
This article is cited in 4 scientific papers (total in 4 papers)
Classification of involutive commutative twovalued groups
V. M. Buchstaber^{abc}, A. P. Veselov^{d}, A. A. Gaifullin^{aebf} ^{a} Steklov Mathematical Institute of Russian Academy of Sciences, Moscow
^{b} Lomonosov Moscow State University
^{c} National Research University "Higher School of Economics"
^{d} Loughborough University, Loughborough, UK
^{e} Skolkovo Institute of Science and Technology
^{f} Institute for Information Transmission Problems of the Russian Academy of Sciences (Kharkevich Institute)
Abstract:
A complete classification of finitely generated involutive commutative twovalued groups is obtained. Three series of such twovalued groups are constructed: a principal series, a unipotent series, and a special series; it is shown that any finitely generated involutive commutative twovalued group is isomorphic to a twovalued group in one of these series. A number of classification results are obtained for topological involutive commutative twovalued groups in the Hausdorff and locally compact cases. The classification of algebraic involutive twovalued groups in the onedimensional case is also discussed.
Bibliography: 45 titles.
Keywords:
classification of multivalued groups, involutive twovalued groups, topological groups, locally compact groups, algebraic twovalued groups.
Received: 12.05.2022
1. Introduction A multivalued group is a generalization of an ordinary group such that the product of any pair of elements is a multiset, that is, an unordered set of elements (possibly with repetitions). The theory of multivalued multiplications, where, given a multivalued group, the cardinality of the result of multiplication of two elements can take various values, has a rich history dating back to the 19th century (see [28]). In this paper we study $n$valued groups. The key to the definition of an $n$valued group is to fix the cardinality $n$ of the multiset that is a product of two elements. The concept of $n$valued group originates from a construction due to Buchstaber and Novikov [11] in the theory of characteristic classes of vector bundles. The foundations of the algebraic theory of $n$valued groups were laid in works of Buchstaber. This theory and its applications were developed by Buchstaber, his students (Kholodov, Yagodovsky, and others), and coauthors (Rees, Veselov, Monastyrsky, Dragović, and others); see references in the surveys [7] and [17]. In this paper we deal only with twovalued groups. We denote by $\operatorname{Sym}^2(X)$ the second symmetric power of the set $X$, that is, the set of all twoelement multisets with elements in $X$. We use square brackets to enumerate elements of multisets. Definition 1.1. A twovalued group is a set $X$ with twovalued multiplication $*\colon X\times X\to \operatorname{Sym}^2(X)$, an identity element $e\in X$, and an operation of taking an inverse $x\mapsto x^{1}$ with the following properties. A twovalued group is called commutative if
$$
\begin{equation*}
x*y=y*x
\end{equation*}
\notag
$$
for all $x,y\in X$. Remark 1.2. The original definition of a twovalued group (see [7]) did not require the uniqueness of an inverse element. This additional condition was used in [15] and [17]. For the present work this condition has turned out to be important. The most important source of $n$valued groups is the construction of coset multivalued groups (see [7], § 6). Coset $n$valued groups are constructed from a pair consisting of a singlevalued group $G$ and a finite subgroup $H$ of its automorphism group. At the same time, the theory of $n$valued groups is not exhausted by coset groups. Firstly, different pairs $(G,H)$ can lead to isomorphic $n$valued groups; secondly, there is a rich class of $n$valued groups that are not coset groups (see [7] and [15]). We will need the following construction of coset twovalued groups. Construction 1.3 (coset twovalued group). Let $G$ be an ordinary (singlevalued) group and $\iota\colon G\to G$ be an involutive automorphism (such that $\iota^2=\mathrm{id}$) of $G$. Then the quotient set $X=G/\iota$ is endowed with the structure of a twovalued group in such a way that
$$
\begin{equation*}
\pi(g)*\pi(h)=\bigl[\pi(gh),\pi(g\iota(h))\bigr]
\end{equation*}
\notag
$$
for all $g,h\in G$, where $\pi\colon G\to X$ is the natural projection. The identity element of this twovalued group is the image of the identity of the group $G$, and the inverse element is defined by the formula $\pi(g)^{1} = \pi(g^{1})$. In the case of an abelian group $G$ there is an important example of an involution, namely, taking the inverse element (the antipodal involution): $\iota_{\mathbf{a}}(g)= g^{1}$. One important feature of the theory of $n$valued groups is the high complexity of classification and enumeration problems. It is due to the fact that the number of isomorphism classes of $n$valued groups on a set of $k$ elements grows extremely rapidly with $k$ and $n$. The same effect holds even if we consider only commutative $n$valued groups. We are aware of attempts to solve the problem of enumerating isomorphism classes of $n$valued groups for given $n$ and $k$ by using highperformance computing technology, which did not lead to significant results because of the computational complexity of the problem. In this connection the problem of identifying those classes of $n$valued groups for which classification problems are amenable, is of natural interest. An important class of this type is the class of involutive twovalued groups introduced by Buchstaber and Veselov in [17] in connection with the Conway topograph. Definition 1.4. An element $x$ of a twovalued group $X$ is called a weak involution if $x^{1}=x$ or, equivalently, if the multiset $x*x$ contains the identity $e$. An element $x\in X$ is called a strong involution if $x*x=[e,e]$. A twovalued group $X$ is called involutive if it consists of weak involutions, that is, $x^{1}=x$ for all $x\in X$. Remark 1.5. Note that in [15] the concept of an involutive $n$valued group has a different meaning: the involutivity of the operation of taking the inverse element $x\mapsto x^{1}.$ It is clear that the class of involutive groups in the sense of [15] contains the class of involutive groups in the sense of [17]. The main result in [15] is the proof that the group algebras of involutive $n$valued groups introduced there coincide with socalled combinatorial algebras (see [1]). In this paper we use the term ‘involutive twovalued group’ in the sense of [17]. The main goal of this paper is to classify involutive commutative twovalued groups. The problem of classifying such twovalued groups was stated in [17]. In that paper it was conjectured that every involutive commutative twovalued group is isomorphic to a coset twovalued group of the form $A/\iota_{\mathbf{a}}$, where $A$ is an abelian group and $\iota_{\mathbf{a}}(g)=g^{1}$ is the antipodal involution on this group. We show that this conjecture is false even for finite twovalued groups. We consider the classification problem for involutive commutative twovalued groups in the following classes: A complete classification will be obtained in the case of finitely generated groups (Theorem 1.10), as well as in the case of compact topological groups without small subgroups (Theorem 15.8). Most of this paper is devoted to the finitely generated case. After that we discuss how our approach extends to other classes of twovalued groups and what results and new questions are obtained along the way. Let us now formulate a classification theorem for finitely generated involutive commutative twovalued groups. We show that, along with the principal series, which consists of the coset twovalued groups of the form $A/\iota_{\mathbf{a}}$, there are two other series of finitely generated involutive commutative twovalued groups; we call them the unipotent and special series. Interestingly, twovalued groups of the unipotent series are also obtained from abelian groups using the coset construction, but with respect to a nonantipodal involution. Twovalued groups in the special series are not coset (see Proposition 4.4 below). In particular, our result gives new constructions of noncoset twovalued groups. Now we present the constructions of these series of involutive commutative twovalued groups. Throughout this article we denote by $C_n$ the cyclic group of order $n$, in particular, $C_{\infty}$ is an infinite cyclic group. It will be convenient for us to use the multiplicative notation for all groups, including abelian ones. 1. The principal series Let $A$ be a (singlevalued) abelian group. Consider the antipodal involution $\iota_{\mathbf{a}}\colon A\to A$ defined by $\iota_{\mathbf{a}}(a)=a^{1}$. Then the coset twovalued group
$$
\begin{equation*}
X^{\mathbf{a}}_A=A/\iota_{\mathbf{a}}
\end{equation*}
\notag
$$
is an involutive commutative group. If $A$ is a finitely generated abelian group, then, as is well known, it can be represented uniquely up to isomorphism as a product $C_{d_1}\times C_{d_2}\times \cdots\times C_{d_k}$, where $2\leqslant d_i\leqslant \infty$ and $d_i$ divides $d_{i+1}$ for all $i=1,\dots, k1$ (see, for example, [41]). Set
$$
\begin{equation*}
X_{d_1,\dots,d_k}^{\mathbf{a}}= X_{C_{d_1}\times\cdots\times C_{d_k}}^{\mathbf{a}}.
\end{equation*}
\notag
$$
It can be verified directly that if the group $A$ is finite, then the cardinality of the twovalued group $X^{\mathbf{a}}_A$ is
$$
\begin{equation*}
\frac{1}{2}(d_1d_2\cdots d_k+2^r),
\end{equation*}
\notag
$$
where $r$ is the number of even integers among $d_1,\dots,d_k$. Remark 1.6. The simplest but important case of involutive twovalued groups in the principal series is the groups $X_{m\times 2}^{\mathbf{a}}=C_2^m/\iota_{\mathbf{a}}$ (where $m\times 2$ denotes the sequence $2,\dots,2$ of $m$ twos). The twovalued group $X_{m\times 2}^{\mathbf{a}}$ is the socalled double of the group $C_2^m$. This means that $X_{m\times 2}^{\mathbf{a}}$ coincides with $C_2^m$ as a set, and the operation in $X_{m\times 2}^{\mathbf{a}}$ is obtained by doubling the operation in $C_2^m$, that is, $x*y=[xy,xy]$. Note that the double can be defined for any group $G$. The result is always a twovalued group, which is involutive in the sense of [15]. It is commutative and involutive in the sense of [17] only if $G$ is a vector space over the twoelement field $\mathbb{F}_2$. Remark 1.7. It is natural to consider twovalued groups $X_{\infty,\dots,\infty}^{\mathbf{a}}$ as free involutive commutative twovalued groups. The twovalued groups $X_{\infty}^{\mathbf{a}}=C_{\infty}/\iota_{\mathbf{a}}$ and $X_{\infty,\infty}^{\mathbf{a}}=(C_{\infty}\times C_{\infty})/\iota_{\mathbf{a}}$ were studied in detail in [17], where they were called the Buchstaber–Novikov twovalued group and the Conway twovalued group and denoted by $\mathbb{X}_1$ and $\mathbb{X}_2$, respectively. 2. The unipotent series We need the notion of a Boolean group. A group is said to be Boolean if each nonidentity element of it has order $2$ (see [21], Chap. 1). It is easy to show that any such group is abelian. Thus, a Boolean group is a vector space over the twoelement field $\mathbb{F}_2$ with the operation written in the multiplicative form. For a Boolean group $V$ we denote its dimension as a vector space by $\dim V$. Let $V$ be a Boolean group. Consider the unipotent involution
$$
\begin{equation*}
\iota_{\mathbf{u}}\colon V\times V\to V\times V
\end{equation*}
\notag
$$
defined by
$$
\begin{equation}
\iota_{\mathbf{u}}(a,b)=(a,ab).
\end{equation}
\tag{1}
$$
Then the coset twovalued group
$$
\begin{equation*}
X^{\mathbf{u}}_V=(V\times V)/\iota_{\mathbf{u}}
\end{equation*}
\notag
$$
is an involutive commutative twovalued group. Remark 1.8. Using the automorphism of the group $V\times V$ defined by $(a,b)\mapsto (ab,b)$, the involution $\iota_{\mathbf{u}}$ is transformed into the permutation involution
$$
\begin{equation*}
\iota_{\mathbf{t}}\colon (a,b)\mapsto (b,a),
\end{equation*}
\notag
$$
but it is more convenient for our purposes to consider the involution in the unipotent form (1). The equivalence of these two involutions is explained by the wellknown isomorphism
$$
\begin{equation*}
\operatorname{GL}(2,\mathbb{F}_2)\cong S_3
\end{equation*}
\notag
$$
with the permutation group of a threeelement set, all elements of order $2$ in which are conjugate. It can be verified directly that if $\dim V=n<\infty$, that is, $V\cong C_2^n$, then the twovalued group
$$
\begin{equation*}
X^{\mathbf{u}}_n=X^{\mathbf{u}}_{C_2^n}
\end{equation*}
\notag
$$
consists of $2^{2n1}+2^{n1}$ elements. If $n=1$, then $X_1^{\mathbf{u}}=X^{\mathbf{u}}_{C_2}$ is a threeelement group $\{e,v,x\}$ with multiplication table $v*v=[e,e]$, $v*x=[x,x]$, $x*x=[e,v]$. It is easy to see that $X^{\mathbf{u}}_{C_2}\cong X^{\mathbf{a}}_{C_4}$. When $n=2$, there is a less obvious isomorphism of $10$element twovalued groups $X^{\mathbf{u}}_2\cong X^{\mathbf{a}}_{4,4}$, that is, $X^{\mathbf{u}}_{C_2\times C_2}\cong X^{\mathbf{a}}_{C_4\times C_4}$ (see Example 1.11 and § 9). Thus, when $\dim V\leqslant 2$, the twovalued groups $X^{\mathbf{u}}_V$ are contained in the principal series of examples. Nevertheless, as we will see below, for $\dim V\geqslant 3$ the twovalued groups $X^{\mathbf{u}}_V$ are not isomorphic to any twovalued groups of the principal series. Remark 1.9. It is easy to see that a coset twovalued group $G/\iota$ is involutive if and only if for any element $g\in G$ one of the two equalities $g^{1}=g$ or $g^{1}=\iota(g)$ holds. The first possibility is realized for twovalued groups of the unipotent series, and the second for twovalued groups of the principal series. 3. The special series Let $V$ be a Boolean group with identity $e$. We add an element to the set $V$, which we denote by $s$, and define a commutative twovalued operation on the set $Y_V=V\cup\{s\}$ by
$$
\begin{equation*}
\begin{alignedat}{2} x*y&=[xy,xy],&\qquad x,y&\in V,\quad x\ne y, \\ x*x&=[e,s],&\qquad x&\in V,\quad x\ne e, \\ s*x&=[x,x],&\qquad x&\in V,\quad x\ne e, \\ e*x&=[x,x],&\qquad x&\in V\cup\{s\}, \\ s*s&=[e,e].&& \end{alignedat}
\end{equation*}
\notag
$$
It can be verified directly that the operation introduced is well defined, associative, and transforms $Y_V$ into an involutive commutative twovalued group. If $\dim V=n<\infty$, that is, $V\cong C_2^n$, then the twovalued group
$$
\begin{equation*}
Y_n=Y_{C_2^n}
\end{equation*}
\notag
$$
consists of $2^n+1$ elements. For $n=1$ the twovalued group $Y_1$ consists of the three elements $e$, $s$, and $x$ and has the multiplication table
$$
\begin{equation*}
s*s=[e,e],\quad s*x=[x,x],\quad x*x=[e,s].
\end{equation*}
\notag
$$
Thus, $Y_1\cong X^{\mathbf{a}}_{C_4} \cong X^{\mathbf{u}}_{C_2}$. Nevertheless, as we will see below, for $\dim V\geqslant 2$ the twovalued group $Y_V$ is not coset and, in particular, it is not isomorphic to any two valued group of the principal or unipotent series. Let us give another interpretation of the twovalued group $Y_V$. The set $V\setminus\{e\}$ can naturally be considered as the projectivization $\mathbb{P}(V)$ of the vector space $V$. Then $Y_V$ is the projective space $\mathbb{P}(V)$ with two additional elements $e$ and $s$. The multiplication rule is as follows: 4. A product with a Boolean group If $X$ is an involutive commutative twovalued group and $W$ is a Boolean group, then the direct product $X\times W$ is again endowed with the structure of an involutive commutative twovalued group, multiplication in which is defined by the formula
$$
\begin{equation*}
(x_1,w_1)*(x_2,w_2)=(x_1*x_2,w_1w_2).
\end{equation*}
\notag
$$
More precisely, if $x_1*x_2=[z_1,z_2]$, then
$$
\begin{equation*}
(x_1,w_1)*(x_2,w_2)=\bigl[(z_1,w_1w_2),(z_2,w_1w_2)\bigr].
\end{equation*}
\notag
$$
If we take involutive commutative groups of the principal series as $X$, then this construction does not lead to new examples, since for any abelian group $A$ and any Boolean group $W$ there is a canonical isomorphism
$$
\begin{equation}
X^{\mathbf{a}}_A\times W\cong X^{\mathbf{a}}_{A\times W}.
\end{equation}
\tag{2}
$$
However, if we take twovalued groups of the unipotent or special series as $X$, then we obtain new examples of involutive twovalued groups. Now we are ready to state a classification theorem for finitely generated involutive commutative twovalued groups. Theorem 1.10. Every finitely generated involutive commutative twovalued group is isomorphic to one of the following twovalued groups: For each $m\geqslant 0$ there are isomorphisms
$$
\begin{equation}
X^{\mathbf{a}}_{m\times 2,4}\cong X^{\mathbf{u}}_1\times C_2^m\cong Y_1\times C_2^m
\end{equation}
\tag{3}
$$
$$
\begin{equation}
\textit{and} \qquad X^{\mathbf{a}}_{m\times 2,4,4}\cong X^{\mathbf{u}}_2\times C_2^m,
\end{equation}
\tag{4}
$$
where $m\times 2$ denotes the sequence $2,\dots,2$ of $m$ twos. The indicated isomorphisms exhaust all isomorphisms between twovalued groups in the above list. Example 1.11. Let us consider $10$element twovalued groups $X^{\mathbf{a}}_{4,4}=C_4^2/\iota_{\mathbf{a}}$ and $X^{\mathbf{u}}_2=(C_2^2\times C_2^2)/\iota_{\mathbf{u}}$ and present an explicit isomorphism between them. Denote by $a_1$ and $a_2$ the generators of the factors in the group $C_4^2$ and denote by $b_1$ and $b_2$ the generators of the factors in $C_2^2$. Define a mapping $f\colon C_4^2\to C_2^2\times C_2^2$ by the formula
$$
\begin{equation*}
f(a_1^ka_2^l)=(b_1^kb_2^l,b_1^{[k/2]}b_2^{[l/2]}),\qquad 0\leqslant k\leqslant 3,\quad 0\leqslant l\leqslant 3,
\end{equation*}
\notag
$$
where $[r]$ denotes the integer part of the number $r$. The mapping $f$ is not a homomorphism. However, it can be verified directly that for any elements $a,c\in C_4^2$ the element $f(ac)$ coincides with one of the four elements $f(a)f(c)$, $f(a)\iota_{\mathbf{u}}(f(c))$, $\iota_{\mathbf{u}}(f(a))f(c)$, and $\iota_{\mathbf{u}}(f(a))\iota_{\mathbf{u}}(f(c))$; moreover, there is a coincidence of multisets
$$
\begin{equation*}
\begin{aligned} \, &\bigl[f(ac),f(ac^{1}),f(a^{1}c),f(a^{1}c^{1})\bigr] \\ &\qquad=\bigl[f(a)f(c),f(a)\iota_{\mathbf{u}}(f(c)), \iota_{\mathbf{u}}(f(a))f(c),\iota_{\mathbf{u}}(f(a)) \iota_{\mathbf{u}}(f(c))\bigr]. \end{aligned}
\end{equation*}
\notag
$$
Therefore, $f$ induces a welldefined isomorphism $X^{\mathbf{a}}_{4,4}\cong X^{\mathbf{u}}_2$. Together with the isomorphism (2), this gives the isomorphism (4). Example 1.12. Consider the $5$element group $Y_2=Y_{C_2^2}$, which is the first group in the special series that is not contained in the principal series. Let $x$, $y$, and $z$ be the three nontrivial elements of $C_2^2$. Then the multiplication table in the twovalued group $Y_2$ has the form
$$
\begin{equation*}
\begin{alignedat}{2} x*x&=y*y=z*z=[e,s],&\quad s*s&=[e,e], \\ x*y&=[z,z],\quad y*z=[x,x],&\quad z*x&=[y,y], \\ s*x&=[x,x],\quad s*y=[y,y],&\quad s*z&=[z,z]. \end{alignedat}
\end{equation*}
\notag
$$
This twovalued group is related to the $8$element nonabelian group of quaternions $Q_8=\{\pm 1,\pm i,\pm j,\pm k\}\subset\mathbb{H}$, where $\mathbb{H}$ is the algebra of quaternions. On $Q_8$ there is a conjugation antiautomorphism $\sigma$: $\sigma(gh)=\sigma(h)\sigma(g)$, which fixes the elements $1$ and $1$ and interchanges $i \leftrightarrow i$, $j\leftrightarrow j$, and $k\leftrightarrow k$. The construction of a coset twovalued group from an involutive automorphism does not carry over to the case of an antiautomorphism as a rule. Nevertheless, the antiautomorphism $\sigma$ of $Q_8$ has the following special property: for any two elements $g,h\in Q_8$ the fourelement multiset $[gh,g\sigma(h),\sigma(g)h,\sigma(g)\sigma(h)]$ is the doubling of some twoelement multiset. This property implies easily that the $5$element quotient set $Q_8/\sigma$ has a welldefined structure of a twovalued group. Since multiplication of quaternion units is commutative up to sign, the twovalued group $Q_8/\sigma$ is commutative. It is easy to see that it is isomorphic to $Y_2$ via the isomorphism $1\mapsto e$, $1\mapsto s$, $\pm i\mapsto x$, $\pm j\mapsto y$, $\pm k\mapsto z$. Example 1.13. Similarly to the previous example, there is a connection between the $9$element twovalued group $Y_3$ and the $16$element set
$$
\begin{equation*}
O_{16}=\{\pm 1,\pm e_0,\pm e_1,\dots,\pm e_6\}\subset\mathbb{O},
\end{equation*}
\notag
$$
where $\mathbb{O}$ is the algebra of octonions and $1,e_0,\dots,e_6$ is its standard basis. Since multiplication of octonions is not associative, the set $O_{16}$ is not even a group, but only a socalled Moufang loop (see [ 18], § 7.1). However, conjugation defines an involutive antiautomorphism of this algebraic structure. It can be verified directly that, again, for any two elements $g,h\in O_{16}$ the fourelement multiset $[gh,g\sigma(h),\sigma(g)h,\sigma(g)\sigma(h)]$ is the doubling of some twoelement multiset. This implies again that there is welldefined twovalued multiplication on the $9$element quotient set $O_{16}/\sigma$. Although multiplication in $O_{16}$ is neither associative nor commutative, it is associative and commutative up to sign. Using this, one can check directly that the resulting twovalued multiplication in $O_{16}/\sigma$ is associative and commutative and defines the structure of a commutative twovalued group on this set. Again, it can be verified directly that this twovalued group is isomorphic to $Y_3$. This work is organized as follows. In § 2 we classify singlegenerated involutive twovalued groups. In particular, we show that all such twovalued groups are commutative. Sections 3–12 are devoted to the classification of involutive commutative twovalued groups in the finitely generated case. Sections 13–15 contain a number of classification results in the topological case. In § 16 the case of algebraic twovalued groups is considered. In the final section, § 17, we discuss some open questions.
2. Singlegenerated involutive twovalued groups The problem of the classification of singlegenerated ordinary groups is very simple: the answer is given by the infinite and all possible finite cyclic groups. In the case of $n$valued groups, the problem of classifying singlegenerated groups (even commutative twovalued groups) has turned out to be difficult and is currently far from a complete solution. A rich source of singlegenerated commutative $n$valued groups is given by multivalued multiplication on the set of irreducible representations of a finite group (see [7], §§ 10 and 11). More precisely, the following theorem holds: if a finite group $G$ has a faithful irreducible representation $\rho$, then the multivalued group on the set of its irreducible representations is singlegenerated. The proof is based on Burnside’s theorem (see [19]), which states that in this case any irreducible representation of the group $G$ is contained in a decomposition of some tensor power of the representation $\rho$. Note also that the problem of singlegenerated multivalued groups is closely related to the problem of the integrability of multivalued dynamical systems with discrete time (see [39] and [16]). Nevertheless, in the class of involutive twovalued groups considered in this paper the problem of classifying singlegenerated groups can be completely solved even without the assumption of commutativity. The answer turns out to be as simple as for ordinary groups (the case of free single generated twovalued groups was considered in [17] in connection with the Conway topograph). Theorem 2.1. Any singlegenerated involutive twovalued group is isomorphic to a coset group of the form $X^{\mathbf{a}}_C=C/\iota_{\mathbf{a}}$, where $C$ is a cyclic group and $\iota_{\mathbf{a}}\colon C\to C$ is an antipodal involution. In particular, every such twovalued group is commutative. The proof is divided into several lemmas, which will be useful to us in their own right. Lemma 2.2. Let $X$ be an involutive twovalued group and $x,y,z\in X$ be elements such that $z$ belongs to the multiset $x*y$. Then $x$ belongs to the multiset $y*z$. Proof. Let $x*y=[z,z']$. Then, in view of the involutivity of the twovalued group $X$, the multiset $(x*y)*z=[z*z,z'*z]$ contains the identity $e$. On the other hand, in view of associativity, this multiset can be written as $x*(y*z)$. From the fact that it contains $e$ and the involutivity of $X$ it follows that $y*z$ contains $x$. $\Box$ Lemma 2.3. Let $X$ be an involutive twovalued group. Then for any element $x\in X$ there exists a unique sequence $x^0,x^1,x^2,\dots$ of elements of $X$ such that $x^0=e$, $x^1=x$, and
$$
\begin{equation}
x*x^k=x^k*x=[x^{k1},x^{k+1}]
\end{equation}
\tag{5}
$$
for all $k\geqslant 1$. Proof. Let us show that (5), together with the initial conditions $x^0=e$ and $x^1=x$, defines a consistent recurrence rule for constructing the sequence $\{x^k\}_{k\geqslant 0}$ uniquely. Assume that the elements of $x^0,x^1,\dots,x^n$, where $n\geqslant 1$, have already been defined so that equalities (5) hold for $1\leqslant k\leqslant n1$. We show that $x^{n+1}$ can uniquely be defined so that (5) holds for $k=n$. From equalities (5) for $1\leqslant k\leqslant n1$ it follows, in particular, that $x$ commutes with each of $x^0,x^1,\dots,x^{n1}$. Let us show that $x$ commutes with $x^n$. This is obvious for $n=1$, so suppose $n\geqslant 2$. We have
$$
\begin{equation*}
x*x*x^{n1}=x*[x^{n2},x^n]=[x*x^{n2},x*x^n]
\end{equation*}
\notag
$$
and
$$
\begin{equation*}
x^{n1}*x*x=[x^{n2},x^n]*x=[x^{n2}*x,x^n*x].
\end{equation*}
\notag
$$
Since $x$ commutes with $x^{n1}$, we have $x*x*x^{n1}=x^{n1}*x*x$. Also, $x*x^{n2}=x^{n2}*x$. Hence $x*x^n=x^n*x$.
Now, applying Lemma 2.2 to the triple $(x^{n1},x,x^n)$ we get that the element $x^{n1}$ lies in the multiset $x*x^n=x^n*x$. Therefore, we can and should take the second element of this multiset as $x^{n+1}$. $\Box$ For $k<0$ we set $x^k=x^{k}$ by definition. Then (5) is satisfied for all integers $k$. It is natural to call the element $x^k$ the $k$th power of $x$. Lemma 2.4. Let $X$ be an involutive twovalued group and $x$ be an arbitrary element of it. Then
$$
\begin{equation}
x^k*x^l=[x^{k+l},x^{kl}]
\end{equation}
\tag{6}
$$
and
$$
\begin{equation}
(x^k)^l=x^{kl}
\end{equation}
\tag{7}
$$
for all integers $k$ and $l$. Proof. Since $x^n=x^{n}$, it suffices to prove equality (6) for nonnegative $k$ and $l$. Let us prove it when $0\leqslant k\leqslant l$; the case $k\geqslant l$ is completely analogous. Since $x^0=e$, (6) is true for $k=0$. For $k=1$ this equality becomes formula (5), which was used to construct the sequence $\{x^k\}$; thus it is also true. Let us prove (6) for $2\leqslant k\leqslant l$ using induction on $k$. By the inductive hypothesis
$$
\begin{equation*}
x^{k1}*x^l=[x^{k+l1},x^{kl1}].
\end{equation*}
\notag
$$
Multiplying both sides of this equation by $x$ on the left and applying (5) we have
$$
\begin{equation*}
x*x^{k1}*x^l=[x^{k+l},x^{k+l2},x^{kl},x^{kl2}].
\end{equation*}
\notag
$$
On the other hand, using the inductive assumption and formula (5) again, we obtain
$$
\begin{equation*}
x*x^{k1}*x^l=[x^{k2}*x^l,x^k*x^l]=[x^{k+l2},x^{kl2},x^k*x^l],
\end{equation*}
\notag
$$
whence it follows that $x^k*x^l = [x^{k+l},x^{kl}]$.
From (6) it follows, in particular, that the sequence of elements $x^{kl}$, where $l = 0,1,2,\dots$ , satisfies the recurrence relation
$$
\begin{equation*}
x^k*x^{kl}=[x^{k(l1)},x^{k(l+1)}],
\end{equation*}
\notag
$$
hence $(x^{k})^l = x^{kl}$. $\Box$ Corollary 2.5. Every singlegenerated involutive twovalued group is commutative. Lemma 2.6. Let $X$ be an involutive twovalued group and $x$ be an arbitrary element of it. Then the set $K\subseteq\mathbb{Z}$ of all integers $k$ such that $x^{k}=e$ is a subgroup. If $k\in K$, then $x^{\pm l+k}=x^{l}$ for all $l\in\mathbb{Z}$. Conversely, if $x^l=x^m$, then either $lm\in K$ or $l+m\in K$. Proof. Obviously, $0\in K$ and $k\in K$ for all $k\in K$. Since $e*x^l=[x^l,x^l]$, it immediately follows from (6) that $x^{\pm l+k}=x^{l}$ for all $k\in K$ and $l\in \mathbb{Z}$. In particular, if $l$ also lies in $K$, then $x^{k+l}=e$, so $k+l\in K$. Thus $K\subseteq\mathbb{Z}$ is a subgroup.
Now suppose that $x^l=x^m$. By involutivity the multiset $x^l*x^m=[x^{l+m},x^{lm}]$ contains $e$. Hence either $x^{l+m}=e$ or $x^{lm}=e$, so either $l+m\in K$ or $lm\in K$. $\Box$ The smallest positive integer $k$ such that $x^k=e$ will be called the order of the element $x$ and denoted by $\operatorname{ord} x$. If $x^k\ne e$ for all positive integers $k$, then we say that the order of element $x$ is equal to infinity. It is obvious that $\operatorname{ord} x=1$ if and only if $x=e$. In what follows elements of orders $2$ and $3$ will be of particular interest. It easily follows from the definition and Lemma 2.6 that a nonidentity element $x$ has order $2$ if and only if $x*x=[e,e]$, and it has order $3$ if and only if $x*x=[e,x]$. In particular, elements of orders $1$ and $2$ are exactly strong involutions; see Definition 1.4. The corollary below follows from Lemma 2.6. Corollary 2.7. If $x$ is an element of an involutive twovalued group, then $x^l=x^m$ if and only if at least one of the numbers $lm$ and $l+m$ is divisible by $\operatorname{ord} x$. In particular, $x^l=e$ if and only if $l$ is divisible by $\operatorname{ord} x$. Proof of Theorem 2.1. Let $X$ be a singlegenerated involutive twovalued group with generator $x$. According to Lemma 2.4, equality (6) holds for any $k,l\geqslant 0$. The subset of $X$ consisting of all powers $x^k$ is closed under multiplication and contains $x$. Since the twovalued group $X$ is generated by $x$, the powers $x^k$, $k\in\mathbb{Z}$, exhaust all elements of $X$.
According to Lemma 2.6, the set $K\subseteq\mathbb{Z}$ of all integers $k$ such that $x^k=e$ is a subgroup, and $x^{\pm l+k}=x^{l}$ for all $k\in K$ and $l\in\mathbb{Z}$. Therefore, the correspondence $k\mapsto x^k$ gives a welldefined surjective map $\varphi\colon(\mathbb{Z}/K)/\iota \to X$. It follows from (6) that $\varphi$ is a homomorphism of twovalued groups, and it follows from the last statement of Lemma 2.6 that $\varphi$ is injective. Thus, $\varphi$ is an isomorphism of twovalued groups. $\Box$
3. Subgroup of elements of order 2 Let $X$ be an involutive commutative twovalued group. Let us introduce a partially defined singlevalued operation of multiplication $\cdot$ on $X$ such that It immediately follows from the definition that $x\cdot y =y\cdot x$ and $x\cdot e=x$. Also, $x\cdot x=e$ if $\operatorname{ord} x\leqslant 2$, and $x\cdot x$ is undefined if $\operatorname{ord} x> 2$. Lemma 3.1. If at least one of the two elements $x$ and $y$ has order $2$, then the multiset $x*y$ consists of two identical elements, and thus the expression $x\cdot y$ is defined. Proof. Let $\operatorname{ord} x=2$ and $z\in x*y$. By Lemma 2.2 we have $y\in x*z$, that is, $x*z=[y,y']$ for some $y'$. Then
$$
\begin{equation*}
(x*x)*z=[e,e]*z=[z,z,z,z]
\end{equation*}
\notag
$$
and
$$
\begin{equation*}
x*(x*z)=x*[y,y']=[x*y,x*y'],
\end{equation*}
\notag
$$
so $x*y=[z,z]$. $\Box$ Proposition 3.2. Let $X$ be an involutive commutative twovalued group and $V\subseteq X$ be its subset consisting of the identity $e$ and all elements of order $2$. Then Proof. Statements (a) and (b) immediately follow from Lemma 3.1 and the associativity, commutativity, and involutivity of the twovalued group $X$.
