b_S).

Proof. Expand each summand on the right by multilinearity:

T(b_S,\ldots,b_S) =\sum_{f:\{1,\ldots,r\}\to S} T(b_{f(1)},\ldots,b_{f(r)}).

For a fixed map $f:\{1,\ldots,r\}\to\{1,\ldots,r\}$ , with image $J$ , its coefficient in the alternating sum is

\sum_{S\supseteq J}(-1)^{r-|S|} =\begin{cases} 1,&J=\{1,\ldots,r\},\\ 0,&J\ne\{1,\ldots,r\}. \end{cases}

Indeed, if the complement of $J$ is nonempty, pairing subsets that differ by one fixed element of that complement cancels the terms. The surviving maps are the bijections of an $r$ -element set. There are $r!$ of them, and symmetry makes every surviving term equal to $T(b_1,\ldots,b_r)$ . ◻

If $A$ is an abelian $p$ -group and $r<p$ , multiplication by $r!$ is injective. To see this, let $y\in A$ have order dividing $p^e$ . As $\gcd(r!,p^e)=1$ , choose integers $u,v$ with $ur!+vp^e=1$ . Then

\tag{12} r!y=0\quad\Longrightarrow\quad y=u(r!y)+v(p^ey)=0.

Thus, in Lemma 4, vanishing on every diagonal forces $T=0$ whenever $A$ is a $p$ -group and $r<p$ . This is the factorial-cancellation step of the metabelian commutator lemma in (1, Lemma 2.1).

Proposition 5. Let $G$ be a finite metabelian $p$ -group of nilpotency class $c<p$ . There exist $a,b,g\in G$ such that $\langle a,b,g\rangle$ has class $c$ .

Proof. If $c=0$ , take $a=b=g=1$ . If $c=1$ , choose $a\ne1$ and take $b=g=1$ . Suppose $c\ge2$ and put $r=c-2$ . Since $\gamma_c(G)\ne1$ , Lemma 2 gives $a,b,g_1,\ldots,g_r\in G$ for which

\tag{13} \delta_{g_1}\cdots\delta_{g_r}([a,b])\ne0 \quad\text{in }A.

If $r=0$ , then $[a,b]\ne1$ , and $\langle a,b\rangle$ has class two; take $g=1$ .

Now let $r\ge1$ . We have $\gamma_{r+3}(G)=\gamma_{c+1}(G)=1$ , so Lemma 3 applies to $x=[a,b]$ . If $\delta_g^r([a,b])=0$ for every $g\in G$ , then $T_x$ vanishes on every diagonal of $B^r$ , because every element of $B$ has a representative in $G$ . Equations (9) and (12), with $r<p$ , would give $T_x=0$ , contradicting (13). Hence there exists $g\in G$ with

0\ne\delta_g^r([a,b])\in \gamma_{r+2}(\langle a,b,g\rangle) =\gamma_c(\langle a,b,g\rangle).

The membership assertion follows by forming the basic commutator inside $\langle a,b,g\rangle$ and then taking its $r$ successive commutators with $g$ . Thus this subgroup has class at least $c$ . The inclusion $\gamma_j(H)\le\gamma_j(G)$ for $H\le G$ , proved inductively from (1), gives the reverse bound. ◻

The finite class and its examples

We first record how nilpotency identities behave under the finite group operations used below. If $H\le G$ and $f:G\twoheadrightarrow Q$ is a surjection, induction on $j$ gives

\tag{15} \gamma_j(H)\le\gamma_j(G),\qquad f(\gamma_j(G))=\gamma_j(Q).

The first step uses subgroup inclusion; the second uses $f([x,y])=[f(x),f(y)]$ and surjectivity to lift both commutator arguments. In particular, subgroups and quotients do not increase nilpotency class. The same argument with the derived series shows that they preserve metabelianity.

Lemma 6. Let $F$ be finite and nilpotent. If every Sylow subgroup of $F$ has class at most $c$ , then $\gamma_{c+1}(F)=1$ .

Proof. Every proper subgroup $H$ of a nilpotent group is properly contained in its normalizer. Indeed, in the upper central series choose the least $i$ for which $Z_i(F)\not\le H$ . Then $Z_{i-1}(F)\le H$ , and any $z\in Z_i(F)\setminus H$ normalizes $H$ because $[z,H]\le Z_{i-1}(F)$ .

