Faithful actions of countable groups on five-generated soluble groups of derived length at most four

Problem

Does every countable group embed in the automorphism group of a finitely generated soluble group?

|C|\le\aleph_0\quad\stackrel{?}{\Longrightarrow}\quad \exists S\text{ finitely generated and soluble},\quad C\hookrightarrow\operatorname{Aut}(S).

The construction below gives the quantitative bounds $d(S)\le5$ and $S^{(4)}=1$ .

Faithful actions with a uniform generator bound

Write $S^{(0)}=S$ and $S^{(n+1)}=[S^{(n)},S^{(n)}]$ . The commutator convention is $[x,y]=xyx^{-1}y^{-1}$ , and products of operators act from right to left.

Theorem 1. For every countable group $C$ , there are a group $S$ , elements $s_1,\ldots,s_5\in S$ , and an injective homomorphism $\iota\colon C\to\operatorname{Aut}(S)$ such that

S=\langle s_1,\ldots,s_5\rangle,\qquad S^{(4)}=1.

Here countable includes finite groups. Theorem 1 answers Problem 5.39 of the Kourovka Notebook (1), posed by P. M. Neumann. Cyclic modules isomorphic to proper direct powers occur in the representation theory of infinite soluble groups; see Neumann (2).

We construct $S$ as $M\rtimes(G\times\mathbb Z)$ , where $M$ is abelian and $G$ is three-generated with $G^{(3)}=1$ . A three-generated overgroup $D$ of $C$ acts faithfully on $M$ and centralizes the action of $G\times\mathbb Z$ ; its action therefore extends to $S$ .

A three-generated overgroup

For groups $H,Q$ , use the unrestricted regular wreath product $H\wr Q=H^Q\rtimes Q$ , where $(q\cdot f)(x)=f(q^{-1}x)$ . Write $\delta_q(h)$ for the base function equal to $h$ at $q$ and to $1$ elsewhere.

Lemma 2. Every countable group embeds in a group generated by three elements.

Proof. Choose a surjection $\mathbb N\to C$ , $n\mapsto c_n$ . In $H=C\wr\mathbb Z$ , let $z$ denote the unit shift and define

h_c(k)=\begin{cases}c&k\ge0,\\1&k<0.\end{cases} \qquad [h_c,z]=\delta_0(c).

The commutator identity follows by evaluating $h_c(k)h_c(k-1)^{-1}$ at every integer $k$ .

In $\mathbb Z^2$ , put $p_n=(n,n^2)$ . Nonzero differences between these points are unique. Indeed, if $p_i-p_j=p_k-p_l$ and $i\ne j$ , then

i-j=k-l\ne0,\qquad (i-j)(i+j)=(k-l)(k+l),

so $i+j=k+l$ , and hence $i=k$ , $j=l$ .

Define $f\in H^{\mathbb Z^2}$ by

f(p_0)=z,\qquad f(p_{n+1})=h_{c_n},\qquad f(x)=1\quad\bigl(x\notin\{p_n:n\ge0\}\bigr).

Let $x,y$ be the two coordinate shifts in $H\wr\mathbb Z^2$ and set $D=\langle f,x,y\rangle$ . For $n\ge0$ , the supports of $f$ and its translate by $p_{n+1}$ can meet only at $p_{n+1}$ . This follows from uniqueness of the difference $p_{n+1}-p_0$ . At this point their values are $h_{c_n}$ and $z$ . Thus

[f,\,p_{n+1}fp_{n+1}^{-1}] =\delta_{p_{n+1}}\bigl(\delta_0(c_n)\bigr).

Here a lattice point denotes the corresponding product of powers of $x,y$ . Conjugation by its inverse moves the displayed base element to the origin. Consequently $D$ contains $\delta_0(\delta_0(c))$ for every $c\in C$ . This double-coordinate map is a homomorphism and is injective by evaluation at the two origins. ◻

A cyclic module isomorphic to eight copies of itself

Let $K=\mathbb F_3$ and $W=\bigoplus_{q\in\mathbb Q}K\varepsilon_q$ . Define invertible $K$ -linear operators by

\tag{6} a\varepsilon_q= \begin{cases}-\varepsilon_q&q\in\mathbb Z,\\ \varepsilon_q&q\notin\mathbb Z,\end{cases} \qquad b\varepsilon_q=\varepsilon_{q+1},\qquad d\varepsilon_q=\varepsilon_{2q}.

More precisely, let $G$ be the subgroup generated by the corresponding three elements in the signed affine group

(K^\times)^{\mathbb Q}\rtimes\operatorname{Aff}(\mathbb Q).

The sign group is abelian. The affine group has abelian translation kernel and abelian linear quotient $\mathbb Q^\times$ , so

\tag{8} G=\langle a,b,d\rangle,\qquad G^{(3)}=1.

