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M. A. Mikheenkoab a Faculty of Mechanics and Mathematics, Lomonosov Moscow State University, Moscow, Russia b Moscow Center of Fundamental and Applied Mathematics, Moscow, Russia
Russian version:
Matematicheskii Sbornik, 2024, Volume 215, Number 6, DOI: https://doi.org/10.4213/sm10009 
Bibliographic databases:
 § 1. IntroductionIn this article $\mathbb Z_p$, for a prime number $p$, denotes the field with $p$ elements and $(\mathbb Z_p)^*$ denotes the group of its invertible elements. A cyclic group of order $n$ is denoted by $\langle a \rangle_n$, $\langle b \rangle_n$, $\langle g \rangle_n$ and so on. An infinite cyclic group is denoted by $\langle g \rangle_\infty$, $\langle x_1\rangle_\infty$ and so on. The Cartesian (or unrestricted direct) product of groups $\{G_i\}_{i \in I}$ is denoted by $\prod_{i \in I} G_i$, whereas their direct (restricted) product is denoted by $\times_{i\in I} G_i$. The group ring of a group $G$ with coefficient ring $R$ is denoted by $R G$. Here we call a group $G$ an extension of a group $A$ by a group $B$ if $G$ has a normal subgroup $A$ and the quotient group $G/A$ is isomorphic to $B$. For elements $g$ and $h$ of a group $G$ the expression $g^h$ denotes $h^{-1}gh$ and the expression $g^{nh}$, where $n$ is a (maybe negative) integer denotes $h^{-1}g^nh$. Also, $[g,h]$ denotes $g^{-1}h^{-1}gh$. The Cartesian (or unrestricted) wreath product $A \mathbin{\overline{\wr}} B$ is viewed as $(\prod_{b \in B} A_b)\leftthreetimes B$ in which the $A_b$ are copies of $A$ and $B$ acts on $\prod_{b \in B} A_b$ so that $((a_b)_{b\in B})^{b_1}=(a_b)_{bb_1\in B}=(a_{bb_1^{-1}})_{b\in B}$ if $b_1 \in B$ and $a_b \in A_b$ for any $b\in B$. This paper is devoted to equations and systems of equations over groups. Definition 1. Let $G$ be a group. An equation in variables $x_1,\dots,x_n$ over $G$ is an expression $w(x_1,\dots,x_n)=1$, where $w$ is an element of the free product ${G*F(x_1,\dots,x_n)}$, where $F(x_1,\dots,x_n)$ is the free group with basis $x_1,\dots,x_n$. The equation $w(x_1,\dots,x_n)=1$ is solvable in the group $\widetilde G$ if $\widetilde G \supset G$ and $\widetilde G$ contains a solution of this equations (that is, there are elements $\widetilde g_1,\dots,\widetilde g_n \in \widetilde G$ such that $w(\widetilde g_1,\dots, \widetilde g_n)=1$). The group $\widetilde G$ is called a solution group. Equivalently, $w=1$ is solvable in $\widetilde G$ if there is a homomorphism $G*F(x_1,\dots,x_n) \to \widetilde G$ that is injective on $G$ and takes $w$ to $1$. If $w=1$ is solvable in a group $\widetilde G$, then we say that the equation $w=1$ is solvable over $G$. The solvability of a (finite or infinite) system of equations (possibly in an infinite set of variables) over a group is defined likewise. It is easy to see that a system $\{w_j=1\}_{j \in J}$ of equations over $G$ in variables $\{x_i\}_{i \in I}=X$ is solvable if and only if $G \cap \langle \langle W \rangle \rangle=\{1\}$ in $G*F(X)$, where $W$ is the set $\{w_j\}_{j \in J}$ and $\langle \langle W \rangle \rangle$ is its normal closure in $G*F(X)$. It follows that if every finite subsystem of a system is solvable, then the systems itself is solvable. Definition 2. Let $\{w_j = 1\}_{j \in J}$ be a system of equations in variables $\{x_i\}_{i\in I}=X$ over a group $G$. Consider the free $\mathbb Z$-module $\sum_{i\in I}\mathbb Z \cdot x_i$ with basis $X$. It can also be viewed as a free abelian group in the additive notation. There is a homomorphism $G*F(X) \to \sum_{i \in I}\mathbb Z \cdot x_i$ which takes $G$ to $0$ and $x_i \in X$ to $x_i$. Let us call this homomorphism a trivialization. Let $m_j$ be the image of $w_j$ under trivialization. Then $m_j$ can be viewed as a finitely supported (that is, having only a finite number of nonzero coordinates) row of sums of the exponents of the variables $x_i$ occurring in $w_j$. The system $\{w_j=1\}$ is called nonsingular, if the elements $m_j$ are independent over $\mathbb Z$, that is, no combination of these rows with coefficients in $\mathbb Z$ is equal to zero, apart from the combination with all coefficients equal to zero (equivalently, if these rows are linearly independent over $\mathbb Q$ as elements of the vector space $\sum_{i \in I}\mathbb Q \cdot x_i$). For a prime number $p$ consider likewise $p$-trivialization, that is, the homomorphism $G*F(X) \to \sum_{i \in I}\mathbb Z_p \cdot x_i$. The system is called $p$-nonsingular if the images of the words $w_j$ under $p$-trivialization (that is, rows of sums modulo $p$ of the exponents of variables occurring in $w_j$) are linearly independent over $\mathbb Z_p$. The system is called unimodular if it is $p$-nonsingular for every prime $p$. In particular, an equation $w(x)=1$ in one variable is nonsingular ($p$-nonsingular, unimodular) if the sum of the exponents of $x$ in $w$ is nonzero (not divisible by $p$, equal to $\pm 1$, respectively). Clearly, if a system of equations is $p$-nonsingular for some prime number $p$, then it is nonsingular as well. Also, a unimodular system is the same as a system $p$-nonsingular for every prime $p$, and therefore, in particular, unimodular systems are nonsingular. Example 1. The system of equations has the following rows of sums of exponents organized into a matrix: The determinant of this matrix is $-5$, so the system is nonsingular, $2$-nonsingular and $3$-nonsingular, but it is $5$-singular, so it is not unimodular. Nonsingular systems of equations over groups in many classes are solvable. For example, one of the classical results on equations over groups concerns finite groups. Theorem ([1]). A finite nonsingular system of equations over a finite group $G$ is solvable. Moreover, the solution group can be chosen