Let $P$ be a Sylow subgroup and set $N=N_F(P)$ . If $N<F$ , choose $x\in N_F(N)\setminus N$ . The groups $P$ and $xPx^{-1}$ are Sylow subgroups of $N$ , so some $n\in N$ satisfies $xPx^{-1}=nPn^{-1}$ . It follows that $n^{-1}x\in N_F(P)=N$ , contrary to $x\notin N$ . Therefore every Sylow subgroup of $F$ is normal.

Distinct Sylow subgroups have trivial intersection and commute: their commutator lies in their intersection. Their orders multiply to $|F|$ , so $F$ is their direct product. Under every coordinate projection, $\gamma_{c+1}(F)$ maps into the trivial subgroup $\gamma_{c+1}(P)$ . The projections separate elements, proving the assertion. ◻

Proposition 7. The class $\mathcal C$ is closed under subgroups, quotients and finite direct products.

Proof. Subgroups inherit all conditions in (2) by (15). For quotient closure, let $f:F\twoheadrightarrow Q$ , with $F\in\mathcal C$ . The group $Q$ is nilpotent and metabelian. A surjection of finite groups maps Sylow $p$ -subgroups onto Sylow $p$ -subgroups: if $K=\ker f$ and $P\in\operatorname{Syl}_p(F)$ , then $P\cap K$ is a Sylow subgroup of the normal subgroup $K$ , and

|f(P)|=\frac{|P|}{|P\cap K|} =\frac{|F|_p}{|K|_p}=|Q|_p.

Every Sylow subgroup of $Q$ is a conjugate of $f(P)$ , hence is also the image of a Sylow subgroup of $F$ , by lifting a conjugating element through $f$ . It has class at most $p-1$ by (15). Each $p$ -subgroup of $Q$ lies in one of these Sylow subgroups, so $Q\in\mathcal C$ .

Let $F=\prod_{i\in I}F_i$ with $I$ finite and $F_i\in\mathcal C$ . If $c$ bounds the classes of the factors, the coordinate images of $\gamma_{c+1}(F)$ are trivial; hence $F$ is nilpotent. Its second derived subgroup is trivial by the same coordinate argument. For a $p$ -subgroup $H\le F$ , each coordinate image $H_i\le F_i$ is a $p$ -group, so $\gamma_p(H_i)=1$ . Consequently every coordinate of $\gamma_p(H)$ is trivial, giving $\gamma_p(H)=1$ . The empty product is the trivial group, which satisfies (2). ◻

We now give finite members of unbounded class. The construction is the unipotent construction over $\mathbb F_p$ occurring in Weichsel’s basic groups (3, Definition 2.1); we use its cyclic subgroup of units directly.

Proposition 8. Let $p$ be prime and $1\le n<p$ . Put

\tag{17} R=\mathbb F_p[t]/(t^n),\qquad U=\langle1+t\rangle\le R^\times,\qquad E_{p,n}=R^+\rtimes U,

where units act by multiplication. Then $E_{p,n}\in\mathcal C$ and $E_{p,n}$ has class exactly $n$ .

Proof. The residue classes $1,t,\ldots,t^{n-1}$ form a basis of $R$ over $\mathbb F_p$ , so $R$ is finite,

\tag{18} t^n=0,\qquad t^j\ne0\quad(0\le j<n).

Characteristic $p$ and $n\le p$ give $(1+t)^p=1+t^p=1$ . Thus $1+t$ is a unit with inverse $(1+t)^{p-1}$ , and every element of $U$ has order dividing $p$ .

Represent an element of $E_{p,n}$ by $(a,v)\in R\times U$ , with

\tag{19} (a,v)(b,w)=(a+vb,vw),\qquad (a,v)^{-1}=(-v^{-1}a,v^{-1}).

The group is finite. The $p$ -th power of any element has linear part one, so it is a translation. Translations have exponent $p$ ; therefore every element has order dividing $p^2$ , and $E_{p,n}$ is a $p$ -group. The translations form an abelian normal subgroup with abelian quotient $U$ , so $E_{p,n}$ is metabelian.