We prove statement (c). If $x^2=v$, then $\operatorname{ord} x=4$, and thus $x^3=x$. Hence $v*x=x^2*x=[x^3,x]=[x,x]$, so $v\cdot x=x$.
Conversely, suppose $v\cdot x=x$, that is, $v*x=[x,x]$. Then it follows from Lemma 2.2 that the multiset $x*x$ contains $v$. However, the fact that the group $X$ is involutive implies that this multiset also contains $e$. Since $v\ne e$, we get that $x*x=[e,v]$, that is, $x^2=v$. $\Box$ The twovalued group $X$ splits into orbits with respect to the ‘$\cdot$’action of the group $V$ on it. We call these orbits $V$orbits; the $V$orbit containing the element $x$ is denoted by $Vx$. Remark 3.3. In what follows we will usually write $xy$ instead of $x\cdot y$, which should not lead to confusion.
4. Special and nonspecial twovalued groups Definition 4.1. A pair $(x,y)$ of elements of an involutive commutative twovalued group is called special if $\operatorname{ord} x>2$, $\operatorname{ord} y>2$, but the multiset $x*y$ consists of two identical elements. An involutive commutative twovalued group is called special if it contains at least one special pair of elements, otherwise it is called nonspecial. Remark 4.2. In [17] nonspecial twovalued groups were called strongly two valued. Thus, if a twovalued group is nonspecial, then $x\cdot y$ is defined if and only if one of the elements $x$ and $y$ has order $\leqslant2$. Proposition 4.3. All twovalued groups of the form $X_A^{\mathbf{a}}$ and $X_V^{\mathbf{u}}\times W$ are non special. Twovalued groups $Y_V\times W$ are special when $\dim V\geqslant 2$. Proof. Let us prove that the twovalued group $X_A^{\mathbf{a}}$ is nonspecial. Let $\pi\colon A\to X_A^{\mathbf{a}}$ be the quotient map by the antipodal involution. Consider elements $x,y\in X_A^{\mathbf{a}}$. If $a,b\in A$ are elements such that $x=\pi(a)$ and $y=\pi(b)$, then $x*y=[\pi(ab),\pi(ab^{1})]$. Assume that the orders of $x$ and $y$ are greater than $2$. Then the orders of $a$ and $b$ are also greater than $2$. Hence the elements $ab$ and $ab^{1}$ are neither equal nor inverse to each other. Therefore, the multiset $x*y$ consists of two different elements. Thus, the twovalued group $X_A^{\mathbf{a}}$ is nonspecial.
Let us prove the nonspeciality of the twovalued group
$$
\begin{equation*}
X_V^{\mathbf{u}}=(V\times V)/\iota_{\mathbf{u}},\quad \text{where } \iota_{\mathbf{u}}(a,b)=(a,ab).
\end{equation*}
\notag
$$
Let $\pi\colon V\times V\to X_V^{\mathbf{u}}$ be the quotient map by the involution $\iota_{\mathbf{u}}$. It can be verified directly that the order of element $\pi(a,b)$ of a twovalued group $X_V^{\mathbf{u}}$ is equal to $4$ if $a\ne e$ and to $2$ if $a=e$ and $b\ne e$. Thus, any pair of elements of order greater than $2$ has the form $x=\pi(a_1,b_1)$, $y=\pi(a_2,b_2)$, where $a_1\ne e$ and $a_2\ne e$. Then
$$
\begin{equation*}
x*y=[\pi(a_1a_2,b_1b_2),\pi(a_1a_2,a_1b_1b_2)].
\end{equation*}
\notag
$$
Since the element $a_1b_1b_2$ is neither $b_1b_2$ nor $a_1a_2b_1b_2$, all elements of the multiset $x*y$ are different. Thus, the twovalued group $X_V^{\mathbf{u}}$ is nonspecial. It is easy to see that this implies that all twovalued groups $X_V^{\mathbf{u}}\times W $ are nonspecial.
Now consider the twovalued group $Y_V$, where $\dim V\geqslant 2$. Let $x,y\in V$ be two distinct nonidentity elements. Then the orders of the elements $x$ and $y$ in the twovalued group $Y_V$ are equal to $4$, since $x*x=y*y=[e,s]$ and $s$ is an element of order $2$. Moreover, $x*y=[xy,xy]$, hence $(x,y)$ is a special pair. Thus, the twovalued group $Y_V$ is special, and therefore all twovalued groups $Y_V\times W$ are special. $\Box$ Proposition 4.4. Let $X$ be a special involutive commutative twovalued group. Then $X$ is not coset, that is, not isomorphic to any twovalued group of the form $G/\iota$, where $G$ is an ordinary group and $\iota$ is an automorphism of $G$ such that $\iota^2=\mathrm{id}$. Proof. Assume the opposite: let $X\cong G/\iota$. Denote by $\pi\colon G\to X$ the quotient map by the automorphism $\iota$. Let $(x,y)$ be a special pair in $X$ and let $g,h\in G$ be elements such that $\pi(g)=x$ and $\pi(h)=y$.
We have $x*x=[\pi(g^2),\pi(g\iota(g))]$. Since $\operatorname{ord} x>2$, the multiset $x*x$ must consist of two different elements, so $g^2\ne g\iota(g)$ and thus $\iota(g)\ne g$. Similarly, $\iota(h)\ne h$.
A multiset $x*y=[\pi(gh),\pi(g\iota(h))]$ consists of two identical elements. Hence either $g\iota(h)=gh$ or $g\iota(h)=\iota(gh)=\iota(g)\iota(h)$, which implies that either $\iota(h)=h$ or ${\iota(g)=g}$. As we have proved, neither of these equalities is possible, which completes the proof. $\Box$ Remark 4.5. Note that, generally speaking, commutative twovalued groups can be obtained from noncommutative singlevalued groups using the coset construction. For example, as shown in [7] (see also [17]), the twovalued Buchstaber–Novikov group $X^{\mathbf{a}}_{C_{\infty}}=C_{\infty}/\iota_{\mathbf{a}}$ is isomorphic to the coset group $(C_2*C_2)/\sigma$, where $*$ denotes the free product and $\sigma$ is the automorphism interchanging the generators of the two factors, $C_2$ and $C_2$. We emphasize that Proposition 4.4 asserts that a special involutive commutative twovalued group cannot be a coset twovalued group of any (not necessarily commutative) group $G$ by an automorphism. Nevertheless, we recall that the special twovalued group $Y_2$ ($Y_3$, respectively) can be obtained using an analogue of the coset construction for the antiautomorphism of the noncommutative group $Q_8$ (the noncommutative and nonassociative Moufang loop $O_{16}$, respectively; see Examples 1.12 and 1.13).
5. Quotient groups of twovalued groups The general question of the existence of quotient groups of twovalued groups requires a separate study. We give a positive answer to this question in our case of involutive commutative twovalued groups. 5.1. The existence of quotient groups A subset $Y$ of an involutive commutative twovalued group $X$ is called a subgroup if for any two elements $y_1,y_2\in Y$ both elements of the multiset $y_1*y_2$ lie in $Y$. To the subgroup $Y$ we assign an equivalence relation $\sim$ on the twovalued group $X$ defined by the following rule: $x\sim x'$ if and only if there exists an element $y\in Y$ such that $x'\in x*y$. Lemma 5.1. (a) The relation $\sim$ introduced above is indeed an equivalence relation. (b) $x\sim e$ if and only if $x\in Y$. (c) If $x_1*x_2=[z_1,z_2]$, $x_1'*x_2'=[z_1',z_2']$, $x_1\sim x_1'$, and $x_2\sim x_2'$, then
$$
\begin{equation*}
[z_1,z_2]\sim [z_1',z_2'],
\end{equation*}
\notag
$$
that is, either $z_1\sim z_1'$ and $z_2\sim z_2'$, or $z_1\sim z_2'$ and $z_2\sim z_1'$. Proof. The relation $\sim$ is reflexive since $x\in x*e$ for all $x$. The symmetry of the relation $\sim$ immediately follows from Lemma 2.2. Let us prove its transitivity. Let $x_1\sim x_2$ and $x_2\sim x_3$. Then there are elements $y_1, y_2\in Y$ such that $x_2\in x_1*y_1$ and $x_3\in x_2*y_2$, which immediately implies that $x_3\in x_1*y_1*y_2$. On the other hand $y_1*y_2=[y_3,y_4]$ for some elements $y_3,y_4\in Y$, so
$$
\begin{equation*}
x_1*y_1*y_2=[x_1*y_3,x_1*y_4].
\end{equation*}
\notag
$$
Therefore, the element $x_3$ belongs to one of the multisets $x_1*y_3$ and $x_1*y_4$. Thus, $x_1\sim x_3$, which completes the proof that $\sim$ is an equivalence relation.
If $x\in Y$, then $e\in x*x$ implies that $x\sim e$. Conversely, if $x\sim e$, then there is an element $y\in Y$ such that $e\in x*y$, so it follows from involutivity that $x=y\in Y$.
Because of the commutativity of the twovalued group $X$ and the fact that $\sim$ is an equivalence relation, it suffices to prove statement (c) in the case when $x_2=x_2'$. If $x_1\sim x_1'$, then there exists $y\in Y$ such that $x_1'\in x_1*y$. Then $x_1*y=[x_1',p]$ for some $p$, and we have
$$
\begin{equation*}
[z_1*y, z_2*y]=(x_1*x_2)*y=(x_1*y)*x_2=[x_1'*x_2,p*x_2]=[z_1',z_2',p*x_2].
\end{equation*}
\notag
$$
Swapping $z_1$ and $z_2$ if necessary, we can assume that $z_1'$ belongs to the multiset $z_1*y$; then $z_1\sim z_1'$. If, in addition, the element $z_2'$ lies in the multiset $z_2*y$, then $z_2\sim z_2'$ and thus the claim follows. It remains to consider the case when $z_1*y=[z_1',z_2']$. Then $z_1'\sim z_1\sim z_2'$. Lemma 2.2 implies that $x_1'*y=[x_1,q]$ for some $q$. Then
$$
\begin{equation*}
[z_1'*y,z_2'*y]=(x_1'*x_2)*y=(x_1'*y)*x_2=[x_1*x_2,q*x_2]=[z_1,z_2,q*x_2],
\end{equation*}
\notag
$$
which shows that $z_2$ belongs to one of the multisets $z_1'*y$ and $z_2'*y$. Hence all four elements $z_1$, $z_2$, $z_1'$, and $z_2'$ are pairwise equivalent one to another, which implies the claim. $\Box$ Lemma 5.1 implies that the set $X/Y=X/{\sim}$ is well defined and twovalued multiplication in the group $X$ induces a welldefined twovalued multiplication on the set $X/Y$. Lemma 5.2. The multiplication introduced transforms $X/Y$ into an involutive commutative twovalued group. Proof. The commutativity and associativity of twovalued multiplication in $X/Y$ immediately follows from the commutativity and associativity of multiplication in $X$. It is also obvious that the equivalence class of the identity element, which, according to statement (b) of Lemma 5.1, coincides with the subgroup $Y$, is the (strong) identity element of the twovalued multiplication introduced; we also denote this identity element by $e$.
Let us prove involutivity (and at the same time the existence and uniqueness of an inverse). The fact that $e\in x*x$ for all $x\in X$ immediately implies that $e\in z*z$ for all $z\in X/Y$. Therefore, it is only necessary to prove that if $z_1$ and $z_2$ are two different elements of $X/Y$, then $e\notin z_1*z_2$. To do this we need to show that if $x_1$ and $x_2$ are representatives of two different equivalence classes $z_1$ and $z_2$, then the multiset $x_1*x_2$ does not contain elements of $Y$. Assume that this is not the case, that is, some $y\in Y$ lies in the multiset $x_1*x_2$. Then by Lemma 2.2, $x_2\in x_1*y$, which means that $x_1\sim x_2$ and so $z_1=z_2$, which is not true. This contradiction completes the proof of the lemma. $\Box$ The involutive commutative twovalued group $X/Y$ defined in this way will be called the quotient group of the twovalued group $X$ by its subgroup $Y$. It follows from part (b) of Lemma 5.1 that the kernel of the natural projection $\pi\colon X\to X/Y$ coincides indeed with $Y$. (As in the case of ordinary groups, the kernel of a homomorphism of twovalued groups is defined to be the preimage of the identity.) Remark 5.3. Let $V\subseteq X$ be the subset of all elements of order $\leqslant 2$. By Proposition 3.2, $V$ is a Boolean group, which acts on the twovalued group $X$. Hence if $W\subseteq V$ is a subgroup, then the notation $X/W$ can be given two meanings: (i) the quotient group of the twovalued group $X$ by the subgroup $W$; (ii) the space of orbits of the action of $W$ on $X$, where $W$ is considered as a singlevalued group with respect to the operation $\cdot$ introduced in § 3. It is easy to see, however, that these two sets coincide. 5.2. The homomorphism theorem The fundamental theorem on the image of a homomorphism carries over to the case of involutive commutative twovalued groups. Proposition 5.4. Let $f\colon X\to Z$ be a homomorphism of involutive commutative twovalued groups. Then its kernel $\ker f$ and image $\operatorname{im} f$ are subgroups of the twovalued groups $X$ and $Z$, respectively, and $f$ induces an isomorphism $X/\ker f\cong\operatorname{im} f$. Proof. Obviously, $\operatorname{im} f$ is a subgroup. If $y_1,y_2\in\ker f$ and $y_1*y_2=[z_1,z_2]$, then
$$
\begin{equation*}
[f(z_1),f(z_2)]=f(y_1)*f(y_2)=e*e=[e,e],
\end{equation*}
\notag
$$
so $f(z_1)=f(z_2)=e$. Hence, $\ker f$ is also a subgroup. Let $\sim$ be the equivalence relation on $X$ defined above and corresponding to this subgroup. To complete the proof of the proposition it suffices to prove that the inverse images of the elements $z\in \operatorname{im} f$ are precisely the equivalence classes with respect to $\sim$. Let us prove this. If $x_1\sim x_2$, then there exists $y\in\ker f$ such that $x_2\in x_1*y$. Then
$$
\begin{equation*}
f(x_2)\in f(x_1)*f(y)=f(x_1)*e=[f(x_1),f(x_1)],
\end{equation*}
\notag
$$
so $f(x_2)=f(x_1)$. Conversely, let $x_1$ and $x_2$ be two elements of $X$ such that $f(x_1)=f(x_2)$. Let $x_1*x_2=[p,q]$. Then the multiset $[f(p),f(q)]=f(x_1)*f(x_2)=f(x_1)*f(x_1)$ must contain the identity element because of involutivity. Hence at least one of $p$ and $q$ lies in $\ker f$. Without loss of generality we can assume that $p\in\ker f$. It follows from Lemma 2.2 that $x_2$ lies in the multiset $x_1*p$, hence $x_1\sim x_2$.$\Box$ Proposition 5.4 allows us to define exact sequences of involutive commutative twovalued groups and work with them in exactly the same way as in the case of ordinary abelian groups. 5.3. $C_2$extensions of twovalued groups Definition 5.5. An (involutive commutative) $C_2$extension of an involutive commutative twovalued group $X$ is an exact sequence of involutive commutative two valued groups
$$
\begin{equation}
1\to 2C_2\to \widehat{X} \xrightarrow{\pi} X\to 1,
\end{equation}
\tag{9}
$$
where $2C_2$ is the twoelement twovalued group obtained by doubling the operation in the group $C_2$. We could also study extensions in which the resulting group $\widehat{X}$ is not involutive or commutative, but they are not of interest to us. In what follows a $C_2$extension is always understood as an involutive commutative $C_2$extension. The image of a twovalued group $2C_2$ in $\widehat{X}$ is a twoelement subgroup, one of whose elements is the identity $e$; we denote its second element, which has order $2$, by $u$. It is easy to see that $X$ is obtained from $\widehat{X}$ by taking the quotient by the action of the group $C_2=\{e,u\}$ given by the operation described in § 3. We denote the identity elements of both $X$ and $\widehat{X}$ by $e$, this should not lead to confusion. Lemma 5.6. (a) If $x\in X$ is an element of order other than $2$, then its preimage $\pi^{1}(x)$ consists of two elements, each of which has order $\operatorname{ord} x$ or $2\operatorname{ord} x$. (b) If $x\in X$ is an element of order $2$, then its preimage $\pi^{1}(x)$ consists of either two elements of order $2$ or one element of order $4$. Proof. First, note that $\pi^{1}(e)=\{e,u\}$, so the claim of the lemma is true for ${x=e}$.
If $\hat{x}$ is an element in $\pi^{1}(x)$, then $\pi^{1}(x)=\{\hat{x},u\hat{x}\}$, so we only need to figure out what orders the elements $\hat{x}$ and $u\hat{x}$ can have and when these elements can coincide.
It is obvious that $\pi(\hat{x}^k)=x^k$ for all $k\in\mathbb{Z}$. So if $\hat{x}^k=e$, then $x^k=e$. Therefore, $\operatorname{ord}\hat{x}$ is divisible by $\operatorname{ord} x$. Now take $k=\operatorname{ord} x$. Then $x^k=e$, which means that $\hat{x}^k$ belongs to the set $\pi^{1}(e)=\{e,u\}$, that is, either $\hat{x}^k=e$ or $\hat{x}^k=u$. In the first case $\operatorname{ord} \hat{x}=k$. In the second case $\operatorname{ord} \hat{x}>k$, but at the same time $\hat{x}^{2k}=u^2=e$, so $\operatorname{ord}\hat{x}=2k$.
By Proposition 3.2, (c), equality $u\hat{x}=\hat{x}$ holds if and only if $\hat{x}^2=u$. If so, then $x^2=e$ and $\operatorname{ord} x=2$, and moreover, $\operatorname{ord}\hat{x}=4$. In this case the preimage $\pi^{1}(x)$ consists of one element $\hat{x}$. In all other cases $u\hat{x}\ne\hat{x}$, that is, the preimage $\pi^{1}(x)$ consists of the two different elements $\hat{x}$ and $u\hat{x}$. It remains to note that if $\operatorname{ord}\hat{x}=\operatorname{ord} x= 2$, then according to Proposition 3.2 $\operatorname{ord}(u\hat{x})=2$; thus, $\pi^{1}(x)$ consists of two elements of order $2$. $\Box$ Recall that, according to Proposition 3.2, (a), the subset $V\subseteq X$ consisting of the identity $e$ and all elements of order $2$ is a Boolean group. Denote by $R\subset V$ the subset of all elements $x$ of order $2$ such that the preimage $\pi^{1}(x)$ consists of one element of order $4$. We call $R$ the branching set of the extension (9). Note that the identity element $e$ never belongs to $R$. Moreover, the following statement is true. Lemma 5.7. The set $V\setminus R$ is a subgroup of the group $V$. Proof. Let $x,y\in V\setminus R$, and let $\hat{x}$ and $\hat{y}$ be elements of the preimages $\pi^{1}(x)$ and $\pi^{1}(y)$, respectively. Then $\operatorname{ord}\hat{x}=\operatorname{ord}\hat{y}=2$. By Proposition 3.2, (a), we obtain $\operatorname{ord}(\hat{x}\hat{y})=2$. Since $\hat{x}\hat{y}\in \pi^{1}(xy)$, it follows that $x y\in V\setminus R$. Thus, $V\setminus R$ is a subgroup. $\Box$
6. Direct product decompositions Recall that in § 1 we defined the direct product of an involutive commutative twovalued group $X$ and a Boolean group $W$; the operation of multiplication in the resulting involutive commutative group $X\times W$ is given by
$$
\begin{equation*}
(x_1,w_1)*(x_2,w_2)=\bigl[(z_1,w_1w_2),(z_2,w_1w_2)\bigr],
\end{equation*}
\notag
$$
where $x_1*x_2=[z_1,z_2]$. Definition 6.1. We say that an involutive commutative twovalued group $X$ contains a singlevalued direct factor if $X$ is isomorphic to a twovalued group of the form $X'\times W$, where $X'$ is an involutive commutative twovalued group and $W$ is a nontrivial Boolean group. Otherwise we say that $X$ does not contain a singlevalued direct factor. We wish to reduce the classification problem for finitely generated involutive commutative twovalued groups to the same problem for twovalued groups that do not contain singlevalued direct factors. To do this we need the following two propositions, the first of which gives a convenient criterion when a twovalued group does not contain a singlevalued direct factor, and the second claims that the largest singlevalued direct factor in a twovalued group is determined in a unique way up to isomorphism. Proposition 6.2. A finitely generated involutive commutative twovalued group $X$ does not contain a singlevalued direct factor if and only if every element $v\in X$ of order $2$ is the square of some element of $X$. Proposition 6.3. Any finitely generated involutive commutative twovalued group $X$ can be decomposed into a direct product of a finitely generated involutive commutative twovalued group without singlevalued direct factors and a finite Boolean group. If $X=X_i\times W_i$, where $i=1,2$, are two such decompositions, then $X_1\cong X_2$ and $W_1\cong W_2$. Before proving Propositions 6.2 and 6.3, we establish two useful properties of involutive commutative twovalued groups. Lemma 6.4. Let $X$ be an involutive commutative twovalued group, and let $x,y\in X$ and $x*y=[z_1,z_2]$. Assume that $x^2\ne y^2$ and $z_1\ne z_2$. Then $z_1*z_2=[x^2,y^2]$. Proof. By Lemma 2.2, the element $y$ belongs to each of the multisets $x*z_1$ and $x*z_2$. Therefore, the identity element $e$ belongs to the multiset
$$
\begin{equation*}
(x*z_1)*(x*z_2)=(x*x)*(z_1*z_2)=[e,x^2]*(z_1*z_2)= [z_1*z_2,z_1*z_2,x^2*(z_1*z_2)].
\end{equation*}
\notag
$$
However, $e\notin z_1*z_2$ since $z_1\ne z_2$. Hence $e\in x^2*(z_1*z_2)$, which implies that $x^2\in z_1*z_2$. Similarly, we have $y^2\in z_1*z_2$. Since $x^2\ne y^2$, this implies the lemma. $\Box$ Lemma 6.5. Let $X$ be an involutive commutative twovalued group and $Q\subseteq X$ be the subset consisting of all squares of elements from $X$. Then $Q$ is a subgroup. Proof. We need to prove that for any $x, y\in X$ the multiset $x^2*y^2$ consists of squares. Let $x*y=[z_1,z_2]$; then
$$
\begin{equation*}
x*x*y*y=(x*x)*(y*y)=[e,x^2]*[e,y^2]=[e,e,x^2,x^2,y^2,y^2,x^2*y^2]
\end{equation*}
\notag
$$
and
$$
\begin{equation*}
x*x*y*y=(x*y)*(x*y)=[z_1,z_2]*[z_1,z_2]=[e,e,z_1^2,z_2^2,z_1*z_2,z_1*z_2].
\end{equation*}
\notag
$$
If $x^2\ne y^2$ and $z_1\ne z_2$, then $z_1*z_2=[x^2,y^2]$, hence $x^2*y^2=[z_1^2,z_2^2]$ and the claim is proved. It remains to analyze the cases when $x^2=y^2$ or $z_1= z_2$.
If $x^2=y^2$, then $x^2*y^2=[e,(x^2)^2]$ in agreement with the claim.
If $z_1=z_2$, then $x*x*y*y=[e,e,e,e,z_1^2,z_1^2,z_1^2,z_1^2]$. Hence the multiset $x^2*y^2$ consists of squares, which completes the proof. $\Box$ Proof of Proposition 6.2. The ‘only if’ part of the statement of the proposition is obvious: indeed, if $X=X'\times W$, where $W$ is a nontrivial Boolean group, then for any element $w\in W\setminus\{e\}$ the pair $(e,w)\in X$ is an element of order $2$ that is not a square.
Conversely, suppose that in the twovalued group $X$ there is an element $w$ of order $2$ that is not a square. We set $W=\{e,w\}$ and prove that the singlevalued group $W\cong C_2$ can be defined as a direct factor in $X$. To do this it suffices to construct a homomorphism of twovalued groups $f\colon X\to W$ such that $f(w)=w$. Indeed, if such a homomorphism is constructed, then together with the natural projection $\pi \colon X\to X/W$, it will induce an isomorphism $X\cong(X/W)\times W$.
If we consider $W$ as a twovalued group, then multiplication is given by the formulae
$$
\begin{equation*}
e*e=[e,e],\qquad e*w=[w,w],\qquad w*w=[e,e].
\end{equation*}
\notag
$$
Therefore, the map $f\colon X\to W$ is a homomorphism of twovalued groups if and only if for any elements $x,y,z\in X$ such that $z\in x*y$ we have the equality
$$
\begin{equation}
f(x)f(y)f(z)=e.
\end{equation}
\tag{10}
$$
When considered for all triples $\{x,y,z\}$ such that $z\in x*y$, these equalities form a system of homogeneous linear equations over the field $\mathbb{F}_2$ (in multiplicative notation) with respect to the variables $f(x)$, $x\in X$. We need to prove that this system of equations has a solution that satisfies the additional condition $f(w)=w$. To do this it suffices to show that the equality $f(w)=e$ does not follow from (10). Assume the opposite. Then the equation $f(w)=e$ is a linear combination over $\mathbb{F}_2$ of equations of the form (10). Since we use the multiplicative notation, this means that $f(w)=e$ is the product of the equalities
$$
\begin{equation*}
f(x_i)f(y_i)f(z_i)=e,\qquad i=1,\dots,m,
\end{equation*}
\notag
$$
for some triples $\{x_i,y_i,z_i\}$ such that $z_i\in x_i*y_i$, taking into account that $f(t)^2=e$ for all $t$. Therefore, the $3m$element multiset $[x_1,y_1,z_1,\dots,x_m,y_m,z_m]$ has the form
$$
\begin{equation*}
[w,t_1,t_1,t_2,t_2,\dots,t_n,t_n]
\end{equation*}
\notag
$$
for some $t_1,\dots,t_n\in X$, where, of course, $n=(3m1)/2$.
For each $i$ it follows from $z_i\in x_i*y_i$ that the multiset $x_i*y_i*z_i$ contains the identity $e$. Thus,
$$
\begin{equation*}
\begin{aligned} \, e\in(x_1*y_1*z_1)*\cdots*(x_m*y_m*z_m)&= w*(t_1*t_1)*(t_2*t_2)*\cdots*(t_n*t_n) \\ &=w*[e,t_1^2]*[e,t_2^2]*\cdots*[e,t_n^2]. \end{aligned}
\end{equation*}
\notag
$$
Since the twovalued group $X$ is involutive, it follows that
$$
\begin{equation*}
w\in [e,t_1^2]*[e,t_2^2]*\cdots*[e,t_n^2].
\end{equation*}
\notag
$$
Hence $w$ lies in a subgroup of the twovalued group $X$ generated by the squares $t_i^2$. Thus, by Lemma 6.5, the element $w$ is itself a square, which is impossible. This contradiction means that the system of linear equations (10) has a solution that satisfies $f(w)=w$. This solution is a twovalued group homomorphism, which induces an isomorphism $X\cong (X/W)\times W$. Proof of Proposition 6.3. If the twovalued group $X$ contains some singlevalued direct factor, then we choose it: $X=X'\times W'$, where $\dim W'>0$. If $X'$ also contains some singlevalued direct factor, then we choose it again: $X'=X''\times \widetilde{W}$, where $\dim \widetilde{W}>0$, and therefore $X=X''\times W''$, where $W''=W'\times \widetilde{W}$, ${\dim W''\geqslant 2}$. Continuing in the same way as long as possible, after the $k$th step we obtain a decomposition $X=X^{(k)}\times W^{(k)}$, where $\dim W^{(k)}\geqslant k$. We need to prove that this process stops at some point, that is, after some step the twovalued group $X^{(k)}$ does not contain a singlevalued direct factor. To do this note that every group $W^{(k)}$ with the doubled operation is the quotient group of the twovalued group $X$. Therefore, if $X$ is generated by $N$ elements, then $W^{(k)}$ is also generated by $N$ elements, hence $k\leqslant N$. Thus, the process under consideration stops no later than at the $N$th step, and we obtain the desired decomposition.
Now let $X=X_1\times W_1=X_2\times W_2$ be two decompositions such that $X_1$ and $X_2$ do not contain singlevalued direct factors. Assume first that $W_1=W_2$. Then each of the projections $X\to X_1$ and $X\to X_2$ is taking the quotient with respect to the action of the same group $W_1=W_2$, which immediately implies that $X_1$ and $X_2$ are isomorphic.
Now suppose that $W_1\ne W_2$. Let $Q\subseteq X$ be the set of squares of elements of $X$. Then $Q\subseteq X_1$ and $Q\subseteq X_2$. By Lemma 6.5 the subset $Q$ is a subgroup. Let $V\subseteq X$ be the subset of elements of order $\leqslant 2$. Then $V$ is a Boolean group and $W_1$ and $W_2$ are its subgroups. From the fact that $Q$ is a subgroup of the twovalued group $X$ it follows that $U=Q\cap V$ is a subgroup of $V$. Since $Q\subseteq X_1$ and $Q\subseteq X_2$, we have $U\cap W_1=U\cap W_2=\{e\}$. Assume that the direct product $U\times W_1$ does not coincide with the whole of $V$. Then $v\in V\setminus(U\times W_1)$ is taken by the projection $X=X_1\times W_1\to X_1$ to an element of order $2$ that is not the square of any element of $X_1$. By Proposition 6.2 this contradicts the fact that $X_1$ does not contain singlevalued direct factors. Hence $V=U\times W_1$. Similarly, $V=U\times W_2$. Thus, $W_1$ and $W_2$ are two subspaces of the vector space $V$ over $\mathbb{F}_2$ which are complementary to the subspace $U$. Therefore, there is a homomorphism $f\colon W_1\to U$ such that the formula $w\mapsto wf(w)$ defines an isomorphism $W_1\cong W_2$. Then the map $F\colon X_1\times W_1\to X_1\times W_1$ defined by $F(x,w)=(x,wf(w))$ is an automorphism of the twovalued group $X$ which leaves $X_1$ invariant and takes $W_1$ to $W_2$. Thus there is a direct product decomposition $X=X_1\times W_2$, which implies that $X_1\cong X_2$. $\Box$
7. Properties of nonspecial twovalued groups Recall that an involutive commutative twovalued group $X$ is called nonspecial if the product $x*y$ consists of two distinct elements for any elements $x,y\in X$ whose orders are greater than $2$. In this section we prove several useful properties of such twovalued groups. In all statements of this section $X$ is a nonspecial involutive commutative twovalued group and $V\subseteq X$ is the subset consisting of the identity $e$ and all elements of order $2$. Recall that, according to Proposition 3.2, the set $V$ is a Boolean group, which acts on $X$. In all statements in this section, lowercase Latin letters $x,y,z,\dots$ denote elements of the twovalued group $X$ unless otherwise stated. Lemma 7.1. The equality $x^2=y^2$ holds if and only if the elements $x$ and $y$ lie in the same $V$orbit. Proof. If $x$ and $y$ are in the same $V$orbit, so that $y=vx$ for some $v\in V$, then the equality $x^2=y^2$ immediately follows from (8).
Let us prove the reverse implication. If $x^2=y^2=e$, then $x$ and $y$ lie in $V$, hence in the same $V$orbit. Assume that $x^2=y^2\ne e$. Then the orders of $x$ and $y$ are greater than $2$. Since the pair $(x,y)$ is not special, it follows that $x*y=[z_1,z_2]$, where $z_1\ne z_2$. We have
$$
\begin{equation*}
x*x*y*y=(x*x)*(y*y)=[e,x^2]*[e,x^2]=[e,e,e,x^2,x^2,x^2,x^2,x^4]
\end{equation*}
\notag
$$
and
$$
\begin{equation*}
x*x*y*y=(x*y)*(x*y)=[z_1,z_2]*[z_1,z_2]=[e,e,z_1^2,z_2^2,z_1*z_2,z_1*z_2].
\end{equation*}
\notag
$$
From the fact that $z_1\ne z_2$ it follows that $e\notin z_1*z_2$. So either $z_1^2=e$, or $z_2^2=e$. Without loss of generality we may assume that $z_1^2=e$, so that $z_1\in V$. Since $z_1\in x*y$, by Lemma 2.2 we have $y\in z_1*x$, so $y=z_1x$. $\Box$ Lemma 7.2. If $x*y=[z_1,z_2]$, then $z_1*z_2=[x^2,y^2]$. Proof. Suppose first that $x$ and $y$ belong to different $V$orbits and neither of them lies in $V$. Then by Lemma 7.1 we have $x^2\ne y^2$. Moreover, from the fact that the twovalued group $X$ under consideration is nonspecial it follows that $z_1\ne z_2$. Therefore, the required equality follows from Lemma 6.4.
It remains to consider the case when either $x$ and $y$ lie in the same $V$orbit, or at least one of them lies in $V$. If $x$ and $y$ are in the same $V$orbit, so that $y=vx$ for some $v\in V$, then
$$
\begin{equation*}
x*y=[v,vx^2] \quad\text{and}\quad v*(vx^2)=[x^2,x^2]=[x^2,y^2].