to be finite. This theorem has repeatedly been generalized: see [12], [8], [13], [11] and [6]. The following conjecture has neither been proved nor disproved yet. Conjecture 1 (Howie’s conjecture [2]). Each nonsingular system of equations over any group is solvable. As a system of equations over a group is solvable if and only if every finite subsystem of this system is solvable, it suffices to consider only finite systems in order to check whether the conjecture is true. For example, a result from [1] states that this conjecture is true for finite groups. However, the solution group for an infinite system may not be finite. Apart from finite groups, this conjecture is also true for locally indicable groups. Definition 3. A locally indicable group is a group each of whose nontrivial finitely generated subgroups admits a surjective homomorphism onto $\langle g \rangle_\infty$. Definition 4. A locally $p$-indicable group for a prime $p$ is a group each of whose nontrivial finitely generated subgroup admits a surjective homomorphism onto $\langle g \rangle_p$. Note that locally indicable groups are locally $p$-indicable for any prime $p$. Also, it is easy to see that if a group is locally $p$-indicable for an arbitrarily large prime number $p$, then it is locally indicable. This can be seen by considering the quotient group of a finitely generated subgroup by the commutator subgroup of this subgroup. Theorem 2 ([2]). A finite nonsingular system of equations over a locally indicable group is solvable. A similar result holds for $p$-nonsingular systems. Theorem 3 ([10]). Let $p$ be a prime number. Then a finite $p$-nonsingular system of equations over a locally $p$-indicable group is solvable. The following result shows that more is known about unimodular equations than about nonsingular ones. Theorem 4 ([4]). Any unimodular equation over a torsion-free group is solvable. This result was generalized in [5]. It is unknown if an analogous result is true for nonsingular equations. Some results are connected not only with the solvability of systems of equations over groups but also with the solvability of systems in groups of the same class. Apart from the already mentioned result from [1] for the class of finite groups, there is an old result for the class of nilpotent groups. Definition 5. Let $p$ be a prime number. A group $G$ is called $p'$-torsion free if each nonidentity element of $G$ has either an infinite order or an order equal to a power of $p$. Theorem 5 ([14]). If This implies that, for example, any nonsingular system over a nilpotent torsion-free group is solvable in the divisible hull of the group, which is a nilpotent group of the same nilpotency class. In [7] a similar question was studied for the class of solvable groups: when does a nonsingular system over a solvable group also have a solution in a solvable group and what can be said about the derived length of the solution group? Proposition 1 ([7]). There exists a metabelian group $G$ with a unimodular equation $w(x)=1$ over it such that this equation has no solutions in metabelian groups. Moreover, $G$ can be chosen to be either Theorem 6 ([7]). Let the group $G$ have a subnormal series in which all factors are abelian and all factors except the last one are torsion free. Then any (finite or infinite) nonsingular system of equations over $G$ has a solution in some group $\widetilde G \supset G$ that also has a subnormal series with abelian factors, and all factors except the last one of the series of $\widetilde G$ are torsion free. Section 3 of the present article contains the proof of a similar result for $p$-nonsingular systems. Theorem 7 (main theorem). Let $p$ be a prime number. Let the group $G$ have a subnormal series with factors $B_1, B_2,\dots,B_n$, where the $B_i=G_i/G_{i+1}$ are abelian groups and all factors $B_i$ except $B_n$ are $p'$-torsion free. Then there is a group $\widehat G \supset G$ with a subnormal series with the same properties (that is, its length is $n$, the factors are abelian, and all factors except the last one are $p'$-torsion free) such that any $p$-nonsingular system of equations over $G$ has a solution in $\widehat G$. Moreover, every $p$-nonsingular system over $\widehat G$ is solvable in $\widehat G$ itself.Using arguments of § 3 we can strengthen a theorem from [7], as mentioned in Remark 2. Lemma 1 from § 2 plays an important role in the proof of the main theorem. Also, the method of the proof is based on methods developed in [11], [6] and [7]. In § 4 the main theorem helps us to prove the following fact. Proposition 2. The metabelian group $\begin{pmatrix} 1 & \mathbb Z_7 \\ 0 & (\mathbb Z_7)^*\end{pmatrix}$ of order $42$ from [7] is a minimal (in order) example of a metabelian group over which there exists a unimodular equation that has no solutions in metabelian groups. Some open questions on systems of equations over solvable groups are formulated in § 5. AcknowledgementsThe author thanks the Theoretical Physics and Mathematics Advancement Foundation “BASIS”. The author also thanks Anton A. Klyachko for his valuable advices and comments. The author thanks an anonymous referee, who, apart from other useful comments, proposed a shorter proof for § 4. § 2. Images of rows of group ring elementsProposition 3. Let $R$ be an associative ring with unity of characteristic $p$. Let $M \in R[x]/(x^{p^k}-1)$ be an element such that $f(M)$ is not a left (right) zero divisor, where $f\colon R[x]/(x^{p^k}-1) \to R$ is the natural ring homomorphism such that $f(x)=1.$ Then $M$ itself is not a left (right) zero divisor. Proof. We consider the ‘left’ case.
Consider $M$ as a polynomial in $(x-1)$ with coefficients from $R$:
$$
\begin{equation*}
M=M_0+M_1(x-1)+M_2(x-1)^2+\dots+M_{p^k-1}(x-1)^{p^k-1}.