Write $\tau(a)=(a,1)$ and $s=(0,1+t)$ . Direct multiplication in (19) gives

\begin{align*} [(a,v),(b,w)]&=\tau\bigl((1-w)a+(v-1)b\bigr),\tag{20a}\\ [s,\tau(a)]&=\tau(ta).\tag{20b} \end{align*}

Every $v\in U$ is congruent to one modulo $tR$ . This follows from the identities

vw-1=(v-1)w+(w-1),\qquad v^{-1}-1=-v^{-1}(v-1),

which show that such units form a subgroup containing $1+t$ . For $j\ge0$ , let $T_j=\{\tau(a):a\in t^jR\}$ . Equation (20a) yields

\tag{22} \gamma_2(E_{p,n})\le T_1,\qquad [T_j,E_{p,n}]\le T_{j+1},\qquad \gamma_{j+1}(E_{p,n})\le T_j\quad(j\ge1).

In particular, $\gamma_{n+1}(E_{p,n})\le T_n=1$ .

Define $w_0=\tau(1)$ and $w_{j+1}=[s,w_j]$ . Equation (20b) proves inductively that

\tag{23} w_j=\tau(t^j)\in\gamma_{j+1}(E_{p,n}).

By (18), $w_{n-1}\ne1$ . Hence the class is exactly $n$ .

Every $p$ -subgroup has class at most $n<p$ . If $q\ne p$ , a $q$ -subgroup of the finite $p$ -group $E_{p,n}$ is trivial, since its order divides a power of $p$ and is a power of $q$ . All conditions in (2) follow. ◻

Passage to profinite groups

For a profinite group $G$ , open normal subgroups form a neighbourhood basis at one. The quotients $G/N$ , with $N$ open and normal, are finite and separate points:

\tag{24} \bigcap_{N\trianglelefteq_{\!o}G}N=1.

These facts follow from the defining representation of a profinite group as an inverse limit of finite groups: kernels of finite families of coordinate maps provide the required basis and separate distinct elements.

Proposition 9. If a nonempty class $\mathcal D$ of finite groups is closed under subgroups, quotients and finite products, then $\operatorname{Pro}(\mathcal D)$ is a profinite variety. Every finite group in $\mathcal D$ , with its discrete topology, belongs to $\operatorname{Pro}(\mathcal D)$ .

Proof. The last assertion follows from quotient closure and proves nonemptiness. If $G\in\operatorname{Pro}(\mathcal D)$ and $f:G\twoheadrightarrow H$ is continuous, compose $f$ with any finite continuous quotient map from $H$ . The composite shows that this finite quotient belongs to $\mathcal D$ .

Let $H\le G$ be closed and let $f:H\twoheadrightarrow Q$ be a finite continuous quotient. Its kernel is open in $H$ . By the subspace topology there is an open neighbourhood $O$ of one in $G$ such that $H\cap O\subseteq\ker f$ . Choose $N\trianglelefteq_{\!o}G$ with $N\subseteq O$ . The map $H\to G/N$ has kernel $H\cap N\subseteq\ker f$ , so $f$ induces a surjection

\tag{25} \operatorname{im}(H\longrightarrow G/N)\twoheadrightarrow Q.

The group $G/N$ belongs to $\mathcal D$ . Its displayed subgroup, and then its quotient $Q$ , belong to $\mathcal D$ . Thus $H\in\operatorname{Pro}(\mathcal D)$ .

Let $G=\prod_{i\in I}G_i$ , where each $G_i\in\operatorname{Pro}(\mathcal D)$ , and let $f:G\twoheadrightarrow Q$ be finite and continuous. The open kernel contains a basic neighbourhood of one, supported on a finite set $J\subseteq I$ . Choose $N_j\trianglelefteq_{\!o}G_j$ in the specified neighbourhood for each $j\in J$ . Then

\tag{26} \left(\prod_{j\in J}N_j\right) \times\left(\prod_{i\notin J}G_i\right) \le\ker f.

The natural map $G\to\prod_{j\in J}G_j/N_j$ is surjective: choose representatives in the finitely many selected coordinates and put one elsewhere. Hence $f$ factors as a surjection from this finite product to $Q$ . Each $G_j/N_j$ lies in $\mathcal D$ ; finite-product and quotient closure give $Q\in\mathcal D$ . This proves closure under arbitrary products. ◻

Lemma 10. If every finite continuous quotient of a profinite group $G$ satisfies $\gamma_{c+1}=1$ , then $\gamma_{c+1}(G)=1$ .