\end{equation*}
\notag
$$
If $x\in V$, then $z_1=z_2=xy$ and $z_1*z_2=[e,y^2]=[x^2,y^2]$; the case when $y\in V$ is similar. Corollary 7.3. If $x*y=[z_1,z_2]$, then $z_2\in x^2*z_1$. Applying Lemma 7.2 twice, we obtain the following statement. Corollary 7.4. If $x*y=[z_1,z_2]$, then $x^2*y^2=[z_1^2,z_2^2]$. Lemma 7.5. Let $x*y=[z_1,z_2]$. Proof. By Corollary 7.4 we have $x^2*y^2=[z_1^2,z_2^2]$. Now let the orders of both $x$ and $y$ divide $4$. Then $x^2$ and $y^2$ belong to the set $V$. Therefore, $z_1^2=z_2^2=x^2y^2$ also belongs to $V$, so that $z_1^4=z_2^4=e$, which proves (a).
Further, using the fact that the twovalued group $X$ under consideration is not special and Lemma 7.1, we obtain the following chain of equivalences, proving the first part of statement (b):
$$
\begin{equation*}
(\operatorname{ord} x\mid4)\vee(\operatorname{ord} y\mid4)\ \ \Longleftrightarrow\ \ (x^2\in V)\vee(y^2\in V)\ \ \Longleftrightarrow\ \ z_1^2=z_2^2\ \ \Longleftrightarrow\ \ Vz_1=Vz_2.
\end{equation*}
\notag
$$
The second part of (b) immediately follows from Corollary 7.3.
Now let all the three elements $x$, $z_1$, and $z_2$ lie in the same $V$orbit, so that $z_1=v_1x$ and $z_2=v_2x$, where $v_1, v_2\in V$. Then
$$
\begin{equation*}
[x^2,y^2]=z_1*z_2=[v_1v_2,v_1v_2x^2].
\end{equation*}
\notag
$$
Hence either $y^2=v_1v_2$ and then $\operatorname{ord} y$ divides $4$, or $x^2=v_1v_2$ and $y^2=v_1v_2x^2=e$ and then $\operatorname{ord} y$ divides $2$. $\Box$ Lemma 7.6. If $x*y=z*t$, then either there exists $v\in V$ such that $z=vx$ and $t=vy$, or there exists $v\in V$ such that $z=vy$ and $t=vx$. Proof. Lemma 7.2 immediately implies that if $x*y=z*t$, then $[x^2,y^2]=[z^2,t^2]$. Suppose $x^2=z^2$ and $y^2=t^2$; the second case is completely similar. Then by Lemma 7.1 there are elements $v_1,v_2\in V$ such that $z=v_1x$ and $t=v_2y$. If $v_1=v_2$, then the statement of the lemma holds.
Suppose $v_1\ne v_2$ and set $w=v_1v_2$; then $w\ne e$. The equality $x*y=z*t$ can be rewritten as $w(x*y)=x*y$. Let $x*y=[s_1,s_2]$. Then the following two cases are possible.
1) $ws_1=s_1$ and $ws_2=s_2$. Then it follows from assertion (c) of Proposition 3.2 that $s_1^2=s_2^2=w$, which means that by Lemma 7.1 we have $s_2=us_1$ for some $u\in V$. Now by Lemma 7.2 we get that
$$
\begin{equation*}
[x^2,y^2]=s_1*s_2=[u,us_1^2]=[u,uw].
\end{equation*}
\notag
$$
Without loss of generality $x^2=u$ and $y^2=uw$. Then $ux=x$ and $uv_1v_2y=y$, hence $z=v_1x=uv_1x$ and $t=v_2y=uv_1y$. Thus, the statement of the lemma holds for $v=uv_1$.
2) $ws_1=s_2$ and $ws_2=s_1$. Then by Lemma 7.2 we obtain
$$
\begin{equation*}
[x^2,y^2]=s_1*s_2=[w,ws_1^2].
\end{equation*}
\notag
$$
Without loss of generality $x^2=w=v_1v_2$. Then $v_1v_2x=x$, hence $z=v_1x=v_2x$. Thus, the statement of the lemma holds for $v=v_2$. $\Box$ Lemma 7.7. Suppose that elements $x_1$, $x_2$, and $x_3$ lie in pairwise distinct $V$orbits and none of them lies in $V$. Then the multiset $x_1*x_2*x_3$ consists of four different elements. Also, if $x_1*x_2=[p_3,q_3]$, $x_2*x_3=[p_1,q_1]$, and $x_3*x_1 = [p_2,q_2]$, then
$$
\begin{equation*}
x_1*x_2*x_3=(x_1*p_1)\cup (x_1*q_1)=(x_2*p_2)\cup (x_2*q_2)= (x_3*p_3)\cup (x_3*q_3)
\end{equation*}
\notag
$$
are all three different partitions of the fourelement set $x_1*x_2*x_3$ into two two element subsets. In other words, $x_1*p_1$, $x_1*q_1$, $x_2*p_2$, $x_2*q_2$, $x_3*p_3$, and $x_3*q_3$ are exactly all six distinct twoelement subsets of the fourelement set $x_1*x_2*x_3$. Proof. Let us prove first that the six multisets $x_1*p_1$, $x_1*q_1$, $x_2*p_2$, $x_2*q_2$, $x_3*p_3$, and $x_3*q_3$ are pairwise distinct. Assume the opposite. Then there are two essentially different cases: $x_1*p_1=x_1*q_1$ or $x_1*p_1=x_2*p_2$. Each of the other cases is completely analogous to one of these two. We consider these cases in turn.
1) $x_1*p_1=x_1*q_1$. Then by Lemma 7.6 there is an element $v\in V$ such that either $p_1=q_1=vx_1$, or $vx_1=x_1$ and $vp_1=q_1$. However, the equality $p_1=q_1$ is impossible because the twovalued group $X$ is not special and $x_2,x_3\notin V$. Therefore, $vx_1=x_1$ and $vp_1=q_1$ and $v\ne e$. It follows from assertion (c) of Proposition 3.2 that $\operatorname{ord} x_1=4$ and $v=x_1^2$, and then it follows from statement (b) of Lemma 7.5 that one of the elements $x_2$ and $x_3$ (without loss of generality, $x_2$) has an order that divides $4$ and $x_2^2p_1=q_1$. Hence $x_1^2x_2^2p_1=p_1$ and $x_1^2x_2^2q_1= q_1$. However, since $x_1$ and $x_2$ lie in different $V$orbits, we see from Lemma 7.1 that $x_1^2\ne x_2^2$, so that $x_1^2x_2^2\ne e$. Now it follows again from statement (c) of Proposition 3.2 that $\operatorname{ord} p_1=\operatorname{ord} q_1=4$ and $p_1^2=q_1^2=x_1^2x_2^2$. Then Corollary 7.4 yields $x_2^2*x_3^2=[x_1^2x_2^2,x_1^2x_2^2]$, so that $x_2^2x_3^2=x_1^2x_2^2$ and $x_3^2=x_1^2$. We obtain a contradiction with the fact that $x_1$ and $x_3$ are in different $V$orbits.
2) $x_1*p_1=x_2*p_2$. Since $x_1$ and $x_2$ are in different $V$orbits, by Lemma 7.6 there is an element $v\in V$ such that $p_2=vx_1$ and $p_1=vx_2$. Then by Lemma 2.2 we have $x_3\in x_1*p_2=[v,vx_1^2]$, so $x_3=vx_1^2$ since $x_3\notin V$. Similarly, $x_3=vx_2^2$. Hence $x_1^2=x_2^2$. We obtain a contradiction with the fact that $x_1$ and $x_2$ are in different $V$orbits.
Thus, the six multisets $x_1*p_1$, $x_1*q_1$, $x_2*p_2$, $x_2*q_2$, $x_3*p_3$, and $x_3*q_3$ are pairwise distinct. Each of them consists of two (possibly identical) elements and is contained in the fourelement multiset $x_1*x_2*x_3$. However, if $x_1*x_2*x_3$ had multiple elements, then the number of different twoelement multisets contained in it would be strictly less than $6$. Therefore, $x_1*x_2*x_3$ consists of four different elements, which completes the proof of the lemma. $\Box$ Lemma 7.8. Suppose that the elements $x_1$, $x_2$, and $x_3$ lie in pairwise distinct $V$orbits and none of them lies in $V$. If a multiset $y_1*y_2$ is contained in the multiset $x_1*x_2*x_3$, then there is a permutation $\lambda,\mu$ of the numbers $1,2$, a permutation $i,j,k$ of the numbers $1,2,3$, and an element $v\in V$ such that $y_{\lambda}=vx_i$ and $y_{\mu}\in v(x_j*x_k)$. Proof. Let $x_i*x_j=[p_k,q_k]$ for each permutation $i,j,k$ of $1,2,3$. From Lemma 7.7 we know that $x_1*p_1$, $x_1*q_1$, $x_2*p_2$, $x_2*q_2$, $x_3*p_3$, and $x_3*q_3$ are exactly all the six twoelement subsets of the fourelement set $x_1*x_2*x_3$, so one of them coincides with the multiset $y_1*y_2$. Without loss of generality we may assume that $y_1*y_2=x_1*p_1$. Then by Lemma 7.6 there is an element $v$ such that $y_{\lambda}=vx_1$ and $y_{\mu}=vp_1\in v(x_2*x_3)$ for some permutation $\lambda,\mu$ of the numbers $1,2$. $\Box$
8. Twovalued groups with an element of order other than 1, 2, or 4 In this section we prove the following result. Theorem 8.1. Let $X$ be a nonspecial involutive commutative twovalued group containing at least one element whose order is neither $1$, nor $2$, nor $4$. Then $X$ is isomorphic to a coset twovalued group of the form $X_A^{\mathbf{a}}=A/\iota_{\mathbf{a}}$, where $A$ is an abelian group and $\iota_{\mathbf{a}}$ is the antipodal involution on $A$. Remark 8.2. This theorem does not require that the twovalued group $X$ be finitely generated. Moreover, it is easy to check that a twovalued group $A/\iota_{\mathbf{a}}$ is finitely generated if and only if the abelian group $A$ is finitely generated. We give the proof of Theorem 8.1 using an explicit construction of the abelian group $A$. This construction depends essentially on the choice of an element $t\in X$ such that $\operatorname{ord} t\notin\{1,2,4\}$. However, as we show in § 12 (Proposition 12.2), given a finitely generated twovalued group $X$, the resulting abelian group $A$ does not depend on the choice of $t$ up to isomorphism. Apparently, the same must be true without the assumption of finite generation, but we do not know the proof of this result. The following proposition describes the explicit construction of an abelian group $A$. The proof, broken down into a series of lemmas, will take up the rest of this section. Proposition 8.3. Let $X$ be a nonspecial involutive commutative twovalued group, $V\subseteq X$ be the subset consisting of all elements of orders $\leqslant2$, and $t\in X$ be an element such that $\operatorname{ord}t\notin\{1,2,4\}$. Consider the set
$$
\begin{equation}
A=\{(x,p)\in X\times X\mid p\in t*x\}.
\end{equation}
\tag{11}
$$
Then the following hold. (a) If $(x,p)\in A$, $(y,q)\in A$, $t*x=[p, p']$, and $t*y=[q,q']$, then there is a unique pair $(z,r)\in A$ such that (b) Define an operation of multiplication $\bullet\colon A\times A\to A$ by the formula
$$
\begin{equation*}
(x,p)\bullet (y,q)=(z,r),
\end{equation*}
\notag
$$
where $(z,r)\in A$ is the only pair that has properties (i), (ii), and (iii). Then $(A,\bullet)$ is a commutative (singlevalued) group with identity element $(e,t)$ and the inverse element defined by the formula $(x,p)^{1}=(x,p')$, where $t*x=[p,p']$. (c) The projection onto the first factor $A\to X$, $(x,p)\mapsto x$, induces an isomorphism of twovalued groups $A/\iota_{\mathbf{a}}\cong X$, where $\iota_{\mathbf{a}}$ is the antipodal involution: $\iota_{\mathbf{a}}(a)=a^{1}$. The following lemma is a direct consequence of Proposition 3.2. Lemma 8.4. The group $V$ acts on the set $A$ by the formula $v(x,p)=(vx,vp)$, where $v\in V$ and $(x,p)\in A$. If statement (a) of Proposition 8.3 is also true for elements $\alpha=(x,p)$ and $\beta=(y,q)$ of $A$, then for any $u,v\in V$ statement (a) of Proposition 8.3 is true for $u\alpha=(ux,up)$ and $v\beta=(vy,vq)$ and
$$
\begin{equation}
(u\alpha)\bullet (v\beta)=uv(\alpha\bullet\beta).
\end{equation}
\tag{12}
$$
Lemma 8.5. Statement (a) of Proposition 8.3 holds in the special cases when $(y,q)$ is one of the pairs $(e,t)$, $(t,e)$, $(t,t^2)$, $(t^2,t)$, $(x,p)$, $(x,p')$, $(p,x)$, $(p',x)$, and $(s,p)$, where $t*p=[x,s]$. The following formulae hold:
$$
\begin{equation}
(x,p)\bullet (e,t) =(x,p),
\end{equation}
\tag{13}
$$
$$
\begin{equation}
(x,p)\bullet (t,e) =(p',x),
\end{equation}
\tag{14}
$$
$$
\begin{equation}
(x,p)\bullet (t,t^2) =(p,s),
\end{equation}
\tag{15}
$$
$$
\begin{equation}
(x,p)\bullet (t^2,t) =(s',p'),
\end{equation}
\tag{16}
$$
$$
\begin{equation}
(x,p)\bullet (x,p) =(x^2,f),
\end{equation}
\tag{17}
$$
$$
\begin{equation}
(x,p)\bullet (x,p') =(e,t),
\end{equation}
\tag{18}
$$
$$
\begin{equation}
(x,p)\bullet (p,x) =(t,e),
\end{equation}
\tag{19}
$$
$$
\begin{equation}
(x,p)\bullet (p',x) =(f',x^2),
\end{equation}
\tag{20}
$$
$$
\begin{equation}
(x,p)\bullet (s,p) =(t^2,t),
\end{equation}
\tag{21}
$$
where $t*p'=[x,s']$, $x*p=[t,f]$, and $x*p'=[t,f']$. Proof. The fact that the pairs $(z,r)$ on the righthand sides of (13)–(21) belong to the set $A$ and have properties (i), (ii), and (iii) can be verified directly using Lemma 2.2 and Corollary 7.3. It remains to prove the uniqueness of such a pair $(z,r)$ in each case.
First of all, note that the pair $(z,r)$ is definitely unique if at least one of the elements $x$, $p$, $p'$, $y$, $q$, and $q'$ lies in $V$. Indeed,  $\bullet$ if $x\in V$, then $z\in x*y=[xy,xy]$ and $r\in x*q=[xq,xq]$, hence $(z,r)=(xy,xq)$;
 $\bullet$ if $p\in V$, then $z\in p*q'=[pq',pq']$ and $r\in p*y=[py,py]$, hence $(z,r)=(pq',py)$;
 $\bullet$ if $p'\in V$, then $z\in p'*q=[p'q,p'q]$ and $r'\in p'*y=[p'y,p'y]$, hence $(z,r')=(p'q,p'y)$, and then the element $r$ is uniquely determined from the condition $t*z=[r,r']$;
 $\bullet$ the cases $y\in V$, $q\in V$ and $q'\in V$ are completely similar.
These arguments cover the cases $(y,q)=(e,t)$, $(t,e)$, and $(t,t^2)$. Consider the remaining cases in turn, assuming that none of $x$, $p$, $p'$, $y$, $q$, and $q'$ lies in $V$.
1. In the case when $(y,q)=(t^2,t)$ we have $r\in x*q=[p,p']$, hence $r=p$ or $r=p'$. In addition, we have $r\in p*y= p*t^2$. Assume first that $p\notin p*t^2$. Then the only possible value is $r=p'$. Moreover, the element $z$ must belong to the multisets $t*r=[x,s']$ and $x*y=x*t^2$. These two conditions define $z$ uniquely if $x\notin x*t^2$. If $x\in x*t^2$, then $t^2\in x*x$, hence $t^2=x^2$, so by Lemma 7.1 we get that $x=ut$ for some $u\in V$. Then $[p,p']=x*t=[u,ut^2]$, which is impossible, because none of the elements $p$ and $p'$ lies in $V$ by assumption.
Suppose now that $p\in p*t^2$, so that $t^2\in p*p$. Then $t^2=p^2$, so by Lemma 7.1 we have $p=ut$ for some $u\in V$. On the other hand $y=t^2$, so $t*y=[t,t^3]$ and $q'=t^3$. The element $z$ must belong to the multisets $x*y=[u,ut^4]$ and $p*q'=[ut^2,ut^4]$. Since $t^2\ne e$, it follows that $z=ut^4$. Now, $r$ must belong to the multisets $t*z=[ut^3,ut^5]$ and $x*q=p*y=[ut,ut^3]$. If $t^5\ne t$, then only $z=ut^3$ satisfies these two conditions. If $t^5=t$, then either $t^4=e$ or $t^6=e$. The first equality is impossible since $\operatorname{ord} t\notin\{1,2,4\}$, and the second is impossible since otherwise $q'=t^3$ would belong to $V$, which contradicts our assumption.
Before going through the remaining cases, we make one general remark. We have already proved that statement (a) is true when $(y,q)$ is one of the pairs $(e,t)$, $(t,e)$, $(t,t^2)$, and $(t^2,t)$, that is, when one of $y$ and $q$ is equal to $t$. According to Lemma 8.4, this implies that (a) is true if one of the elements $y$ and $q$ lies in the orbit $Vt$. Similarly, (a) is true if one of $x$ and $p$ lies in the orbit $Vt$. Hence in the proof of statement (a) we can assume that none of $x$, $p$, $y$, and $q$ lies in $Vt$. According to Lemma 7.1, this is equivalent to the assumption that none of $x^2$, $p^2$, $y^2$, and $q^2$ is equal to $t^2$. Now we continue to consider all cases under this additional assumption.
2. In the case when $(y,q)=(x,p)$ we have $z\in x*x=[e,x^2]$ and $z\in p*p'$. If $z=e$, then $p=p'$. Since $x\notin V$ and $t\notin V$, this is impossible because the twovalued group $X$ is nonspecial. So the only possible value is $z=x^2$. Now we have $r\in x*p=[t,f]$ and $r\in x^2*t$. If $t\in x^2*t$, then $x^2\in t*t=[e,t^2]$, which is impossible because $x\notin V$ and $x^2\ne t^2$ by assumption. So $t\notin x^2*t$, and the only possible value is $r=f$.
3. In the case when $(y,q)=(x,p')$ the element $z$ must belong to each of the multisets $x*x=[e,x^2]$, $p*q'=p*p=[e,p^2]$, and $p'*q=p'*p'=[e,{p'}^2]$. If $x^2$, $p^2$, and ${p'}^2$ are not all equal, then $z=e$ and $r=t$. If $x^2=p^2={p'}^2$, then from Lemma 7.1 we get that $p=ux$ and $p'=u'x$, where $u,u'\in V$. Then, by Lemma 7.2 the equalities $[x^2,t^2]=p*p'=[uu',uu'x^2]$ hold. So either $t^2=uu',$ or $x^2=uu'$ and $t^2=uu'x^2=e$. Neither conclusion is possible since $\operatorname{ord} t\notin\{1,2,4\}$. Hence the case $x^2=p^2={p'}^2$ is impossible.
4. In the case when $(y,q)=(p,x)$ we have $r\in x*x=[e,x^2]$ and $r\in p*p=[e,p^2]$. If $x^2\ne p^2$, then $r=e$ and so $z=t$. Let $x^2=p^2$. Then by Lemma 7.1 we get that $p=ux$ for some $u\in V$, so $t\in x*p=[u,ux^2]$. Since $t\notin V$, it follows that $t=ux^2$. Then $t*x=[ux,ux^3]$ and $p'=ux^3$. Therefore, the element $z$ belongs to the multisets $x*y=[u,ux^2]$ and $p'*q=[ux^2,ux^4]$. Since $x^4=t^2\ne e$, the only possible value is $z=ux^2=t$. Thus $r\in z*t=[e,x^4]$. However, it has already been noted that $r\in [e,x^2]$. Moreover, by assumption we have $x^4=t^2\ne x^2$. So the only possible value is $r=e$.
5. In the case when $(y,q)=(p',x)$ we have $r\in x*q=[e,x^2]$ and $r\in p*y=p*p'$. However, since $x\notin V$ and $t\notin V$, and as $X$ is nonspecial, it follows that $p\ne p'$. Therefore, $e\notin p*p'$ and the only possible value is $r=x^2$. Now we have $z\in x*y=[t,f']$ and $z\in t*r=t*x^2$. If the inclusion $t\in t*x^2$ were true, then we would get that $x^2\in t*t=[e,t^2]$, which is impossible since $x\notin V$ and $x^2\ne t^2$. So $t\notin t*x^2$ and the only possible value is $z=f'$.
6. In the case when $(y,q)=(s,p)$ we have $z\in p'*q=p'*p=[t^2,x^2]$ (by Lemma 7.2) and $z\in x*s$.
Suppose that $x^2\notin x*s$. Then the only possible value is $z=t^2$, hence $r=t$ or $r=t^3$. However, $r$ must lie in the multiset $x*q=x*p=[t,f]$. If $t^3=f$ were true, then according to Lemma 7.2 we would have $[x^2,p^2]=t*f=[t^2,t^4]$, which is impossible since $x^2\ne t^2$ and $p^2\ne t^2$ by assumption. So $t^3\ne f$, and therefore the only possible value is $r=t$.
Now suppose that $x^2\in x*s$. Then $s\in x^2*x=[x,x^3]$. However, $x\ne s$, since $t,p\notin V$ and the twovalued group $X$ is not special. Therefore, $s=x^3$. By Lemma 7.2 we have $[t^2,p^2]=x*s=[x^2,x^4]$. Since $x^2\ne t^2$, it follows that $t^2=x^4$ and $p^2=x^2$. Hence $p=ux$ and $t=vx^2$ for some $u,v\in V$. Then $[p,p']=t*x=[vx,vx^3]$. If the equality $p'=vx$ were true, then from Lemma 7.2 we would get that $[t^2,x^2]=p*p'=[uv,uvx^2]$, which would imply that either $t^2=uv$, or $x^2=uv$ and $t^2=uvx^2=e$, which is impossible since $\operatorname{ord} t\notin\{1,2,4\}$. Hence $p=vx$ and $p'=vx^3$. Further, $[q,q']=t*s=[vx,vx^5]$, hence $q'=vx^5$. The element $z$ must belong to the multisets $x*y=[x^2,x^4]$ and $p*q'=[x^4,x^6]$. If the equality $x^2=x^6$ were true, then it would follow that $x^8=e$, hence $t^4=(vx^2)^4=e$, which is not valid. Therefore, $x^2\ne x^6$ and the only possible value is $z=x^4$. Now, the element $r$ must belong to the multisets $x*q=[v,vx^2]$ and $p*y=[vx^2,vx^4]$. The equality $x^4=e$ is impossible, because it would imply that $t^2=(vx^2)^2=e$. Therefore, the only possible value is $r=vx^2$. $\Box$ Proof of statement (a) of Proposition 8.3. Lemmas 8.4 and 8.5 imply that statement (a) holds for pairs $(x,p)$ and $(y,q)$ if at least one of the following conditions is satisfied:
It remains to consider the case of ‘general position’, when neither of these two conditions is satisfied. Then by Lemma 7.7 the multiset $t*x*y$ consists of four different elements and $(x*q)\cup (x*q')$ and $(p*y)\cup (p'*y)$ are two different partitions of this fourelement set into two twoelement subsets. Therefore, the sets $x*q$ and $p*y$ have a single common element, which we denote by $r$. Also, $t*x*y=(t*z)\cup (t*z')$, where $x*y=[z,z']$, which means that $r$ belongs to exactly one of the two twoelement sets $t*z$ and $t*z'$. Renaming the elements $z$ and $z'$ if necessary, we can assume that $r\in t*z$ and $r\notin t*z'$. Then $(z,r)$ is the only pair in $A$ that can have properties (i), (ii), and (iii). It remains to show that it does indeed have these properties. Property (ii) and the inclusion $z\in x*y$ hold by construction.
Let us prove property (iii). By Lemma 7.7
$$
\begin{equation*}
t*x*y=(x*q)\cup (x*q')=(p*y)\cup (p'*y)=(t*z)\cup (t*z')
\end{equation*}
\notag
$$
are all the three different partitions of the fourelement set $t*x*y$ into two twoelement subsets. Therefore, since $r$ belongs to the subsets $x*q$, $p*y$, and $t*z$, the element $r'$ of $t*z$ that is different from $r$ belongs to $x*q'$ and $p'*y$, which is property (iii).
Now we prove that $z\in p*q'$. The element $z$ belongs to the multiset
$$
\begin{equation*}
x*y\subseteq (p*t)*y=p*(t*y)=(p*q)\cup(p*q').
\end{equation*}
\notag
$$
Therefore, if $z\notin p*q'$, then $z$ must lie in $p*q$. Assume that this is indeed the case. By Lemma 7.7 for the triple $(t,x,y)$, the multiset $t*x*y$ consists of four different elements and $x*q$, $p*y$, and $t*z$ are three pairwise distinct twoelement subsets of it which intersect in the common element $r$. Hence the union of these three subsets (without taking account of the multiplicities of elements) coincides with the whole of $t*x*y$. On the other hand, since $x\in t*p$, $y\in t*q$, and $z\in p*q$, each of the multisets $x*q$, $p*y$, and $t*z$ is contained in the fourelement multiset $t*p*q$. Hence $t*p*q=t*x*y$. Therefore, the $32$element multiset $(t*p*q)*(t*x*y)$ must contain the identity $e$ with multiplicity at least $4$. On the other hand, we have $x*p = [t,f]$ and $y*q=[t,g]$ for some $f,g\in X$; then
$$
\begin{equation*}
\begin{aligned} \, (t*p*q)*(t*x*y)&=(t*t)*(x*p)*(y*q)=(t*t)*[t,f]*[t,g] \\ &=[t*t*t*t,t*t*t*f,t*t*t*g,t*t*f*g]. \end{aligned}
\end{equation*}
\notag
$$
At the same time the multiset
$$
\begin{equation*}
t*t*t*t=[e,t^2]*[e,t^2]=[e,e,e,t^2,t^2,t^2,t^2,t^4]
\end{equation*}
\notag
$$
contains $e$ only with multiplicity $3$ since $\operatorname{ord} t\ne \{1,2,4\}$. Therefore, $e$ must belong to at least one of the three multisets $t*t*t*f$, $t*t*t*g$, and $t*t*f*g$.
If $e\in t*t*t*f$, then $f\in t*t*t=[t,t,t,t^3]$. Hence either $f=t$ or $f=t^3$. By Lemma 7.2
$$
\begin{equation*}
[x^2,p^2]=t*f=\begin{cases} [e,t^2]&\text{if}\ f=t, \\ [t^2,t^4]&\text{if}\ f=t^3. \end{cases}
\end{equation*}
\notag
$$
Hence $x^2=t^2$ or $p^2=t^2$, so it follows from Lemma 7.1 that $Vx=Vt$ or $Vp=Vt$, which contradicts the assumption that condition 1) fails. Similarly, we arrive at a contradiction by assuming that $e\in t*t*t*g$. If $e\in t*t*f*g$, then by Lemma 7.2 we have
$$
\begin{equation*}
t*t*f*g=(t*f)*(t*g)=[x^2,p^2]*[y^2,q^2].
\end{equation*}
\notag
$$
Hence one of the equalities $x^2=y^2$, $x^2=q^2$, $p^2=y^2$, and $p^2=q^2$ holds, which implies by Lemma 7.1 that condition 2) holds. This leads to a contradiction again.
Thus, $z\notin p*q$, so $z\in p*q'$ and, similarly, $z\in p'*q$. Therefore, the pair $(z,r)$ has indeed properties (i), (ii), and (iii). $\Box$ By construction the operation $\bullet$ is commutative. It follows from formula (13) that $\varepsilon=(e,t)$ is the identity element for multiplication $\bullet$, that is, $\varepsilon\bullet\alpha=\alpha$ for all $\alpha\in A$. Further, it follows from (18) that $(x,p)\bullet (x,p')=\varepsilon$ if $t*x=[p,p']$. If $\alpha=(x,p)\in A$, then we denote the element $(x,p')$ by $\alpha^{1}$. Then $(\alpha^{1})^{1}=\alpha$ and $\varepsilon^{1}=\varepsilon$. Note, however, that we have not yet proved the uniqueness of the inverse, that is, we have not shown that the equality $\alpha\bullet\beta=\varepsilon$ implies that $\beta=\alpha^{1}$. In order to prove that $(A,\bullet)$ is an abelian group, we need to prove the uniqueness of an inverse and the associativity of multiplication. The proof is divided into several lemmas. The next lemma immediately follows from the fact that properties (i), (ii), and (iii) do not change when we interchange simultaneously $p\leftrightarrow p'$, $q\leftrightarrow q'$, and $r\leftrightarrow r'$. Lemma 8.6. For any two elements $\alpha,\beta\in A$ the equality
$$
\begin{equation*}
(\alpha\bullet\beta)^{1}=\alpha^{1}\bullet \beta^{1}
\end{equation*}
\notag
$$
holds. Lemma 8.7. For any two elements $\alpha,\beta\in A$, the equality
$$
\begin{equation*}
\alpha^{1}\bullet(\alpha\bullet\beta)=\beta
\end{equation*}
\notag
$$
holds. Proof. Let $\alpha=(x,p)$, $\beta=(y,q)$, and $\alpha\bullet\beta=(z,r)$; also let $t*x=[p,p']$, $t*y=[q,q']$, and $t*z=[r,r']$. Then the triples $(x,p,p')$, $(y,q,q')$, and $(z,r,r')$ satisfy properties (i), (ii), and (iii). We need to prove that $(x,p')\bullet (z,r)=(y,q)$. This means that we have to prove that properties (i), (ii), and (iii) still hold if we substitute the triples $(x,p',p)$, $(z,r,r')$, and $(y,q,q')$ instead of $(x,p,p')$, $(y,q,q')$, and $(z,r,r')$, respectively. Let us write down the required properties: The fact that the combination of these three properties is equivalent to the combination of the three original properties (i), (ii), and (iii) immediately follows from Lemma 2.2. $\Box$ Corollary 8.8. For any element $\alpha\in A$ the operation of multiplication by $\alpha$ defines a bijection of the set $A$ onto itself. In particular, if $\alpha\bullet\beta=\varepsilon$, then $\beta=\alpha^{1}$, that is, the inverse element is unique. We introduce the notation $\tau=(t,e)$ and $\tau^k=(t^k,t^{k1})$ for all $k\in\mathbb{Z}$. Denote by $\mathcal{T}\subseteq A$ the subset consisting of all elements $\tau^k$, $k\in\mathbb{Z}$. By verifying properties (i), (ii), and (iii) directly, we easily obtain the following. Lemma 8.9. The equalities $\tau^k\bullet\tau^m=\tau^{k+m}$ hold for all $k,m\in\mathbb{Z}$. Thus, the set $\mathcal{T}$ with the operation $\bullet$ is a cyclic group generated by the element $\tau$. Lemma 8.10. For any two elements $\alpha,\beta\in A$ the equality
$$
\begin{equation*}
\tau\bullet(\alpha\bullet\beta)=(\tau\bullet\alpha)\bullet\beta
\end{equation*}
\notag
$$
holds. Proof. Let $\alpha=(x,p)$, $\beta=(y,q)$, $\alpha\bullet\beta=(z,r)$, and let $t*x=[p,p']$, $t*y=[q,q']$, and $t*z=[r,r']$. Then the triples $(x,p,p')$, $(y,q,q')$, and $(z,r,r')$ satisfy properties (i), (ii), and (iii). According to formula (14) we have $\tau\bullet\alpha=(p',x)$ and $\tau\bullet(\alpha\bullet\beta)=(r',z)$. Thus, we need to prove that
$$
\begin{equation}
(p',x)\bullet (y,q)=(r',z).
\end{equation}
\tag{22}
$$
Let $(p',x)\bullet (y,q)=(\zeta,\rho)$.
First we prove that $\rho=z$. Indeed, according to the construction of the multiplication $\bullet$, the element $\rho$ must belong to the multisets $p'*q$ and $x*y$. However, $z$ also belongs to these two multisets. Assume that $\rho\ne z$. Then $p'*q=x*y=[z,\rho]$. By Lemma 7.6 this implies that there is an element $u\in V$ such that either $p'=ux$ and $q=uy$, or $p'=uy$ and $q=ux$. Consider these cases separately.
1. Let $p'=ux$ and $q=uy$. Then the element $t$ belongs to each of the multisets $x*p'=[u,ux^2]$ and $y*q=[u,uy^2]$. Since $t\notin V$, we get that $t=ux^2=uy^2$, so $x^2=y^2$. By Lemma 7.1 this implies that $y=vx$ for some element $v\in V$. We have $t*x=[ux,ux^3]$, hence $p=ux^3$. Thus, $\alpha=(x,ux^3)$ and $\beta=(vx,uvx)$. Now it follows from (18) that $\alpha\bullet\beta=(v,vt)$, that is, $z=v$ and $r=r'=vt$. The equality (22), which we need to prove, takes the form $(ux,x)\bullet (vx,uvx)=(vt,v)$ and immediately follows from (19).
2. Let $p'=uy$ and $q=ux$. It follows from (20) and (17) that
$$
\begin{equation*}
\alpha\bullet\beta=(x,p)\bullet (up',ux)=(uf',ux^2)
\end{equation*}
\notag
$$
and
$$
\begin{equation*}
(\tau\bullet\alpha)\bullet\beta=(p',x)\bullet (up',x)=(u{p'}^2,uf'),
\end{equation*}
\notag
$$
where $x*p'=[t,f']$. By Lemma 7.2 we have $t*f'=[x^2,{p'}^2]$, so (22) becomes
$$
\begin{equation*}
\tau\bullet (uf',ux^2)= (u{p'}^2,uf')
\end{equation*}
\notag
$$
and follows from (14).
