Abstract. In 1958, Sierpiński asked whether there exists a linear order X that is isomorphic to its lexicographically ordered cube but is not isomorphic to its square. The main result of this paper is that the answer is negative. More generally, if X is isomorphic to any one of its finite powers X n , n > 1, it is isomorphic to all of them.
We construct non-isomorphic linear orders X and Y that are both left-hand and right-hand divisors of one another, answering positively a question of Sierpiński.2010 Mathematics Subject Classification: 06A05, 03E05. Key words and phrases: linear order, lexicographical product, Schroeder-Bernstein problem 1 2 GARRETT ERVIN Kaplansky test properties. The first of these, sometimes individually referred to as the Schroeder-Bernstein property, asserts that whenever two structures X, Y ∈ K are each isomorphic to a divisor of the other, then X and Y must themselves be isomorphic. That is, ifKaplansky himself was interested in determining whether these properties held for various classes of infinite abelian groups. He considered their failure a heuristic indication that a given class admitted no useful structure theorem, a la the classification theorem for finitely generated abelian groups. In his book [7] in which he recorded them, he said "I believe their defeat is convincing evidence that no reasonable invariants exist." The two properties had been considered previously for other classes of structures by several authors, including Tarski [13] and Hanf [5].A third property, related to the two Kaplansky properties, is the cube property. Whenever a structure X from a class K is isomorphic to its own square X 2 , it is isomorphic to all of its finite powers X n . In particular, it is isomorphic to its cube. The cube property asserts that the converse holds: for all X ∈ K, if X ∼ = X 3 then it is already the case that X ∼ = X 2 . If K contains a counterexample to the cube property, namely, a structure X that is isomorphic to its cube but not to its square, then the pair X and Y = X 2 witnesses the failure of both Kaplansky properties for K.
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