\end{equation*}
\notag
$$
Note that $(x-1)^{p^k}=x^{p^k}-1=0$ because the characteristic of $R$ is $p$. By the definition of $f$, $M_0$ is equal to $f(M)$, hence $M_0$ is not a left zero divisor.
Now assume that $M$ is a left zero divisor. This means that there is a nonzero polynomial such that $MB=0$. Considering $MB$ also as a polynomial in $(x-1)$, we look at the constant term of $MB$. On the one hand it is equal to $M_0B_0$, on the other it must be zero. The term $M_0$ is not a left zero divisor, and therefore $B_0=0$. Now we look at the linear term of $MB$. As $B_0$ equals $0$, this linear term is equal to $M_0B_1(x-1)$, while it must be equal to $0$ as well as the constant term, and so on. Eventually, we see that all coefficients $B_i$ are equal to $0$. This contradicts the condition $B\neq 0$. Thus, the assumption that $M$ is a left zero divisor is false.The ‘right’ case can be treated similarly. The proposition is proved. Corollary 1. Let $R$ be an associative ring with unity of characteristic $p$. Let $M \in R[x_1,\dots,x_l]/(x_1^{p^{k_1}}-1,\dots, x_l^{p^{k_l}}-1)$ be an element such that $f(M)$ is not a left (right) zero divisor, where $f\colon R[x_1,\dots,x_l]/(x_1^{p^{k_1}}-1,\dots, x_l^{p^{k_l}}-1) \to R$ is the natural ring homomorphism with $f(x_i)=1.$ Then $M$ itself is not a left (right) zero divisor. Proof. We use induction on $l$.
If $l=1$, then this is exactly Proposition 3. Now we show that if the corollary is true for $l$ variables, then it is also true for $l+1$ variables. Note that
$$
\begin{equation*}
\begin{aligned} \, &R[x_1,\dots,x_l,x_{+1}]/(x_1^{p^{k_1}}-1,\dots, x_l^{p^{k_l}}-1, x_{l+1}^{p^{k_{l+1}}}-1) \\ &\qquad =\bigl(R[x_1,\dots,x_l]/(x_1^{p^{k_1}}-1,\dots, x_l^{p^{k_l}}-1)\bigr) [x_{l+1}]/(x_{l+1}^{p^{k_{l+1}}}-1). \end{aligned}
\end{equation*}
\notag
$$
We denote $R[x_1,\dots,x_l]/(x_1^{p^{k_1}}-1,\dots, x_l^{p^{k_l}}-1)$ by $Q$. The ring $Q$ also has characteristic $p$. We look at the image $h(M)$ of $M$ under the natural homomorphism $h\colon Q[x_{l+1}]/(x_{l+1}^{p^{k_{l+1}}}-1) \to Q$, $h(x_{l+1})=1$. This image is not a left (right) zero divisor by the induction assumption. Then by Proposition 3 the element $M$ is not a left (right) zero divisor either. Now let $P$ be a finitely generated abelian $p$-group. It can be represented as $P= \langle g_1\rangle_{p^{k_1}}\times\dots\times \langle g_n\rangle_{p^{k_l}}$. Then the group algebra $\mathbb Z_p P$ can be represented as $\mathbb Z_p P=\mathbb Z_p [x_1,\dots,x_l]/(x_1^{p^{k_1}}-1,\dots, x_l^{p^{k_l}}-1)$. And the ring of $n\times n$ matrices over $\mathbb Z_p P$ is $M_n(\mathbb Z_p) [x_1,\dots,x_l]/(x_1^{p^{k_1}}-1,\dots, x_l^{p^{k_l}}-1)$. As we know that a nonsingular matrix over a field is not a zero divisor, we obtain the following result. Corollary 2. Let $M\in M_n(\mathbb Z_p P)$ be an $n \times n$ matrix over the group algebra, where $P$ is a finitely generated abelian $p$-group. Suppose that the image of $M$ under the natural ring homomorphism $\varepsilon\colon M_n(\mathbb Z_p P) \to M_n(\mathbb Z_p)$ which takes the elements of $P$ to the unity of $\mathbb Z_p$ is a nonsingular matrix. Then $M$ is not a zero divisor. In particular, the rows of $M$ are independent over $\mathbb Z_p P$. The statement of Corollary 2 is also true for $P\times A$, where $P$ and $A$ are finitely generated abelian groups, $P$ is a $p$-group and $A$ is torsion free. Corollary 3. Let $M\in M_n(\mathbb Z_p (P\times A))$ be an $n \times n$ matrix over a group algebra, where $P$ is a finitely generated abelian $p$-group and $A$ is a finitely generated torsion-free abelian group. Suppose that the image of $M$ under the natural ring homomorphism $\varepsilon\colon M_n(\mathbb Z_p (P\times A)) \to M_n(\mathbb Z_p)$, which takes elements of $P\times A$ to the unity of $\mathbb Z_p$, is a nonsingular matrix. Then $M$ is not a zero divisor. In particular, the rows of $M$ are independent over $\mathbb Z_p (P\times A)$. Proof. We look at the image of $M$ under the ring homomorphism which takes the elements of $P$ to the unity of $\mathbb Z_p$ and does not change the elements of $A$. The matrix $f(M)$ is not a zero divisor.
Indeed, $\mathbb Z_p A$ can be embedded into its field of fractions, as $\mathbb Z_p A$ is an integral domain. Therefore, the matrix $f(M)$ not being a zero divisor is equivalent to the nonsingularity of $f(M)$ over the field of fractions, that is, to it having a nonzero determinant. On the other hand the determinant of $f(M)$ is not equal to zero, as by assumption the determinant of $\varepsilon(M)=h(f(M))$ is not equal to zero, where $h\colon M_n(\mathbb Z_p A) \to M_n(\mathbb Z_p)$ is the natural ring homomorphism which takes the elements of $A$ to the unity of $\mathbb Z_p$. Further, note that $M_n(\mathbb Z_p (P\times A))$ is naturally isomorphic to
$$
\begin{equation*}
M_n(\mathbb Z_p A) [x_1,\dots,x_l]/(x_1^{p^{k_1}}-1,\dots, x_l^{p^{k_l}}-1).