Thus we have proved that $\rho=z$ in all cases. We need to prove that $\zeta=r'$. Assume the opposite. According to the construction of the multiplication $\bullet$ the element $\zeta$ must belong to the multisets $p'*y$ and $x*q'$. However, $r'$ also belongs to these two sets. Hence $p'*y=x*q'=[\zeta,r']$. According to Lemma 7.6 this implies that there is an element $u\in V$ such that either $p'=ux$ and $q'=uy$, or $p'=uy$ and $q'=ux$. Consider these cases separately.
1. Let $p'=ux$ and $q'=uy$. Just like in case 1 above, we get that $t=ux^2=uy^2$ and $y=vx$ for some element $v\in V$. We have $t*x=[ux,ux^3]$, hence $p=ux^3$. Similarly, $q=uy^3=uvx^3$. Thus, $\alpha=(x,ux^3)$ and $\beta=(vx,uvx^3)$. Now it follows from (17) that $\alpha\bullet\beta=(vx^2,vf)$, where $x*p=[t,f]$. However, $x*p=[ux^2,ux^4]$, so $f=ux^4$. Thus $z=vx^2=uvt$ and $r=uvx^4$. We have $t*z=[uv,uvx^4]$, hence $r'=uv$. Equality (22) takes the form $(ux,x)\bullet (vx,uvx^3)=(uv,uvt)$ and immediately follows from (18).
2. Let $p'=uy$ and $q'=ux$. It follows from (19) that
$$
\begin{equation*}
\alpha^{1}\bullet\beta^{1}=(x,p')\bullet (up',ux)=(ut,u)=u\tau.
\end{equation*}
\notag
$$
Hence, by Lemmas 8.6 and 8.9
$$
\begin{equation*}
\begin{aligned} \, \alpha\bullet\beta&=u\tau^{1}, \\ \tau\bullet(\alpha\bullet\beta)&=u\varepsilon, \\ (\tau\bullet(\alpha\bullet\beta))\bullet\beta^{1}&= u\beta^{1}=(p',x)=\tau\bullet\alpha, \end{aligned}
\end{equation*}
\notag
$$
so that in view of Corollary 8.8 we have the required equality (22).
Thus, $\zeta=r'$ in all cases, which completes the proof of the lemma. $\Box$ Corollary 8.11. The operation $\bullet$, as restricted to $\mathcal{T}\times A\subseteq A\times A$, defines an action of the cyclic group $\mathcal{T}$ on the set $A$ such that for any elements $\alpha,\beta\in A$ and any integer $k$ the equality $\tau^k\bullet(\alpha\bullet\beta)=(\tau^k\bullet\alpha)\bullet\beta$ is satisfied. Recall that the formula $v(x,p)=(vx,vp)$, $v\in V$, $(x,p)\in A$, defines an action of the commutative group $V$ on the set $A$. According to (12) this action has the property $v(\alpha\bullet\beta)=(v\alpha)\bullet\beta$ for all $v\in V$ and $\alpha,\beta\in A$. It is easy to see that the actions of the groups $V$ and $\mathcal{T}$ on $A$ thus constructed commute. That is, we obtain an action of the direct product $G=V\times \mathcal{T}$ on $A$, which has the property
$$
\begin{equation}
g(\alpha\bullet\beta)=(g\alpha)\bullet\beta
\end{equation}
\tag{23}
$$
for all $g\in G$ and $\alpha,\beta\in A$. It is easy to see that the two lemmas below follow from this property and the uniqueness of the inverse with respect to the operation $\bullet$. Lemma 8.12. The operation $\bullet$ induces a welldefined commutative operation on the set of $G$orbits $A/G$, with respect to which the orbit $G\varepsilon$ is the unit element, and the element inverse to any orbit $G\alpha$ is unique and coincides with the orbit $G\alpha^{1}$. Lemma 8.13. Let the elements $\alpha_1$, $\alpha_2$, $\alpha_3$, $\beta_1$, $\beta_2$, and $\beta_3$ of $A$ be such that for each $i$ the elements $\alpha_i$ and $\beta_i$ lie in the same $G$orbit. Then $(\alpha_1\bullet\alpha_2)\bullet\alpha_3=\alpha_1\bullet(\alpha_2\bullet\alpha_3)$ if and only if
$$
\begin{equation*}
(\beta_1\bullet\beta_2)\bullet\beta_3=\beta_1\bullet(\beta_2\bullet\beta_3).
\end{equation*}
\notag
$$
The following statement easily follows from Lemmas 8.13 and 8.7. Corollary 8.14. If the elements $\alpha_1$, $\alpha_2$, and $\alpha_3$ of $A$ are such that some $\alpha_i$ lies in the $G$orbit $G\varepsilon$ or some $\alpha_i$ and $\alpha_j^{1}$, $i\ne j$, lie in the same $G$orbit, then
$$
\begin{equation*}
(\alpha_1\bullet\alpha_2)\bullet\alpha_3= \alpha_1\bullet(\alpha_2\bullet\alpha_3).
\end{equation*}
\notag
$$
Lemma 8.15. Let the elements $\alpha=(x,p)$ and $\beta=(y,q)$ of $A$ be such that $\beta$ does not lie in the same $G$orbit as $\alpha$ or $\alpha^{1}$, and let $t*x=[p,p']$ and $t*y=[q,q']$. Then none of the elements $x$, $p$, and $p'$ lies in the same $V$orbit as any of $y$, $q$, and $q'$. Proof. It follows from equalities (14) and (15) that the elements $(x,p)$, $(p',x)$, and $(p,s)$ (where $t*p=[x,s]$) lie in the same $G$orbit. Similarly, $(y,q)$, $(q',y)$, and $(q,z)$ (where $t*q=[y,z]$) also lie in the same $G$orbit. Therefore, if one of the elements $x$, $p$, and $p'$ lies in the same $V$orbit as one of $y$, $q$, and $q'$, then in the $G$orbits of $\alpha$ and $\beta$ there are elements $\widetilde{\alpha}=(\widetilde{x},\widetilde{p}\,)$ and $\widetilde{\beta}=(\widetilde{y},\widetilde{q}\,)$, respectively, such that $\widetilde{y}=v\widetilde{x}$ for some $v\in V$. Then either $\widetilde{q}=v\widetilde{p}$ or $\widetilde{q}=v\widetilde{p}\,'$, where $t*\widetilde{x}=[\widetilde{p},\widetilde{p}\,']$. Hence $\widetilde{\beta}$ lies in the same $G$orbit as one of $\widetilde{\alpha}$ and $\widetilde{\alpha}^{1}$, so by Lemma 8.12 the element $\beta$ lies in the same $G$orbit as one of $\alpha$ and $\alpha^{1}$. The next lemma follows directly from the construction of the operation $\bullet$. Lemma 8.16. Let $\alpha_i=(x_i,p_i)$, $i=1,2,3$, be elements of the set $A$, and let $t*x_i=[p_i,p_i']$. Let $(z,r)=(\alpha_1\bullet \alpha_2) \bullet\alpha_3$ and $t*z=[r,r']$. Then Lemma 8.17. Let the elements $\alpha_1$, $\alpha_2$, and $\alpha_3$ of $A$ lie in pairwise distinct $G$orbits. Then
$$
\begin{equation*}
(\alpha_1\bullet\alpha_2)\bullet\alpha_3= \alpha_1\bullet(\alpha_2\bullet\alpha_3)= (\alpha_1\bullet\alpha_3)\bullet\alpha_2.
\end{equation*}
\notag
$$
Proof. If some $\alpha_i$ lies in the $G$orbit $G\varepsilon$ or some $\alpha_i$ and $\alpha_j^{1}$, $i\ne j$, lie in the same $G$orbit, then the required equality holds by Corollary 8.14. Therefore, we can assume that no $\alpha_i$ lies in the same $G$orbit as $\varepsilon$ or any of the elements $\alpha_j^{\pm 1}$ for $j\ne i$. Then it follows from Lemma 8.15 that, first, none of the elements $x_i$, $p_i$ and $p_i'$, where $i=1,2,3$, lies in $V$ or $Vt$, and second, for $i\ne j$ none of $x_i$, $p_i$, and $p_i'$ lies in the same $V$orbit as $x_j$, $p_j$, or $p_j'$.
Assume that the statement of the lemma is false, so that taking the products of $\alpha_1$, $\alpha_2$ and $\alpha_3$ in two of the three indicated orders gives different results $(z_1,r_1)\ne (z_2,r_2)$. Let $t*z_i=[r_i,r_i']$, $i=1,2$. Note that properties (a)–(c) in Lemma 8.16 are invariant under permutations of $\alpha_1$, $\alpha_2$, $\alpha_3$, so that these properties hold for both $(z_1,r_1,r_1')$ and $(z_2,r_2,r_2')$.
Let $(y_i,q_i)=\alpha_j\bullet\alpha_k$ for each permutation $i,j,k$ of $1,2,3$. Then $y_i\in x_j*x_k$ and $q_i\in t*y_i$. Let $x_j*x_k=[y_i,y_i']$ and $t*y_i=[q_i,q_i']$.
Assume first that $z_1\ne z_2$. According to property (a) in Lemma 8.16, both $z_1$ and $z_2$ belong to the multiset $x_1*x_2*x_3$. However, since $x_1$, $x_2$, and $x_3$ lie in pairwise distinct $V$orbits, it follows from Lemma 7.7 that $x_1*x_2*x_3$ consists of four distinct elements and $x_1*y_1$, $x_1*y'_1$, $x_2*y_2$, $x_2*y'_2$, $x_3*y_3$, and $x_3*y'_3$ are all the six twoelement subsets of it. Therefore, one of these six subsets coincides with $[z_1,z_2]$. Renumbering the elements $\alpha_1$, $\alpha_2$, and $\alpha_3$ we can assume that $[z_1,z_2]=x_1*h$, where $h$ is one of the elements $y_1$ and $y_1'$.
Now consider the multisets $x_i*p_j*p_k'$, where $i,j,k$ are permutations of $1,2,3$. According to property (a) in Lemma 8.16, each of these multisets contains the multiset $[z_1,z_2]=x_1*h$. Therefore, by Lemma 7.8, for each triple $(x_i,p_j,p_k')$ there is an element $v\in V$ such that one element of the triple is equal to $vx_1$ or $vh$, and the product of the other two elements of the triple contains the remaining element of the pair $vx_1$, $vh$. Moreover, by what we have proved, the elements $x_2$, $p_2$, $p_2'$, $x_3$, $p_3$, and $p_3'$ cannot lie in the orbit $Vx_1$. In addition, $p_1$ and $p_1'$ cannot both lie in $Vx_1$, since by Lemma 7.5, (c), this would imply that $\operatorname{ord} t$ divides $4$. Now consider three cases.
1. Both $p_1$ and $p_1'$ lie in the orbit $Vh$. Since $p_1$ and $p_1'$ cannot simultaneously lie in the orbit $Vx_1$, we have $Vx_1\ne Vh$. Since $\operatorname{ord} t\notin\{1,2,4\}$, by Lemma 7.5, (b), we have $\operatorname{ord} x_1\in\{1,2,4\}$. None of the elements $x_2$, $p_2$, $p_2'$, $x_3$, $p_3$, and $p_3'$ lies in $Vx_1$ or $Vp_1=Vh$. Thus, it follows from Lemma 7.8 for the triples $(x_i,p_j',p_k)$, $i=2,3$, that each of the multisets $x_2*p_3$, $x_2*p_3'$, $x_3*p_2$, and $x_3*p_2'$ intersects the orbit $Vx_1$. Hence, by Lemma 2.2 the multiset $x_1*x_2$ intersects both $Vp_3$ and $Vp_3'$, and the multiset $x_1*x_3$ intersects both $Vp_2$ and $Vp_2'$. However, since the order of $x_1$ divides $4$, it follows from Lemma 7.5, (b), that each of the multisets $x_1*x_2$ and $x_1*x_3$ consists of two elements which lie in the same $V$orbit. Therefore, $p_2'=w_2p_2$ and $p_3'=w_3p_3$ for some elements $w_2,w_3\in V$. Since $t*x_i=[p_i,p_i']$, it now follows from Lemma 7.5, (b), that the orders of $x_2$ and $x_3$ also divide $4$. Note that $t$ belongs to the subgroup of the twovalued group $X$ generated by $V$ and the elements $x_1$, $x_2$, and $x_3$. Indeed, $t\in x_3*p_3$, and $p_3$ lies in the same $V$orbit as the elements of the multiset $x_1*x_2$. Hence, by Lemma 7.5, (a), we get that the order of $t$ divides $4$, which is a contradiction.
2. One of $p_1$ and $p_1'$ lies in the orbit $Vx_1$ and the other belongs to the orbit $Vh$. Since $p_1$ and $p_1'$ cannot simultaneously lie in the orbit $Vx_1$, we have $Vx_1\ne Vh$. Consider the case when $p_1\in Vx_1$ and $p_1'\in Vh$; the second case is completely similar. From the fact that $p_1=ux_1$ for some $u\in V$, it follows that $t\in x_1*p_1=[u,ux_1^2]$, hence $t=ux_1^2$ since $t\notin V$. Also, since $t\notin V$, we have $x_1^2\notin V$, which means that the order of $x_1$ does not divide $4$. Then, according to statement (c) of Proposition 3.2, all elements $vx_1$, where $v$ runs through the group $V$, are pairwise distinct. In particular, $vx_1\ne x_1$ for $v\ne e$ and $vx_1\ne ux_1$ for $v\ne u$. Moreover, none of the elements $x_2$, $p_2$, $p_2'$, $x_3$, $p_3$, and $p_3'$ lies in $Vx_1$ or $Vp_1=Vh$. Therefore, from Lemma 7.8 for the triples $(x_1,p_2,p_3')$, $(x_1,p_3,p_2')$, $(x_2,p_1,p_3')$, and $(x_3,p_1,p_2')$ it follows that $h$ lies in both $p_2*p_3'$ and $p_3*p_2'$, and $uh$ lies in both $x_2*p_3'$ and $x_3*p_2'$. Using Lemma 2.2, we obtain from here that the multiset $h*p_2'$ contains $ux_3$ and $p_3$, and $h*p_3'$ contains $ux_2$ and $p_2$. If the equality $p_2=ux_2$ were true, then we would get that $t\in x_2*p_2=[u,ux_2^2]$, so $t=ux_2^2$ since $t\notin V$. But then it would follow from the equality $x_2^2=ut=x_1^2$ and Lemma 7.1 that the elements $x_1$ and $x_2$ lie in the same $V$orbit, which is not the case. Hence $ux_2\ne p_2$ and thus $h*p_3'=[ux_2,p_2]$. Likewise, $ux_3\ne p_3$ and $h*p_2'=[ux_3,p_3]$. Then from Lemma 7.2 we get that $h^2$ belongs to the multisets $ux_2*p_2$ and $ux_3*p_3$. However, the element $ut$ also belongs to the same two multisets. Moreover, it follows from Lemma 7.6 that $ux_2*p_2\ne ux_3*p_3$, since none of $x_2$ and $p_2$ lies in the same $V$orbit as any of $x_3$ and $p_3$. Therefore, $h^2=ut=x_1^2$, which contradicts the inequality $Vh\ne Vx_1$ in view of Lemma 7.1.
3. At least one of $p_1$ and $p_1'$ belongs to neither $Vx_1$ nor $Vh$. We assume that $p_1'\notin Vx_1\cup Vh$; the second case is similar. No element of the triples $(x_2,p_3,p_1')$ and $(x_3,p_2,p_1')$ lies in the orbit $Vx_1$; moreover, $p_1'$ does not lie in the orbit $Vh$. By Lemma 7.8 for these triples, at least one of $x_2$ and $p_3$ lies in $Vh$ and at least one of $x_3$ and $p_2$ lies in $Vh$. However, none of $x_2$ and $p_2$ lies in the same $V$orbit as $x_3$ or $p_3$. Therefore, either $x_2,p_2\in Vh$ and $x_3,p_3\notin Vh$, or $x_2,p_2\notin Vh$ and $x_3,p_3\in Vh$. These cases are similar, so we consider the first. Let $x_2=uh$ and $p_2=vh$, where $u,v\in V$. It follows from the equality $p_2=uvx_2$ that $t\in x_2*p_2=[uv,uvx_2^2]$, so $t=uvx_2^2$ since $t\notin V$. Now recall that $h$ is either $y_1$ or $y_1'$, hence $h\in x_2*x_3$. Therefore, $x_3\in x_2*h=[u,ux_2^2]=[u,vt]$, which contradicts the fact that $x_3$ lies in neither $V$ nor $Vt$.
The contradiction obtained proves that $z_1=z_2$. Denote this element simply by $z$. Consider the triple of elements $\beta_1=\tau\bullet\alpha_1$, $\beta_2=\alpha_2$, $\beta_3=\alpha_3$. Lemma 8.10 implies that
$$
\begin{equation*}
\beta_i\bullet(\beta_j\bullet\beta_k)= \tau\bullet(\alpha_i\bullet(\alpha_j\bullet\alpha_k))
\end{equation*}
\notag
$$
for all permutations $i,j,k$ of $1,2,3$. Therefore, it follows from the fact that the products of $\alpha_1$, $\alpha_2$, and $\alpha_3$ in two different orders are equal to $(z,r_1)$ and $(z,r_2)$ and from formula (14) that the products of $\beta_1$, $\beta_2$, and $\beta_3$ in the same two orders are equal to $\tau\bullet (z,r_1)=(r_1',z)$ and $\tau\bullet (z,r_2)=(r_2',z)$ respectively, where $t*z=[r_1,r_1']=[r_2,r_2']$. The elements $\beta_1$, $\beta_2$, and $\beta_3$ also lie in pairwise different $G$orbits. Repeating for them the arguments made for the elements $\alpha_1$, $\alpha_2$, and $\alpha_3$ in the above proof of the equality $z_1=z_2$, we obtain $r_1'=r_2'$. Hence $r_1=r_2$. $\Box$ Lemma 8.18. For any element $\alpha$ of the set $A$ we have the following equality:
$$
\begin{equation*}
(\alpha\bullet\alpha)\bullet(\alpha\bullet\alpha)= \alpha\bullet(\alpha\bullet(\alpha\bullet\alpha)).
\end{equation*}
\notag
$$
Proof. Consider the three elements $\alpha$, $\beta=\alpha\bullet(\alpha\bullet\alpha)$, and $\gamma=(\alpha\bullet\alpha)^{1}$. Suppose first that $\alpha$ and $\beta$ lie in different $G$orbits and $\beta$ and $\gamma$ lie in different $G$orbits. Then
$$
\begin{equation}
(\alpha\bullet\beta)\bullet\gamma=\alpha\bullet(\beta\bullet\gamma).
\end{equation}
\tag{24}
$$
Indeed, if $\alpha$, $\beta$, and $\gamma$ belong to three pairwise distinct $G$orbits, then (24) follows from Lemma 8.17, and if $\alpha$ and $\gamma$ lie in the same $G$orbit, then by Lemma 8.13 equality (24) is equivalent to the valid equality $(\alpha\bullet\beta)\bullet\alpha=\alpha\bullet(\beta\bullet\alpha)$. Using equality (24) and Lemma 8.7 we obtain
$$
\begin{equation*}
\bigl((\alpha\bullet\alpha)\bullet(\alpha\bullet\alpha)\bigr)\bullet(\alpha\bullet\alpha)^{1}=\alpha\bullet\alpha
\end{equation*}
\notag
$$
and
$$
\begin{equation*}
\begin{aligned} \, \bigl(\alpha\bullet(\alpha\bullet(\alpha\bullet\alpha))\bigr)\bullet(\alpha\bullet\alpha)^{1} &=(\alpha\bullet\beta)\bullet\gamma=\alpha\bullet(\beta\bullet\gamma) \\ &=\alpha\bullet\bigl((\alpha\bullet(\alpha\bullet\alpha))\bullet (\alpha\bullet\alpha)^{1}\bigr)=\alpha\bullet\alpha. \end{aligned}
\end{equation*}
\notag
$$
Applying Corollary 8.8 we arrive at the statement of the lemma.
It remains to consider the case when one of the elements $\alpha$ and $\gamma$ lies in the same $G$orbit as $\beta$.
Assume that $\alpha$ and $\beta=\alpha\bullet(\alpha\bullet\alpha)$ lie in the same $G$orbit. Then by Lemma 8.12 the element $\alpha\bullet\alpha$ lies in the same $G$orbit as $\varepsilon$, hence $\alpha$ and $\alpha^{1}$ lie in the same $G$orbit. In this case the equality in question follows from Corollary 8.14 applied to the triple $\alpha$, $\alpha$, $\alpha\bullet\alpha$.
Suppose now that $\beta=\alpha\bullet(\alpha\bullet\alpha)$ and $\gamma=(\alpha\bullet\alpha)^{1}$ lie in the same $G$orbit. Then $\beta=v\tau^k\bullet \gamma$ for some $v\in V$, $k\in \mathbb{Z}$. Hence by Lemma 8.6
$$
\begin{equation*}
\alpha\bullet\alpha =\gamma^{1}=v\tau^k\bullet\beta^{1}= v\tau^k\bullet \bigl(\alpha^{1}\bullet(\alpha\bullet\alpha)^{1}\bigr).
\end{equation*}
\notag
$$
Applying Lemmas 8.6 and 8.7, and Corollary 8.11 we obtain
$$
\begin{equation*}
\begin{aligned} \, (\alpha\bullet\alpha)\bullet(\alpha\bullet\alpha)&= (\alpha\bullet\alpha)\bullet\bigl(v\tau^k\bullet(\alpha^{1}\bullet (\alpha\bullet\alpha)^{1})\bigr) \\ &=v\tau^k\bullet\bigl((\alpha\bullet\alpha)\bullet (\alpha^{1}\bullet(\alpha\bullet\alpha)^{1})\bigr)= v\tau^k\bullet\alpha^{1}, \\ \alpha\bullet(\alpha\bullet(\alpha\bullet\alpha))&= \alpha\bullet(v\tau^k\bullet(\alpha\bullet\alpha)^{1})= v\tau^k\bullet(\alpha\bullet(\alpha^{1}\bullet \alpha^{1}))=v\tau^k\bullet\alpha^{1}. \ \square \end{aligned}
\end{equation*}
\notag
$$
Lemma 8.19. For any elements $\alpha$ and $\beta$ of the set $A$ the equality
$$
\begin{equation*}
(\alpha\bullet\alpha)\bullet\beta=\alpha\bullet(\alpha\bullet\beta)
\end{equation*}
\notag
$$
holds. Proof. Suppose first that none of $\alpha$ and $\beta^{1}$ lies in the same $G$orbit as $\alpha\bullet\beta$. Using Lemmas 8.17, 8.13, and 8.7 we obtain
$$
\begin{equation*}
(\alpha\bullet(\alpha\bullet\beta))\bullet \beta^{1}= \alpha\bullet((\alpha\bullet\beta)\bullet \beta^{1})=\alpha\bullet\alpha.
\end{equation*}
\notag
$$
Multiplying both sides of this equality by $\beta$ and using Lemma 8.7 again we obtain the required equality $\alpha\bullet(\alpha\bullet\beta)=(\alpha\bullet\alpha)\bullet\beta$.
It remains to consider the case when either $\alpha$ or $\beta^{1}$ lies in the same $G$orbit as $\alpha\bullet\beta$.
Assume that $\alpha$ and $\alpha\bullet\beta$ lie in the same $G$orbit. Then it follows from (23) and Lemma 8.7 that $\varepsilon=\alpha^{1}\bullet\alpha$ and $\beta=\alpha^{1}\bullet(\alpha\bullet\beta)$ also lie in the same $G$orbit. In this case the required statement immediately follows from Corollary 8.14.
Suppose now that $\alpha\bullet\beta$ and $\beta^{1}$ lie in the same $G$orbit. Then it follows from Lemmas 8.12 and 8.7 that $\alpha$ lies in the same $G$orbit as $\beta^{1}\bullet\beta^{1}$. Therefore, it follows from Lemma 8.13 that the required associativity property for the triple $(\alpha,\alpha,\beta)$ is equivalent to the associativity property for the triple $(\beta^{1}\bullet\beta^{1},\beta^{1}\bullet\beta^{1},\beta)$. However, using Lemmas 8.18 and 8.7 we get that
$$
\begin{equation*}
\begin{aligned} \, &\bigl((\beta^{1}\bullet\beta^{1})\bullet(\beta^{1}\bullet\beta^{1})\bigr) \bullet\beta=\bigl(((\beta^{1}\bullet\beta^{1})\bullet\beta^{1})\bullet \beta^{1}\bigr)\bullet\beta \\ &\qquad=(\beta^{1}\bullet\beta^{1})\bullet\beta^{1}= (\beta^{1}\bullet\beta^{1})\bullet \bigl((\beta^{1}\bullet\beta^{1})\bullet\beta\bigr). \qquad\square \end{aligned}
\end{equation*}
\notag
$$
Now we can finally fully prove the associativity of the multiplication $\bullet$. Lemma 8.20. The equality $(\alpha_1\bullet\alpha_2)\bullet\alpha_3=\alpha_1\bullet(\alpha_2\bullet\alpha_3)$ holds for any elements $\alpha_1$, $\alpha_2$, and $\alpha_3$ of $A$. Proof. The required equality follows from Lemma 8.17 if the elements $\alpha_1$, $\alpha_2$, and $\alpha_3$ lie in pairwise distinct $G$orbits, and it follows from Lemmas 8.19 and 8.13 if two of these three elements lie in the same $G$orbit. $\Box$ Statement (b) of Proposition 8.3 follows from Corollary 8.8 and Lemma 8.20; statement (c) immediately follows from statements (a) and (b).
9. Nonspecial twovalued groups consisting of elements of orders $1$, $2$, and $4$ Theorem 9.1. Let $X$ be a finitely generated nonspecial involutive commutative twovalued group that does not contain a singlevalued direct factor and consists entirely of elements of orders $1$, $2$, and $4$. Then $X$ is isomorphic to one of the groups in the following two series: Each of the twovalued groups $X_{n\times 4}^{\mathbf{a}}$ and $X_n^{\mathbf{u}}$ consists of $2^{2n1}+2^{n1}$ elements. When $n\leqslant 2$, these two twovalued groups are isomorphic, while for $n\geqslant 3$ they are not isomorphic. Let $X$ be a twovalued group satisfying the assumptions of Theorem 9.1 and $V\subseteq X$ be the subgroup of all elements of orders $1$ and $2$. The order of each element $x\in X\setminus V$ is $4$, so the order of its square $x^2$ is $2$, that is, $x^2\in V\setminus\{e\}$. For each element $v\in V$ denote by $X_v\subseteq X$ the subset of elements $x$ such that $x^2=v$. Then $X_e=V$, while for $v\ne e$ $X_v$ consists of elements of order $4$. According to Proposition 6.2 the fact that $X$ does not contain a singlevalued direct factor implies that all subsets $X_v$ are nonempty. Moreover, Lemma 7.1, Corollary 7.4, and Lemma 7.5, (b), immediately imply the following statement. Lemma 9.2. Each set $X_v$ is a $V$orbit. For $v\ne e$ the stabilizer of the elements of this orbit is the twoelement group $\{e,v\}$. If $x\in X_u$, $y\in X_{v}$, and $x*y=[z_1,z_2]$, then
$$
\begin{equation*}
z_1,z_2\in X_{uv}\quad\textit{and}\quad z_2=uz_1=vz_1.
\end{equation*}
\notag
$$
Corollary 9.3. If the twovalued group $X$ is finitely generated, then the Boolean group $V$ is finite. Proof. Let $X$ be generated by elements $x_1,\dots,x_N$ and let $x_i\in X_{v_i}$, $i=1,\dots,N$. Then any multiset of the form $x_{i_1}*\cdots*x_{i_k}$ consists of elements lying in the orbits $X_v$, where $v=v_{i_1}^{\pm1}\cdots v_{i_k}^{\pm 1}$. Hence the union of the orbits $X_v$, where $v$ runs through the subgroup generated by $v_1,\dots,v_N$, is the whole twovalued group $X$. Therefore, $v_1,\dots,v_N$ generate the whole group $V$, so $\dim V \leqslant N$. Corollary 9.4. If $\dim V=n<\infty$, then the twovalued group $X$ consists of exactly $2^{2n1}+2^{n1}$ elements. In each subset $X_v$, $v\in V$, we choose some representative $x_v$, and as $x_e$ we take the identity $e$. Lemma 9.2 implies that
$$
\begin{equation}
x_u*x_v=\bigl[\varphi(u,v)x_{uv},u\varphi(u,v)x_{uv}\bigr]= \varphi(u,v)[x_{uv},ux_{uv}]
\end{equation}
\tag{25}
$$
for some element $\varphi(u,v)\in V$ satisfying
$$
\begin{equation}
\varphi(u,v) \equiv \varphi(v,u) \pmod{\!\langle u,v\rangle},
\end{equation}
\tag{26}
$$
$$
\begin{equation}
\varphi(u,u) \equiv e\pmod{\!\langle u\rangle},
\end{equation}
\tag{27}
$$
$$
\begin{equation}
\varphi(e,u) \equiv\varphi(u,e)\equiv e\pmod{\!\langle u\rangle},
\end{equation}
\tag{28}
$$
where we let $\langle v_1,\dots,v_k\rangle$ denote the subgroup generated by $v_1,\dots,v_k$. Note that condition (25) defines the element $\varphi(u,v)$ not uniquely, but only up to multiplication by an element of the subgroup $\langle u,v\rangle$ generated by $u$ and $v$. Indeed, multiplication by $u$ or $v$ simply rearranges the elements of the multiset $x_u*x_v$. It is easy to see that multiplication in the twovalued group $X$ can completely be recovered from the map $\varphi\colon V\times V\to V$ we have constructed. Let us now write down the condition that a map $\varphi$ satisfying (26)–(28) defines indeed an associative multiplication on $X$. For arbitrary $u,v,w\in V$ we have
$$
\begin{equation*}
\begin{aligned} \, (x_u*x_v)*x_w&=\varphi(u,v)[x_{uv},ux_{uv}]*x_w \\ &=\varphi(u,v)\varphi(uv,w)[x_{uvw},ux_{uvw},vx_{uvw},wx_{uvw}] \end{aligned}
\end{equation*}
\notag
$$
and
$$
\begin{equation*}
\begin{aligned} \, x_u*(x_v*x_w)&=x_u*\varphi(v,w)[x_{vw},vx_{vw}] \\ &=\varphi(v,w)\varphi(u,vw)[x_{uvw},ux_{uvw},vx_{uvw},wx_{uvw}]. \end{aligned}
\end{equation*}
\notag
$$
Since the element $x_{uvw}$ is stabilized by $e$ and $uvw$, but it is not stabilized by any other elements of $V$, the condition that these two multisets coincide can be written as an inclusion:
$$
\begin{equation}
\varphi(u,v)\varphi(uv,w)\varphi(u,vw)\varphi(v,w)\in\langle u,v,w\rangle.
\end{equation}
\tag{29}
$$
If instead of the set $\{x_v\}$ of representatives of the sets $X_v$ we choose another set of representatives $\{x_v'\}$, where $x_e'=e$ again, then $x_v'=\chi(v)x_v$ for some $\chi(v)\in V$ such that $\chi(e)=e$. Then the map $\varphi$ will be replaced by the map given by the formula
$$
\begin{equation*}
\varphi'(u,v)=\varphi(u,v)\chi(u)\chi(v)\chi(uv).
\end{equation*}
\notag
$$
In view of the fact that, as noted above, $\varphi(u,v)$ is defined up to multiplication by an element of the group $\langle u,v\rangle$, it would be more correct to say that $\varphi$ will be replaced by a map $\varphi'$ such that
$$
\begin{equation}
\varphi'(u,v)\equiv\varphi(u,v)\chi(u)\chi(v)\chi(uv) \pmod{\!\langle u,v\rangle}.
