Finite-generation obstructions for classes of group isotopes

Problem

If a variety, quasivariety, or pseudovariety $\mathcal M$ is generated by one finite group, is its isotope closure $I\mathcal M$ generated by one finite quasigroup?

\mathcal M=\mathsf C(G),\quad |G|<\infty \quad\stackrel{?}{\Longrightarrow}\quad I\mathcal M=\mathsf C(A),\quad |A|<\infty.

The proof distinguishes modern pseudovarieties of finite algebras from the original convention using disjunctive identities.

Setup and result

A quasigroup is a nonempty algebra $(A;\cdot,\backslash,/)$ satisfying

\tag{1} x\backslash(x\cdot y)=y,\quad x\cdot(x\backslash y)=y,\quad (x\cdot y)/y=x,\quad (x/y)\cdot y=x.

An isotopy from a quasigroup $A$ to a group $H$ consists of bijections $\alpha,\beta,\gamma:A\to H$ such that

\alpha(x)\beta(y)=\gamma(x\cdot y)\qquad(x,y\in A).

For a class $\mathcal M$ of groups, let $I\mathcal M$ be the class of all quasigroups isotopic to members of $\mathcal M$ .

Problem 9.4 of the Kourovka Notebook asks whether $I\mathcal M$ is generated by a single finite quasigroup whenever $\mathcal M$ is generated by a single finite group (3). The question is posed for varieties, quasivarieties and pseudovarieties. The last term requires care: Gvaramiya’s original paper (2, p. 1048) uses classes defined by disjunctive identities, rather than pseudovarieties of finite algebras. We treat both conventions.

Write $\mathsf{V}(A)$ and $\mathsf{Q}(A)$ for the classes defined by all identities and all quasiidentities, respectively, valid in $A$ . A quasiidentity has finitely many equational premises and one equational conclusion. For finite $A$ , write $\mathsf{PV}(A)$ for its closure under homomorphic images, subalgebras and finite direct products. All homomorphisms preserve the three operations in (1). For a group these operations are

x\cdot y=xy,\qquad x\backslash y=x^{-1}y,\qquad x/y=xy^{-1}.

When the generating algebra is a group, the corresponding group class is understood within groups.

Theorem 1. Let $G$ be a nontrivial finite group. For every finite quasigroup $A$ of order $m$ , there is an isotope $B$ of $G^{m!+1}$ such that

B\notin\mathsf{V}(A),\qquad B\notin\mathsf{Q}(A),\qquad B\notin\mathsf{PV}(A).

Consequently, for $\mathsf C\in\{\mathsf{V},\mathsf{Q},\mathsf{PV}\}$ , the class $I\mathsf C(G)$ is not generated under $\mathsf C$ by any single finite quasigroup.

A disjunctive identity is a universally quantified formula

\tag{5} \bigvee_{j=1}^{r}s_j(\mathbf x)=t_j(\mathbf x),

where the $s_j,t_j$ are terms. Let $\mathsf{D}(A)$ be the class of algebras satisfying every such formula valid in $A$ .

Theorem 2. Let $C_3$ be the cyclic group of order three. There is no finite quasigroup $A$ such that

\mathsf{D}(A)=I\mathsf{D}(C_3).

Falconer (1, Theorem 5.2) constructs an isotope of a countable elementary abelian $p$ -group with an infinite one-generated subquasigroup. Her result already yields the variety and quasivariety obstructions in Theorem 1. We do not claim those conclusions as new. The proof below replaces the infinite construction by a cyclic coordinate shift on a finite direct power and applies equally to $\mathsf{PV}$ . Theorem 2 uses a separate obstruction on quasigroups of order at most three.

A finite separating isotope

Define constant-free terms recursively by

\tag{7} t_0(x,y)=x,\qquad t_{n+1}(x,y)=t_n(x,y)\cdot y.

If $R_y(x)=x\cdot y$ , then $t_n(x,y)=R_y^n(x)$ .

Lemma 3. Every quasigroup $A$ of finite order $m$ satisfies

\tag{8} t_{m!}(x,y)=x.

For a fixed $n$ , the identity $t_n(x,y)=x$ is preserved under homomorphic images, subalgebras and finite direct products.

Proof. The last two identities in (1) make $R_y$ a permutation, with inverse $z\mapsto z/y$ . Every permutation of an $m$ -element set has its $m!$ -th power equal to the identity, which proves (8).

If $f:A\to B$ is a homomorphism, induction on $n$ gives

f(t_n^A(x,y))=t_n^B(f(x),f(y)).

For surjective $f$ , lifting $x,y\in B$ proves preservation under images. For injective $f$ , cancellation proves preservation under subalgebras. In a direct product, the identity holds in each coordinate. ◻

Lemma 4. Let $G\neq1$ be a group and $n\geq1$ . There is an isotope $B_n$ of $G^{n+1}$ which does not satisfy $t_n(x,y)=x$ .