Use the same letters for their represented operators. Let $R$ be the unital subring of $\operatorname{End}_K(W)$ generated by the image of $G$ . For $g\in G$ , the operator representing $g^{-1}$ also lies in $R$ , so the representation takes values in $R^\times$ .

Lemma 3. Put $e=a-1\in R$ and, for $0\le i<8$ , put

u_i=b^i d^3e,\qquad v_i=e d^{-3}b^{-i}.

Then $e^2=e\ne0$ and

\tag{10} u_i e=u_i,\qquad v_i e=v_i,\qquad v_i u_j=\delta_{ij}e,\qquad \sum_{i=0}^7u_i v_i=e.

Proof. In characteristic three, $-1-1=1$ , so $e$ is the projection onto the integer coordinates of $W$ . In particular $e\varepsilon_0= \varepsilon_0\ne0$ . On basis vectors,

\begin{align*} u_i\varepsilon_q&= \begin{cases}\varepsilon_{8q+i}&q\in\mathbb Z,\\0&q\notin\mathbb Z,\end{cases}\\ v_i\varepsilon_q&= \begin{cases}\varepsilon_{(q-i)/8}&(q-i)/8\in\mathbb Z,\\0&\text{otherwise}.\end{cases} \end{align*}

The first two relations follow, since $(q-i)/8\in\mathbb Z$ implies $q\in\mathbb Z$ . For integer $n$ and $0\le i,j<8$ , $(8n+j-i)/8$ is integral exactly when $i=j$ . This proves the third relation. Each integer has a unique expression $8n+i$ with $0\le i<8$ , and no noninteger has such an expression. Hence the operators $u_i v_i$ are the eight disjoint residue-class projections whose sum is $e$ . ◻

Proposition 4. The left $R$ -module $V=Re$ is nonzero and cyclic, and

\tag{12} \Phi\colon V\longrightarrow V^8,\quad v\longmapsto(vu_i)_{i=0}^7

is an $R$ -linear isomorphism, with inverse $\Psi((w_i))=\sum_i w_i v_i$ .

Proof. The left ideal is $Re=\{v\in R:ve=v\}$ , with nonzero generator $e$ . Since $u_i e=u_i$ and $v_i e=v_i$ , right multiplication by $u_i$ or $v_i$ maps $Re$ into $Re$ . Right multiplication commutes with the left $R$ -action, so the maps are $R$ -linear. The remaining relations give

\Psi\Phi(v)=v\sum_i u_i v_i=ve=v,\qquad (\Phi\Psi((w_i)))_j=\sum_i w_i v_i u_j=w_j e=w_j.

◻

Transport and a faithful quotient

Fix $D=\langle r_1,r_2,r_3\rangle$ containing $C$ , as in Lemma 2. Let

(\sigma_0,\ldots,\sigma_7) =(1,r_1,r_1^{-1},r_2,r_2^{-1},r_3,r_3^{-1},1).

These are eight labelled entries; coincidences among their values are allowed. Write $\pi_i\colon V\to V$ for the $i$ th component of $\Phi$ and $j_i\colon V\to V$ for the inverse image under $\Phi$ of the $i$ th coordinate inclusion. Thus $\pi_k j_i=\delta_{ki}1_V$ .

On the algebraic direct sum $A=\bigoplus_{q\in D}V$ , define

\tag{15} T(\delta_q(v))=\sum_{i=0}^7\delta_{q\sigma_i}(\pi_i(v)), \qquad L_c\delta_q(v)=\delta_{cq}(v),\qquad \xi=\delta_1(e).

Both maps are $R$ -linear, and $T L_c=L_c T$ because $(cq)\sigma_i=c(q\sigma_i)$ .

Lemma 5. The map $T$ is surjective. The only $R$ -submodule of $A$ containing $\xi$ and preserved by $T$ is $A$ . There is $r\in R$ such that

\tag{16} rT\xi=\xi.

Proof. The component relations give, even when some labels coincide,

\tag{17} T\bigl(\delta_q(j_i(v))\bigr)=\delta_{q\sigma_i}(v).

Taking $i=0$ proves surjectivity on every summand and hence on $A$ . If a submodule $N$ contains $\xi$ , cyclicity of $V$ gives $\delta_1(V)\subseteq N$ . Equation (17) and $T$ -invariance then give $\delta_{q\sigma_i}(V)\subseteq N$ whenever $\delta_q(V)\subseteq N$ . Every element of $D$ is a word in the labels, so $N$ contains every summand. Finally choose $r\in R$ with $re=j_0(e)$ . By $R$ -linearity of $T$ ,

rT\xi=T(r\xi)=T\bigl(\delta_1(j_0(e))\bigr)=\xi.

◻

Put

\tag{19} N=\bigcup_{n\ge0}\ker T^n,\qquad M=A/N,\qquad \mu=\xi+N.