\end{equation}
\tag{30}
$$
We call maps $\varphi\colon V\times V\to V$ satisfying conditions (28) and (29) quasicocycles on the Boolean group $V$. A quasicocycle $\varphi$ is called symmetric if it satisfies (26) and involutive if it satisfies (27). We call two quasicocycles $\varphi$ and $\varphi'$ cohomologous if there exists a map $\chi\colon V\to V$ such that $\chi(e)=e$ and for all $u,v\in V$ the congruence (30) holds. In particular, a quasicocycle is called cohomologically trivial if it is cohomologous to the quasicocycle identically equal to one. Since for the quasicocycle $\varphi$ in (25) the values $\varphi(u,v)$ are defined only up to multiplication by elements of $\langle u,v\rangle$, it will be convenient to say that two quasicocycles $\varphi$ and $\varphi'$ on $V$ are equivalent if for all $u,v\in V$ there is a congruence $\varphi(u,v)\equiv \varphi'(u,v)\pmod{\!\langle u,v\rangle}$ The following proposition immediately follows from the above. Proposition 9.5. A finitely generated nonspecial involutive commutative two valued group $X$ that does not contain a singlevalued direct factor and consists of elements of orders $1$, $2$, and $4$, is completely determined by a Boolean group $V$ and an involutive symmetric quasicocycle $\varphi\colon V\times V\to V$. The twovalued groups corresponding to pairs $(V,\varphi)$ and $(V',\varphi')$ are isomorphic if and only if there is an isomorphism of the groups $V$ and $V'$ that takes the quasicocycle $\varphi$ to a quasicocycle cohomologous to the quasicocycle $\varphi'$. The terminology introduced above has the following origin. Let $G$ be a group and $A$ be an abelian group in additive notation. Recall the definition of the cohomology groups $H^*(G;A)$ (see [2], Chap. III, § 1). For each $k\geqslant 0$ consider the group $\mathcal{C}^k(G;A)$ consisting of the maps
$$
\begin{equation*}
\varphi\colon \underbrace{G\times\cdots\times G}_{k}\to A
\end{equation*}
\notag
$$
such that $\varphi(g_1,\dots,g_k)=0$ if at least one of the $g_i$ is equal to $e$. The differential $\delta\colon \mathcal{C}^k(G;A)\to \mathcal{C}^{k+1}(G;A)$ is defined by
$$
\begin{equation*}
\begin{aligned} \, (\delta\varphi)(g_1,g_2,\dots,g_{k+1})&=\varphi(g_2,\dots,g_{k+1})+ \sum_{i=1}^k(1)^i\varphi(g_1,\dots,g_ig_{i+1},\dots,g_{k+1}) \\ &\qquad+(1)^{k+1}\varphi(g_1,\dots,g_k). \end{aligned}
\end{equation*}
\notag
$$
The maps $\varphi$ that lie in the kernel of this mapping are called $k$cocycles on the group $G$ with values in $A$. The cohomology group $H^k(G;A)$ is by definition the quotient group of the group of $k$cocycles on $G$ with values in $A$ by the subgroup $\delta\mathcal{C}^{k1}(G;A)$. If the group $A$ is Boolean and we use for it the multiplicative notation, then the equation of a $2$cocycle can be written as
$$
\begin{equation}
\varphi(u,v)\varphi(uv,w)\varphi(u,vw)\varphi(v,w)=e
\end{equation}
\tag{31}
$$
for all $u,v,w\in G$. Moreover, two cocycles $\varphi$ and $\varphi'$ are cohomologous if and only if there exists a map $\chi \colon G\to A$ such that $\chi(e)=e$ and
$$
\begin{equation*}
\varphi'(u,v)=\varphi(u,v)\chi(u)\chi(v)\chi(uv)
\end{equation*}
\notag
$$
for all $u,v\in G$. Thus, the above definitions for quasicocycles are weaker versions of these conditions. Proposition 9.5 reduces our classification problem for finitely generated nonspecial involutive commutative twovalued groups that do not contain singlevalued direct factors and consist of elements of orders $1$, $2$, and $4$ to the problem of the classification of involutive symmetric quasicocycles on Boolean groups up to cohomology. On the set $\mathcal{H}(V)$ of cohomology classes of involutive symmetric quasicocycles on $V$ the operation of multiplication defines the structure of an abelian group. Our task is to calculate this group. Before solving this problem, we note that for genuine cocycles satisfying equation (31) exactly, a similar problem is trivial. Lemma 9.6. Let $V$ and $W$ be Boolean groups and $\varphi\colon V\times V\to W$ be an involutive symmetric cocycle, that is, a map satisfying cocycle equation (31) and the equations
$$
\begin{equation}
\varphi(u,v)=\varphi(v,u)
\end{equation}
\tag{32}
$$
and
$$
\begin{equation}
\varphi(u,u)=\varphi(e,u)=\varphi(u,e)=e.
\end{equation}
\tag{33}
$$
Then $\varphi$ is cohomologically trivial, that is, $\varphi=\delta\chi$ for some $\chi\colon V\to W$ such that $\chi(e)=e$. Proof. It obviously suffices to prove the lemma in the case when $W=\mathbb{F}_2$. (Of course, in this case we write $\mathbb{F}_2$ in the additive notation.) Let $\dim V=n$ and $b_1,\dots,b_n$ be a basis of the Boolean group $V$ as a vector space over $\mathbb{F}_2$. Then
$$
\begin{equation*}
H^*(V;\mathbb{F}_2)=H^*\bigl((\mathbb{RP}^{\infty})^n;\mathbb{F}_2\bigr)= \mathbb{F}_2[t_1,\dots,t_n],
\end{equation*}
\notag
$$
where $t_1,\dots,t_n$ is the basis of the space $H^1(V;\mathbb{F}_2)=\operatorname{Hom}(V,\mathbb{F}_2)$ that is dual to the basis $b_1,\dots,b_n$. Therefore, the cohomology class of the cocycle $\varphi$ has the form $\sum\lambda_{i}t_i^2+\sum_{i< j}\mu_{ij}t_it_j$ for some $\lambda_i,\mu_{ij}\in\mathbb{F}_2$. It immediately follows from the definition of multiplication in cohomology that $\lambda_i=\varphi(b_i,b_i)$ and $\mu_{ij}=\varphi(b_i,b_j)+\varphi(b_j,b_i)$. Thus, it follows from equations (32) and (33) that the cohomology class of the cocycle $\varphi$ is zero. Now let us give a nontrivial example of an involutive symmetric quasicocycle. Let $\mathcal{B}=\{b_i\}_{i\in\Lambda}$ be a basis of a Boolean group $V$ (as a vector space over $\mathbb{F}_2$). Then each element of $V$ can uniquely be written in the form $b_I=\prod_{i\in I}b_i$, where $I\subseteq \Lambda$ is a finite subset; in particular, $b_{\varnothing}=e$. Put
$$
\begin{equation}
\varphi_{\mathcal{B}}(b_I,b_J)=b_{I\cap J}.
\end{equation}
\tag{34}
$$
Then conditions (26)–(29) can be verified directly, so that $\varphi_{\mathcal{B}}$ is indeed an involutive symmetric quasicocycle on $V$. Our main result about the group $\mathcal{H}(V)$ is as follows. Proposition 9.7. (a) If $\dim V\leqslant 2$, then every involutive symmetric quasicocycle on $V$ is trivial, and thus the group $\mathcal{H}(V)$ is trivial. (b) If $3\leqslant\dim V<\infty$, then the quasicocycle $\varphi_{\mathcal{B}}$ defined by formula (34) is cohomologically nontrivial. In this case every involutive symmetric quasicocycle on $V$ is cohomologous to either the trivial quasicocycle or the quasicocycle $\varphi_{\mathcal{B}}$. (In particular, the quasicocycles $\varphi_{\mathcal{B}}$ corresponding to different bases $\mathcal{B}$ are cohomologous one to another.) Thus, $\mathcal{H}(V)$ is the cyclic group of order $2$. We defer the proof of this proposition to the next section, and now we show how Theorem 9.1 follows from it. To do this, we calculate the quasicocycles $\varphi$ corresponding to the twovalued groups $X^{\mathbf{a}}_{n\times 4}$ and $X^{\mathbf{u}}_n$. 1. Consider the twovalued group $X=X^{\mathbf{u}}_{n}=(C_2^{n}\times C_2^{n})/\iota_{\mathbf{u}}$, where $\iota_{\mathbf{u}}(a,b)=(a,ab)$. The group of elements of order $2$ in $X$ is exactly the subgroup $V=\{e\}\times C_2^n\subset C_2^n\times C_2^n$ which consists of the elements fixed by the involution $\iota_{\mathbf{u}}$. Identifying $v$ with $(e,v)$ we identify $C_2^n$ with $V$. For any $v\in V=C_2^n$ the subset $X_v\subset X$ consists of the images of all elements of the form $(v,w)$, where $w\in C_2^n$, under the projection $\pi\colon C_2^n\times C_2^n\to X$. We choose the representative $x_v=\pi(v,e)=\pi(v,v)$ of this subset. Then
$$
\begin{equation*}
x_u*x_v=[x_{uv},ux_{uv}],
\end{equation*}
\notag
$$
and therefore the twovalued group $X^{\mathbf{u}}_n$ corresponds to the trivial quasicocycle on the group $C_2^n$. 2. Consider the twovalued group $X=X^{\mathbf{a}}_{n\times 4}=C_4^n/\iota_{\mathbf{a}}$, where $\iota_{\mathbf{a}}(a)=a^{1}$. Let $\{a_i\}_{i\in\Lambda}$ be the system of standard generators of the direct product $C_4^n$. The group of ordertwo elements of $X$ is exactly the subgroup $V= C_2^n\subset C_4^n$ of elements fixed by the involution $\iota_{\mathbf{a}}$. The basis $\mathcal{B}=\{b_i\}_{i\in\Lambda}$ of the Boolean group $V$ consists of the elements $b_i=a_i^2$. An arbitrary element of $V$ has the form $b_I=\prod_{i\in I}b_i$, where $I\subseteq \Lambda$ is a finite subset. As a representative $x_{b_I}$ of the subset $X_{b_I}\subset X$ we take the image of the element $a_I=\prod_{i\in I}a_i$ under the projection $\pi\colon C_4^n\to X$. We have $a_Ia_J=b_{I\cap J}a_{I\bigtriangleup J}$ and $a_I^{1}a_J=b_{I\setminus J}a_{I\bigtriangleup J}$, where $I\bigtriangleup J$ is the symmetric difference between the sets $I$ and $J$. Hence
$$
\begin{equation*}
x_{b_I}*x_{b_J}=[b_{I\cap J}x_{b_{I\bigtriangleup J}},b_{I\setminus J} x_{b_{I\bigtriangleup J}}]=b_{I\cap J}[x_{b_{I\bigtriangleup J}}, b_Ix_{b_{I\bigtriangleup J}}].
\end{equation*}
\notag
$$
Thus, the twovalued group $X_{n\times 4}^{\mathbf{a}}$ corresponds to the quasicocycle $\varphi_{\mathcal{B}}$ on the group $C_2^n$. Theorem 9.1 immediately follows from the above calculation and Propositions 9.5 and 9.7. Recall also that an explicit isomorphism between twovalued groups $X^{\mathbf{a}}_{4,4}$ and $X^{\mathbf{u}}_2$ was described in Example 1.11.
10. Quasicocycles on Boolean groups In this section we prove Proposition 9.7. When $\dim V\leqslant 2$, it immediately follows from the fact that $\langle u,v\rangle =V$ for any pair of distinct nontrivial elements $u,v\in V$, that every involutive symmetric quasicocycle on $V$ is equivalent, and therefore cohomologous, to the trivial one. Thus, statement (a) of Proposition 9.7 is proved. We divide the proof of statement (b) into several lemmas. Lemma 10.1. Let $V$ be the Boolean group of eight elements, that is, a threedimensional vector space over $\mathbb{F}_2$. Then $\mathcal{H}(V)$ is a cyclic group of order $2$. If $\mathcal{B}$ is a basis of the Boolean group $V$, then the cohomology class of the quasicocycle $\varphi_{\mathcal{B}}$ is a generator of $\mathcal{H}(V)$. Proof. Let $\varphi$ be an involutive symmetric quasicocycle on $V$. If $u$ and $v$ are distinct nontrivial elements of $V$, then there are only two cosets of the group $V$ by the subgroup $\langle u,v\rangle$. Set $\lambda(u,v)=0$ if $\varphi(u,v)\in \langle u,v\rangle$ and $\lambda(u,v)=1$ if $\varphi(u,v)\notin \langle u,v\rangle$. Then $\lambda(u,v)=\lambda(v,u)$. The cocycle condition (29) for the triple $(u,v,uv)$ will be written as $\lambda(u,v)=\lambda(v,uv)$, which implies that the value $\lambda(u,v)$ depends only on the twodimensional subspace $\langle u,v\rangle\subset V$ and not on the specific choice of a basis $u$, $v$ in it. Moreover, twodimensional subspaces of the vector space $V$ are in a natural onetoone correspondence with nontrivial elements of the dual vector space $V^*=\operatorname{Hom}(V,\mathbb{F}_2)$. Therefore, we get a welldefined map $\lambda\colon V^*\setminus\{e\}\to \mathbb{F}_2$. It is easy to see that equivalent quasicocycles $\varphi$ lead to identical maps $\lambda$ and a quasicocycle $\varphi$ can uniquely be recovered from the corresponding map $\lambda$ up to equivalence.
Let us now prove that corresponding to any mapping $\lambda\colon V^*\setminus\{e\}\to \mathbb{F}_2$ there is an equivalence class of quasicocycles. To do this, first we calculate the mapping $\lambda_{\mathcal{B}}$ corresponding to the quasicocycle $\varphi_{\mathcal{B}}$, where $\mathcal{B}=\{b_1,b_2,b_3\}$ is some basis of the Boolean group $V$. Direct calculation shows that $\lambda_{\mathcal{B}}(\xi_{\mathcal{B}})=1$, where $\xi_{\mathcal{B}}\in V^*$ is an element such that $\xi_{\mathcal{B}}(b_1)=\xi_{\mathcal{B}}(b_2)=\xi_{\mathcal{B}}(b_3)=1$, and $\lambda_{\mathcal{B}}(\xi)=0$ for all $\xi\ne\xi_{\mathcal{B}}$. The quasicocycles on $V$ form a group with respect to pointwise multiplication, and the product of quasicocycles corresponds to pointwise addition of the corresponding maps $\lambda$. Thus, it is easy to see that by multiplying quasicocycles $\varphi_{\mathcal{B}}$ for different $\mathcal{B}$, we can obtain any map $\lambda\colon V^*\setminus\{e\}\to \mathbb{F}_2$ as $\lambda$. Since $V^*\setminus\{e\}$ contains seven elements, there are exactly $128$ equivalence classes of quasicocycles on $V$.
Now we find out how the cohomological equivalence of two cocycles can be rewritten in terms of the maps $\lambda$. Given a map $\lambda\colon V^*\setminus\{e\}\to \mathbb{F}_2$, we set
$$
\begin{equation*}
\sigma(\lambda)=\sum_{\xi\in V^*\setminus\{e\}}\lambda(\xi).
\end{equation*}
\notag
$$
Let us prove that the quasicocycles corresponding to two maps $\lambda$ and $\lambda'$ are cohomologous if and only if $\sigma(\lambda)=\sigma(\lambda')$.
For two elements $a,b\in V\setminus\{e\}$ consider the map $\chi_{a,b}\colon V\to V$ such that $\chi_{a,b}(a)=b$ and $\chi_{a,b}(u)=e$ for $u\ne a$. Consider two quasicocycles that are cohomologous to each other by means of $\chi_{a,b}$:
$$
\begin{equation*}
\varphi'(u,v)=\varphi(u,v)\chi_{a,b}(u)\chi_{a,b}(v)\chi_{a,b}(uv),
\end{equation*}
\notag
$$
and compare the maps $\lambda,\lambda'\colon V^*\setminus\{e\}\to \mathbb{F}_2$ corresponding to them. Direct calculation shows that
$$
\begin{equation*}
\lambda'(\xi)=\begin{cases} \lambda(\xi)+1&\text{if}\ \xi(a)=0\text{ and }\xi(b)=1, \\ \lambda(\xi) &\text{otherwise}. \end{cases}
\end{equation*}
\notag
$$
Thus, if $a=b$, then the quasicocycles $\varphi$ and $\varphi'$ are equivalent and $\lambda=\lambda'$, and if $a\ne b$, then $\lambda'(\xi)\ne \lambda(\xi)$ for exactly two elements $\xi\in V^*\setminus\{e\}$. Therefore, firstly, $\sigma(\lambda')=\sigma(\lambda)$, and secondly, by choosing a pair $(a,b)$ we can change the map $\lambda$ at any two prescribed elements of $V^*\setminus\{e\}$. Since any map $\chi\colon V\to V$ can be represented as a pointwise product of mappings of the form $\chi_{a,b}$ for different pairs $(a,b)$, we obtain that the equivalence classes of the quasicocycles corresponding to the maps $\lambda,\lambda'\colon V^*\setminus\{e\}\to\mathbb{F}_2$ lie in the same cohomology class if and only if $\sigma(\lambda)=\sigma(\lambda')$. Therefore, $\mathcal{H}(V)$ is the cyclic group of order $2$. Since $\sigma(\lambda_{\mathcal{B}})=1$ for any basis $\mathcal{B}$ of the group $V$, all quasicocycles $\varphi_{\mathcal{B}}$ are cohomologous and their cohomology class generates the group $\mathcal{H}(V)$. $\Box$ Let $V$ be a Boolean group and $\varphi\colon V\times V\to V$ be a quasicocycle. If $U\subset V$ is a subgroup and $\Pi\colon V\to U$ is a projection, then we call the map
$$
\begin{equation*}
U\times U\subset V\times V\xrightarrow{\varphi}V\xrightarrow{\Pi} U
\end{equation*}
\notag
$$
the restriction of the quasicocycle $\varphi$ to the subgroup $U$ along the projection $\Pi$. It is easy to see that this map is a quasicocycle on $U$. Moreover, if the original quasicocycle is involutive and symmetric, then the same holds for the resulting quasi cocycle on $U$. The operation of taking the restriction along the projection $\Pi$ induces a homomorphism $\Pi_*\colon\mathcal{H}(V)\to\mathcal{H}(U)$. Lemma 10.2. Let $V$ be a finite Boolean group such that $\dim V> 3$ and $\Pi\colon V\to U$ be a projection onto some subgroup $U\subset V$ with $\dim U=3$. Then the homomorphism $\Pi_*\colon \mathcal{H}(V)\to\mathcal{H}(U)$ is surjective. Proof. Let $\mathcal{B}$ be a basis of the Boolean group $V$ part of which is a basis $\mathcal{A}$ of the subgroup $U$. Then the restriction of the quasicocycle $\varphi_{\mathcal{B}}$ along $\Pi$ coincides with the quasicocycle $\varphi_{\mathcal{A}}$. This immediately implies the surjectivity of the homomorphism $\Pi_*$, since according to Lemma 10.1 the group $\mathcal{H}(U)$ is generated by the cohomology class of the quasicocycle $\varphi_{\mathcal{A}}$. $\Box$ Lemma 10.3. Let $V$ be a finite Boolean group and $\Pi\colon V\to U$ be a projection onto some subgroup $U\subset V$ with $\dim U\geqslant 3$. Then the homomorphism $\Pi_*\colon \mathcal{H}(V)\to\mathcal{H}(U)$ is injective. Proof. Note that it suffices to prove the lemma in the case when the subgroup $U$ has codimension $1$ in $V$ (if we consider $V$ as a vector space over $\mathbb{F}_2$); then the general case will automatically follow by induction. Denote by $c$ the generator of the onedimensional kernel of the projection $\Pi$. Then $V=U\times\langle c\rangle$.
Let $\varphi$ be an involutive symmetric quasicocycle on $V$ and $\psi$ be its restriction to $U$ along $\Pi$. We need to prove that if $\psi$ is cohomologically trivial, then $\varphi$ is also cohomologically trivial.
It follows from the cohomological triviality of the quasicocycle $\psi$ that there exists a map $\xi\colon U\to U$ such that $\xi(e)=e$ and $\varphi(u_1,u_2)\xi(u_1)\xi(u_2)\xi(u_1u_2)\in \langle u_1,u_2,c\rangle$ for arbitrary $u_1,u_2\in U$. We set $\chi(u)=\xi(u)$ and $\chi(uc)=\xi(u)\varphi(u,c)$ for all $u\in U$ and consider the involutive symmetric quasicocycle $\varphi'$ on $V$ given by the formula $\varphi'(v,w)=\varphi(v,w)\chi(v)\chi(w)\chi(vw)$. Then the quasicocycle $\varphi'$ is cohomologous to $\varphi$ and
$$
\begin{equation}
\varphi'(u_1,u_2)\equiv c^{\lambda(u_1,u_2)}\pmod{\!\langle u_1,u_2\rangle},\qquad u_1,u_2\in U,
\end{equation}
\tag{35}
$$
where $\lambda\colon U\times U\to\mathbb{F}_2$ is some map satisfying $\lambda(u_1,u_2)=\lambda(u_2,u_1)$ and $\lambda(u,u)=\lambda(u,e)=0$. Moreover, $\varphi'(u,c)\in\langle u,c\rangle$ for all $u\in U$.
It follows from condition (29) for triples $(c,c,u)$, where $u\in U$, that $\varphi'(uc,c)\in\langle u,c\rangle$.
Now, from condition (29) for triples $(c,u_1,u_2)$, where $u_1,u_2\in U$, it follows that $\varphi'(u_1c,u_2)\in\langle u_1,u_2,c\rangle$. Consequently,
$$
\begin{equation}
\varphi'(u_1c,u_2)\equiv c^{\mu(u_1,u_2)}\pmod{\!\langle u_1c,u_2\rangle},
\end{equation}
\tag{36}
$$
where $\mu\colon U\times U\to\mathbb{F}_2$ is a map satisfying $\mu(e,u)=\mu(u,e)=0$.
Finally, from condition (29) for triples $(u_1,c,u_2c)$, where $u_1,u_2\in U$, it follows that $\varphi'(u_1c,u_2c)\in\langle u_1,u_2,c\rangle$. Consequently,
$$
\begin{equation}
\varphi'(u_1c,u_2c)\equiv c^{\nu(u_1,u_2)}\pmod{\!\langle u_1c,u_2c\rangle},
\end{equation}
\tag{37}
$$
where $\nu\colon U\times U\to\mathbb{F}_2$ is some map satisfying $\nu(u_1,u_2)=\nu(u_2,u_1)$ and $\nu(u,u)=\nu(u,e)=0$.
Condition (29) for triples $(u_1,u_2,u_3)$, $(u_1c,u_2,u_3)$, and $(u_1c,u_2c,u_3)$, where $u_1,u_2, u_3\in U$, can now be written as the system of equations
$$
\begin{equation}
\lambda(u_2,u_3)+\lambda(u_1u_2,u_3)+\lambda(u_1,u_2u_3)+\lambda(u_1,u_2) =0,
\end{equation}
\tag{38}
$$
$$
\begin{equation}
\lambda(u_2,u_3)+\mu(u_1u_2,u_3)+\mu(u_1,u_2u_3)+\mu(u_1,u_2) =0 \quad\text{for } u_1\notin\langle u_2,u_3\rangle,
\end{equation}
\tag{39}
$$
$$
\begin{equation}
\mu(u_2,u_3)+\lambda(u_1u_2,u_3)+\nu(u_1,u_2u_3)+\nu(u_1,u_2) =0 \quad\text{for } u_1\notin\langle u_3\rangle,\ u_2\notin\langle u_3\rangle.
\end{equation}
\tag{40}
$$
The restrictions $u_1\notin\langle u_2,u_3\rangle$ in (39) and $u_1\notin\langle u_3\rangle$, $u_2\notin\langle u_3\rangle$ in (40) are essential. They are related to the fact that exactly under these restrictions the element $c$ does not belong to the subgroup generated by the triple of elements under consideration, and so we can write down the condition that the degree of the occurrence of $c$ on the lefthand side of the corresponding congruence (29) is zero.
Equation (38) is exactly the usual cocycle condition for $\lambda$. Lemma 9.6 implies that there is a map $\eta\colon U\to\mathbb{F}_2$ such that $\eta(e)=0$ and
$$
\begin{equation}
\lambda(u_1,u_2)=\eta(u_1)+\eta(u_2)+\eta(u_1u_2).
\end{equation}
\tag{41}
$$
If $u,u'\in U$ are distinct elements none of which is equal to $e$, then equation (40) for the triple $(u_1,u_2,u_3)=(u,u,u')$ gives
$$
\begin{equation}
\mu(u,u')=\nu(u,uu').
\end{equation}
\tag{42}
$$
Note that if one of $u$ and $u'$ is equal to $e$, then (42) also holds, because $\mu(u,e)=\mu(e,u)=e$ and $\nu(u,u)=\nu(u,e)=\nu(e,u)=e$. However, the condition $u\ne u'$ is essential.
Now suppose that elements $u_1,u_2,u_3\in U$ are such that $u_1\notin\langle u_2,u_3\rangle$. Then each of the pairs $(u_1u_2,u_3)$, $(u_1,u_2u_3)$, and $(u_1,u_2)$ consists of distinct elements. Therefore, substituting (41) and (42) into (39) we obtain the equation
$$
\begin{equation*}
\nu(u_1u_2,u_1u_2u_3)+\nu(u_1,u_1u_2u_3)+\nu(u_1,u_1u_2)= \eta(u_2)+\eta(u_3)+\eta(u_2u_3).
\end{equation*}
\notag
$$
Making the change of variables $a_1=u_1$, $a_2=u_1u_2$, $a_3=u_1u_2u_3$ in this equation and introducing the new function
$$
\begin{equation}
\rho(u,u')=\nu(u,u')+\eta(uu'),
\end{equation}
\tag{43}
$$
we get that
$$
\begin{equation}
\rho(a_1,a_2)+\rho(a_2,a_3)+\rho(a_3,a_1)=0
\end{equation}
\tag{44}
$$
provided that $a_1$ is not equal to any of the elements $e$, $a_1a_2$, $a_2a_3$, and $a_3a_1$. The last condition means exactly that none of $a_1$, $a_2$, $a_3$, and $a_1a_2a_3$ is equal to $e$. Note that the symmetry of the function $\nu$ immediately implies the symmetry of $\rho$. Moreover, $\rho(u,u)=\nu(u,u)+\eta(e)=0$ for all $u\in U$.
Until now we have never used the key condition $\dim U\geqslant 3$, without which the lemma is false. Let us prove that under this condition equality (44) holds for all triples $(a_1,a_2,a_3)$ of elements of the set $U\setminus\{e\}$ without the assumption that $a_1a_2a_3\ne e$. Indeed, let $a_1$, $a_2$, and $a_3$ be elements of $U\setminus\{e\}$ such that $a_1a_2a_3= e$. Then $\dim\langle a_1,a_2,a_3\rangle=2$. Therefore, it follows from the condition $\dim U\geqslant 3$ that there is an element $b\in U$ not belonging to the subgroup $\langle a_1,a_2,a_3\rangle$. Then, as we have proved, equation (44) holds for each of the triples $(a_1,a_2,b)$, $(a_2,a_3,b)$, and $(a_3,a_1,b)$. Adding these equations we obtain (44) for the initial triple $(a_1,a_2,a_3)$.
From the fact that equation (44) holds for all triples $(a_1,a_2,a_3)$ of elements of the set $U\setminus\{e\}$ it immediately follows that there is a function $\zeta\colon U\setminus\{e\}\to\mathbb{F}_2$ such that
$$
\begin{equation}
\rho(u,u')=\zeta(u)+\zeta(u'), \qquad u,u'\ne e.
\end{equation}
\tag{45}
$$
Indeed, choosing an arbitrary element $u_0\in U\setminus\{e\}$ we can put $\zeta(u)=\rho(u_0,u)$ for all $u\in U\setminus\{e\}$. Then equality (45) follows from (44). From (42), (43), and (45) we obtain
$$
\begin{equation}
\mu(u,u') =\zeta(u)+\eta(u')+\zeta(uu'), \qquad u \ne u',
\end{equation}
\tag{46}
$$
$$
\begin{equation}
\nu(u,u') =\zeta(u)+\zeta(u')+\eta(uu'), \qquad u,u' \ne e.
\end{equation}
\tag{47}
$$
We extend the function $\zeta$ to the whole of $U$ by setting $\zeta(e)=0$. Note, however, that formulae (45) and (47) are not necessarily true if $u$ or $u'$ is equal to $e$, and the same holds for formula (46) if $u=u'$.
Consider the map $\theta\colon V\to V$ and the involutive symmetric quasicocycle $\varphi''\colon V\times V\to V$ defined by the formulae
$$
\begin{equation*}
\begin{alignedat}{2} \theta(u)&=c^{\eta(u)},&\qquad u&\in U, \\ \theta(uc)&=c^{\zeta(u)},&\qquad u&\in U, \\ \varphi''(v,w)&=\varphi'(v,w)\theta(v)\theta(w)\theta(vw),&\qquad v,w&\in V. \end{alignedat}
\end{equation*}
\notag
$$
It follows from (35)– (37), (41), (46), and (47) that
$$
\begin{equation}
\varphi''(v,w)\in \langle v,w\rangle
\end{equation}
\tag{48}
$$
for all $v,w\in V$, that is, the quasicocycle $\varphi''$ is equivalent to the trivial one. Note that in the exceptional cases $(v,w)=(uc,u)$ and $(v,w)=(uc,c)$, when formulae (46) and (47) may not be true, the inclusion (48) still holds, because in these cases $c\in \langle v,w\rangle$. Since $\varphi''$ is cohomologous to $\varphi$ by construction, we obtain the statement of the lemma. $\Box$ Statement (b) of Proposition 9.7 follows from Lemmas 10.1–10.3.
11. Special twovalued groups In this section we prove the following theorem. Theorem 11.1. Any special involutive commutative twovalued group that does not contain singlevalued direct factors is isomorphic to a twovalued group $Y_V$, where $V$ is a Boolean group with $\dim V\geqslant 2$. Lemma 11.2. If $(x,y)$ is a special pair in an involutive commutative twovalued group $X$ and $x*y=[z,z]$, then $x^2=y^2=z^2$ and the five elements $e$, $x$, $y$, $z$, and $s=x^2$ are pairwise distinct and form a subgroup of the twovalued group $X$ with multiplication table
$$
\begin{equation*}
\begin{alignedat}{2} x*x&=y*y=z*z=[e,s], &\quad s*s&=[e,e], \\ x*y&=[z,z],\quad y*z=[x,x],&\quad z*x&=[y,y], \\ s*x&=[x,x],\quad s*y=[y,y],&\quad s*z&=[z,z], \end{alignedat}
\end{equation*}
\notag
$$
that is, they form a subgroup isomorphic to the twovalued group $Y_2$. In particular, $x$, $y$, and $z$ are elements of order $4$. Proof. Let $x*y=[z,z]$. By Lemma 2.2 we have $y*z=[x,x']$ for some $x'$. Then
$$
\begin{equation*}
x*y*z=(x*y)*z=[z,z]*z=[e,e,z^2,z^2]
\end{equation*}
\notag
$$
and
$$
\begin{equation*}
x*y*z=x*(y*z)=x*[x,x']=[e,x^2,x*x'].
\end{equation*}
\notag
$$
However, we know that $x^2\ne e$ because $\operatorname{ord} x>2$. Hence $x^2=z^2$. In addition, we see that the identity $e$ is contained in the multiset $x*x'$, which implies that $x'=x$. Thus, $y*z=[x,x]$. Similarly, we have $y^2=x^2$ and $x*z=[y,y]$. Now set $s=x^2=y^2=z^2$, so that $x*x=y*y=z*z=[e,s]$.
We have
$$
\begin{equation*}
x*y*y=(x*y)*y=[z,z]*y=[x,x,x,x]
\end{equation*}
\notag
$$
and
$$
\begin{equation*}
x*y*y=x*(y*y)=x*[e,s]=[x,x,x*s],
\end{equation*}
\notag
$$
so $x*s=[x,x]$ and $x^3=x$. Since $\operatorname{ord} x>2$, it follows that $\operatorname{ord} x=4$. Hence $\operatorname{ord} s=2$, that is, $s*s=[e,e]$. Similarly to the equality $x*s=[x,x]$ we prove that $y*s=[y,y]$ and $z*s=[z,z]$. Thus the five elements $e$, $s$, $x$, $y$, and $z$ form a subgroup of the twovalued group $X$ which is isomorphic to $Y_2$. $\Box$ It follows from this lemma that if $(x,y)$ is a special pair and $x*y=[z,z]$, then $(x,z)$ and $(y,z)$ are also special pairs and $x*z=[y,y]$ and $y*z=[x,x]$. We call such triples of elements $(x,y,z)$ special triples. Elements of a special triple always have order $4$ and $x^2=y^2=z^2$ holds for them. Proposition 11.3. Let $X$ be a special involutive commutative twovalued group. Then (a) all elements of $X$ have order $1$, $2$, or $4$; (b) all elements of order $4$ of $X$ have the same square, that is, there exists an element $s\in X$ such that $x^2=s$ for all elements $x\in X$ of order $4$. The proof of this proposition will be divided into several lemmas. Lemma 11.4. Given an involutive commutative twovalued group $X$, assume that an element $x$ has order $4$, while the order of an element $q$ is neither $1$, nor $2$, nor $4$. Set $s=x^2$. Then there is a sequence $\{x_n\}_{n\in\mathbb{Z}}$ of elements of $X$ such that $x_0=x$ and
$$
\begin{equation}
q^k*x_n =[x_{nk},x_{n+k}], \qquad k,n \in\mathbb{Z},
\end{equation}
\tag{49}
$$
$$
\begin{equation}
s x_n =x_{n}, \qquad n \in\mathbb{Z},
\end{equation}
\tag{50}
$$
$$
\begin{equation}
x*x_1 =[q,s q],
\end{equation}
\tag{51}
$$
$$
\begin{equation}
x_1*x_1 =[e,s q^2].
\end{equation}
\tag{52}
$$
Moreover, such a sequence is unique up to changing the sign of the index $n$, that is, up to interchanging $x_n\leftrightarrow x_{n}$ simultaneously for all $n$. At the same time $x_1\ne x_{1}$ and neither of $x_1$ and $x_{1}$ has order $1$, $2$, or $4$. Proof. Set $x_0=x$. Since $\operatorname{ord} q\ne 4$, Lemma 11.2 implies that the pair $(q,x)$ is not special. Since the orders of $q$ and $x$ are greater than $2$, it follows that $q*x=[x_1,x_{1}]$ for some elements $x_1,x_{1}\in X$ such that $x_1\ne x_{1}$.
The element $s=x^2$ has order $2$, and we have the following equalities:
$$
\begin{equation*}
[x*x_1,x*x_{1}]=x*(x*q)=(x*x)*q=[e,s]*q=[q,q,sq,sq].