\end{equation*}
\notag
$$
Then from Corollary 1 for $R$ equal to $M_n(\mathbb Z_p A)$ we get that $M$ is not a zero divisor since $f(M)$ is not. The structure $P\times A$ (where $P$ and $A$ are from Corollary 3) is the structure of every finitely generated $p'$-torsion-free abelian group. We can now go from the finitely generated case to the infinitely generated one. Corollary 4. Let $M\in M_n(\mathbb Z_p D)$ be an $n \times n$ matrix over a group algebra, where $D$ is a $p'$-torsion-free abelian group. Suppose that the image of $M$ under the natural ring homomorphism $\varepsilon\colon M_n(\mathbb Z_p D) \to M_n(\mathbb Z_p)$ is a nonsingular matrix. Then $M$ is not a zero divisor. In particular, the rows of $M$ are independent over $\mathbb Z_p D$. Proof. Assume that this is not true. This means that there is a nonzero matrix $B$ over $\mathbb Z_p D$ such that either $BM=0$ or $MB=0$.
We look at all the elements of $B$ and $M$. There is a finite number of them. Each of them contains a finite number of elements of $D$ with nonzero coefficients. Hence there is only a finite number of elements of $D$ which are contained in these matrices with nonzero coefficient. Consider the subgroup $H$ generated by these elements. It is a finitely generated $p'$-torsion-free abelian group. The elements of $B$ and $M$ are contained in the corresponding subalgebra $\mathbb Z_p H$. Therefore, $M$ is a zero divisor in a matrix ring over the group algebra of a finitely generated $p'$-torsion-free abelian group, which is prohibited by Corollary 3. This means that the assumption that $M$ is a zero divisor is false. Corollary 4 can be generalized to the case of infinite rows. Lemma 1. Let $D$ be a $p'$-torsion-free abelian group and $\{m_i\}_{i\in I}$ be elements of a free module over $\mathbb Z_p D$ (that is, finitely supported rows of elements of this group algebra). Consider the natural ring homomorphism $f \colon \mathbb Z_p D \to \mathbb Z_p$ which takes the elements of $D$ to the unity of $\mathbb Z_p$. It defines naturally a mapping from a free module over $\mathbb Z_p D$ to a vector space over $\mathbb Z_p$. Now let the system of rows $\{f(m_i)\}_{i\in I}$ be linearly independent over $\mathbb Z_p$. Then the system $\{m_i\}_{i\in I}$ is independent over $\mathbb Z_p D$. Proof. We argue by contradiction.
Assume that $\{m_i\}_{i\in I}$ is dependent over $\mathbb Z_p D$. This means that some finite subsystem of these rows is dependent. Without loss of generality suppose that this subsystem consists of $m_1, m_2,\dots, m_n$. By the hypotheses of the lemma $f(m_1),f(m_2),\dots, f(m_n)$ are linearly independent over $\mathbb Z_p$. So if we write the rows $f(m_1),f(m_2),\dots, f(m_n)$ one over another and remove all zero columns (after which only a finite number of columns remains), we obtain a matrix with linearly independent rows. In particular, this matrix has a nonsingular square submatrix of size $n\times n$. Now consider the corresponding submatrix for $m_1,m_2,\dots, m_n$. By Corollary 4 the rows of this submatrix are independent over $\mathbb Z_p D$. So $m_1,m_2,\dots, m_n$ are independent over $\mathbb Z_p D$. We arrive at a contradiction, thus, the assumption of the dependence of $\{m_i\}_{i\in I}$ is false. The lemma is proved. § 3. The proof of the main theoremLet us use induction on $n$. If $n=1$ (that is, the group $G$ is abelian), then the divisible hull of $G$ can be taken as $\widehat G$. Indeed, suppose that $\{w_j=1\}_{j \in J}$ is a nonsingular (but not even necessarily $p$-nonsingular) system of equations over a divisible abelian group $H$ in variables $\{x_i\}_{i \in I}$. The group $H$ is naturally embedded into
$$
\begin{equation*}
R=( H \times (\times_{i\in I}\langle x_i \rangle_\infty) ) \big/\langle \{w_j\}_{j\in J}\rangle.