Proof. For every $N\trianglelefteq_{\!o}G$ , the quotient map sends $\gamma_{c+1}(G)$ into $\gamma_{c+1}(G/N)=1$ . Thus $\gamma_{c+1}(G)$ lies in the intersection in (24), and is trivial. ◻

Lemma 11. Let $\mathcal W$ be a locally nilpotent profinite variety. For each $d\ge1$ , there is an integer $c_d\ge0$ such that every member of $\mathcal W$ topologically generated by $d$ elements has class at most $c_d$ .

Proof. Suppose no bound exists for a fixed $d$ . For each $m\ge0$ , choose $G_m\in\mathcal W$ and a topological generating tuple $g_{1,m},\ldots,g_{d,m}$ such that $\gamma_{m+1}(G_m)\ne1$ . In $P=\prod_{m\ge0}G_m$ , form the diagonal elements

\tag{27} a_j=(g_{j,m})_{m\ge0}\quad(1\le j\le d),\qquad D=\overline{\langle a_1,\ldots,a_d\rangle}\le P.

Product and closed-subgroup closure give $D\in\mathcal W$ . The tuple $a_1,\ldots,a_d$ , regarded as elements of $D$ , generates $D$ topologically: its closure in the subspace $D$ is its ambient closure intersected with $D$ , hence is $D$ itself.

The coordinate projection $D\to G_m$ is continuous. Its image is compact, hence closed in the Hausdorff group $G_m$ , and contains every $g_{j,m}$ . It is therefore surjective. Local nilpotence gives $\gamma_{c+1}(D)=1$ for some $c$ . By (15), the surjection $D\to G_c$ gives $\gamma_{c+1}(G_c)=1$ , contradicting the choice of $G_c$ . ◻

Proof of the counterexample

Proof of Theorem 1. Proposition 7 and Proposition 9 show that $\mathcal V$ is a profinite variety. It is nonempty by Proposition 8. Its finite continuous quotients are nilpotent by (2); hence every member of $\mathcal V$ is pronilpotent.

For each $m\ge1$ , choose a prime $p_m>m$ and put $F_m=E_{p_m,m}$ . Such primes exist by the infinitude of primes. Each $F_m$ belongs to $\mathcal C$ and has class $m$ . Consequently

\tag{28} P=\prod_{m\ge1}F_m\in\mathcal V.

If $P$ had class at most $c$ , its quotient $F_{c+1}$ would have class at most $c$ , contradicting Proposition 8. Therefore $\mathcal V$ is not nilpotent.

Let $\mathcal W\le\mathcal V$ be locally nilpotent, and let $c$ be its rank-three bound from Lemma 11. Take a finite member $F\in\mathcal W$ . Its identity map is a finite continuous quotient, so $F\in\mathcal C$ . For each prime $p$ and $S\in\operatorname{Syl}_p(F)$ , the subgroup $S$ is closed and belongs to $\mathcal W$ ; it is metabelian and has class less than $p$ . Proposition 5 gives a subgroup $H=\langle a,b,g\rangle\le S$ with

\operatorname{cl}(S)=\operatorname{cl}(H).

Since $F$ is finite, $H$ is closed and belongs to $\mathcal W$ . Its three generators also generate it topologically, and thus $\operatorname{cl}(H)\le c$ . Every Sylow subgroup of $F$ has class at most $c$ . Lemma 6 gives

\tag{30} \gamma_{c+1}(F)=1\qquad (F\in\mathcal W,\ F\text{ finite}).

Finally, let $G\in\mathcal W$ . Every finite continuous quotient of $G$ remains in $\mathcal W$ , so it satisfies (30). Lemma 10 gives $\gamma_{c+1}(G)=1$ . Thus all members of $\mathcal W$ satisfy the same nilpotency identity, and $\mathcal W$ is nilpotent. ◻

References

Preprint · Lean (GitHub)

N. D. Gupta and M. F. Newman, On metabelian groups, J. Austral. Math. Soc. 6 (1966), 362–368. doi:10.1017/S1446788700004316.

E. I. Khukhro and V. D. Mazurov (eds.), Unsolved Problems in Group Theory: The Kourovka Notebook, 21st ed., 2026, Problem 7.38(a), p. 24. arXiv:1401.0300.

P. M. Weichsel, On metabelian $p$ -groups, J. Austral. Math. Soc. 7 (1967), 55–63. doi:10.1017/S1446788700005097. Corrigendum, 8 (1968), 128. doi:10.1017/S1446788700004675.