\end{equation*}
\notag
$$
In addition,
$$
\begin{equation*}
s(x*x_1)=(s x)*x_1=x*x_1.
\end{equation*}
\notag
$$
Hence
$$
\begin{equation*}
x*x_1=x*x_{1}=[q,s q].
\end{equation*}
\notag
$$
From the fact that $q\in x*x_1$ it follows that $q^2$ belongs to the multiset
$$
\begin{equation*}
(x*x_1)*(x*x_1)=(x*x)*(x_1*x_1)=[e,s]*[e,x_1^2]= [e,e,s,s,x_1^2,x_1^2,s x_1^2,s x_1^2].
\end{equation*}
\notag
$$
If $x_1$ had order $1$, $2$ or $4$, then the order of the element $x_1^2$ would be $1$ or $2$ and the multiset $(x*x_1)*(x*x_1)$ would consist of elements of orders $1$ and $2$. But this cannot be true since the order of $q^2$ is neither $1$ nor $2$. So $\operatorname{ord} x_1\notin\{1,2,4\}$. Similarly, $\operatorname{ord} x_{1}\notin\{1,2,4\}$.
Now it follows from Proposition 3.2, (c), that $s x_1\ne x_1$ and $s x_{1}\ne x_{1}$. However, at the same time we have
$$
\begin{equation*}
[s x_1,s x_{1}]=s [x_1,x_{1}]=s(x*q)=(sx)*q=x*q=[x_1,x_{1}].
\end{equation*}
\notag
$$
Hence $s x_1=x_{1}$ and $s x_{1}=x_1$.
Equalities (49) for $k=1$ have the form
$$
\begin{equation}
q*x_n=[x_{n1},x_{n+1}].
\end{equation}
\tag{53}
$$
These equalities allow us to define uniquely the elements of the sequence $\{x_n\}$ using recursion, starting from the already defined elements $x_0$, $x_1$, and $x_{1}$. Indeed, the elements $x_n$ with positive indices are defined consequently in the ascending order of $n$. By construction, for $n\geqslant 1$ the element $x_{n}$ lies in the multiset $q*x_{n1}$, and therefore, by Lemma 2.2 the multiset $q*x_n$ does indeed contain $x_{n1}$ and we can and must take the second element of this multiset as $x_{n+1}$. Similarly, the elements $x_n$ with negative indices are uniquely defined in the descending order of $n$.
Let us prove that $q^k*x_n=[x_{n+k},x_{nk}]$ using induction on $k$. For $k=0$ and $k=1$ this is true by construction. For $k\geqslant 2$ we prove this by assuming that the statement is true for $k2$ and $k1$. We have
$$
\begin{equation*}
\begin{aligned} \, q*q^{k1}*x_n&=q*(q^{k1}*x_n)=q*[x_{n+k1},x_{nk+1}] \\ &=[x_{n+k},x_{n+k2},x_{nk+2},x_{nk}] \end{aligned}
\end{equation*}
\notag
$$
and
$$
\begin{equation*}
q*q^{k1}*x_n=(q*q^{k1})*x_n=[q^{k2},q^k]*x_n= [x_{n+k2},x_{nk+2},q^k*x_n],
\end{equation*}
\notag
$$
so $q^k*x_n=[x_{n+k}, x_{nk}]$.
It suffices to prove equality (50) for $n\geqslant 0$, and this is also done using induction on $n$. Indeed, the basis of induction for $n=0$ and $n=1$ is true. The inductive step is obtained by the action of the element $s$ on the equality (53): if $s x_{n1}=x_{n+1}$ and $s x_n=x_{n}$, then
$$
\begin{equation*}
[x_{n+1},x_{n1}]=q*x_{n}=q*(s x_n)=s(q*x_n)= s [x_{n+1},x_{n1}]=[s x_{n+1},x_{n+1}],
\end{equation*}
\notag
$$
so $s x_{n+1}=x_{n1}$.
Now we prove (51) and (52).
Since $q*x=[x_1,s x_1]$, $x_1$ lies in the multisets $q*x$ and $(s q)*x$. By Lemma 2.2 this implies that the elements $q$ and $sq$ lie in $x*x_1$. However, $\operatorname{ord} q\ne 4$, so $q\ne sq$ by Proposition 3.2, (c). Hence $x*x_1=[q,sq]$.
Finally, we obtain
$$
\begin{equation*}
(s q^2)*x_1=q^2*(s x_1)=q^2*x_{1}=[x_1,x_{3}],
\end{equation*}
\notag
$$
so, it follows from Lemma 2.2 that $s q^2$ lies in the multiset $x_1*x_1$. Since $\operatorname{ord} q\ne 4$, we have $\operatorname{ord} q^2\ne 2$, hence $s q^2\ne e$. Therefore, $x_1*x_1=[e,s q^2]$. $\Box$ Lemma 11.5. Let $(x,y,z)$ be a special triple in $X$, let $s=x^2=y^2=z^2$, and let $q\in X$ be an element whose order is neither $1$, nor $2$, nor $4$. Let $\{x_n\}_{n\in\mathbb{Z}}$, $\{y_n\}_{n\in\mathbb{Z}}$, and $\{z_n\}_{n\in\mathbb{Z}}$ be sequences satisfying the assumptions of Lemma 11.4 for the pairs $(x,q)$, $(y,q)$, and $(z,q)$, respectively. Then for all $n\in\mathbb{Z}$ the following equalities hold:
$$
\begin{equation*}
\begin{alignedat}{3} x_n*y&=[z_n,s z_n],&\qquad y_n*z&=[x_n,s x_n],&\qquad z_n*x&=[y_n,s y_n], \\ x*y_n&=[z_n,s z_n],&\qquad y*z_n&=[x_n,s x_n],&\qquad z*x_n&=[y_n,s y_n]. \end{alignedat}
\end{equation*}
\notag
$$
In addition, making the change $z_{n}\leftrightarrow z_{n}$ for all $n\in\mathbb{Z}$ simultaneously if necessary, the following equalities can be ensured:
$$
\begin{equation*}
\begin{alignedat}{2} x_1*y_1&=[z,z_2],&\qquad x_2*y_1=x_1*y_2&=[z_1,z_3], \\ y_1*z_1&=[x,x_2],&\qquad y_2*z_1=y_1*z_2&=[x_1,x_3], \\ z_1*x_1&=[y,y_2],&\qquad z_2*x_1=z_1*x_2&=[y_1,y_3]. \end{alignedat}
\end{equation*}
\notag
$$
Proof. We have
$$
\begin{equation*}
q^n*x*y=(q^n*x)*y=[x_n,sx_n]*y=[x_n*y,s(x_n*y)]
\end{equation*}
\notag
$$
and
$$
\begin{equation*}
q^n*x*y=q^n*(x*y)=q^n*[z,z]=[z_n,s z_n,z_n,s z_n],
\end{equation*}
\notag
$$
so $x_n*y=[z_n,s z_n]$. The formulae for $y_n*z$, $z_n*x$, $x*y_n$, $y*z_n$, and $z*x_n$ are proved similarly.
We have
$$
\begin{equation*}
(x*y)*(q*q)=[z,z]*[e,q^2]=[z,z,z,z,z_2,z_2,sz_2,sz_2]
\end{equation*}
\notag
$$
and
$$
\begin{equation*}
(q*x)*(q*y)=[x_1,s x_1]*[y_1,s y_1]= [x_1*y_1,x_1*y_1,s (x_1*y_1),s (x_1*y_1)].
\end{equation*}
\notag
$$
Since $s z=z$, it follows that either $x_1*y_1=[z,z_2]$ or $x_1*y_1=[z,s z_2]=[z,z_{2}]$. In the second case we need to make the change of $z_n\leftrightarrow z_{n}$ simultaneously for all $n$. Thus, we achieve the equality $x_1*y_1=[z,z_2]$. Then
$$
\begin{equation*}
q*x_1*y_1=(q*x_1)*y_1=[x,x_2]*y_1=[z_1,s z_1,x_2*y_1]
\end{equation*}
\notag
$$
and
$$
\begin{equation*}
q*x_1*y_1=q*(x_1*y_1)=q*[z,z_2]=[z_1,s z_1,z_1,z_3],
\end{equation*}
\notag
$$
so $x_2*y_1=[z_1,z_3]$. Similarly, $x_1*y_2=[z_1,z_3]$.
In a similar way we get that either $y_1*z_1=[x,x_2]$, or $y_{1}*z_1=[x,x_{2}]$. However, now we can no longer make the permutation $x_{n}\leftrightarrow x_{n}$ since it would spoil the already proved equality $x_1*y_1=[z,z_2]$. Therefore, we need to prove the equality $y_1*z_1=[x,x_2]$ in another way. Note that in the case when $x_2=x_{2}$ this equality has already been proved. Therefore, we only need to consider the case $x_2\ne x_{2}$. In this case $x_2\ne x$. Indeed, if the equality $x=x_2$ were true, then the equality $x_{2}=s x_2=s x=x=x_2$ would hold. Now note that it follows from the already proved equalities $x*y_1=[z_1,s z_1]$ and $x_2*y_1=[z_1,z_3]$ and Lemma 2.2 that $x$ and $x_2$ lie in the multiset $y_1*z_1$. Since $x_2\ne x$, this implies the required equality $y_1*z_1=[x,x_2]$. In the same way we have $z_1*x_1=[y,y_2]$.
The formulae for $y_2*z_1$, $y_1*z_2$, $z_2*x_1$, and $z_1*x_2$ are now proved similarly to the ones for $x_2*y_1$ and $x_1*y_2$. $\Box$ Lemma 11.6. Let $(x,y)$ be a special pair in an involutive commutative group $X$. Then there is no element $q$ in $X$ such that $q^2=x$. Proof. Assume that such an element $q$ exists. Let $x*y=[z,z]$ and $s=x^2=y^2=z^2$, and let $\{x_n\}_{n\in\mathbb{Z}}$, $\{y_n\}_{n\in\mathbb{Z}}$, and $\{z_n\}_{n\in\mathbb{Z}}$ be sequences satisfying the assumptons of Lemma 11.5.
We have
$$
\begin{equation*}
[x_2,s x_2]=q^2*x=x*x=[e,s],
\end{equation*}
\notag
$$
so either $x_2=e$ or $x_2=s$. Therefore, either $x_2*y_1=[y_1,y_1]$ or $x_2*y_1=[s y_1,s y_1]$. However, by Lemma 11.5 we have $x_2*y_1=[z_1,z_3]$, which means that either $y_1=z_1$, or $s y_1=z_1$, that is, $y_1=z_{1}$. In either case we get that the multiset $q*y_1$ contains $y$ and $z$. Since $y\ne z$, we have $q*y_1=[y,z]$. Finally,
$$
\begin{equation*}
\begin{aligned} \, [y_1,y_1,x*y_1]&=[e,x]*y_1=(q*q)*y_1=q*(q*y_1)=q*[y,z] \\ &=[y_1,y_{1},z_1,z_{1}]=[y_1,y_{1},y_1,y_{1}], \end{aligned}
\end{equation*}
\notag
$$
so $x*y_1=[y_{1},y_{1}]$. However, by a formula in Lemma 11.5 we have $x*y_1=[y_{1},y_{1}]$. We obtain a contradiction since $y_1\ne y_{1}$, which completes the proof of the lemma. $\Box$ Lemma 11.7. A special involutive commutative twovalued group $X$ cannot contain elements of order $3$. Proof. Assume the opposite: let $q\in X$ be an element of order $3$. Then $q^2=q$, that is, $q*q=[e,q]$. Let $(x,y,z)$ be some special triple in $X$, let $s=x^2=y^2=z^2$, and let $\{x_n\}_{n\in\mathbb{Z}}$, $\{y_n\}_{n\in\mathbb{Z}}$, and $\{z_n\}_{n\in\mathbb{Z}}$ be sequences satisfying the assumptions of Lemma 11.5. Then we have
$$
\begin{equation*}
[x_2,x_{2}]=q^2*x=q*x=[x_1,x_{1}].
\end{equation*}
\notag
$$
So either $x_2=x_1$ or $x_2=x_{1}=s x_1$. If the equality $x_2=x_1$ were true, then we would get that $q*x_1=[x_1,x]$, and therefore
$$
\begin{equation*}
[x_1,x_1,x_1,x]=[e,q]*x_1=(q*q)*x_1=q*(q*x_1)=q*[x,x_1]=[x_1,x_{1},x_1,x].
\end{equation*}
\notag
$$
This is not true since $x_{1}\ne x_{1}$. Hence $x_2=s x_1$. Likewise, $y_2=s y_1$ and $z_2=s z_1$.
Now, using the equalities $x_1*x_1=[e,s q]$, $x_1*z=[y_1,s y_1]$, $x_1*y_1=[z,z_2]$, and $x_1*z_1=[y,y_2]$ we obtain
$$
\begin{equation*}
(x_1*x_1)*y_1=[e,s q]*y_1=[y_1,y_1,s y_2,s y]=[y_1,y_1,y_1,y]
\end{equation*}
\notag
$$
and
$$
\begin{equation*}
x_1*(x_1*y_1)=x_1*[z,z_2]=x_1*[z,s z_1]= [y_1,s y_1,s y,s y_2]=[y_1,y_{1},y,y_1],
\end{equation*}
\notag
$$
which is a contradiction since $y_1\ne y_{1}$. $\Box$ Proof of Proposition 11.3, (a). Assume that the twovalued group $X$ contains an element $q$ whose order is neither $1$, nor $2$, nor $4$. According to Lemma 11.7, the order of $q$ is not equal to $3$; thus $\operatorname{ord} q>4$. Let $(x,y,z)$ be some special triple in $X$, $s=x^2=y^2=z^2$, and let $\{x_n\}_{n\in\mathbb{Z}}$, $\{y_n\}_{n\in\mathbb{Z}}$, and $\{z_n\}_{n\in\mathbb{Z}}$ be sequences satisfying the assumptions of Lemma 11.5. Using the properties of these sequences we obtain
$$
\begin{equation*}
q*y_1*x_2=q*(y_1*x_2)=q*[z_1,z_3]=[z,z_2,z_2,z_4]
\end{equation*}
\notag
$$
and
$$
\begin{equation*}
q*y_1*x_2=(q*y_1)*x_2=[y,y_2]*x_2=[z_2,s z_2,x_2*y_2],
\end{equation*}
\notag
$$
which implies that $s z_2$ coincides with one of the three elements $z$, $z_2$, and $z_4$.
If the equality $s z_2=z_2$ were true, then we would get that $q^2*z=[z_2,z_2]$. Since $\operatorname{ord} q^2>2$ and $\operatorname{ord} z=4$, it would follow that $(q^2,z)$ is a special pair, which is impossible by Lemma 11.6. Therefore, $s z_2\ne z_2$.
If the equality $s z_2=z$ were true, then we would again get that
$$
\begin{equation*}
s z_2=z=s z=z_2,
\end{equation*}
\notag
$$
which is impossible as we have already shown. So $s z_2\ne z$.
Thus, $s z_2=z_4$. Then
$$
\begin{equation*}
[z_3,z_5]=q*z_4=q*(s z_2)=s (q*z_2)=s [z_1,z_3]=[s z_1,s z_3].
\end{equation*}
\notag
$$
If $z_3=s z_3$, then $q^3*z=[z_3,z_3]$. Hence either $\operatorname{ord} q^3\leqslant 2$ or $(q^3,z)$ is a special pair, and then the order of $q^3$ is $4$ by Lemma 11.2. Thus, in any case, the order of $q^3$ divides $4$, hence the order of $q$ divides $12$. On the other hand the order of $q$ is larger than $4$, so either $\operatorname{ord} q=6$ or $\operatorname{ord} q=12$. Then either $q^2$ or $q^4$ has order $3$, which is impossible by Lemma 11.7. Hence $z_3\ne s z_3$, and therefore $z_3= s z_1$. Then
$$
\begin{equation*}
[z_2,z_4]=q*z_3=q*(s z_1)=s(q*z_1)=s[z,z_2]=[z,sz_2]=[z,z_4],
\end{equation*}
\notag
$$
so that $z=z_2$. But then again we obtain $s z_2=z_2$, which is impossible as we have already proved. The resulting contradiction completes the proof of part (a) of the proposition. $\Box$ Lemma 11.8. Let $x$ and $p$ be two elements of order $4$ in an involutive commutative twovalued group $X$ such that $x^2\ne p^2$. Then there is an element $a\in X$ of order $4$ such that $a^2=x^2 p^2$ and
$$
\begin{equation*}
x*p=[a,x^2 a],\qquad x*a=[p,x^2 p],\quad\textit{and}\quad p*a=[x,p^2 x].
\end{equation*}
\notag
$$
Proof. Since $x^2\ne p^2$, it follows from Lemma 11.2 that the pair $(x,p)$ is not special. So $x*p=[a,b]$ for some $a\ne b$. Lemma 2.2 implies now that $p\in x*a$. At the same time, on the one hand
$$
\begin{equation*}
x^2(x*a)=(x^2x)*a=x*a,
\end{equation*}
\notag
$$
and on the other hand $x^2 p\ne p$ (by Proposition 3.2, (c), since $p^2\ne x^2$). Hence $x*a=[p,x^2 p]$. Similarly, $p*a=[x,p^2 x]$.
Now we have
$$
\begin{equation*}
(x*x)*(a*a)=[e,x^2]*[e,a^2]=[e,e,x^2,x^2,a^2,a^2,x^2 a^2,x^2 a^2]
\end{equation*}
\notag
$$
and
$$
\begin{equation*}
(x*a)*(x*a)=[p,x^2 p]*[p,x^2 p]=[e,e,x^2,x^2,p^2,p^2,x^2p^2,x^2 p^2],
\end{equation*}
\notag
$$
so either $a^2=p^2$ or $a^2=x^2p^2$. From the consideration of the product $p*p*a*a$ it follows similarly that either $a^2=x^2$ or $a^2=x^2 p^2$. As $x^2\ne p^2$, we have $a^2=x^2 p^2$.
Since $x^2$ and $p^2$ are different elements of order $2$, we get that $\operatorname{ord} a^2=2$, hence $\operatorname{ord} a=4$. Moreover, we see that $a^2\ne x^2$ since $p^2\ne e$. Hence $x^2 a\ne a$ by Proposition 3.2, (c). However,
$$
\begin{equation*}
[x^2a,x^2 b]=x^2(x*p)=(x^2 x)*p=x*p=[a,b],
\end{equation*}
\notag
$$
hence $b=x^2 a$. $\Box$ Proof of Proposition 11.3, (b). Let $(x,y,z)$ be some special triple in $X$ and let $s=x^2=y^2=z^2$. Assume that there is an element $p\in X$ of order $4$ such that $p^2=t\ne s$. Applying Lemma 11.8 to the pairs $(x,p)$, $(y,p)$, and $(z,p)$ we get that there are elements $a$, $b$, and $c$ of order $4$ in $X$ such that
$$
\begin{equation*}
\begin{alignedat}{3} &&a^2=b^2&=c^2=st,&& \\ x*p&=[a,s a],&\qquad x*a&=[p,s p],&\qquad p*a&=[x,t x], \\ y*p&=[b,s b],&\qquad y*b&=[p,s p],&\qquad p*b&=[y,t y], \\ z*p&=[c,s c],&\qquad z*c&=[p,s p],&\qquad p*c&=[z,t z]. \end{alignedat}
\end{equation*}
\notag
$$
We have
$$
\begin{equation*}
x*y*p=(x*y)*p=[z,z]*p=[c,c,s c,s c]
\end{equation*}
\notag
$$
and
$$
\begin{equation*}
x*y*p=x*(y*p)=x*[b,s b]=[x*b,s(x*b)].
\end{equation*}
\notag
$$
Moreover, by Lemma 11.2 the pair $(x,b)$ is not special since $x^2\ne b^2$. Hence the multiset $x*b$ consists of two different elements. Therefore, $x*b=[c,s c]$. Now it follows from Lemma 2.2 that $x$ belongs to the multiset $b*c$. Hence the element $t x=st x=b^2 x$ also belongs to
$$
\begin{equation*}
b^2(b*c)=(b^2 b)*c=b*c.
\end{equation*}
\notag
$$
At the same time $t x\ne x$ by Proposition 3.2, (c), since $x^2\ne t$. Thus, $b*c=[x,t x]$. Finally, we get that
$$
\begin{equation*}
\begin{aligned} \, [c,c,c,c]&=[e,c^2]*c=[e,b^2]*c=(b*b)*c=b*(b*c) \\ &=b*[x,t x]=[c,s c,t c,st c]=[c,s c,s c,c], \end{aligned}
\end{equation*}
\notag
$$
so $s c=c$, which is impossible according to Proposition 3.2, (c), since $c^2\ne s$. Thus, $p^2=s$ for all elements of $p\in X$ of order $4$. $\Box$ Proof of Theorem 11.1. According to Proposition 11.3 all elements of the two valued group $X$ have orders $1$, $2$, or $4$ and, moreover, there is an element $s$ of order $2$ in $X$ such that $s=x^2$ for all elements $x\in X$ of order $4$. Since $X$ does not contain singlevalued direct factors, it follows from Proposition 6.2 that $s$ is the only element of order $2$ in $X$. Thus, $x^2=s$ for all elements $x\in X$, except $e$ and $s$. The elements $e$ and $s$ form a twoelement subgroup of $X$. Consider the twovalued quotient group $V=X/\{e,s\}$. We have $sx=x$ for all elements $x$, except $e$ and $s$, so $V$ is simply the set $X$ in which $e$ and $s$ are identified into one element $e$ and no other identifications are made. In the twovalued group $V$ the equality $x^2=s$ turns into $x^2=e$. Thus, all nontrivial elements of the twovalued group $V$ have order $2$. According to Proposition 3.2, (a), $V$ is a Boolean group with doubled multiplication, so $X\cong Y_V$.
12. End of classification in the finitely generated case In this section we collect together the results in §§ 2–11 and prove Theorem 1.10. It follows from Proposition 6.3 and Theorems 8.1, 9.1, and 11.1 that every finitely generated involutive commutative twovalued group $X$ is isomorphic to one of the following twovalued groups: $X^{\mathbf{a}}_A\times C_2^m$, where $A$ is a finitely generated abelian group, $X^{\mathbf{u}}_n\times C_2^m$, or $Y_n\times C_2^m.$ Taking the isomorphism $X_A^{\mathbf{a}}\times C_2^m\cong X_{A\times C_2^m}$ into account we see that $X$ is isomorphic to one of the twovalued groups listed in Theorem 1.10. Now we need to show that all these twovalued groups are pairwise nonisomorphic, with the following exceptions:
$$
\begin{equation*}
X^{\mathbf{a}}_{m\times 2,4}\cong X^{\mathbf{u}}_1\times C_2^m\cong Y_1\times C_2^m
\end{equation*}
\notag
$$
and
$$
\begin{equation*}
X^{\mathbf{a}}_{m\times 2,4,4}\cong X^{\mathbf{u}}_2\times C_2^m.
\end{equation*}
\notag
$$
The first of these exceptional isomorphisms is obvious; the second follows from the isomorphism $X^{\mathbf{a}}_{4,4}\cong X^{\mathbf{u}}_2$ (see Example 1.11 and Theorem 9.1). Twovalued groups are divided into two types, special and nonspecial ones, and twovalued groups of different types cannot be isomorphic one to another. According to Proposition 4.3, twovalued groups $Y_n\times C_2^m$ with $n\geqslant 2$ are special, and all other twovalued groups listed in Theorem 1.10 are nonspecial. The special twovalued group $Y_n\times C_2^m$ consists of $2^m(2^n+1)$ elements, so for different pairs $(n,m)$ these groups are nonisomorphic. The twovalued group $X_n^{\mathbf{u}}\times C_2^m$ consists of $2^m(2^{2n1}+2^{n1})$ elements, so for different pairs $(n,m)$ these groups are also nonisomorphic. To complete the proof of Theorem 1.10 it remains to prove the following two propositions. Proposition 12.1. If $n\geqslant 3$ and $m\geqslant 0$, then the twovalued group $X_n^{\mathbf{u}}\times C_2^m$ is not isomorphic to any twovalued group $X^{\mathbf{a}}_{A}$ of the principal series. Proposition 12.2. Let $A$ and $B$ be finitely generated abelian groups such that the twovalued groups $X_A^{\mathbf{a}}$ and $X_B^{\mathbf{a}}$ are isomorphic. Then $A$ and $B$ are also isomorphic. We start with the following simple lemma, which will be used in the proofs of both propositions. Lemma 12.3. Let $A$ be an abelian group and $\pi\colon A\to X_A^{\mathbf{a}}$ be the quotient map by the antipodal involution $\iota_{\mathbf{a}}$. Then $\operatorname{ord} \pi(a)=\operatorname{ord} a$ for all $a\in A$. Proof. By the definition of a coset twovalued group we have
$$
\begin{equation*}
\pi(a^k)*\pi(a)=\bigl[\pi(a^{k1}),\pi(a^{k+1})\bigr]
\end{equation*}
\notag
$$
for all $k$. It follows easily by induction from this equality that $\pi(a)^k=\pi(a^k)$ for all $a\in A$ and $k>0$. Since $\operatorname{ord} \pi(a)$ is by definition the least $k$ such that $\pi(a)^k=e$ and the inverse image of the identity element of the twovalued group $X_A^{\mathbf{a}}$ under the map $\pi$ is precisely the identity element of $A$, we have $\operatorname{ord} \pi(a)=\operatorname{ord} a$. $\Box$ Proof of Proposition 12.1. The twovalued group $X_n^{\mathbf{u}}\times C_2^m$ consists of elements of orders $1$, $2$, and $4$. Therefore, the isomorphism $X_n^{\mathbf{u}}\times C_2^m\cong X^{\mathbf{a}}_A$ can hold only if all elements of the twovalued group $X^{\mathbf{a}}_A$ have order $1$, $2$, or $4$. According to Lemma 12.3, this is equivalent to the fact that all elements of the abelian group $A$ have order $1$, $2$, or $4$, that is, $A\cong C_4^p\times C_2^q$ for some $p,q\geqslant 0$. Then the twovalued group $X^{\mathbf{a}}_A$ consists of $2^q(2^{2p1}+2^{p1})$ elements, and so $p=n$ and $q=m$. Thus, the only twovalued group of the principal series that could be isomorphic to $X_n^{\mathbf{u}}\times C_2^m$ is $X^{\mathbf{a}}_{C_4^n\times C_2^m}\cong X^{\mathbf{a}}_{n\times 4}\times C_2^m$. It is easy to see that all elements of order $2$ in $X^{\mathbf{a}}_{n\times 4}$ and $X^{\mathbf{u}}_n$ are squares. Therefore, by Proposition 6.2 these two twovalued groups do not contain singlevalued direct factors. Hence, by Proposition 6.3 the isomorphism $X_n^{\mathbf{u}}\times C_2^m\cong X^{\mathbf{a}}_{n\times 4}\times C_2^m$ would imply the isomorphism $X_n^{\mathbf{u}}\cong X^{\mathbf{a}}_{n\times 4}$. It remains to note that by Theorem 9.1 the twovalued groups $X_n^{\mathbf{u}}$ and $X^{\mathbf{a}}_{n\times 4}$ are not isomorphic for $n\geqslant 3$. $\Box$ Denote by $N_k(A)$ ($N_k(X)$) the number of elements of order $k$ in the abelian group $A$ (in the involutive commutative twovalued group $X$, respectively). The following wellknown property of finite abelian groups is easily deduced from their classification theorem. Lemma 12.4. If $A$ and $B$ are finite abelian groups such that $N_k(A)=N_k(B)$ for all $k$, then $A$ and $B$ are isomorphic. Corollary 12.5. If $A$ and $B$ are finitely generated abelian groups such that $N_k(A)=N_k(B)$ for all $k$ and the ranks of the free parts of $A$ and $B$ are equal, then $A$ and $B$ are isomorphic. Lemma 12.6. If $A$ is a finitely generated abelian group, then
$$
\begin{equation*}
N_2(X_A^{\mathbf{a}})=N_2(A)\quad\textit{and}\quad N_k(X_A^{\mathbf{a}})=\frac{N_k(A)}{2}\quad\textit{for all } k>2.
\end{equation*}
\notag
$$
Proof. Under the projection $\pi\colon A\to X^{\mathbf{a}}_A$ each pair $\{a,a^{1}\}$ of mutually inverse elements of $A$ is glued into one element of the twovalued group $X^{\mathbf{a}}_A$. In addition, $a=a^{1}$ if $\operatorname{ord}a=2$, and $a\ne a^{1}$ if $\operatorname{ord}a>2$. Therefore, the claim immediately follows from Lemma 12.3. $\Box$ Lemma 12.7. Let $A$ be a finitely generated abelian group. Then the rank of the free part of $A$ is equal to the largest cardinality of a system of elements $x_1,\dots,x_r\in X_A^{\mathbf{a}}$ such that no multiset $x_1^{n_1}*\cdots*x_r^{n_r}$, where $n_1,\dots,n_r\in\mathbb{Z}$ and at least one of the numbers $n_i$ is nonzero, contains the identity element $e$. Proof. If $a_1,\dots,a_r$ are elements of the group $A$ and $x_i=\pi(a_i)$, $i=1,\dots,r$, then the multiset $x_1^{n_1}*\cdots*x_r^{n_r}$ consists of the $2^{r1}$ elements $\pi(a_1^{n_1}a_2^{\pm n_2}\cdots a_r^{\pm n_r})$ corresponding to all possible choices of signs $\pm$. Therefore, the condition that if at least one of the numbers $n_i$ is nonzero, then these elements are distinct from $e$, is equivalent to the condition of the absence of a nontrivial relation $a_1^{m_1}\cdots a_r^{m_r}=e$. $\Box$ Proof of Proposition 12.2. According to Lemmas 12.6 and 12.7, the numbers $N_k(A)$ and the rank of the free part of a finitely generated abelian group $A$ are uniquely determined by the structure of the twovalued group $X_A^{\mathbf{a}}$. Therefore, Proposition 12.2 follows from Corollary 12.5. $\Box$ Thus, Theorem 1.10 is proved.
13. Nonfinitely generated twovalued groups Let us now discuss to what extent our classification theorem can be extended to the case of nonfinitely generated involutive commutative twovalued groups. There are two natural formulations of the problem: we can consider twovalued groups without topology or topological twovalued groups. We consider the first problem in this section and the second in §§ 14 and 15. Of course, in neither case can we expect a complete classification theorem such as Theorem 1.10 in the finitely generated case. Indeed, in the nonfinitely generated case there is no classification theorem even for singlevalued commutative groups. Therefore, the best result we can hope for is the solution of the classification problem for involutive commutative twovalued groups ‘modulo’ the classification of singlevalued commutative groups. The main result of this section is the following theorem. Theorem 13.1. Every involutive commutative twovalued group is isomorphic to one of the following twovalued groups: Twovalued groups from different series are never isomorphic to each other. Two twovalued groups that both belong to the series 2) or both belong to the series 3) are isomorphic to each other if and only if the corresponding Boolean groups $V$ are isomorphic and the corresponding Boolean groups $W$ are also isomorphic. In order to give Theorem 13.1 a complete form we would like to answer the following question. Question 13.2. Is it possible that abelian groups $A$ and $B$ are not isomorphic, but the twovalued groups $X_A^{\mathbf{a}}$ and $X_B^{\mathbf{a}}$ are isomorphic? We do not know the answer to this question for arbitrary abelian groups. Recall that in the case of finitely generated abelian groups a negative answer was obtained in Proposition 12.2. This was proved by counting the elements of different orders; of course, this proof cannot be generalized to the nonfinitely generated case. The proof of Theorem 13.1 follows the same scheme as the proof of Theorem 1.10. Let $X$ be an arbitrary involutive commutative twovalued group. Just as in the proof of Theorem 1.10, we consider three cases: In the first case, by Theorem 8.1 a twovalued group $X$ is isomorphic to a coset twovalued group of the form $X_A^{\mathbf{a}}$, where $A$ is an abelian group. Recall that in the proof of Theorem 8.1 we never used the finite generation property (see Remark 8.2). In the second and third cases we note first of all that Propositions 6.2 and 6.3 can easily be carried over to the nonfinitely generated case. The only difference is that in the proof of Proposition 6.3 the usual induction on dimension must be replaced by transfinite induction in the standard way. Thus from an arbitrary involutive commutative twovalued group a maximal singlevalued direct factor, which is a Boolean group, is split off in a unique way up to isomorphism. This allows us to reduce the classification theorem to the case of twovalued groups $X$ that do not contain singlevalued direct factors, that is (by Proposition 6.2) to twovalued groups $X$ in which every element of order $2$ is the square of an element of order $4$. If such a twovalued group $X$ is special, then $X$ is isomorphic to a group of the form $Y_V$ by Theorem 11.1, in the proof of which the finite generation property was not used either. It remains to consider the case when $X$ is a nonspecial twovalued group consisting entirely of elements of orders $1$, $2$, and $4$, in which each element of order $2$ is the square of an element of order $4$, and to prove that in this case $X$ is isomorphic either to a group of the form $X_{C_4^{\omega}}^{\mathbf{a}}$, where $\omega$ is some (possibly infinite) cardinal, or to a group of the form $X_V^{\mathbf{u}}$, where $V\cong C_2^{\omega}$ is a (possibly infinite) Boolean group. Here $A^{\omega}$ denotes the direct sum (not the direct product) of $\omega$ copies of the abelian group $A$. This is the most interesting case which requires some considerations in addition to those made in §§ 9 and 10. More precisely, for a not necessarily finitely generated twovalued group $X$ (and therefore the not necessarily finite Boolean group $V$ of its elements of order $2$) all arguments in §§ 9 and 10 except for the proof of Lemma 10.3 carry over word for word. Thus, we only need to prove the following statement. Lemma 13.3. Let $V$ be an infinite Boolean group and $\Pi\colon V\to U$ be a projection onto some finite subgroup $U\subset V$ with $\dim U\geqslant 3$. Then the homomorphism $\Pi_*\colon \mathcal{H}(V)\to\mathcal{H}(U)$ is injective. We will need the following two technical lemmas. Lemma 13.4. Let $\varphi$ be an involutive symmetric quasicocycle on a Boolean group $W$ for which the conditions
$$
\begin{equation}
\varphi(u,v)=\varphi(v,u)
\end{equation}
\tag{54}
$$
and
$$
\begin{equation}
\varphi(u,u)=\varphi(e,u)=\varphi(u,e)=e
\end{equation}
\tag{55}
$$
hold exactly. Consider the map $\delta\varphi\colon W\times W\times W\to W$ given by
$$
\begin{equation}
(\delta\varphi)(u,v,w)=\varphi(v,w)\varphi(uv,w)\varphi(u,vw)\varphi(u,v).