\end{equation*}
\notag
$$
In fact, to show that the natural mapping is injective it suffices to show that $H \cap \langle \{w_j\}_{j\in J}\rangle=\{1\}$ in $H \times (\times_{i\in I}\langle x_i \rangle_\infty)$. Consider an element $w_j \in H \times (\times_{i\in I}\langle x_i \rangle_\infty)$ and its component in $\times_{i\in I}\langle x_i \rangle_\infty$. The group $\times_{i\in I}\langle x_i \rangle_\infty$ can be rewritten in the additive notation to obtain the $\mathbb Z$-module $\sum_{i \in I}\mathbb Z \cdot x_i$. In this notation the component of $w_j$ is equal to $m_j$, the image of $w_j$ under trivialization. Then the $(\times_{i\in I}\langle x_i \rangle_\infty)$-component of an arbitrary element $w_{j_1}^{n_1}\dotsb w_{j_s}^{n_s}$ of $\langle \{w_j\}_{j\in J}\rangle$ in the additive notation is $n_1m_{j_1} + \dots + n_s m_{j_s}$. If the element $w_{j_1}^{n_1}\dotsb w_{j_s}^{n_s}$ lies in $H$, then its $(\times_{i\in I}\langle x_i \rangle_\infty)$-component is $1$ in the multiplicative notation, that is, $0$ in the additive notation. However, as the system $\{w_j=1\}$ is nonsingular, the $m_j$ are independent over $\mathbb Z$, hence $n_1m_{j_1} + \dots + n_s m_{j_s}=0$ only if $n_1=\dots=n_s=0$. This means that if the element $w_{j_1}^{n_1}\dotsb w_{j_s}^{n_s}$ lies in $H$, then it must be equal to $1$ (in the multiplicative notation). This shows that $H \cap \langle \{w_j\}_{j\in J}\rangle=\{1\}$ in $H \times (\times_{i\in I}\langle x_i \rangle_\infty)$ and also shows the injectivity of the natural mapping $H\to R$. Clearly, $R$ contains a solution $\{x_i\}_{i \in I}$ of $\{w_j=1\}$. Since $H$ is divisible, it is a direct factor of $R$. The images of $\{x_i\}_{i \in I}$ under the projection onto $H$ form a solution of the system in $H$. Now assume that $n>1$. We embed $G$ into the Cartesian wreath product $G_2 \mathbin{\overline{\wr}} B_1$ by the Kaloujnine–Krasner theorem [9]. After this we can embed $B_1$ into its divisible hull $\widehat B_1$ and embed $G_2$ (using induction) into a group $\widetilde G_2$ with a required subnormal series of length $n-1$ and such that every $p$-nonsingular system of equations over $\widetilde G_2$ has a solution in $\widetilde G_2$ itself. Thus, we embed $G_2 \mathbin{\overline{\wr}} B_1$ into the wreath product $\widetilde G_2 \mathbin{\overline{\wr}} \widehat B_1$. Note that the divisible hull $\widehat B_1$ of the $p'$-torsion-free abelian group $B_1$ is $p'$-torsion free too (to understand this, it suffices to consider the torsion-free component and the $p$-primary component of $\widehat B_1$ and to see that the projection of $\widehat B_1$ onto the product of these two components is injective on $B_1\subset \widehat B_1$). Now we need the following lemma. Lemma 2. Suppose that $\widehat H$ is a group such that every $p$-nonsingular system over it has a solution in $\widehat H$ itself. Also suppose that $\widehat B$ is a divisible $p'$-torsion-free abelian group. Then every $p$-nonsingular system of equations over the Cartesian wreath product $\widehat H \mathbin{\overline{\wr}} \widehat B=(\prod_{b \in \widehat B}\widehat H_b)\leftthreetimes \widehat B$ has a solution in $\widehat H \mathbin{\overline{\wr}} \widehat B$ itself. Proof. Suppose that $\{v_j=1\}_{j\in J}$ is a $p$-nonsingular system of equations over $\widehat H \mathbin{\overline{\wr}} \widehat B$ in variables $\{x_i\}_{i\in I}$. Consider the image of this system under the natural mapping of coefficients $\widehat H \mathbin{\overline{\wr}} \widehat B \to \widehat B$. This image is a $p$-nonsingular system of equations over the divisible abelian group $\widehat B$, and therefore it has a solution in $\widehat B$ itself.
This means that we can change the variables in the original system $\{v_j=1\}_{j\in J}$ over $\widehat H \mathbin{\overline{\wr}} \widehat B$ so that all words $v_j$ become words in the alphabet $\{x_i^{\pm b}\}_{i\in I,\, b\in \widehat B} \sqcup \prod_{b \in \widehat B}\widehat H_b$. Now we assume that all words $v_j$ were originally words in this alphabet. We denote $\prod_{b \in \widehat B}\widehat H_b$ by $C$. Let us introduce new variables $\{x_{ib}\}_{i\in I,\, b\in \widehat B}$ and rewrite the words $v_j$ in the coefficients from $C$ and the variables $\{x^{\pm 1}_i\}$ as words $w_j$ in the coefficients from $C$ and the variables $\{x^{\pm 1}_{ib}\}$ such that for every $j \in J$, $v_j$ is $w_j$ in which each $x^{\pm 1}_{ib}$ is replaced by $x_i^{\pm b}$ for all $i\in I$ and $b \in \widehat B$. In other words,
$$
\begin{equation*}
v_j=w_j\Bigl(c_{(j,1)},\dots, c_{(j,k_j)}, x_{i_{(j,1)}}^{b_{(j,1)}}, \dots, x_{i_{(j,l_j)}}^{b_{(j,l_j)}}\Bigr).
\end{equation*}
\notag
$$
We search for a solution of $\{v_j=1\}_{j \in J}$ among the elements of $C$. Let $\{\widetilde x_i\}_{i\in I}$ be the set of elements of $C$. Then we let $[\widetilde x_i]_b$ denote the coordinate of $\widetilde x_i$ corresponding to the factor $\widehat H_b$. Note that for every $d \in \widehat B$ For every $i \in I$ we replace $x_i$ in $v_j$ by $\widetilde x_i$. We denote the result (which is an element of $C$) by $\widetilde v_j$ and find its coordinates:
$$
\begin{equation*}
\begin{aligned} \, [\widetilde v_j]_b &=w_j \Bigl( [c_{(j,1)}]_b,\dots, [c_{(j,k_j)}]_b, [\widetilde x_{i_{(j,1)}}^{b_{(j,1)}}]_b, \dots, [\widetilde x_{i_{(j,l_j)}}^{b_{(j,l_j)}}]_b \Bigr) \\ &=w_j \bigl( [c_{(j,1)}]_b,\dots, [c_{(j,k_j)}]_b, [\widetilde x_{i_{(j,1)}}]_{bb_{(j,1)}^{-1}}, \dots, [\widetilde x_{i_{(j,l_j)}}]_{bb_{(j,l_j)}^{-1}} \bigr). \end{aligned}
\end{equation*}
\notag
$$
Thus, finding a solution $\{\widetilde x_i\}_{i\in I}$ of the equation $\{v_j=1\}_{j\in J}$ in $C=\prod_{b \in \widehat B}\widehat H_b$ is the same as finding a set $\{[\widetilde x_i]_b\}_{i\in I,\, b\in B}$ of elements of $\widehat H$ such that
$$
\begin{equation*}
w_j \bigl( [c_{(j,1)}]_b,\dots, [c_{(j,k_j)}]_b, [\widetilde x_{i_{(j,1)}}]_{bb_{(j,1)}^{-1}}, \dots, [\widetilde x_{i_{(j,l_j)}}]_{bb_{(j,l_j)}^{-1}} \bigr) =1
\end{equation*}
\notag
$$
for all $j\in J$ and $b \in \widehat B$.
