Two formula values, an infinite subgroup

Nilradical

A parameter-free first-order formula $\varphi(x)$ is concise in a group $G$ if finiteness of its value set $G_\varphi$ implies finiteness of $\langle G_\varphi\rangle$ . Conte and Petschick ask whether every such formula is concise in residually finite groups. Their question became Notebook Problem 21.106. Conciseness of first-order formulae, Question 1.

Theorem. A positive, parameter-free formula defines exactly two elements in the integral Heisenberg group, and those elements generate an infinite cyclic subgroup. Thus the formula is not concise in residually finite groups.

This answers the formula question. The separate word-conciseness question, Problem 21.105, is outside its scope.

The formula

Use $[a,b]=a^{-1}b^{-1}ab$ and put

\begin{aligned} \varphi(x):\quad \exists a,b\;\forall g,h\;\exists u,v\;\bigl(&x=[a,b]\land[u,a]=1\land[v,b]=1\\ &{}\land[u,b]=[g,h]\land[a,v]=[g,h]\bigr). \end{aligned}

Every variable except $x$ is bound. The formula uses only multiplication, inversion, the identity and equality; its matrix is a conjunction of equations. In the Heisenberg group, the commuting conditions restrict the witnesses so that $[a,b]$ must be a primitive central element.

A determinant calculation

Write the integral Heisenberg group $H$ as triples with multiplication

(r,s,t)(r',s',t')=(r+r',s+s',t+t'+rs').

The identity is $(0,0,0)$ and the inverse of $(r,s,t)$ is $(-r,-s,rs-t)$ . Put $c(n)=(0,0,n)$ and, for group elements $a,b$ , let $D(a,b)$ be the determinant of their first two coordinates. Direct calculation gives

[a,b]=c(D(a,b)),\qquad c(m)c(n)=c(m+n).

Suppose $x$ satisfies $\varphi$ , with initial witnesses $a,b$ . Apply its universal condition to $g=(1,0,0)$ and $h=(0,1,0)$ , whose commutator is $c(1)$ . The resulting $u,v$ satisfy

D(u,a)=D(v,b)=0,\qquad D(u,b)=D(a,v)=1.

The two-dimensional determinant identity

D(u,a)D(v,b)-D(u,v)D(a,b)+D(u,b)D(a,v)=0

therefore gives $D(u,v)D(a,b)=1$ . Both factors are integers, so $D(a,b)=\pm1$ . Consequently $x=c(1)$ or $c(-1)$ .

Conversely, choose $\varepsilon\in\{1,-1\}$ and take $a=(1,0,0)$ , $b=(0,\varepsilon,0)$ . For arbitrary $g,h$ , write $d=D(g,h)$ and choose

u=(\varepsilon d,0,0),\qquad v=(0,d,0).

The required zero determinants vanish because the corresponding vectors are parallel, and

D(u,b)=\varepsilon^2d=d=D(a,v), \qquad D(a,b)=\varepsilon.

These witnesses satisfy the formula. Thus its entire value set is

H_\varphi=\{c(1),c(-1)\}.

The two elements are distinct and generate $\{c(n):n\in\mathbb Z\}\cong\mathbb Z$ .

Finite quotients and the scope of the answer

Reduce all three coordinates modulo an integer $m\ge2$ . The multiplication formula makes this a homomorphism to the finite Heisenberg group over $\mathbb Z/m\mathbb Z$ . If a triple is nonidentity, choose a nonzero coordinate $n$ and take $m=|n|+1$ . That coordinate remains nonzero after reduction. Every nonidentity element is therefore separated in a finite quotient, proving residual finiteness and completing the counterexample.

Conte and Petschick prove that existential formulae are concise in torsion-free class-two nilpotent groups: Theorem 1.2 in their arXiv version and Theorem 1.1 in the published paper, Monatshefte für Mathematik 209 (2026), 215–240. The universal quantifiers in the displayed formula are essential to its distinction from that positive result.

The question and that earlier theorem are credited to Conte and Petschick. The Heisenberg group and determinant identity are standard mathematics; Nilradical supplies the displayed formula, its exact value-set argument and the counterexample formalization. The complete formalization uses Lean and mathlib and includes the parameter-free syntax as well as its interpretation. The formula's quantifier complexity offers a concrete direction for further study.