\end{equation}
\tag{56}
$$
Then the quasicocycle $\varphi$ is cohomologically trivial if and only if there exists a map $\lambda\colon W\times W\to\mathbb{F}_2$ such that $\lambda(u,u)=\lambda(e,u)=\lambda(u,e)=0$ for all $u\in W$ and
$$
\begin{equation}
\begin{aligned} \, \nonumber (\delta\varphi)(u,v,w)&=u^{\lambda(uv,w)+\lambda(u,vw)+\lambda(u,v)} v^{\lambda(v,w)+\lambda(uv,w)+\lambda(vw,u)+\lambda(v,u)} \\ &\qquad\times w^{\lambda(w,v)+\lambda(w,uv)+\lambda(wv,u)} \end{aligned}
\end{equation}
\tag{57}
$$
for all $u,v,w\in W$. Proof. It is easy to see that any quasicocycle cohomologous to $\varphi$ and satisfying conditions (54) and (55) has the form
$$
\begin{equation*}
\varphi'(u,v)= \varphi(u,v)\chi(u)\chi(v)\chi(uv)u^{\lambda(u,v)}v^{\lambda(v,u)}
\end{equation*}
\notag
$$
for some maps $\chi\colon W\to W$ and $\lambda\colon W\times W\to\mathbb{F}_2$ such that $\chi(e)=e$ and $\lambda(u,u)=\lambda(e,u)=\lambda(u,e)=0$ for all $u\in W$. Then
$$
\begin{equation*}
\begin{aligned} \, (\delta\varphi')(u,v,w)&=(\delta\varphi)(u,v,w) u^{\lambda(uv,w)+\lambda(u,vw)+\lambda(u,v)} \\ &\qquad\times v^{\lambda(v,w)+\lambda(uv,w)+\lambda(vw,u)+\lambda(v,u)} w^{\lambda(w,v)+\lambda(w,uv)+\lambda(wv,u)}. \end{aligned}
\end{equation*}
\notag
$$
If $\varphi$ is cohomologically trivial, then as $\varphi'$ we can take the trivial quasicocycle, and we obtain exactly equality (57).
Conversely, if there is a map $\lambda$ satisfying (57), then the quasicocycle $\varphi$ is equivalent to a genuine involutive symmetric cocycle
$$
\begin{equation*}
\varphi'(u,v)=\varphi(u,v)u^{\lambda(u,v)}v^{\lambda(v,u)},
\end{equation*}
\notag
$$
for which the condition $(\delta\varphi')(u,v,w)=1$ and conditions (54) and (55) are satisfied exactly. By Lemma 9.6, $\varphi'$ is cohomologically trivial, hence the quasicocycle $\varphi$ is also cohomologically trivial. Lemma 13.5. An involutive symmetric quasicocycle on an infinite Boolean group $V$ is cohomologically trivial if its restrictions to all finite subgroups along all possible projections are cohomologically trivial. Proof. Let $\varphi$ be an involutive symmetric quasicocycle on $V$ whose restrictions along all possible projections onto finite subgroups are cohomologically trivial. Replacing the quasicocycle $\varphi$ with an equivalent one we can assume that conditions (54) and (55) are satisfied exactly. Consider the map $\delta\varphi\colon V\times V\times V\to V$ given by formula (56).
Denote the set of finite subgroups of $V$ by $\mathcal{F}$. By Lemma 13.4, for each $U\in\mathcal{F}$ there is a map $\lambda\colon U\times U\to\mathbb{F}_2$ such that $\lambda(u,u)=\lambda(u,e)=\lambda(e,u)=0$ for all $u\in U$ and equality (57) holds for all $u,v,w\in U$. Denote by $\Lambda_U$ the set of maps $\lambda$ that have these properties. Then $\Lambda_U$ is a nonempty finite set. Obviously, if $U'\subset U$, then $\lambda_{U'}\in\Lambda_{U'}$ for every $\lambda\in\Lambda_U$. (For simplicity, in what follows we write $\lambda\big_{U'}$ instead of $\lambda\big_{U'\times U'}$.)
Let us prove that it is possible to choose representatives $\lambda_U\in\Lambda_U$, $U\in\mathcal{F}$, so that for any two subgroups $U_1,U_2\in\mathcal{F}$ the restrictions of $\lambda_{U_1}$ and $\lambda_{U_2}$ to $U_1\cap U_2$ coincide. This choice is made in the following way. By Zermelo’s theorem the set $\mathcal{F}$ has a wellordering $\prec$. Moreover, we can assume that the trivial subgroup $\{e\}$ is minimal with respect to this ordering. We select the representatives of $\lambda_U$ in the order given by the ordering $\prec$ using transfinite recursion. Each representative $\lambda_U$ will be chosen so that the following property holds:
Recursion starts with the trivial map $\lambda_{\{e\}}$ on the trivial subgroup $\{e\}$.
Let us prove that a representative $\lambda_U$ can indeed be chosen in accordance with condition $(*)$ if the previous representatives $\lambda_{U'}$, $U'\prec U$, were chosen so that condition $(*)$ is satisfied for each of them. Assume the contrary: no map in the set $\Lambda_U$, chosen as $\lambda_U$, satisfies condition $(*)$. Let $\lambda_1,\dots,\lambda_N$ be all the elements of the finite set $\Lambda_U$. Then for each map $\lambda_j$ there is a set of subgroups $U_1^{(j)},\dots,U_{k_j}^{(j)}\in\mathcal{F}$ such that $U_1^{(j)}\prec\cdots\prec U_{k_j}^{(j)}=U$, and there is a subgroup $W_j\in \mathcal{F}$ containing $U_1^{(j)},\dots,U_{k_j}^{(j)}$ such that the set of maps $\lambda_{U_1^{(j)}},\dots,\lambda_{U_{k_j1}^{(j)}},\lambda_j$ cannot be extended to a map in the set $\Lambda_{W_j}$. We combine the sets of subgroups $U_1^{(j)},\dots,U_{k_j}^{(j)}$, $j=1,\dots,N$, by removing the repeating subgroups and ordering the resulting finite set of subgroups using the relation $\prec$. As a result, we obtain a finite sequence of subgroups $U_1\prec \cdots\prec U_k=U$. Let $W$ be the finite subgroup of $V$ generated by the subgroups $W_1,\dots,W_N$. Then for no $j$ can the set of maps $\lambda_{U_1},\dots,\lambda_{U_{k1}},\lambda_j$ be extended to a map in the set $\Lambda_{W}$. However, it follows from condition $(*)$ for the subgroup $U_{k1}\prec U$ that there is a map $\mu\in\Lambda_W$ such that $\mu\big_{U_i}=\lambda_{U_i}$, $i=1,\dots,k1$. Since $\mu\big_U$ is one of the mappings $\lambda_1,\dots,\lambda_N$, we obtain a contradiction, which completes the proof of the possibility of constructing maps $\lambda_U$.
From condition $(*)$ it follows, in particular, that for any two subgroups $U_1, U_2\in \mathcal{F}$ the restrictions of $\lambda_{U_1}$ and $\lambda_{U_2}$ to the intersection $U_1\cap U_2$ coincide. Therefore, all maps $\lambda_U$ combine into a single map $\lambda\colon V\times V \to\mathbb{F}_2$ such that $\lambda(u,u)=\lambda(u,e)=\lambda(e,u)=0$ for all $u\in V$ and equality (57) holds for all $u,v,w\in V$. Thus, by Lemma 13.4 the quasicocycle $\varphi$ is cohomologically trivial. $\Box$ Lemma 13.3 follows from Lemmas 10.2, 10.3, and 13.5. Thus we have proved that any involutive commutative twovalued group belongs to one of the three series listed in Theorem 13.1. Twovalued groups of the third series are special and therefore not isomorphic to twovalued groups of the first two series. Twovalued groups of the first and second series (in the case when they contain only elements of orders $1$, $2$, and $4$) are different since the corresponding quasicocycles are not cohomologous. To complete the proof of Theorem 13.1 we need to establish that a pair of Boolean groups $(V,W)$ can be recovered from a twovalued group $X$ of the second or third series uniquely up to isomorphism. This is indeed true: the Boolean group $W$ is the maximal singlevalued direct factor in the decomposition $X\cong X'\times W$, which is unique up to isomorphism by the analogue of Proposition 6.3, and $V$ is the Boolean group of elements of order $2$ in $X'$ for the second series and the Boolean group of elements of order $4$ in $X'$ for the third.
14. Topological twovalued groups If $Y$ is a topological space, then we endow the symmetric square $\operatorname{Sym}^2(Y)$ with the quotient topology of the direct product topology on $Y\times Y$. Definition 14.1. A topological twovalued group is a topological space $X$ with two valued group structure such that the operation of twovalued multiplication $X\times X\to \operatorname{Sym}^2(X)$ and the operation of taking the inverse $X\to X$ are continuous. We consider only Hausdorff topological twovalued groups; the nonHausdorff case is apparently much more complicated. It is easy to see that if $A$ is a single valued commutative Hausdorff topological group and $\iota$ is an involutive continuous automorphism of $A$, then the coset twovalued group $A/\iota$ has the natural structure of a Hausdorff topological twovalued group. Thus, the topological twovalued groups $X_A^{\mathbf{a}}$, for an arbitrary Hausdorff topological abelian group $A$, and $X_V^{\mathbf{u}}$, for a Hausdorff topological Boolean group $V$, are well defined. (Here and below, by a topological Boolean group we mean a topological group all of whose nontrivial elements have order $2$.) Again, we have three essentially different cases: Let us consider them separately. In cases 1) and 3) we obtain analogues of our results, Theorems 8.1 and 11.1. Case 2) is more complicated. In this case the classification problem remains open. We will construct a nontrivial example showing that a direct analogue of Theorem 9.1 on classification is false. 14.1. Topological twovalued groups $X$ containing an element $t$ such that $\operatorname{ord} t\notin\{1,2,4\}$ In this case there is a direct analogue of Theorem 8.1, but its proof requires additional considerations. Theorem 14.2. Let $X$ be an involutive commutative Hausdorff topological two valued group containing at least one element whose order is neither $1$, nor $2$, nor $4$. Then $X$ is isomorphic to a coset topological twovalued group of the form $X_A^{\mathbf{a}}=A/\iota_{\mathbf{a}}$, where $A$ is a commutative Hausdorff topological group and $\iota_{\mathbf{a}}$ is the antipodal involution on $A$. As in § 8, we define the desired topological group as the set
$$
\begin{equation}
A=\{(x,p)\in X\times X\mid p\in t*x\}
\end{equation}
\tag{58}
$$
with the operations of multiplication and taking the inverse defined in Proposition 8.3. We endow $A$ with the topology induced by the direct product topology on $X\times X$. The fact that the resulting structure of a commutative group is well defined was proved in § 8, so we only need to prove that the operations of multiplication and taking inverse introduced in the group $A$ are continuous. (It is obvious that the isomorphism $A/\iota_{\mathbf{a}}\to X$ given by the projection onto the first factor is continuous in both directions.) To prove the continuity of operations we need the notion of a Smith–Dold ramified covering (see [36] and [20]). Definition 14.3. Let $\pi\colon \widetilde{Y}\to Y$ be a surjective continuous mapping of topological spaces with finite inverse images of points. A map $\pi$ is called $d$fold ramified covering in the sense of Smith–Dold if there is a continuous map $\tau\colon Y\to \operatorname{Sym}^d(\widetilde{Y})$ such that Remark 14.4. In the above definition we have followed [14]. The equivalent original definition due to Smith is as follows. A surjective continuous mapping $\pi\colon \widetilde{Y}\to Y$ is called a ramified covering if there is a map $\mu\colon \widetilde{Y}\to \mathbb{Z}_{>0}$, called the multiplicity map, such that the following two conditions are satisfied. The multiplicity map $\mu$ is recovered from the map $\tau$ in Definition 14.3 as the multiplicity of the occurrence of the point $\tilde y$ in the multiset $\tau\circ\pi(\tilde y)$. Informally speaking, a $d$fold ramified covering is a continuous map having a continuous $d$valued inverse. Let $\pi\colon A\to X$ be the projection onto the first factor. Since multiplication in the twovalued group $X$ is continuous, the map $x\mapsto (x,t*x)$ is a continuous twovalued inverse of $\pi$. Therefore, the projection $\pi$ is a twofold ramified Smith–Dold covering. As noted by Dold ([20], § 1), in the Hausdorff case the partially defined operation of removing one element of a multiset
$$
\begin{equation*}
\operatorname{Sym}^d(Y)\times Y \dashrightarrow \operatorname{Sym}^{d1}(Y),\qquad (\sigma,y)\mapsto \sigmay,
\end{equation*}
\notag
$$
is continuous on its domain. This immediately implies the continuity of the operation of taking the inverse in the group $A$, which was defined by the formula $(x,p)^{1}=(x,p')$, where $[p,p']=x*t$. Remark 14.5. Note the specifics of twofold Smith–Dold ramified coverings: on the total space of such a covering there is always a continuous involution that swaps the sheets. In our case this is the antipodal involution on the group $A$. In the case of a $d$fold Smith–Dold ramified covering $\widetilde{Y}\to Y$ with $d>2$ the situation is more complicated: generally speaking, there is no action of a group of order $d$ on the space $\widetilde{Y}$ itself. Nevertheless, Buchstaber and Rees [14], using results of Dold [20], found a construction of an associated $d!$fold ramified covering $E$ over $Y$ with an action of the permutation group $S_d$ such that $\widetilde{Y}=E\times_{S_d}\{1,\dots,d\}$. To prove the continuity of the product operation we need the following technical lemma. Lemma 14.6. Let $\pi\colon\widetilde{Y}\to Y$ be a Smith–Dold ramified covering of Hausdorff topological spaces and $s\colon Y\to \widetilde{Y}$ be its settheoretic section, that is, a map such that $\pi\circ s=\mathrm{id}_Y$. Assume that the graph $\Gamma_s\subset Y\times \widetilde{Y}$ of $s$ is closed. Then $s$ is continuous. Proof. Identifying $\Gamma_s$ with its preimage under the embedding $\pi\times\mathrm{id}\colon \widetilde{Y}\to Y\times\widetilde{Y}$, we may assume that $\Gamma_s\subset\widetilde{Y}$. The subset $\Gamma_s$ is closed in $\widetilde{Y}$ since it is closed in $Y\times\widetilde{Y}$. Let us prove the continuity of the section $s$ at an arbitrary point $y\in Y$. Consider an arbitrary neighbourhood $U$ of the point $s(y)$. Let $y_1=s(y),y_2,\dots,y_k$ be all the different inverse images of the point $y$ under the map $\pi$ without regard to multiplicities. The points $y_2,\dots,y_k$ do not belong to the graph $\Gamma_s$. Therefore, since the graph is closed, they have open neighbourhoods $U_2,\dots,U_k$, respectively, disjoint from $\Gamma_s$. Now, from the continuity of the multivalued inverse to the projection $\pi$ it follows that there is an open neighbourhood $V$ of the point $y$ in $Y$ such that the set $\pi^{1}(V)$ is contained in the union of neighbourhoods $U\cup U_2\cup\cdots\cup U_k$. Therefore, $s(V)$ is contained in the intersection of this union of neighbourhoods with the graph $\Gamma_s$, and therefore in the original neighbourhood $U$. Hence the map $s$ is continuous at $y$. $\Box$ Let us deduce from Lemma 14.6 the continuity of the operation of multiplication $\bullet$ constructed in Proposition 8.3. Consider the closed subset $B\subset A\times A\times A$ of all triples $((x,p),(y,q),(z,r))$ such that $z\in x*y$, and look at the projection $\sigma\colon B\to A\times A$ onto the product of the first two factors. It is easy to see that $\sigma$ is a fourfold Smith–Dold ramified covering. Indeed, the continuous fourvalued inverse mapping assigns to each pair $((x,p),(y,q))$ four triples $((x,p),(y,q),(z,r))$ (counting multiplicities) such that $z\in x*y$ and $r\in t*z$. According to the construction in Proposition 8.3, as the product $(x,p)\bullet (y,q)$ we must take the one of the four points $(z,r)$ such that $((x,p),(y,q),(z,r))\in\sigma^{1}((x,p),(y,q))$ that satisfies additionally the conditions $z\in x*y$, $z\in p*q'$, $z\in p'*q$, $r\in x*q$, $r\in p*y$, $r'\in x*q'$, and $r'\in p'*y$, where $t*x=[p,p']$, $t*y=[q,q']$, and $t*z=[r,r']$, and these conditions determine the required point $(z,r)$ uniquely. In this way,
$$
\begin{equation*}
s\colon\bigl((x,p),(y,q)\bigr)\mapsto \bigl((x,p),(y,q),(x,p)\bullet (y,q)\bigr)
\end{equation*}
\notag
$$
is a settheoretic section of the ramified covering $\sigma$. As noted before, it follows from the Hausdorff property that the points $p'$, $q'$, and $r'$ depend continuously on the pairs $(x,p)$, $(y,q)$, and $(z,r)$, respectively. Therefore, each of the above conditions cuts out a closed subset of $A\times A\times A$. The graph of the section $s$ is the intersection of all these subsets and the closed set $B$. Hence it is closed, and therefore, by Lemma 14.6, the section $s$ is continuous, which completes the proof of Theorem 14.2. 14.2. Special topological twovalued groups Let $V$ be a Boolean group with discrete topology and $\dim V\geqslant 2$, let $U$ be an arbitrary Hausdorff topological Boolean group and $s\in U$ be an arbitrary element other than $e$. Consider the quotient space $W=U/\langle s\rangle$ with the quotient topology (which is always Hausdorff). Let $\pi\colon U\to W$ be the projection. Now consider the topological space
$$
\begin{equation}
Y_{V,U,s}=(\{e\}\times U)\sqcup \bigl((V\setminus\{e\})\times W\bigr),
\end{equation}
\tag{59}
$$
where $\sqcup$ denotes the disjoint union, and define commutative twovalued multiplication on it by the formulae
$$
\begin{equation*}
\begin{alignedat}{2} (e,u_1)*(e,u_2)&=\bigl[(e,u_1u_2),(e,u_1u_2)\bigr]&&\quad\text{if } u_1,u_2\in U, \\ (e,u)*(v,w)&=\bigl[(v,\pi(u)w),(v,\pi(u)w)\bigr]&&\quad\text{if } u\in U,\ v\in V\setminus\{e\},\\ &&&\qquad w\in W, \\ (v,w_1)*(v,w_2)&=\bigl(e,\pi^{1}(w_1w_2)\bigr)&&\quad\text{if } v\in V\setminus\{e\},\\ &&&\qquad w_1,w_2\in W, \\ (v_1,w_1)*(v_2,w_2)&= \bigl[(v_1v_2,w_1w_2),(v_1v_2,w_1w_2)\bigr]&&\quad\text{if } v_1,v_2\in V\setminus\{e\},\\ &&&\qquad v_1\ne v_2, \ w_1,w_2\in W. \end{alignedat}
\end{equation*}
\notag
$$
It can be verified directly that $Y_{V,U,s}$ is a special Hausdorff topological twovalued group. Moreover, if we forget about topology, then this twovalued group becomes isomorphic to the twovalued group $Y_V\times W$. Theorem 14.7. Any special involutive commutative Hausdorff topological two valued group $X$ is isomorphic to one of the topological twovalued groups of the form $Y_{V,U,s}$. Two topological twovalued groups $Y_{V_i,U_i,s_i}$, $i=1,2$, are isomorphic to each other if and only if $V_1\cong V_2$ as abstract groups and there exists an isomorphism of topological groups $U_1\cong U_2$ taking $s_1$ to $s_2$. Proof. Theorem 13.1 immediately implies that, if we forget about the topology, then $X$ is isomorphic to some twovalued group of the form $Y_{V,U,s}$. Therefore, the first statement of Theorem 14.7 is a consequence of the following proposition.
Proposition 14.8. Any Hausdorff topology on the set $Y_{V,U,s}$ with respect to which the above operation of twovalued multiplication is continuous, has the form of the disconnected union of some topology on $\{e\}\times U=U$ with respect to which $U$ is a Hausdorff topological group and the product topology on $(V\setminus\{e\})\times W$, where the set $V\setminus\{e\}$ is endowed with the discrete topology and the set $W$ is equipped with the quotient topology of the topology chosen on $U$. Proof. Consider an arbitrary Hausdorff topology $\mathcal{T}$ on the twovalued group $Y=Y_{V,U,s}$ with respect to which the operation of twovalued multiplication is continuous. We identify the subset $\{e\}\times U$ with $U$ and denote an element $(e,u)$ of the twovalued group $Y$ simply by $u$. Note that $s\in x*x$ if and only if the element $x$ lies in the subset $(V\setminus\{e\})\times W$ of $Y$. Since the operation of multiplication in $Y$ is continuous, the condition $s\in x*x$ distinguishes a closed subset of $Y$. Hence $\{e\}\times U$ is open in $Y$. Denote by $\mathcal{T}_U$ the restriction of the topology $\mathcal{T}$ to $U$; then $(U,\mathcal{T}_U)$ is a topological group. Denote the quotient topology of the topology $\mathcal{T}_U$ under the projection $\pi\colon U\to W$ by $\mathcal{T}_W$.
It remains to prove that each set $\{v\}\times W$ is also open in $Y$ and the topology $\mathcal{T}_{\{v\}\times W}$ coincides with the quotient topology $\mathcal{T}_W$ under the standard identification of $\{v\}\times W$ with $W$. Let $y = (v,e)$; then the formula $x\mapsto x*y$ defines a continuous map $\mu_y\colon Y\to\operatorname{Sym}^2(Y)$. First, note that $\{v\}\times W$ coincides with the inverse image of the open set $\operatorname{Sym}^2(U)$ under $\mu_y$, so that $\{v\}\times W$ is open. Second, if $\mu_y$ is restricted to $\{e\}\times U$, then its two branches coincide and it defines a continuous mapping $\nu_y\colon \{e\}\times U\to\{v\}\times W$, coinciding with $(e,u)\mapsto \bigl(v,\pi(u)\bigr)$. Third, the mapping $\mu_y_{\{v\}\times W}$ is a continuous twovalued inverse of the mapping $\nu_y$. Therefore, $\nu_y$ is a twofold Smith–Dold ramified covering, which immediately implies that the topology on the set $\{v\}\times W$ coincides with the quotient topology $\mathcal{T}_W$. The second statement of Theorem 14.7 follows from the fact that the topological Boolean group $U$, its element $s$, and the discrete Boolean group $V$ are uniquely recovered up to isomorphism from the topological twovalued group $Y=Y_{V,U,s}$. Indeed, $U\subset Y$ is the subgroup consisting of the identity element and all elements of order $2$, with the topology of a subset, $s$ is the element that is the square of all elements of order $4$, and $V\cong Y/U$. $\Box$ 14.3. Nonspecial topological twovalued groups consisting of elements of orders 1, 2, and 4 Here we give an example showing that fundamentally new effects arise in this case and that the direct analogue of Theorem 9.1 is false. First of all, we note that for topological twovalued groups there are no analogues of direct factorization results such as Propositions 6.2 and 6.3. This is because even for ordinary commutative Hausdorff topological groups there is no welldefined operation of taking a quotient group that would leave us in the category of Hausdorff topological groups (for example, when a subgroup is dense in a group). Therefore, in the topological case it is natural to generalize the definition of the unipotent series $X_V^{\mathbf{u}}\times W$ as follows. We will need the following category $\mathcal{P}$, which we call the category of pairs of Hausdorff topological Boolean groups. An object of this category is a pair $(V,U)$, where Note that we do not require the topology $\mathcal{T}_U$ to be the same as the restriction of the topology $\mathcal{T}_V$ to $U$; the definition only implies that the topology of $\mathcal{T}_U$ either coincides with the restriction of the topology $\mathcal{T}_V$ or is finer than the latter. A morphism between objects $(V_1,U_1)$ and $(V_2,U_2)$ in the category $\mathcal{P}$ is a commutative diagram where $\alpha$ is a homomorphism continuous with respect to $\mathcal{T}_{U_1}$ and $\mathcal{T}_{U_2}$ and $\beta$ is a homomorphism continuous with respect to the topologies $\mathcal{T}_{V_1}$ and $\mathcal{T}_{V_2}$. We call objects of the category $\mathcal{P}$ pairs of Hausdorff topological Boolean groups. Let $(V,U)$ be a pair in the category $\mathcal{P}$. Consider the continuous unipotent involution $\iota_{\mathbf{u}}\colon U\times V\to U\times V$ defined by
$$
\begin{equation*}
\iota_{\mathbf{u}}(u,v)=(u,uv).
\end{equation*}
\notag
$$
Then the coset twovalued group
$$
\begin{equation}
X^{\mathbf{u}}_{V,U}=(U\times V)/\iota_{\mathbf{u}}
\end{equation}
\tag{60}
$$
is an involutive commutative Hausdorff topological twovalued group. In the particular case when the subgroup $U$ is distinguished as a topological direct factor: $V=U\times W$, and the topology $\mathcal{T}_V$ is the product topology, this construction yields the topological twovalued groups $X^{\mathbf{u}}_{V,U}\cong X^{\mathbf{u}}_{U}\times W$. Denote by $\mathcal{C}$ the category of nonspecial involutive commutative Hausdorff topological twovalued groups consisting of elements of orders $1$, $2$, and $4$ and of their continuous homomorphisms. A natural analogue of Theorem 9.1 would be a statement that any topological twovalued group in the category $\mathcal{C}$ is isomorphic either to a twovalued group of the form $X^{\mathbf{a}}_A$, where $A$ is a Hausdorff topological group consisting of elements of orders $1$, $2$, and $4$, or to a twovalued group of the form $X^{\mathbf{u}}_{V,U}$, where $(V,U)$ is a pair of Hausdorff topological Boolean groups. However, below we give an example showing that this is not true. Before building this example, let us discuss how the cohomological classification approach from §§ 9 and 10 is adapted to the topological case under consideration and what problems arise along the way. Let $X$ be a topological twovalued group in the category $\mathcal{C}$. Denote by $V$ the Boolean group of all elements of orders $1$ and $2$ in $X$, with the Hausdorff topology $\mathcal{T}_V$ induced by the inclusion $V\subset X$. Let $U\subset V$ denote the subgroup of elements $x^2$, where $x\in X$. Similarly to Lemma 9.2, for every element $u\in U$ the set $X_u$ of all $x$ such that $x^2=u$ is a $V$orbit. Moreover, the stabilizer of the elements of this orbit is the subgroup $\{e,u\}\subset V$, and if $x\in X_u$, $y\in X_{w}$, and $x*y=[z_1,z_2]$, then $z_1,z_2\in X_{uw}$ and $z_2=uz_1=wz_1$. Thus, there is a canonical bijection $X/V\leftrightarrow U$, and the (singlevalued) multiplication in $U$ is induced by the twovalued multiplication in $X$. We endow $U$ with the quotient topology $\mathcal{T}_U$. It is not obvious a priori that this topology is Hausdorff. Note, however, that the continuity of the operation of squaring in $X$ implies the continuity of the embedding $X/V=U\subset V$, where $U$ is endowed with the quotient topology $\mathcal{T}_U$ and $V$ with the subset topology $\mathcal{T}_V$. Therefore, the topology $\mathcal{T}_U$ is Hausdorff and the pair $(V,U)$ is an object of the category $\mathcal{P}$. Thus, the correspondence
$$
\begin{equation*}
X\mapsto (V,U)
\end{equation*}
\notag
$$
we have constructed defines a functor from the category $\mathcal{C}$ to the category $\mathcal{P}$, which we denote by $\Phi$. It can be verified directly that $\Phi(X_{V,U}^{\mathbf{u}})=(V,U)$ for any pair $(V,U)$ in the category $\mathcal{P}$. This implies, in particular, that the topological twovalued groups $X^{\mathbf{u}}_{V_1,U_1}$ and $X^{\mathbf{u}}_{V_2,U_2}$ are isomorphic if and only if the pairs of topological Boolean groups $(V_1,U_1)$ and $(V_2,U_2)$ are isomorphic. The functor $\Phi$ allows us to split the problem of classifying the topological twovalued groups $X$ in the category $\mathcal{C}$ into separate problems of classifying all $X$ with an arbitrary fixed (up to isomorphism) pair $\Phi(X)=(V,U)$. If in each $V$orbit $X_u$, where $u\in U$, we choose a representative $x_u$ (so that $x_e=e$), then multiplication in the twovalued group $X$ is uniquely defined by the rule of multiplication for the representatives $x_u$, which looks like
$$
\begin{equation}
x_u*x_v=\varphi(u,v)[x_{uv},ux_{uv}],\qquad u,v\in U,
\end{equation}
\tag{61}
$$
for some involutive symmetric quasicocycle $\varphi\colon U\times U\to V$, that is, a map such that
$$
\begin{equation}
\varphi(u,v)\equiv \varphi(v,u) \pmod{\!\langle u,v\rangle},
\end{equation}
\tag{62}
$$
$$
\begin{equation}
\varphi(u,u)\equiv\varphi(e,u)\equiv\varphi(u,e)\equiv e\pmod{\!\langle u\rangle},
\end{equation}
\tag{63}
$$
$$
\begin{equation}
\varphi(u,v)\varphi(uv,w)\varphi(u,vw)\varphi(v,w)\in\langle u,v,w\rangle
\end{equation}
\tag{64}
$$
for all $u,v,w\in U$. So far, the only difference from § 9 is that the group $U$ is not necessarily the same as $V$. However, now we must ensure that multiplication given by (61) is continuous. Of course, to follow the continuity property of multiplication we would like the map $U\to X$, $u\mapsto x_u$, indicating the representatives of $V$orbits, to be continuous (with respect to the topology $\mathcal{T}_U$ on $U$). We call a topological group $X$ in the category $\mathcal{C}$ tame if it there exists a continuous mapping describing a choice of representatives of $V$orbits $u\mapsto x_u$, and we call it wild if there is no such a continuous mapping. This is where the first fundamental problem arises. Question 14.9. Does the category $\mathcal{C}$ contain wild topological twovalued groups? The authors do not know the answer to this question at the moment. On the one hand there are no reasonable arguments in favour of the negative answer, especially if we take into account that the topological space $U$ is not simply connected. On the other hand, so far we have failed to construct an example of a wild twovalued group. In what follows we focus on the study of tame twovalued groups in the category $\mathcal{C}$ and show that even among them there is a topological group that belongs to neither the principal nor the unipotent series. First of all, we prove that, as a topological space, the twovalued group $X$ is homeomorphic to the unipotent twovalued group $X^{\mathbf{u}}_{V,U}=(U\times V)/\iota_{\mathbf{u}}$. Proposition 14.10. Let $X$ be a tame topological twovalued group in the category $\mathcal{C}$ and let $\Phi(X)=(V,U)$. Then the twovalued group $X$ is homeomorphic as a topological space to the unipotent twovalued group $X^{\mathbf{u}}_{V,U}=(U\times V)/\iota_{\mathbf{u}}$. More precisely, if $u\mapsto x_u$ is a continuous mapping describing a choice of representatives in the $V$orbits of the twovalued group $X$, then the mapping $g\colon U\times V\to X$ defined by $g(u,v)= vx_u$ induces a homeomorphism $\bar g\colon (U\times V)/\iota_{\mathbf{u}}\approx X$. Proof. From the fact that the stabilizer of each element of the orbit $Vx_{u}$ under the action of the group $V$ is equal to $\{e,u\}$ it follows that $\bar g$ is a bijection. The continuity of the map $g$ (and therefore, of $\bar g$) follows from the continuity of the mapping $u\mapsto x_u$ and the continuity of multiplication in the twovalued group $X$. Denote the (continuous) projection $X\to X/V=U$ by $\pi$. The mapping $g$ admits a twovalued continuous inverse mapping $X\to \operatorname{Sym}^2(U\times V)$, which is defined by $x\mapsto [\pi(x),x*x_{\pi(x)}]$. (Thus, $g$ is a twofold Smith–Dold ramified covering.) This immediately implies the continuity of $\bar g^{1}$. Thus, an arbitrary tame twovalued group $X$ with $\Phi(X)=(V,U)$ can be considered as a twovalued group of the unipotent series $X_{V,U}^{\mathbf{u}}$ with multiplication deformed by an involutive symmetric quasicocycle $\varphi$ in accordance with formula (61). (Recall that, as shown in § 9, the trivial quasicocycle corresponds to the twovalued group $X_{V,U}^{\mathbf{u}}$ itself.) We will use the notation $X_{V,U,\varphi}$ for the twovalued group with multiplication (61). It is easy to check that for multiplication in the twovalued group $X_{V,U,\varphi}$ to be continuous it is necessary and sufficient that the quasicocycle $\varphi$ is quasicontinuous in the following sense. Recall that, in fact, the values $\varphi(u,v)$ are not uniquely defined, but only up to multiplication by an element of $\langle u,v\rangle$. We say that a quasi cocycle $\varphi\colon U\times U\to V$ is quasicontinuous if the fourvalued mapping $U\times U\to \operatorname{Sym}^4(V)$ defined by
$$
\begin{equation*}
(u,v)\mapsto \bigl[\varphi(u,v),u\varphi(u,v),v\varphi(u,v), uv\varphi(u,v)\bigr],
\end{equation*}
\notag
$$
is continuous. If instead of a set of representatives of $V$orbits $x_u$ we choose another set of representatives $x_u'$, where $x_e'=e$ again, then $x_u'=\chi(u)x_u$ for some mapping $\chi\colon U\to V$ such that $\chi(e)=e$. Then the quasicocycle $\varphi$ is replaced by a quasicocycle $\varphi'$ such that
$$
\begin{equation}
\varphi'(u,v)\equiv\varphi(u,v)\chi(u)\chi(v)\chi(uv) \pmod{\!\langle u,v\rangle}.