The kernels form an increasing chain of $R$ -submodules, so $N$ is a submodule. Moreover $Tx\in N$ if and only if $x\in N$ .

Proposition 6. The map $T$ induces an $R$ -linear automorphism $t$ of $M$ . Left translation induces a faithful homomorphism $D\to\operatorname{Aut}_R(M)$ commuting with $t$ , and $\mu\ne0$ . Every $t$ -invariant $R$ -submodule of $M$ containing $\mu$ equals $M$ .

Proof. The equivalence $Tx\in N\Longleftrightarrow x\in N$ gives injectivity of the induced map; surjectivity follows from that of $T$ . Each $L_c$ and its inverse commute with $T$ , and therefore preserve $N$ , giving an action by $R$ -linear automorphisms of $M$ . Its commutation with $t$ descends from (15).

The recovery identity iterates to

\tag{20} r^nT^n\xi=\xi\qquad(n\ge0).

This follows inductively from $R$ -linearity of $T$ ; commutativity of $R$ is not required. Since $\xi\ne0$ , no $T^n\xi$ is zero, and therefore $\mu\ne0$ .

Suppose $L_c\mu=\mu$ . Then $T^n(L_c\xi-\xi)=0$ for some $n$ . The map $L_c$ is $R$ -linear and commutes with $T$ , so multiplying this equality by $r^n$ and using (20) gives

L_c\xi-\xi=0.

For $c\ne1$ , evaluation at $c$ would instead give $e\ne0$ . Thus $c=1$ , proving faithfulness. Finally the preimage in $A$ of a $t$ -invariant submodule containing $\mu$ contains $\xi$ and is $T$ -invariant. Lemma 5 makes that preimage all of $A$ . ◻

The soluble group and its automorphisms

The group $G$ acts on $M$ through the units of $R$ , and $t$ commutes with that action by $R$ -linearity. Thus $G\times\mathbb Z$ acts by

(g,n)\cdot m=\rho(g)t^n(m),

where $\rho\colon G\to R^\times$ is the given representation. Form the semidirect product

\tag{23} S=M\rtimes(G\times\mathbb Z).

We identify $\mu\in M$ , $a,b,d\in G$ and the unit generator $t\in\mathbb Z$ with their natural images in $S$ .

Proposition 7. The group $S$ satisfies

S=\langle\mu,a,b,d,t\rangle,\qquad S^{(4)}=1.

The action of $D$ on $M$ extends faithfully to automorphisms of $S$ that fix $G\times\mathbb Z$ pointwise.

Proof. Let $H=\langle\mu,a,b,d,t\rangle\le S$ . It contains $G\times\mathbb Z$ . Its intersection with the additive normal subgroup $M$ contains $\mu$ and is closed under conjugation by $G$ and $t$ . Thus $H\cap M$ is closed under the operators $\rho(g)$ and under $t$ . Since it is an additive subgroup, it is also closed under integer multiples, sums, differences and compositions of the operators $\rho(g)$ . These generate $R$ , so $H\cap M$ is an $R$ -submodule. Proposition 6 gives $H\cap M=M$ , and hence $H=S$ .

By (8), $(G\times\mathbb Z)^{(3)}=1$ . The image of $S^{(3)}$ in $S/M$ is therefore trivial. Thus $S^{(3)}\le M$ , and $S^{(4)}=1$ since $M$ is abelian.

For $c\in D$ , define

\alpha_c\bigl(m,(g,n)\bigr)=\bigl(L_c m,(g,n)\bigr).

The $R$ -linearity of $L_c$ and $L_ct=tL_c$ give $L_c((g,n)\cdot m)=(g,n)\cdot L_c(m)$ , so $\alpha_c$ preserves the semidirect-product multiplication. Its inverse is $\alpha_{c^{-1}}$ , and $\alpha_c\alpha_d=\alpha_{cd}$ . If $\alpha_c=1$ , its restriction to $M$ is the identity; Proposition 6 then gives $c=1$ . ◻

Proof of Theorem 1. Compose the embedding $C\hookrightarrow D$ from Lemma 2 with the faithful homomorphism $D\to\operatorname{Aut}(S)$ of Proposition 7. The generating set and the equality $S^{(4)}=1$ also follow from Proposition 7. ◻

References

Preprint · Lean (GitHub)

E. I. Khukhro and V. D. Mazurov (eds.), Unsolved Problems in Group Theory: The Kourovka Notebook, 21st ed., 2026, Problem 5.39. arXiv:1401.0300.

P. M. Neumann, Pathology in the representation theory of infinite soluble groups, in A. C. Kim and B. H. Neumann (eds.), Groups–Korea 1988, Lecture Notes in Mathematics 1398, Springer, Berlin, 1989, 124–139. doi:10.1007/BFb0086249.