| Russian Mathematical Surveys |
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This article is cited in 1 scientific paper (total in 1 paper) Brief communications Commutativity of involutive two-valued groups A. A. Gaifullinabcd a Steklov Mathematical Institute of Russian Academy of Sciences b Skolkovo Institute of Science and Technology c Lomonosov Moscow State University d Institute for Information Transmission Problems of the Russian Academy of Sciences (Kharkevich Institute)
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     Over the past decades, the theory of $n$-valued groups has been enriched by numerous new results and applications; see [1] and [2]. One of its features is the complexity of classification problems even in the commutative case. In [2] Buchstaber and Veselov, in connection with the Conway topograph, introduced the class of involutive two-valued groups, for which the classification problems seem amenable: see [3]. Following [2] and [3], we give the following definition. Definition. An involutive two-valued group is a set $X$ endowed with a multiplication $*\colon X\times X\to\operatorname{Sym}^2(X)$ (where $\operatorname{Sym}^2(X)$ is the symmetric square of $X$) and an identity element $e\in X$ that have the following properties: (1) (associativity) for any $x,y,z\in X$ there is an equality of $4$-element multisets $(x*y)*z=x*(y*z)$, (2) (strong identity) $x*e=e*x=[x,x]$ for all $x\in X$, (3) (involutivity) the multiset $x*y$ contains the identity $e$ if and only if $x=y$. A two-valued group is said to be commutative if it, in addition, has the following property: (4) (commutativity) $x*y=y*x$ for all $x,y\in X$. Note that in [4] the concept of an involutive $n$-valued group was used in another, not equivalent meaning: see Remark 1.5 in [3] for details. In [3] a complete classification of finitely generated commutative involutive two- valued groups and partial classification results in the non-finitely generated and topological cases were obtained. The main result of the present note is the following theorem, which implies that all classification results in [3] are also valid without the assumption of commutativity. For single-generated two-valued groups this theorem was proved in [3]. The following lemma is Lemma 2.2 in [3]. Lemma 1. Suppose that $x$, $y$, and $z$ are elements of an involutive two-valued group $X$. Then $z\in x*y$ if and only if $y\in z*x$. Of key importance for us will be the concepts of powers and order of an element of an involutive two-valued group, which were introduced in [3]. Namely, for each element $x\in X$ there is a unique sequence $x^0=e$, $x^1=x$, $x^2,x^3,\ldots$ of elements of $X$ such that $x^k*x^m=[x^{|k-m|},x^{k+m}]$ for all $k$ and $m$. The order of $x$ (denoted by $\operatorname{ord} x$) is the smallest positive integer $k$ such that $x^k=e$. In particular, the order of a non-identity element $x$ is equal to $2$ if and only if $x*x=[e,e]$. Lemma 2. Suppose that $x$ and $y$ are elements of an involutive two-valued group $X$ such that $\operatorname{ord} x =2$. Then $x*y=y*x=[z,z]$ for some element $z\in X$. Proof. Let $z$ be an element of $x*y$. By Lemma 1 the multiset $z*x$ contains $y$, that is, $z*x=[y,y']$ for some $y'$. Then $[y*x,y'*x]=(z*x)*x=z*(x*x)=z*[e,e]=[z,z,z,z]$. Hence $y*x=[z,z]$. Applying Lemma 1 again, we obtain $x*z=[y,y'']$ for some $y''$. Then $[x*y,x*y'']=x*(x*z)=(x*x)*z=[e,e]*z=[z,z,z,z]$. Therefore, $x*y=[z,z]$.
Proof of the theorem. We need to prove that $x*y=y*x$ for all $x$ and $y$. If one of the two elements $x$ and $y$ has order $2$, then the required equality follows from Lemma 2. So we may assume that $\operatorname{ord} x>2$ and $\operatorname{ord} y>2$, that is, $x^2\ne e$ and $y^2\ne e$. Put $x*y=[z_1,z_2]$ and $y*x=[w_1,w_2]$. Then we have Hence either at least two of the four multisets $z_i*w_j$ contain $e$ or one of these four multisets is equal to $[e,e]$. From involutivity it follows that $e\in z_i*w_j$ if and only if $z_i=w_j$. Up to the permutations $z_1\leftrightarrow z_2$ and $w_1\leftrightarrow w_2$ and the reversal of the roles of the elements $x$ and $y$, there are three substantially different cases.
Case 1: $z_1=w_1$ and $z_2=w_2$. Then $x*y=y*x$, as required. Case 2: $z_1=z_2=w_1\ne w_2$. We have Since $\operatorname{ord} x>2$, we have $x*x=[e,x^2]$, where $x^2\ne e$. Moreover, by involutivity $e\notin x*x'$ unless $x'=x$. Hence it follows from (2) that $y*z_1=[x,x]$ and $x^2=z_1^2$. It is proved similarly that $z_1*x=[y,y]$ and $y^2=z_1^2$. Therefore, $x^2=y^2$. Consequently, $z_1*w_1=z_2*w_1=z_1*z_1=[e,z_1^2]=[e,x^2]$ and $x*y^2*x=x*x^2*x=[x,x^3]*x=[e,x^2,x^2,x^4]$. Thus, (1) reads $[e,e,x^2,x^2,z_1*w_2,z_1*w_2]=[e,e,e,x^2,x^2,x^2,x^2,x^4]$. Since $x^2\ne e$, we obtain $e\in z_1*w_2$, which leads to a contradiction since $z_1\ne w_2$. Therefore, the case under consideration is impossibleCase 3: $z_1*w_1=[e,e]$, that is, $z_1=w_1$ and $\operatorname{ord} z_1=2$. We may assume that the elements $z_1$, $z_2$, and $w_2$ are pairwise different, since otherwise we arrive at one of the two cases already considered (Case 1 or 2). By Lemma 1 we obtain $y\in z_1*x$. Since $\operatorname{ord} z_1=2$, it follows from Lemma 2 that $x*z_1=z_1*x=[y,y]$. We have
$$
\begin{equation*}
(x*z_1)*(z_1*x)=[y,y]*[y,y]=[e,e,e,e,y^2,y^2,y^2,y^2]
\end{equation*}
\notag
$$
and Hence $x^2=y^2$. As in Case 2, it follows that $e\in x*y^2*x$, and therefore $e$ occurs in the multiset (1) with multiplicity at least three. Hence $e$ is contained in at least one of the three multisets $z_1*w_2$, $z_2*w_1=z_2*z_1$, and $z_2*w_2$, which is impossible, since the elements $z_1$, $z_2$, and $w_2$ are pairwise different. The contradiction obtained completes the proof of the theorem. The author is grateful to V. M. Buchstaber and A. P. Veselov for multiple fruitful discussions. 
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\Bibitem{Gai24} |
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