A set \(X\) with a binary operation \(\cdot\) is called a cycle set if \(y\mapsto x\cdot y\) is invertible and \((x\cdot y)\cdot(x\cdot z)=(y\cdot x)\cdot(y\cdot z)\) for all \(x,y,z\in X\). We say \(X\) is non-degenerate if \(x\mapsto x\cdot x\) is bijective. Non-degenerate cycle sets are known to be in one-to-one correspondence with set-theoretical solutions of the quantum Yang-Baxter equation which are non-degenerate and unitary. In the case where \(X=A\) an Abelian group one can consider the free Abelian group on \(A\). If \(A\) has a multiplication \(\cdot\) which makes \(A\) a cycle set such that \(a\cdot(b+c)=a\cdot b+a\cdot c\) and \((a+b)\cdot c=(a\cdot b)\cdot(a\cdot c)\), then \(A\) is a linear cycle set. A non-degenerate cycle set can be embedded into a linear cycle set.
Let \(X\) be a square-free cycle set, that is \(x\cdot x=x\) for all \(x\). (Such an \(X\) is necessarily non-degenerate.) Let \(Y\) and \(Z\) be two finite sub-cycle-sets such that \(Y\) and \(Z\) operate transitively on each other. The main theorem in this paper is that \(y\cdot z=y'\cdot z\) for all \(y,y'\in Y\) and \(z\in Z\). As a result, if \(X\) is finite and has a decomposition \(X=Y\sqcup Z\) such that \(Y\) and \(Z\) operate transitively on each other, then \(X\) is a bi-cycle. As another consequence, if \(X\) is a nonempty finite square-free cycle set operating transitively on itself then \(X\) consists of a single point.
The key to the results in this paper is the relationship between linear cycle sets and braces. Here an Abelian group with a right distributive multiplication if and only if the operation \(a\circ b=ab+a+b\) (in the free Abelian group on \(A\)) gives a group structure on \(A\). An explicit one-to-one correspondence is given between braces and linear cycle sets.