Every linear order isomorphic to its cube is isomorphic to its square

TL;DR

This paper proves that any linear order isomorphic to its cube must also be isomorphic to its square, resolving a long-standing question and showing such orders are highly rigid.

Contribution

It establishes that linear orders isomorphic to any of their finite powers are actually isomorphic to all such powers, answering Sierpinski's 1958 question.

Findings

Linear orders isomorphic to their cube are also isomorphic to their square.

Orders isomorphic to any finite power are isomorphic to all powers.

The result confirms the rigidity of such linear orders.

Abstract

In 1958, Sierpinski 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.