In other words, if the system of equations
$$
\begin{equation*}
\bigl\{w_j\bigl([c_{(j,1)}]_b,\dots, [c_{(j,k_j)}]_b, y_{i_{(j,1)},bb_{(j,1)}^{-1}},\dots, y_{i_{(j,l_j)},bb_{(j,l_j)}^{-1}}\bigr)=1 \mid j \in J,\, b \in \widehat B \bigr\}
\end{equation*}
\notag
$$
in the variables $\{y_{i,b} \mid i \in I, b \in \widehat B\}$ is solvable in $\widehat H$, then $\{v_j = 1\}$ has a solution consisting of elements of $C$, so that it is solvable in $\widehat H \mathbin{\overline{\wr}} \widehat B$, which proves the lemma. We denote
$$
\begin{equation*}
w_j \bigl( [c_{(j,1)}]_b,\dots, [c_{(j,k_j)}]_b, y_{i_{(j,1)},bb_{(j,1)}^{-1}}, \dots, y_{i_{(j,l_j)},bb_{(j,l_j)}^{-1}} \bigr)
\end{equation*}
\notag
$$
by $f_{j,b}$. By the assumption on the group $\widehat H$, if the system $\{f_{j,b}=1\}$ is $p$-nonsingular, then it is solvable in $\widehat H$ (as required). Let us show this.
Consider $p$-trivialization. Let $m_{j,b}$ be the image of $f_{j,b}$ under $p$-trivialization. There is a $\mathbb Z_p \widehat B$-module structure on $\sum_{i\in I,\, b \in \widehat B} \mathbb Z_p \cdot y_{i,b}$, which is defined by $d \cdot y_{i,b} = y_{i,db}$ for $d \in \widehat B$. Thus, we obtain a free $\mathbb Z_p \widehat B$-module $\sum_{i \in I}\mathbb Z_p\widehat B \cdot y_{i,1}$. Note that $m_{j,b}=b\cdot m_{j,1}$. Consider a linear combination of the rows $m_{j,b}$ that have the same index $j \in J$. A linear combination $n_1m_{j,b_1} + \dots + n_sm_{j,b_s}$ is equal to $(n_1b_1+\dots+n_sb_s)\cdot m_{j,1}$, that is, to $m_{j,1}$ times an element of the group algebra $\mathbb Z_p\widehat B$. Now consider a linear combination $\lambda$ of the rows $m_{j,b}$, where $j \in J$ can be different. It is a sum of linear combinations of the rows having the same index $j$. Each of these combinations looks like $(n_1b_1+\dots+n_tb_t)m_{j,1}$. So $\lambda$ is a combination of the rows $m_{j,1}$ with coefficients from $\mathbb Z_p \widehat B$ for some indices $j \in J$. In particular, the rows $\{m_{j,b}\}_{i \in J,\, b \in \widehat B}$ are linearly independent over $\mathbb Z_p$ if and only if the rows $\{m_{j,1}\}_{j \in J}$ are independent over $\mathbb Z_p \widehat B$. Consider the images of the rows $m_{j,1}$ under the natural ring homomorphism $\varepsilon\colon \mathbb Z_p \widehat B \to \mathbb Z_p$ which takes elements of $\widehat B$ to the unity. The image of the element at the $i$th position in the row $m_{j,1}$ (that is, of the coefficient of $y_{i,1}$ in $m_{j,1}$) is equal to the sum modulo $p$ of the sums of the exponents of the $y_{i,b}$ in $f_{j,1}$ for all $b\in\widehat B$ where the index $i$ is fixed. In other words, it is the sum over all $b\in\widehat B$ of the sums of the exponents of the $x_{i,b}$ (the index $i$ is fixed) in $w_j([c_{(j,1)}]_1,\dots, [c_{(j,k_j)}]_1,y_{i_{(j,1)},b_{(j,1)}^{-1}}, \dots, y_{i_{(j,l_j)},b_{(j,l_j)}^{-1}})$ modulo $p$. And this is the same as the sum of the exponents of $x_i$ in the word $v_j = w_j(c_{(j,1)},\dots, c_{(j,k_j)},x_{i_{(j,1)}}^{b_{(j,1)}}, \dots, x_{i_{(j,l_j)}}^{b_{(j,l_j)}})$ modulo $p$. So the image of the row $m_{j,1}$ is the same as the row of the sums of the exponents of the variables $x_i$ in the word $v_j$ modulo $p$. In particular, the images of the rows $m_{j,1}$ are linearly independent over $\mathbb Z_p$ since the system $\{v_j=1\}$ is $p$-nonsingular. Hence by Lemma 1 the rows $m_{j,1}$ are independent over $\mathbb Z_p \widehat B$. Thus, $\{f_{j,b}=1\}$ is $p$-nonsingular. This means that this system is solvable in $\widehat H$, which finishes the proof of the lemma. Once Lemma 2 has been proved, it can be applied to $\widetilde G_2 \mathbin{\overline{\wr}} \widehat B_1$ to show that any $p$-nonsingular system of equations over this group is solvable in $\widetilde G_2 \mathbin{\overline{\wr}} \widehat B_1$ itself. As this group contains $G$, we can take $\widetilde G_2 \mathbin{\overline{\wr}} \widehat B_1$ as $\widehat G$. The conclusion of the theorem is true for $\widehat G$: it has a required subnormal