Geodesic
Ternary $\ast$-bands are globally determined
Ural mathematical journal, Tome 9 (2023) no. 1, pp. 64-77 Cet article a éte moissonné depuis la source Math-Net.Ru

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A non-empty set $S$ together with the ternary operation denoted by juxtaposition is said to be ternary semigroup if it satisfies the associativity property $ab(cde)=a(bcd)e=(abc)de$ for all $a,b,c,d,e\in S$. The global set of a ternary semigroup $S$ is the set of all non empty subsets of $S$ and it is denoted by $P(S)$. If $S$ is a ternary semigroup then $P(S)$ is also a ternary semigroup with a naturally defined ternary multiplication. A natural question arises: "Do all properties of $S$ remain the same in $P(S)$?" The global determinism problem is a part of this question. A class $K$ of ternary semigroups is said to be globally determined if for any two ternary semigroups $S_1$ and $S_2$ of $K$, $P(S_1)\cong P(S_2)$ implies that $S_1\cong S_2$. So it is interesting to find the class of ternary semigroups which are globally determined. Here we will study the global determinism of ternary $\ast$-band.
Keywords: rectangular ternary band, involution ternary semigroup, involution ternary band, ternary $\ast$-band, ternary projection. 
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Indrani Dutta; Sukhendu Kar. Ternary $\ast$-bands are globally determined. Ural mathematical journal, Tome 9 (2023) no. 1, pp. 64-77. http://geodesic.mathdoc.fr/item/UMJ_2023_9_1_a4/

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