\end{equation}
\tag{65}
$$
Moreover, the transition to the new set of representatives $x_u'$ depends on $u$ continuously if and only if the mapping $\chi$ is quasicontinuous in the sense that the twovalued mapping $U\to \operatorname{Sym}^2(V)$ given by the formula
$$
\begin{equation*}
u\mapsto \bigl[\chi(u),u\chi(u)\bigr]
\end{equation*}
\notag
$$
is continuous. Thus, it is natural to call two quasicontinuous involutive symmetric quasi cocycles $\varphi$ and $\varphi'$ cohomologous if they are related by (65) for some quasicontinuous mapping $\chi\colon U\to V$ such that $\chi(e)=e$. Then the following analogue of Proposition 9.5 holds. Proposition 14.11. Topological twovalued groups $X_{V,U,\varphi}$ and $X_{V',U',\varphi'}$ are isomorphic if and only if there is an isomorphism of pairs of topological Boolean groups $(V,U)$ and $(V',U')$ that takes the quasi continuous quasicocycle $\varphi$ to a quasi continuous quasicocycle cohomologous to $\varphi'$. In particular, $X_{V,U,\varphi}$ is isomorphic to $X_{V,U}^{\mathbf{u}}$ if and only if the quasicontinuous quasicocycle $\varphi$ is cohomologically trivial. Denote by $\mathcal{H}_{\mathrm{qc}}(U,V)$ the Boolean group of cohomology classes of quasicontinuous quasicocycles $U\times U\to V$. Thus, the classification of tame topological twovalued groups $X$ such that $\Phi(X)=(V,U)$ reduces to the following problem. Problem 14.12. Describe the group $\mathcal{H}_{\mathrm{qc}}(U,V)$. Of course, there is a topologyforgetting homomorphism
$$
\begin{equation*}
f\colon\mathcal{H}_{\mathrm{qc}}(U,V)\to\mathcal{H}(U,V),
\end{equation*}
\notag
$$
where $\mathcal{H}(U,V)$ is the Boolean group of all quasicocycles (without the requirement of quasi continuity) up to the cohomology defined by an arbitrary mapping $\chi$ (also without the requirement of quasicontinuity). Using Lemma 9.6 it is easy to verify that the isomorphism $\mathcal{H}(U,V)\cong \mathcal{H}(U)$ always holds, hence the group $\mathcal{H}(U,V)$ is trivial when $\dim U\leqslant 2$, and $\mathcal{H}(U,V)\cong C_2$ when $\dim U\geqslant 3$. In the general situation the homomorphism $f$ is apparently neither injective, nor surjective. Now we give an example of a certain element in the kernel of the homomorphism $f$ which provides us with the construction of a tame topological twovalued group in the category $\mathcal{C}$ that belongs neither to the antipodal family $X_A^{\mathbf{a}}$ nor to the unipotent family $X_{V,U}^{\mathbf{u}}$. In our example, $U=V$, so from now on we write $\mathcal{H}_{\mathrm{qc}}(V)$ instead of $\mathcal{H}_{\mathrm{qc}}(V,V)$ and $X_{V,\varphi}$ instead of $X_{V,V,\varphi}$. To construct the above example we need the notion of a free topological Boolean group. Let $Y$ be a completely regular Hausdorff topological space. Markov [29] introduced the concepts of a free topological group and a free abelian topological group with basis $Y$. We need a similar concept of a free Boolean group $B(Y)$ with basis $Y$ (see [35]). Algebraically, $B(Y)$ is just the Boolean group $C_2^Y$, that is, the vector space over the field $\mathbb{F}_2$ whose basis is the set $Y$. As a topology on $B(Y)$ we take a topology that is consistent with the group structure and satisfies the following two conditions: Such a topology exists, is unique, and is simultaneously the finest one among all topologies compatible with the group structure and satisfying condition 1), and the coarsest one among all topologies compatible with the group structure and satisfying condition 2). The proofs of analogues of these statements for free topological groups and free abelian topological groups are contained in [29] and [23]; the necessary extension of these results to the Boolean case can be found in [35]. Elements of $B(Y)$ can be interpreted as finite subsets of the space $Y$ (in the usual sense, that is, without multiplicities). Then the group operation is the symmetric difference of subsets, and the identity element is the empty set. For convenience we denote the symmetric difference of subsets $u$ and $v$ by $uv$. We are interested in the case when $Y$ is a compact metric space with metric $d$. Then the topology of a free topological Boolean group on $B(Y)$ can be described as follows (see [23] and [35]). For each subset $v\in B(Y)$ define its norm $\v\$ with values in $\mathbb{R}_{\geqslant 0}\cup\{+\infty\}$ as follows: The function constructed is a norm in the following sense: This norm is an analogue for topological Boolean groups of the seminorms considered by Markov [29] and Graev [23]; see also [35], § 2, for specific formulae. So the function
$$
\begin{equation*}
\rho(u,v)=\uv\
\end{equation*}
\notag
$$
defines a welldefined $\infty$metric on $B(Y)$, that is, a metric with values in $\mathbb{R}_{\geqslant 0}\cup\{+\infty\}$. We could look at the corresponding metric topology in $B(Y)$, but this would not be the topology we want. The topology we need is finer, it is obtained as follows. For each $k\geqslant 0$ we denote by $B_k(Y)$ the subset of the group $B(Y)$ consisting of all subsets $v\subset Y$ with at most $k$ elements. We endow each $B_k(Y)$ with the metric topology induced by the metric $\rho$. The desired topology of the free topological Boolean group on $B(Y)$ is the topology of the direct limit of these metric topologies. This means that a set $U$ is open in $B(Y)$ if and only if the intersection $U\cap B_k(Y)$ is open in $(B_k(Y),\rho)$ for every $k$. Informally, $B(Y)$ can be thought of as the configuration space of finite collections of points on $Y$ with the condition that when a pair of points collides, both of them vanish. In particular, it is easy to see that if $Y$ is pathconnected, then $B(Y)$ consists of two pathconnected components, $B_{\rm even}(Y)$ and $B_{\rm odd}(Y)$, formed by the subsets of even and odd cardinality, respectively. Example 14.13. Let $K\subset \mathbb{R}^d$ be a compact convex body (that is, a convex set with nonempty interior) in a finitedimensional Euclidean space of dimension $d\geqslant 1$. Consider a free topological Boolean group $B(K)$. We define a quasicocycle $\varphi\colon B(K)\times B(K)\to B(K)$ by the formula
$$
\begin{equation}
\varphi(u,v)=\prod_{p\in u}\,\prod_{q\in v}\biggl\{\frac{p+q}2\biggr\},
\end{equation}
\tag{66}
$$
where $\{\,\cdot\,\}$ denotes a onepoint subset. It can be verified directly that the map $\varphi$ satisfies conditions (62)– (64), that is, it is indeed an involutive symmetric quasi cocycle. Moreover, it is obvious that $\varphi$ is continuous and, a fortiori, quasi continuous. Proposition 14.14. Let $\varphi\colon B(K)\times B(K)\to B(K)$ be a quasicocycle given by formula (66). Then Proof. Let us prove statement (a). Define a discontinuous map $\chi_0\colon B(K)\to B(K)$ by the formula
$$
\begin{equation}
\chi_0(v)=\prod_{\{p,q\}\subset v,\ p\ne q}\biggl\{\frac{p+q}{2}\biggr\},
\end{equation}
\tag{67}
$$
where the product is taken over all unordered subsets $\{p,q\}$ of the set $v$. It is easy to see that
$$
\begin{equation}
\varphi(u,v)=\chi_0(u)\chi_0(v)\chi_0(uv)
\end{equation}
\tag{68}
$$
for all $u,v\in B(K)$, from which (a) immediately follows. Note that the map $\chi_0$ defined by (67) is not quasicontinuous, so (68) does not imply the cohomological triviality of the quasicocycle $\varphi$ in the group $\mathcal{H}_{\mathrm{qc}}(B(K))$.
We prove statement (b) by contradiction. Suppose that $\varphi$ is cohomologically trivial as a quasicontinuous quasicocycle. Then there is a quasicontinuous mapping $\chi\colon B(K)\to B(K)$ such that $\chi(e)=e$ and
$$
\begin{equation*}
\varphi(u,v)\equiv \chi(u)\chi(v)\chi(uv)\pmod{\!\langle u,v\rangle}
\end{equation*}
\notag
$$
for all $u,v\in B(K)$.
Restricting $\chi$ to oneelement subsets of $K$ we obtain the mapping $K\to B(K)$, $q\mapsto\chi(\{q\})$. It is quasicontinuous in the following sense: the twovalued mapping
$$
\begin{equation}
q\mapsto\bigl[\chi(\{q\}),\{q\}\chi(\{q\})\bigr]
\end{equation}
\tag{69}
$$
is continuous. Now note that of the two elements $\chi(\{q\})$ and $\{q\}\chi(\{q\})$ exactly one always lies in the connected component $B_{\rm even}(K)$ and the other one lies in $B_{\rm odd}(K)$. Therefore, if of these two elements we always choose the one that lies in $B_{\rm even}(K)$, then we obtain a continuous onevalued branch of the twovalued mapping (69). Set
$$
\begin{equation*}
\psi(\{q\})=\begin{cases} \chi(\{q\})&\text{if}\ \chi(\{q\})\in B_{\rm even}(K), \\ \{q\}\chi(\{q\})&\text{if}\ \chi(\{q\})\in B_{\rm odd}(K). \end{cases}
\end{equation*}
\notag
$$
Then the map $q\mapsto\psi(\{q\})$ is continuous. We extend the map $\psi$ to a continuous homomorphism $\psi\colon B(K)\to B_{\rm even}(K)$ by setting
$$
\begin{equation*}
\psi(v)=\prod_{q\in v}\psi(\{q\}).
\end{equation*}
\notag
$$
Let us replace $\chi$ by the new map $\widetilde{\chi}\colon B(K)\to B(K)$ such that
$$
\begin{equation*}
\widetilde\chi(v)=\chi(v)\psi(v)
\end{equation*}
\notag
$$
for all $v\in B(K)$. Since $\psi$ is a homomorphism, we still have $\widetilde\chi(e)=e$ and
$$
\begin{equation}
\varphi(u,v)\equiv \widetilde\chi(u)\widetilde\chi(v) \widetilde\chi(uv)\pmod{\!\langle u,v\rangle}
\end{equation}
\tag{70}
$$
for all $u,v\in B(K)$. Moreover, it follows from the continuity of $\psi$ that $\widetilde\chi$ is still quasicontinuous.
By construction, for any point $q\in K$ we have either $\widetilde\chi(\{q\})=e$ or $\widetilde\chi(\{q\})=\{q\}$. Let $p$ and $q$ be two different points of $K$. Congruence (70) for the pair $(u,v)=(\{p\},\{q\})$ implies that
$$
\begin{equation*}
\biggl\{\frac{p+q}2\biggr\}\subseteq \widetilde\chi(\{p,q\})\subseteq \biggl\{\frac{p+q}{2}\,,p,q\biggr\}.
\end{equation*}
\notag
$$
In addition, note that the set $\{p,q\}$ belongs to the same connected component $B_{\rm even}(K)$ of the space $B(K)$ as the empty set $e$. The quasicontinuity of $\widetilde\chi$ means that the unordered pair of sets $\bigl[\widetilde\chi(w),w\widetilde\chi(w)\bigr]$ depends on $w$ continuously. If $w=e$, then both sets in this pair are empty and therefore lie in $B_{\rm even}(K)$. Hence both $\widetilde\chi(\{p,q\})$ and $\{p,q\}\widetilde\chi(\{p,q\})$ also lie in $B_{\rm even}(K)$. So $\widetilde\chi(\{p,q\})$ is one of the two sets $\{(p+q)/2,p\}$ and $\{(p+q)/2,q\}$.
Now let $p$, $q$, and $r$ be three points in $K$ such that the six points $p$, $q$, $r$, $(p+q)/2$, $(q+r)/2$, and $(r+p)/2$ are pairwise distinct. Then it follows from congruence (70) for the pair $(u,v)=(\{p,q\},\{r\})$ that the set $\widetilde\chi(\{p,q,r\})$ contains exactly one of the two points $p$ and $q$. However, in a similar way this set contains exactly one of the two points $p$ and $r$ and exactly one of $q$ and $r$. In combination, these three statements lead to a contradiction, which completes the proof of the proposition. $\Box$ Corollary 14.15. The tame topological twovalued group $X_{B(K),\varphi}$ is neither isomorphic (as a topological group) to any twovalued group $X_A^{\mathbf{a}}$ of the principal series, nor to any twovalued group $X_{V,U}^{\mathbf{u}}$ of the unipotent series. Proof. It follows from statement (a) of Proposition 14.14 that, as an abstract two valued group (that is, after forgetting the topology), $X_{B(K),\varphi}$ is isomorphic to the twovalued group $X^{\mathbf{u}}_{B(K)}$ of the unipotent series. From here it immediately follows (by Theorem 13.1) that $X_{B(K),\varphi}$ is not isomorphic to any group $X_A^{\mathbf{a}}$ of the principal series even without taking the topology into account.
We have $\Phi(X_{B(K),\varphi})=(B(K),B(K))$. Therefore, the topological twovalued group $X_{B(K),\varphi}$ certainly cannot be isomorphic to any topological twovalued group of the unipotent series, except for $X_{B(K),B(K)}^{\mathbf{u}}= X_{B(K)}^{\mathbf{u}}$. That the topological groups $X_{B(K)}^{\mathbf{u}}$ and $X_{B(K),\varphi}$ are not isomorphic follows from the cohomological nontriviality of the quasicocycle $\varphi$ in the group $\mathcal{H}_{\mathrm{qc}}(B(K))$ and Proposition 14.11. Remark 14.16. The construction of the quasicocycle $\varphi$ can be modified quite strongly so that Proposition 14.14 and Corollary 14.15 remain true (and their proofs are verbatim the same). Namely, we can consider any quasicocycle of the form
$$
\begin{equation*}
\varphi(u,v)=\prod_{p\in u}\,\prod_{q\in v}F(p,q),
\end{equation*}
\notag
$$
where $F\colon K\times K\to B_{\rm odd}(K)$ is an arbitrary symmetric continuous mapping such that $F(q,q)=\{q\}$ for all $q\in K$. In this case only the pathwise connectivity of the set $K$ is necessary and the convexity is not important: we needed it only for the particular map $F(p,q)=\{(p+q)/2\}$, which we used, to be well defined.
15. Locally compact twovalued groups The most important class of topological groups is the class of locally compact (Hausdorff) topological groups. There is a rich structural theory for this class of groups, whose development has primarily been associated with the solution of Hilbert’s fifth problem. The formulation of this problem in its most common interpretation is as follows: is it true that every topological group that is a topological manifold is isomorphic to a Lie group? In connection with the questions considered in this paper, we are primarily interested in commutative locally compact groups. The foundations of their structural theory were laid by Pontryagin [33], [34]; in particular, he gave a positive solution to Hilbert’s fifth problem for commutative groups. A positive solution to Hilbert’s fifth problem in the general case was obtained by Gleason [22] and Montgomery and Zippin [30]. Subsequently, even stronger structure theorems were obtained for locally compact groups. The most important for us is the following theorem on locally compact groups without small subgroups. By definition, a topological group has no small subgroups if it contains a neighbourhood of identity that does not contain nontrivial subgroups. Theorem 15.1 (Gleason–Yamabe theorem). Let $G$ be a locally compact Hausdorff topological group without small subgroups. Then $G$ is isomorphic to a Lie group. Remark 15.2. The statement of this theorem was proved by Gleason [22] under the additional assumption that the group is finitedimensional and by Yamabe [45] in the general case (see also the monograph [37], Corollary 1.5.8). It is well known that any commutative Lie group can be represented as a direct product of a compact torus $(S^1)^m$, a vector group $\mathbb{R}^n$, and a discrete abelian group. Thus, in the commutative case the Gleason–Yamabe theorem takes the following form. Corollary 15.3. Let $G$ be a commutative locally compact Hausdorff topological group without small subgroups. Then $G\cong (S^1)^m\times\mathbb{R}^n\times A$, where $m\geqslant 0$, $n\geqslant 0$, and $A$ is a discrete abelian group. The natural question is how this result can be carried over to the case of involutive commutative twovalued groups. As in the case of ordinary groups, we say that a topological twovalued group has no small subgroups if it contains a neighbourhood of the identity that does not contain nontrivial twovalued subgroups. Theorem 15.4. Let $X$ be an involutive commutative locally compact Hausdorff topological twovalued group without small subgroups. Then one of two possibilities holds: Remark 15.5. Discrete twovalued groups are, of course, always locally compact and do not contain small subgroups. A classification of discrete involutive commutative twovalued groups is presented in Theorem 13.1. To prove Theorem 15.4 we need two simple lemmas. Lemma 15.6. Let $X$ be a Hausdorff topological twovalued group with identity $e$. Assume that the oneelement set $\{e\}$ is open. Then $X$ is discrete. Proof. Consider the subset $W$ of $\operatorname{Sym}^2(X)$ consisting of all multisets of the form $[e,x]$, where $x\in X$. The fact that the set $\{e\}$ is open in $X$ immediately implies that the set $W$ is open in $\operatorname{Sym}^2(X)$. For any element $y\in X$ multiplication by $y$ defines a continuous map $\mu_y\colon X\to\operatorname{Sym}^2(X)$. Since $e\in x*y$ if and only if $x=y$, we have $\mu_y^{1}(W)=\{y\}$. Thus, all oneelement sets $\{y\}$ are open, hence the topology on $X$ is discrete. $\Box$ Lemma 15.7. Let $X$ satisfy the conditions of Theorem 15.4. Then there is a neighbourhood of the identity in $X$ that does not contain elements of orders $2$ and $4$. Proof. Let $U$ be an open neighbourhood of the identity element of the twovalued group $X$ that does not contain nontrivial twovalued subgroups. Let $U'$ be the set consisting of all elements of $x\in X$ such that $x^2\in U$. Since the map $x\mapsto x^2$ is continuous, $U'$ is also an open neighbourhood of the identity.
Note that $U$ does not contain elements of order $2$. Indeed, if there were an element of order $2$ in $U$, that is, an element $x\ne e$ such that $x*x=[e,e]$, then the set $\{e,x\}$ would be a nontrivial twovalued subgroup contained in $U$. The square of any element of order $4$ is an element of order $2$. Moreover, the squares of elements in $U'$ lie in $U$. Therefore, $U'$ does not contain elements of order $4$. Thus, we can take the set $U\cap U'$ as the desired neighbourhood. $\Box$ Proof of Theorem 15.4. Let $V$ be an open neighbourhood of the identity without elements of order $2$ and $4$ given by Lemma 15.7. Assume that the twovalued group $X$ is not discrete. Then by Lemma 15.6 the oneelement set $\{e\}$ is not open. Hence $V$ contains at least one element $t$ different from $e$. Then $\operatorname{ord} t\notin \{1,2,4\}$. By Theorem 14.2 there is an isomorphism $X\cong G/\iota_{\mathbf{a}}$, where $G$ is a commutative Hausdorff topological group and $\iota_{\mathbf{a}}$ is the antipodal involution. Using the fact that the projection $G\to X$ is a twofold Smith–Dold ramified covering, it is easy to deduce the local compactness of $G$ from the local compactness of $X$. Now, by the Gleason–Yamabe theorem there is an isomorphism $G\cong (S^1)^m\times\mathbb{R}^n\times A$, where $m+n>0$ since $X$ is not discrete. $\Box$ For compact twovalued groups without small subgroups, Theorem 15.4 leads to a complete classification similar to Theorem 1.10. Theorem 15.8. Let $X$ be an involutive commutative compact Hausdorff topological twovalued group without small subgroups. Then $X$ is isomorphic to one of the following twovalued groups: All isomorphisms between twovalued groups in this list are exhausted by the isomorphisms (3) and (4) in Theorem 1.10. Proof. The theorem follows almost immediately from Theorems 15.4 and 1.10. We only need to verify that the topological twovalued groups $X^{\mathbf{a}}_{d_1,\dots,d_k}(m)$ corresponding to different tuples $(m;d_1,\dots,d_k)$ are pairwise nonisomorphic and, for $m\geqslant 1$, are not isomorphic to twovalued groups of the unipotent and special series. To do this, note that the connected component of the identity in a twovalued group $X=X^{\mathbf{a}}_{d_1,\dots,d_k}(m)$ is the group $X_0=X^{\mathbf{a}}_{\varnothing}(m) = (S^1)^m/\iota_{\mathbf{a}}$, and the quotient group $X/X_0$ is isomorphic to the twovalued group $X^{\mathbf{a}}_{d_1,\dots,d_k}$. Moreover, the topological twovalued groups $X^{\mathbf{a}}_{\varnothing}(m)$ are pairwise nonisomorphic since they have different dimensions, and the twovalued groups $X^{\mathbf{a}}_{d_1,\dots,d_k}$ are pairwise nonisomorphic by Lemma 1.10. This completes the proof of the theorem. $\Box$
16. Algebraic twovalued groups We now discuss the algebraic version of commutative involutive groups following [6], [7], and [17]. For simplicity we work over the field of complex numbers $\mathbb C$. In the simplest setting an algebraic twovalued multiplication law
$$
\begin{equation*}
(x,y)\mapsto z=x*y
\end{equation*}
\notag
$$
is given in local coordinates by an equation $F(x,y,z)=0$, where $F(x,y,z)$ is a polynomial which is symmetric in the variables $x,y,z$ and has degree 2 in each of them. The twovalued multiplication law must satisfy the condition of associativity
$$
\begin{equation*}
(x*y)*z=x*(y*z)
\end{equation*}
\notag
$$
in the sense of the theory of correspondences: after the elimination of $u$ and $v$ the systems of equations $F(x,y,u)=0$, $F(u,z,w)=0$ and $F(y,z,v)=0$, $F(x,v,w)=0$ define the same set $(x,y,z,w) \in \mathbb C^4$. We also assume that
$$
\begin{equation*}
F(0,y,z)=(zy)^2
\end{equation*}
\notag
$$
which means that 0 is the (strong) identity element of the group. Note that by symmetry we have the equality $F(x,y,0)=(xy)^2,$ which means that the corresponding group is automatically involutive. And vice versa, the addition law of an involutive twovalued group must be symmetric in all three variables (see Lemma 2.2). Such groups were classified in [6] and [17], where it was shown that the corresponding polynomial must be of the form
$$
\begin{equation}
F(x,y,z)=(x+y+za_2 xyz)^24(1+a_3 xyz)(xy+xz+yz+a_1 xyz),
\end{equation}
\tag{71}
$$
where $a_1$, $a_2$, and $a_3$ are arbitrary parameters. In general, the equation $F=0$ for such a polynomial defines the coset group $\mathcal E/\iota_{\mathbf{a}}$, where $\mathcal E$ is an elliptic curve (considered as an abelian group) and $\iota_{\mathbf{a}}\colon \mathcal E \to \mathcal E$ is the antipodal involution. More precisely, consider an elliptic curve in the standard Weierstrass form $v^2=4u^3g_2ug_3$ and a point $\alpha$ on it, and set the corresponding parameters to be
$$
\begin{equation}
a_1=3\wp(\alpha),\quad a_2=3\wp(\alpha)^2\frac{g_2}{4}\,,\quad\text{and}\quad a_3=\frac{1}{4}(4\wp(\alpha)^3g_2\wp(\alpha)g_3),
\end{equation}
\tag{72}
$$
where $\wp$ is the classical Weierstrass elliptic function satisfying the equation
$$
\begin{equation*}
(\wp')^2=4\wp^3g_2\wpg_3.
\end{equation*}
\notag
$$
Theorem 16.1 ([6], [17]). The equation $F(x,y,z)=0$ with the symmetric polynomial $F(x,y,z)$ satisfying the condition $F(0,y,z)=(zy)^2$ defines the structure of an algebraic twovalued group if and only if $F$ has the form (71). The multiplication law $F(x,y,z)=0$ for $F$ of the form (71) with parameters (72) is reduced to the form $X\pm Y\pm Z=0$ by the change of variables
$$
\begin{equation}
x=(\wp(X)+\wp(\alpha))^{1},\quad y=(\wp(Y)+\wp(\alpha))^{1},\quad z=(\wp(Z)+\wp(\alpha))^{1}.
\end{equation}
\tag{73}
$$
The corresponding twovalued group is isomorphic to the involutive coset group $\mathcal E/\iota_{\mathbf{a}}=\mathbb{CP}^1$, where $\mathcal E$ is the elliptic curve given by the equation
$$
\begin{equation}
v^2=u^3+a_1u^2+a_2u+a_3,
\end{equation}
\tag{74}
$$
and $\iota_{\mathbf{a}}$ is its involution $v\mapsto v.$ In the case where the roots of the polynomial on the righthand side coincide, the elliptic curve degenerates into a rational one, and the corresponding twovalued involutive groups are isomorphic to the coset groups $\mathbb C^*/\iota_{\mathbf{a}}$ and $\mathbb C/\iota_{\mathbf{a}}$, where $\iota_{\mathbf{a}}(z)= z^{1}$ and $\iota_{\mathbf{a}}(z)=z$, respectively. Interestingly, the general family (71) includes all cases of the formal twovalued groups that first appeared in topology in the works [11] and [3]–[5] by Buchstaber and Novikov: the case $a_1=0$ corresponds to elliptic cohomology, the case $a_2=a_3=0$ to $K$theory, and the most degenerate case $a_1=a_2=a_3=0$ to the ordinary cohomology (for details, see [17]). Note that, as follows from the parametrization (73), from the algebrogeometric point of view the general equation (71) defines the (affine part) of the special Kummer surface $\mathcal E\times\mathcal E/{\pm I}$. General Kummer surfaces corresponding to Jacobi varieties of curves of genus 2 were discussed by Buchstaber and Dragović [8] in connection with integrable billiards. The corresponding twovalued groups on Kummer surfaces embedded in $\mathbb{CP}^3$ are explicitly described in terms of the addition laws for Klein $\wp$functions [9]. It is natural to conjecture that general algebraic twovalued commutative involutive groups are obtained by the coset construction from abelian varieties and their degenerations, but the details of a precise formulation need to be clarified. Results due to Pillay [32] which use Hrushovski’s theorem from model theory, can be useful for the proof. The general theory of onedimensional twovalued formal groups was actually completed by Buchstaber in [4] and [5]. In particular, the concept of canonical invariant operators $d_n$ was introduced there, and it was shown that for the twovalued formal addition laws given by the relation
$$
\begin{equation*}
z^2\Theta_1(x,y)z+\Theta_2(x,y)=0,
\end{equation*}
\notag
$$
the canonical operator $d_1$ has the form
$$
\begin{equation}
d_1=\frac{1}{2}\varphi_1(x)\frac{d}{dx}+ \frac{1}{8}\varphi_2(x) \frac{d^2}{dx^2}\,,
\end{equation}
\tag{75}
$$
where
$$
\begin{equation*}
\begin{gathered} \, \varphi_1(x)=\frac{\partial\Theta_1(x,y)}{\partial y}\bigg_{y=0}\quad\text{and} \quad \varphi_2(x)=\frac{\partial\sigma(x,y)}{\partial y}\bigg_{y=0}, \end{gathered}
\end{equation*}
\notag
$$
and
$$
\begin{equation*}
\sigma(x,y)=\Theta_1^2(x,y)4\Theta_2(x,y).
\end{equation*}
\notag
$$
It was also shown there that the algebra of invariant operators on the group is freely generated by the canonical operator $d_1$, and a general formula for the exponential on such groups was found. It is easy to check that for algebraic twovalued groups (71) the canonical operator (75) in the parametrization (73) is simply the second derivative operator $d^2/dX^2$ in the elliptic coordinate $X$ (or the special case of the Lamé operator $\mathcal L=d^2/dX^2m(m+1)\wp(X)$ for $m=0$), which generates the algebra of differential operators that are invariant under shifts on the elliptic curve (74) and the involution $\iota_{\mathbf{a}}$. The corresponding exponential was written explicitly in [6] in terms of elliptic functions.
17. Concluding remarks The structural theory of $n$valued groups is at the very beginning of its development and most questions are open. In our paper we have completed the classification of twovalued commutative involutive groups in the finitely generated case and in some important topological classes. In the general case the most promising direction seems to be the development of the theory of $n$valued Lie groups. In particular, in the case of an $n$valued group $G/H$, $H \subset \operatorname{Aut} G$, obtained from the Lie group $G$ using the coset construction, it is natural to study the algebra of differential operators on $G$ that are invariant under (say, left) shifts on $G$ and actions of the automorphism group $H$. The corresponding algebra can be considered as an analogue of the universal enveloping algebra for the $n$valued Lie group. The study of this algebra in the general case seems to be an important and interesting problem. The example of an elliptic twovalued group shows the existence of the structure of a twovalued Lie group on the twodimensional sphere, which does not admit the ordinary group structure. The question of topological obstructions to the $n$valued group structure is of great interest. As shown by Hopf [26], if a topological space admits (singlevalued) multiplication with identity element, then the cohomology ring of this space has the structure of a Hopf algebra. The nonexistence of such a structure is an obstruction to the existence of multiplication on a topological space. It was shown by Buchstaber and Rees [12], [13] (see also [14]) that the existence of $n$valued multiplication on a space leads to a similar structure, which they called a Hopf $n$algebra. In the case of fourdimensional manifolds the corresponding obstructions were studied by Panov [31]. He obtained an explicit list of the homotopy types of fourdimensional manifolds whose cohomology admits the structure of a Hopf $2$algebra. As a result, only manifolds in this list may admit twovalued multiplication. Note that the simplyconnected singular Kummer variety $\mathcal{E}\times\mathcal{E}/\pm I\approx T^4/\iota_{\mathbf{a}}$ is a twovalued group that realizes the case of second Betti number $6$ pointed out by Panov. A number of results on multivalued multiplications on spheres were obtained by Gugnin [25]. In the theory of formal twovalued groups significant advances were made in the 1970s by Buchstaber [3], [4], who was motivated by applications to cobordism theory [11]. These works, in particular, revealed a deep connection with the Delsarte–Levitan theory of generalized shift operators [27]. The methods in these papers, as well as their connections with the theory of quantum groups and Hopf algebras [12], [13], may be useful in the more general context of our problems. In the theory of $n$valued groups there are important constructions arising from various branches of mathematics. One of such constructions is the $n$valued group structure on the set of irreducible representations of a noncommutative group (see [7]). It is natural to consider this construction as an interpretation of Tanaka–Krein duality (see [1]), which is an analogue of Pontryagin duality for non commutative groups, in group theoretic terms. Duality theory for $n$valued groups is discussed in [10], [43], and [44]. Let us mention one more construction. The group algebra of a coset $n$valued group is the subalgebra of the group algebra of the original group that is generated by the elements obtained as the sums of the points on orbits of the action of the automorphism group. A more general example of group origin is given by socalled Schur rings (see [42]). In connection with the problem of Schur rings some examples of noncoset $n$valued groups were constructed in [15] using numbertheoretic methods. Finally, there are important connections with the theory of integrable systems [16], [39], [40], where many questions also remain open (see, for example, the discussion and references in [17]). In particular, Veselov [38], [40] proposed an approach to the integrability of multivalued mappings in terms of the growth of the number of images of their iterations. The connection of this approach with the theory of multivalued groups was discussed in [16]. It is interesting to note that the original proof of Gromov’s remarkable theorem stating that finitely generated groups of polynomial growth are almost nilpotent, uses results of Gleason, Montgomery, and Zippin (see [24]).



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Citation:
V. M. Buchstaber, A. P. Veselov, A. A. Gaifullin, “Classification of involutive commutative twovalued groups”, Uspekhi Mat. Nauk, 77:4(466) (2022), 91–172; Russian Math. Surveys, 77:4 (2022), 651–727
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https://www.mathnet.ru/eng/rm10064https://doi.org/10.4213/rm10064e https://www.mathnet.ru/eng/rm/v77/i4/p91

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