series of length $n$ and any $p$-nonsingular system of equations over $\widehat G$ is solvable in $\widehat G$ itself. Now the main theorem (Theorem 7) is proved. Remark 1. If $n$ (the length of a subnormal series) in the hypotheses of the main theorem is also equal to the derived length of $G$, then every $p$-nonsingular system of equations over $G$ has a solution in a solvable group of the same derived length as $G$. Remark 2. An analogue of Lemma 1 in the case when $D$ is a torsion-free abelian group and $\mathbb Z_p$ is replaced by $\mathbb Q$ is also true. Therefore, if we repeat the proof of the main theorem, then we can strengthen the result in [7]: for any group $G$ with a subnormal series having torsion-free abelian factors (where the last factor can have torsion) there is a group $\widetilde G \supset G$ with a similar subnormal series of the same length such that $\widetilde G$ contains a solution of every nonsingular system of equations over $G$ (and even over $\widetilde G$). In [7] different nonsingular systems over $G$ could have different solution groups. Remark 3. An analogous result for two primes $p$ and $q$ (that is, the main theorem for $\{p,q\}'$-torsion-free groups — these are groups in which the finite orders of elements are not divisible by any primes distinct from $\{p,q\}$ — instead of $p'$-torsion-free ones, and $\{p,q\}$-nonsingular systems — these are systems which are simultaneously $p$-nonsingular and $q$-nonsingular — instead of $p$-nonsingular ones) is not true. This is shown by the example of a group of order $42$ in [7], Proposition 1, a: the first factor of the subnormal series of this group is $\{2,3\}'$-torsion free and the equation over the group is unimodular, but this equation has no solutions in metabelian groups. Thus, the main theorem does not extend to the case of two or more prime numbers. Moreover, for every two prime numbers $p$ and $q$ we can construct an example of a metabelian group with $\{p,q\}'$-torsion-free first factor such that some unimodular equation over this group has no solutions in metabelian groups. Example 2. Let $p$ and $q$ be two different prime numbers. Consider the metabelian group $G=\langle c \rangle_2 \wr (\langle a \rangle_p \times \langle b \rangle_q)$. Its order is $2^{pq}pq$. Note that $c c^{ab} = [c,ab]$ belongs to the commutator subgroup of $G$. As $p$ and $q$ are coprime, there are integers $n$ and $m$ such that $np+mq=1$. Then consider the equation
$$
\begin{equation*}
x^n\cdot x^{na}\dotsb x^{na^{p-1}}\cdot x^m\cdot x^{mb}\dotsb x^{mb^{q-1}}=c c^{ab}
\end{equation*}
\notag
$$
over $G$. Since $np+mq=1$, this equation is unimodular, and its right-hand side is an element of the commutator subgroup of $G$. Assume that there is a metabelian group $\widetilde G \supset G$ containing a solution $\widetilde x \in \widetilde G$ of this equation. On taking the quotient $\widetilde G/\widetilde G'$, the equality
$$
\begin{equation*}
\widetilde x^n\cdot \widetilde x^{na}\dotsb \widetilde x^{na^{p-1}}\cdot \widetilde x^m\cdot \widetilde x^{mb}\dotsb \widetilde x^{mb^{q-1}}=c c^{ab}
\end{equation*}
\notag
$$
transforms into $\widetilde x = 1$. This means that $\widetilde x$ belongs to the commutator subgroup of $\widetilde G$. As $\widetilde G$ is metabelian, this implies that $\widetilde x$ commutes with its conjugates. Using the notation $\widetilde x^g \widetilde x^h=\widetilde x^{g+h}$ for simplicity, we obtain
$$
\begin{equation*}
\begin{aligned} \, &(\widetilde x^n\cdot \widetilde x^{na}\dotsb \widetilde x^{na^{p-1}}\cdot \widetilde x^m\cdot \widetilde x^{mb}\dotsb \widetilde x^{mb^{q-1}})^{1+ab} \\ &\qquad =\widetilde x^{(n(1+b)(1+a+\dots+a^{p-1})+m(1+a)(1+b+\dots+b^{q-1}) )} \\ &\qquad =(\widetilde x^n\cdot \widetilde x^{na}\dotsb \widetilde x^{na^{p-1}}\cdot \widetilde x^m\cdot \widetilde x^{mb}\dotsb \widetilde x^{mb^{q-1}})^{a+b}. \end{aligned}
\end{equation*}
\notag
$$
However,
$$
\begin{equation*}
(c c^{ab})^{1+ab}=c^{1+2ab+a^2b^2}\neq c^{a+b+a^2b+ab^2}=(c c^{ab})^{a+b}.
