Groups Universally Equivalent to the Solvable Baumslag–Soliter Group

Abstract

We describe the finitely generated groups that are universally equivalent to the solvable Baumslag–Soliter group with parameters $(1,n)$, $|n|>1$.

1. Introduction

In 1995, Chapuis [1] proved that the universal theory of a free metabelian group is solvable, and then in [2] he characterized the groups that possess the same universal theory as the free metabelian group of rank $\geq 2$. The author established in [3] that the universal theory of the free solvable group of class $\geq 4$ is algorithmically unsolvable. Recall also that more than ten years ago the author gave a definition of a rigid solvable group, and then, in his and joint works with Myasnikov, he studied the properties of rigid groups, the algebraic geometry over them, and the model-theoretical aspects of the theory of divisible rigid groups. Using the notion of rigid group, the assertion of the second result of Chapuis looks as follows: The groups that are universally equivalent to the free two-step solvable group are exactly the rigid two-step solvable groups.

A wider class $\mathcal{R}_{2}$ of metabelian groups was defined in [4]. By definition, a two-step solvable group $G$ belongs to $\mathcal{R}_{2}$ provided that the following holds: There is a normal abelian subgroup $\rho_{2}(G)$ of $G$ such that $G/\rho_{2}(G)=A$ is a nontrivial torsion-free abelian group. The action by conjugations of $G$ on $\rho_{2}(G)$ equips the last subgroup with the structure of a (right) module over the group ring $𝕑A$. Let $R$ stand for the quotient ring of $𝕑A$ by the module annihilator. In these circumstances, $\rho_{2}(G)$ may be considered as an $R$-module. It is required that $\rho_{2}(G)$ has no module torsion, and $A$ is embeddable into the group $R^{\ast}$ of the invertible elements in $R$.

It was noticed that $\rho_{2}(G)$ is uniquely defined by these conditions, and the class $\mathcal{R}_{2}$ is closed under the universal equivalence of groups. As above, the so-called $r$-pair $(A,R)$ is put into correspondence with every group in $\mathcal{R}_{2}$, where $R$ is a commutative domain, and $A$ is a nontrivial torsion-free subgroup of $R^{\ast}$ which generates $R$ as a ring. Conversely, given an $r$-pair $(A,R)$, we may define the group of matrices

$$ M(A,R)=\begin{pmatrix}A&0\\ R&1\end{pmatrix} $$

in the class $\mathcal{R}_{2}$ which may be considered as an extension of the additive group of $R$ by $A$. In [4], the notion of universal theory of an $r$-pair $(A,R)$ was defined, and it was proved that the universal theories of an $r$-pair $(A,R)$ and the group $M(A,R)$ are equivalent (Theorem 1); i.e., each one of them is interpreted in the other.

It was established as well that the universal theories of two split groups $G_{1},G_{2}\in\mathcal{R}_{2}$ coincide if and only if the universal theories of the corresponding $r$-pairs $(A_{1},R_{1})$ and $(A_{2},R_{2})$ coincide (Theorem 2).

Consider a solvable Baumslag–Soliter group with parameters $(1,n)$, where $|n|>1$. It may be identified with the group

$$ BS(1,n)=M(C,𝕈_{n})=\begin{pmatrix}C&0\\ 𝕈_{n}&1\end{pmatrix}; $$

here $𝕈_{n}=𝕑[1/n]=𝕈_{-n}$ is the ring of rationals whose denominators are some powers of $|n|$, and $C$ is the multiplicative group generated by $c=n$. Let us present the main results enabling us to obtain the description of the finitely generated groups with the same universal theory as $BS(1,n)$.

Theorem 1

The universal theory of a group $G$ coincides with that of $BS(1,n)$ if and only if $G\in\mathcal{R}_{2}$ and the universal theory of the $r$-pair $(A,R)$, corresponding to $G$, coincides with one of the $r$-pair $(C,𝕈_{n})$ that corresponds to $BS(1,n)$.

Theorem 2

The universal theory of a finitely generated $r$-pair $(A,R)$ coincides with one of the $r$-pair $(C,𝕈_{n})$ if and only if the abelian group $A$ is freely generated by $c,y_{1},\dots,y_{m}$, where $y_{1},\dots,y_{m}$ are some algebraically independent over $𝕑$ elements in $R$. Furthermore, $R$ is the ring of Laurent polynomials in $y_{1},\dots,y_{m}$ over $𝕈_{n}$.

