4 Green’s Relations in Finite Semigroups
Important lemma: (2).
In a finite semigroup, the \(\mathcal{D}\) and \(\mathcal{J}\) relations are equivalent. That is, for any two elements \(x, y \in S\), \(x \mathcal{D} y\) if and only if \(x \mathcal{J} y\).
The forward direction, that \(x \mathcal{D} y\) implies \(x \mathcal{J} y\), holds in any semigroup, not just finite ones. This is because if \(x \mathcal{D} y\), there exists \(z\) such that \(x \mathcal{R} z\) and \(z \mathcal{L} y\). These relations imply \(x \leq _{\mathcal{J}} z\) and \(z \leq _{\mathcal{J}} y\), and by transitivity, \(x \leq _{\mathcal{J}} y\). A symmetric argument shows \(y \leq _{\mathcal{J}} x\), so \(x \mathcal{J} y\).
The reverse direction relies on the semigroup being finite. Suppose \(x \mathcal{J} y\). Then there exist \(s, t, u, v\) such that \(x = syt\) and \(y = uxv\). Substituting, we get \(x = s \cdot u \cdot x \cdot v \cdot t\), showing that \(x\) satisfies a sandwich equation. By 2, there exist positive integers \(k\) and \(\ell \) such that \((su)^k x = x\) and \(x(vt)^\ell = x\).
We use \(z = xv\) as the intermediate element to show \(x \mathcal{D} y\). We need to prove \(x \mathcal{R} xv\) and \(xv \mathcal{L} y\).
For \(x \mathcal{R} xv\): clearly \(xv \leq _{\mathcal{R}} x\) since \(xv = x \cdot v\). For the reverse, we have \(x = x(vt)^\ell = xv \cdot t(vt)^{\ell -1}\), so \(x \leq _{\mathcal{R}} xv\). Thus \(x \mathcal{R} xv\).
\(xv \mathcal{L} y\) follows similarly.
Therefore \(x \mathcal{R} xv\) and \(xv \mathcal{L} y\), which means \(x \mathcal{D} y\).
In a finite semigroup, if two elements are \(\mathcal{J}\)-equivalent, then a one-sided preorder implies the corresponding one-sided equivalence. Specifically, if \(x \mathcal{J} y\) and \(x \leq _{\mathcal{R}} y\), then \(x \mathcal{R} y\). Similarly, if \(x \mathcal{J} y\) and \(x \leq _{\mathcal{L}} y\), then \(x \mathcal{L} y\).
Suppose \(x \mathcal{J} y\) and \(x \leq _{\mathcal{R}} y\). The preorder \(x \leq _{\mathcal{R}} y\) means there exists \(z\) such that \(x = yz\). If \(z = 1\) (in the monoid extension), then \(x = y\) and we are done. Otherwise, \(z\) is a proper element and \(x = yz\).
We need to show \(y \leq _{\mathcal{R}} x\), i.e., that \(y = xt\) for some \(t\). From \(x \mathcal{J} y\), there exist \(u, v\) such that \(y = uxv\). Substituting \(x = yz\), we get \(y = u \cdot y \cdot zv\), so \(y\) satisfies a sandwich equation. By 2, there exists a positive integer \(n\) such that \(y(zv)^n = y\).
Now we have:
Thus \(y \leq _{\mathcal{R}} x\), and combined with \(x \leq _{\mathcal{R}} y\), we get \(x \mathcal{R} y\).
The argument for \(\mathcal{L}\) is symmetric.
In a finite semigroup, if an element \(x\) can be written as \(x = uxv\) for some \(u, v \in S\), then \(x\) is \(\mathcal{H}\)-equivalent to both \(ux\) and \(xv\).
Recall that \(x \mathcal{H} y\) if and only if \(x \mathcal{R} y\) and \(x \mathcal{L} y\). We prove both \(x \mathcal{H} ux\) and \(x \mathcal{H} xv\).
Showing \(x \mathcal{H} ux\): We need \(x \mathcal{R} ux\) and \(x \mathcal{L} ux\).
For \(x \mathcal{R} ux\): We have \(ux \leq _{\mathcal{R}} x\) trivially (since \(ux = u \cdot x\)). From \(x = uxv\), we see \(x \leq _{\mathcal{R}} ux\) (via \(v\)). Also, \(x \mathcal{J} ux\) since \(x = ux \cdot v\) (so \(x \leq _{\mathcal{J}} ux\)) and \(ux = u \cdot x \cdot 1\) gives \(ux \leq _{\mathcal{J}} x\). By 10, the \(\mathcal{R}\)-preorder strengthens to \(x \mathcal{R} ux\).
For \(x \mathcal{L} ux\): We have \(x \mathcal{J} ux\) as shown above. Since \(ux = u \cdot x\), we have \(ux \leq _{\mathcal{L}} x\). By 10, this strengthens to \(ux \mathcal{L} x\).
Showing \(x \mathcal{H} xv\): We need \(x \mathcal{R} xv\) and \(x \mathcal{L} xv\).
For \(x \mathcal{R} xv\): We have \(xv \leq _{\mathcal{R}} x\) trivially. Also \(x \mathcal{J} xv\) since \(x = u \cdot xv\) (so \(x \leq _{\mathcal{J}} xv\)) and \(xv \leq _{\mathcal{J}} x\). By 10, this strengthens to \(x \mathcal{R} xv\).
For \(x \mathcal{L} xv\): We have \(x = uxv = u \cdot xv\), so \(x \leq _{\mathcal{L}} xv\). Also, \(xv = x \cdot v\) gives \(xv \leq _{\mathcal{L}} x\). With \(x \mathcal{J} xv\), by 10, we get \(x \mathcal{L} xv\).
Therefore \(x \mathcal{H} ux\) and \(x \mathcal{H} xv\).