Proof. Put $d=n+1$ and $H=G^d$ , with coordinates indexed by $\mathbb Z/d\mathbb Z$ . Let

(\sigma x)_i=x_{i+1}.

On the set $H$ , define

\tag{11} x*y=\sigma(x)y,\qquad x\backslash_*z=\sigma(x)^{-1}z,\qquad z/_*y=\sigma^{-1}(zy^{-1}).

These operations satisfy (1), and $(\sigma,1,1)$ is an isotopy from $B_n=(H;*,\backslash_*,/_*)$ to $H$ .

Let $e$ be the identity of $H$ , choose $g\neq1$ in $G$ , and put

a_0=g,\qquad a_i=1\quad(i\neq0).

Since $x*e=\sigma(x)$ , induction gives

t_n^{B_n}(a,e)=\sigma^n(a).

Its zeroth coordinate is $a_n=1$ , whereas $a_0=g\neq1$ . Thus $t_n^{B_n}(a,e)\neq a$ . ◻

Proof of Theorem 1. Take $n=m!$ . Lemma 3 gives an identity of $A$ , and hence of every member of $\mathsf{V}(A)$ and $\mathsf{Q}(A)$ . The preservation assertion gives the same identity throughout $\mathsf{PV}(A)$ . The isotope $B_n$ in Lemma 4 violates it.

Each of $\mathsf{V}(G)$ , $\mathsf{Q}(G)$ and $\mathsf{PV}(G)$ contains the finite direct power $G^{m!+1}$ . Thus $B_n\in I\mathsf C(G)$ but $B_n\notin\mathsf C(A)$ for all three operators. The argument applies to every finite proposed generator $A$ . ◻

Disjunctive identities

The argument in this section uses only a cardinality bound on the possible generator.

Lemma 5. Every nonempty quasigroup of order at most three satisfies at least one of the following identities:

\begin{align*} (x\cdot y)\cdot z&=x\cdot(y\cdot z),\tag{14a}\\ (x\cdot y)\cdot y&=x,\tag{14b}\\ x\cdot x&=y\cdot y.\tag{14c} \end{align*}

Proof. For orders one and two the Latin square condition gives an associative operation. For order three, identify the underlying set with $\mathbb F_3$ . The multiplication has the form

\tag{15} x\cdot y=ax+by+c, \qquad a,b\in\{1,-1\},\quad c\in\mathbb F_3.

To see the exhaustion, permute rows and columns to make the first row and first column $0,1,2$ . The remaining entries are then forced, giving the table of addition modulo three. Every permutation of $\mathbb F_3$ is affine, so undoing the permutations gives (15).

If $a=-1$ , then $(x\cdot y)\cdot y=x$ . If $a=1$ and $b=-1$ , then $x\cdot x=c$ . In the remaining case $a=b=1$ , the operation $x+y+c$ is associative. These are (14b), (14c), and (14a), respectively. ◻

Proof of Theorem 2. The disjunctive identity

\tag{16} \bigvee_{0\leq i<j\leq3}x_i=x_j

holds in $C_3$ . Every group in $\mathsf{D}(C_3)$ therefore has at most three elements, and isotopy preserves cardinality.

Suppose $\mathsf{D}(A)=I\mathsf{D}(C_3)$ . Since $A\in\mathsf{D}(A)$ , it follows that $|A|\leq3$ . Moreover, $C_3\in\mathsf{D}(C_3)$ , so the target class contains both addition on $\mathbb F_3$ and the isotope

x\circ y=-x+y.

Ordinary identities are special cases of disjunctive identities. Each identity in Lemma 5 is incompatible with containing both these quasigroups:

\begin{align*} (0+1)+1&=2\neq0,\\ 0+0&=0\neq2=1+1,\\ (1\circ0)\circ0&=1\neq-1=1\circ(0\circ0). \end{align*}

Thus (14b) and (14c) exclude addition on $\mathbb F_3$ , while (14a) excludes $\circ$ . Every possible $A$ is excluded. ◻

References

Preprint · Lean (GitHub)

E. Falconer, Isotopy invariants in quasigroups, Trans. Amer. Math. Soc. 151 (1970), 511–526, doi:10.1090/S0002-9947-1970-0272932-4.

A. A. Gvaramiya, Quasigroup classes that are invariant under isotopy, and abstract classes of invertible automata, Dokl. Akad. Nauk SSSR 282 (1985), no. 5, 1047–1051 (Russian), https://www.mathnet.ru/eng/dan9110.

E. I. Khukhro and V. D. Mazurov (eds.), The Kourovka Notebook: Unsolved problems in group theory, 21st ed., September 2026 update, Problem 9.4, arXiv:1401.0300v46.