Note that the solvability of the universal theory of $BS(1,n)$ follows from Theorem 8 of [5], where it is stated that a finitely generated metabelian split group possesses the solvable universal theory.

We now pose the following question: Is it possible to generalize our assertions and to characterize the groups universally equivalent to a given group in $\mathcal{R}_{2}$ with a corresponding $r$-pair $(A,R)$, where $A$ is an (infinite) cyclic group?

In [4], there was also formulated the general problem of classification of the groups in the class $\mathcal{R}_{2}$ by their universal theories.

2. Auxiliary Assertions and Proof of Theorem 1

2.1. Recall some definitions from [4].

An $r$-pair $(A,R)$ is finitely generated if $A$ is a finitely generated group.

A subpair of an $r$-pair $(A,R)$ is an $r$-pair $(A_{1},R_{1})$ such that $R_{1}$ is a subring of $R$, and $A_{1}$ is a subgroup of $A$.

A morphism $(A,R)\rightarrow(A_{1},R_{1})$ of two $r$-pairs is a ring homomorphism $R\rightarrow R_{1}$ which maps $A$ into $A_{1}$. We say that a morphism is essential if the image of $A$ in $A_{1}$ is a nontrivial subgroup.

A homomorphism $G_{1}\rightarrow G_{2}$ of some groups in the class $\mathcal{R}_{2}$ is essential provided that the image of $G_{1}$ is a nonabelian group.

Let us recall a series of known facts:

Lemma 1 [6]

The universal theories or two equationally Noetherian groups coincide if and only if each of the groups is locally discriminated by the other.

Lemma 2 [7]

Every group in $\mathcal{R}_{2}$ is an equationally Noetherian group.

Lemma 3 [4]

The universal theories of two $r$-pairs coincide if and only if each of the pairs is locally discriminated by the other by essential morphisms.

Lemma 4 [4]

Every essential homomorphism of groups in $\mathcal{R}_{2}$ induces an essential morphism of the corresponding $r$-pairs.

We need two more lemmas.

Lemma 5

Let $G_{1},G_{2}\in\mathcal{R}_{2}$. Assume that there exists a set of essential group homomorphisms $G_{1}\rightarrow G_{2}$ which discriminate $G_{1}$. Then the set of the induced essential morphisms of the $r$-pairs $(A_{1},R_{1})\rightarrow(A_{2},R_{2})$ discriminates $(A_{1},R_{1})$.

Proof

Consider a set $\{u_{1},\dots,u_{s}\}$ of nonzero elements in $R_{1}$ which we want to discriminate by a morphism $(A_{1},R_{1})\rightarrow(A_{2},R_{2})$ of $r$-pairs. Choose a nontrivial element $t$ in $\rho_{2}(G_{1})$ and take a set $\{tu_{1},\dots,tu_{s}\}$ of nontrivial elements in $\rho_{2}(G_{1})$. By hypothesis, there is an essential group homomorphism $G_{1}\rightarrow G_{2}$ which discriminates this set. Let $\overline{t}$ denote the image of $t$. Note that $\overline{t}$ is nontrivial. The images of $tu_{1},\dots,tu_{s}$ are of the shape $\overline{t}\overline{u}_{1},\dots,\overline{t}\overline{u}_{s}$, where $\overline{u}_{1},\dots,\overline{u}_{s}\in R_{2}$ are some images of $u_{1},\dots,u_{s}$ under the induced morphism of $r$-pairs $(A_{1},R_{1})\rightarrow(A_{2},R_{2})$. Obviously, they should be nonzero. The lemma is proved.

Lemma 6

Given an $r$-pair $(A,R)$, consider an $r$-pair $(B,R(y_{1},\dots,y_{m}))$, where $R(y_{1},\dots,y_{m})$ is the Laurent polynomial ring in $y_{1},\dots,y_{m}$ over $R$, and $B=A\times\langle y_{1},\dots,y_{m}\rangle$. Then the universal theories of the $r$-pairs $(A,R)$ and $(B,R(y_{1},\dots,y_{m}))$ coincide.