\end{equation*}
\notag
$$
Thus, we arrive at a contradiction: some combinations of conjugates to the left-hand side of the equation are equal, though the same combinations of conjugates to the right-hand side are not. Therefore, the assumption that a solution exists in a metabelian group is false. Remark 4. From the proof of the main theorem (Theorem 7) we can deduce that $\widehat G_n/\widehat G_{n+1}$ is a Cartesian power of the divisible hull of the group $B_n=G_n/G_{n+1}$. Therefore, if $B_n$ is $p'$-torsion free (like the other factors of the series of $G$), then $\widehat G_n / \widehat G_{n+1}$ is $p'$-torsion free too. More generally, $\widehat G_n / \widehat G_{n+1}$ has the same set of prime orders of elements as $B_n$. For example, if $B_n$ is $\langle b \rangle_{30}$, then $\widehat G_n / \widehat G_{n+1}$ has elements of prime orders $2$, $3$ and $5$ and has no elements of prime orders $7$, $13$ or $19$. § 4. Proof of minimalityIn this section we call a solvable group, over which there is a unimodular equation that has no solutions in solvable groups of the same derived length, a counterexample. ‘Counterexample’ here means a counterexample to the following statement: any unimodular equation over a solvable group has a solution in solvable groups with the same derived length. The main theorem (Theorem 7) shows that if a metabelian group $G$ is an extension of an abelian group by an abelian $p$-group for some prime number $p$, then any unimodular equation over $G$ has a solution in some metabelian group, as unimodular equations are $p$-nonsingular for every prime $p$. If, moreover, $G$ is abelian, then any unimodular equation has a solution in $G$ itself. So in this case $G$ cannot be a counterexample. This helps us to prove that every metabelian group of order at most $41$ is not a counterexample and thus the group from Proposition 2 is indeed a minimal counterexample with respect to order. Consider metabelian groups of order at most $41$. First note that, if the order of $G$ is $p^k$, where $p$ is a prime number, then $G$ is nilpotent, and unimodular equations over a nilpotent group are solvable in this group itself by Shmel’kin’s theorem [14]. Hence groups of order $p^k$ (even not necessarily metabelian) are not counterexamples. In our case these are groups of orders $1$, $2$, $3$, $4$, $5$, $7$, $8$, $9$, $11$, $13$, $16$, $17$, $19$, $23$, $25$, $27$, $29$, $31$, $32$, $37$ and $41$. Further, note that a nonabelian group of order $pq$, where $p<q$ are two distinct primes, is (if it exists) an extension of $\langle g \rangle_q$ by $\langle h \rangle_p$ (see [3], for example). This is why groups of orders $6$, $10$, $14$, $15$, $21$, $22$, $26$, $33$, $34$, $35$, $38$ and $39$ are not counterexamples either. Also, the following fact holds: in a group of order $p^2 q$, where $p$ and $q$ are distinct primes, one of Sylow subgroups is normal (this fact can also be seen from [3]). This means that this group is either an extension of a group of order $p^2$ (which is an abelian $p$-group) by $\langle g \rangle_q$ or an extension of $\langle g \rangle_q$ by an abelian $p$-group. In both cases this group is not a counterexample. Thus, orders $12$, $18$, $20$ and $28$ need not be considered. The remaining orders are framed in Table 1. These are $24$, $30$, $36$ and $40$. Table 1.Considered and unconsidered orders of groups
Suppose that a metabelian group $G$ has one of these orders. Consider a maximal abelian group $H$ containing the commutator subgroup of $G$. Then $H$ is normal with abelian factor $G/H$ (as $H$ contains the commutator subgroup). Note that the centralizer of $H$ in $G$ is precisely $H$ itself because, if an element $g$ lies in the centralizer of $H$ but does not lie in $H$, then the subgroup $\langle g, H \rangle$ is also abelian and strictly contains $H$, which is a contradiction to the maximality of $H$. In other words, if we consider the action $G$ on $H$ by conjugation, then we obtain an embedding of $G/H$ into $\operatorname{Aut}(H)$. Now consider each of these orders separately. Order $24$: $H$ cannot have order $1$ or $2$ (as subgroups of these orders are central). If $H$ has order $3$, $6$, $8$, $12$ or $24$, then $G/H$ is an abelian $p$-group (or $G$ itself is abelian). In this case $G$ is not a counterexample. The remaining case is $|H| = 4$. In this case $H$ is either $\langle h \rangle_4$ or $V_4$. Hence $\operatorname{Aut}(H)$ is either $\langle a \rangle_2$ or $S_3$. As $G/H$ embeds into $\operatorname{Aut}(H)$ and has order $6$, $\operatorname{Aut}(H)\cong S_3$. But $G/H$ is a cyclic group of order $6$ (as it is abelian) and cannot be embedded into $S_3$. This is why the case $|H|=4$ is impossible. We see that if $|G| = 24$, then $G$ is not a counterexample. Order $30$: again, $H$ cannot have order $1$ or $2$. If $H$ is of order $6$, $10$ or $15$, then $G/H$ is an abelian $p$-group (or $G$ itself is abelian). If $|H|=3$ or $5$, then $\operatorname{Aut}(H)$ has order either $2$ or $4$, and $G/H$ (which has order $10$ or $6$) cannot be embedded into $\operatorname{Aut}(H)$. So $|H|$ cannot be $3$ or $5$. Therefore, if $|G|=30$, then $G$ is not a counterexample either. Order $36$: If $H$ has order $4$, $9$, $12$, $18$ or $36$, then $G/H$ is an abelian $p$-group (or $G$ is abelian itself). The group $H$ cannot have order $1$, $2$ (as subgroups of these orders are central), $3$ (because then $|\operatorname{Aut}(H)| = 2$ and $|G/H| = 18$) or $6$ (because then $|\operatorname{Aut}(H)| = 2$ and $|G/H| = 6$). Hence $G$ is not a counterexample in this case. Order $40$: If $H$ has order $5$, $8$, $10$, $20$ or $40$, then $G/H$ is an abelian $p$-group (or $G$ is abelian itself). The group $H$ cannot have order $1$, $2$ (as subgroups of these orders are central) or $4$ (because then $|\operatorname{Aut}(H)|=2$ or $6$, while $|G/H| = 10$). So in this case $G$ is not a counterexample either. As a result, a metabelian group of order at most $41$ cannot be a counterexample. Combining this with a counterexample of order $42$ we obtain the proof of Proposition 2. § 5. Open questionsQuestion 1. Is there any example of a solvable group of derived length greater than $2$ such that not every unimodular equation over this group has a solution in solvable groups of the same derived length? Question 2. In particular, is there an example of a solvable group of order less than $42$ but of derived length greater than $2$ such that not every unimodular equation over this group has a solution in solvable groups of the same derived length? For example, is $S_4$ (which is a group of order $24$ and derived length $3$) such an example? This group has no subnormal series satisfying the assumptions of the main theorem. Question 3. Does every unimodular equation over the group from Proposition 2 have a solution in a solvable group? 
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\Bibitem{Mik24} |
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