Proof

The pair $(A,R)$ is embedded into $(B,R(y_{1},\dots,y_{m}))$. Hence, the first pair is discriminated by the second. Note that the second pair is discriminated by the first. Consider some set $\{f_{1}(y_{1},\dots,y_{m}),\dots,f_{s}(y_{1},\dots,y_{m})\}$ of nonzero elements in $R(y_{1},\dots,y_{m})$, i.e., the Laurent polynomials, which we want to discriminate. Fix a nontrivial $a\in A$. It is easy to understand that there is a set of values

$$ y_{1}=a^{k_{1}},\dots,y_{m}=a^{k_{m}}\quad(0\neq k_{i}\in 𝕑) $$

such that $f_{j}(a^{k_{1}},\dots,a^{k_{m}})\neq 0$. It means that the $r$-pair $(B,R(y_{1},\dots,y_{m}))$ is discriminated by its subpair $(A,R)$ by essential morphisms, and then the universal theories of $(A,R)$ and $(B,R(y_{1},\dots,y_{m}))$ coincide by Lemma 3. The lemma is proved.

2.2. We proceed now to the proof of Theorem 1. Let $G$ possess the same universal theory as $BS(1,n)$. Then $G$ belongs to $\mathcal{R}_{2}$, and let $(A,R)$ be its corresponding $r$-pair. By Lemma 1 we may assert that each of the groups $G$ and $BS(1,n)$ is locally discriminated by the other, and the corresponding group homomorphisms are essential. By Lemma 5 each of the $r$-pairs $(A,R)$ and $(C,𝕈_{n})$ is locally discriminated by the other by essential morphisms, whence the universal theories of these $r$-pairs coincide by Lemma 3.

Conversely, assume that $G\in\mathcal{R}_{2}$ and the universal theory of the $r$-pair $(A,R)$, which corresponds to $G$, coincide with one of the $r$-pair $(C,𝕈_{n})$, corresponding to $BS(1,n)$. By Lemma 3 each of these pairs is locally discriminated by the other by essential morphisms. In particular, there is an essential morphism of pairs $(C,𝕈_{n})\rightarrow(A,R)$. Obviously, such morphism is an embedding, whence $(C,𝕈_{n})$ is a subpair of the $r$-pair $(A,R)$. We need to prove that each of the groups $G$ and $BS(1,n)$ is locally discriminated by the other. It suffices to consider the case when $G$ is a finitely generated group. By Theorem 2 of [8] $G$ is embedded into a split finitely generated group $H\in\mathcal{R}_{2}$, which has the same $r$-pair $(A,R)$. By Theorem 2 of [4] $H$ has the same universal theory as $BS(1,n)$. Therefore, $H$ and, hence, $G$ are discriminated by $BS(1,n)$. It remains to notice that $BS(1,n)$ is discriminated by $G$. There is $a\in G$ such that the conjugation by $a$ corresponds to the multiplication by $n$ in the module $\rho_{2}(G)$. Choose a nontrivial $b\in\rho_{2}(G)$ as well. Then the subgroup of $G$, generated by $a$ and $b$, is isomorphic to $BS(1,n)$, which is sufficient. Theorem 1 is proved.

3. Proof of Theorem 2

In one direction, everything follows from Lemma 6: if $R$ is the Laurent polynomial ring in $y_{1},\dots,y_{m}$ over $𝕈_{n}$ then the universal theories of the $r$-pairs $(C,𝕈_{n})$ and $(A,R)$ coincide.

Prove the converse statement. Assume that the universal theories of the $r$-pairs $(A,R)$ and $(C,𝕈_{n})$ coincide. Since the $r$-pair $(A,R)$ is discriminated by its subpair $(C,𝕈_{n})$ by essential morphisms. Therefore, there is a group epimorphism $A\rightarrow C$, which is identical on $C$, whence $C$ is a direct factor of $A$. Let $A=C\times\langle y_{1},\dots,y_{m}\rangle$ be the free abelian group with the base $\{c,y_{1},\dots,y_{m}\}$. Assume the contrary: There is an algebraic dependence among $y_{1},\dots,y_{m}$, i.e., there is a nontrivial integer polynomial $f(x_{1},\dots,x_{m})$ such that $f(y_{1},\dots,y_{m})=0$. Given a set $\varepsilon=(\varepsilon_{1},\dots,\varepsilon_{m})$ with $\varepsilon_{i}=\pm 1$, we denote by $f_{\varepsilon}$ the integer polynomial that results from $f\bigl{(}x_{1}^{\varepsilon_{1}},\dots,x_{m}^{\varepsilon_{m}}\bigr{)}$ by multiplying by a proper denominator of the shape $x_{1}^{k_{1}}\dots x_{m}^{k_{m}}$ $(k_{i}\geq 0)$. We have $f_{\varepsilon}\bigl{(}y_{1}^{\varepsilon_{1}},\dots,y_{m}^{\varepsilon_{m}}\bigr{)}=0$.

First, we consider the case $n>0$. Distinguish the positive and negative parts of $f$; i.e., consider a representation $f=f^{\prime}-f^{\prime\prime}$, where $f^{\prime}$ and $f^{\prime\prime}$ are some linear combinations of distinct monomials with positive integer coefficients, and one of the parts may be trivial. From here we get the corresponding representation $f_{\varepsilon}=f^{\prime}_{\varepsilon}-f^{\prime\prime}_{\varepsilon}$. Let a natural $k$ bound from above the power of every variable $x_{i}$ in all polynomials $f_{\varepsilon}$. Furthermore, we assume that all coefficients in $f^{\prime}$ and $f^{\prime\prime}$ are less than $n^{k+1}$. Write these coefficients in the $n$-adic representation, i.e., decompose them by powers of $n$. As a result, we may assert that $f^{\prime}_{\varepsilon}$ and $f^{\prime\prime}_{\varepsilon}$ are some linear combinations of monomials of the shape $n^{k_{0}}x_{1}^{k_{1}}\dots x_{m}^{k_{m}}$ $(0\leq k_{i}\leq k)$ with the coefficients in $\{0,\dots,n-1\}$. Under a fixed $\varepsilon$ and distinct sets $(k_{0},k_{1},\dots,k_{m})$, all elements $n^{k_{0}}y_{1}^{\varepsilon_{1}k_{1}}\dots y_{m}^{\varepsilon_{m}k_{m}}$ $(0\leq k_{i}\leq k)$ in $A$ are different. Form the set $D_{\varepsilon}$ from these elements. By hypothesis, there is a morphism of the $r$-pairs $\varphi:(A,R)\rightarrow(C,𝕈_{n})$, for which the images of all elements in every $D_{\varepsilon}$ remain distinct. Let $y_{1}\varphi=n^{l_{1}},\dots,y_{m}\varphi=n^{l_{m}}$. We may assume that the set $(l_{1},\dots,l_{m})$ consists of nonzero integers. Let $\varepsilon=(\varepsilon_{1},\dots,\varepsilon_{m})$ be the corresponding set of the signs of these numbers. We have

$$ \begin{gathered}\displaystyle f_{\varepsilon}\bigl{(}y_{1}^{\varepsilon_{1}},\dots,y_{m}^{\varepsilon_{m}}\bigr{)}\varphi=f_{\varepsilon}(n^{\varepsilon_{1}l_{1}},\dots,n^{\varepsilon_{m}l_{m}})\\ \displaystyle=f^{\prime}_{\varepsilon}(n^{\varepsilon_{1}l_{1}},\dots,n^{\varepsilon_{m}l_{m}})-f^{\prime\prime}_{\varepsilon}(n^{\varepsilon_{1}l_{1}},\dots,n^{\varepsilon_{m}l_{m}}).\end{gathered} $$

By construction, the nonnegative integer $f^{\prime}_{\varepsilon}(n^{\varepsilon_{1}l_{1}},\dots,n^{\varepsilon_{m}l_{m}})$ is given by a decomposition by different powers of $n$ (an analogous statement may be said about $f^{\prime\prime}_{\varepsilon}(n^{\varepsilon_{1}l_{1}},\dots,n^{\varepsilon_{m}l_{m}})$), and distinct powers of $n$ take part in both decompositions. Then the difference

$$ f^{\prime}_{\varepsilon}(n^{\varepsilon_{1}l_{1}},\dots,n^{\varepsilon_{m}l_{m}})-f^{\prime\prime}_{\varepsilon}(n^{\varepsilon_{1}l_{1}},\dots,n^{\varepsilon_{m}l_{m}}) $$

should be nonzero; a contradiction with the equality $f_{\varepsilon}\bigl{(}y_{1}^{\varepsilon_{1}},\dots,y_{m}^{\varepsilon_{m}}\bigr{)}=0$.

In the case $n<0$, we need to consider an algebraical dependence among $y_{1}^{2},\dots,y_{m}^{2}$. Then changing $n$ by $n^{2}$, the proof, which was given above for the case $n>0$, is completely valid. Theorem 2 is proved.