finalized

4 Green’s Relations in Finite Semigroups

Lemma 9 Equivalence of D and J Relations in Finite Semigroups

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 \(x \mathcal{J} y\).

Proof

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. If \(x \mathcal{J} y\), then \(x \leq _{\mathcal{J}} y\) and \(y \leq _{\mathcal{J}} x\). This means there exist \(s, t, u, v \in S^1\) such that \(x = syt\) and \(y=uxv\). Substituting these into each other shows that \(x\) is of the form \(axb\) for some \(a,b \in S\). In a finite semigroup, this implies that some power of \(a\) and \(b\) will lead to an idempotent element related to \(x\), which can be used to construct the intermediate element for the \(\mathcal{D}\) relation. This relies on the property that for any element \(a\) in a finite semigroup, the sequence \(a, a^2, a^3, \dots \) must contain an idempotent.

Lemma 10 J-Equivalence Strengthening Preorders

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 \(x \mathcal{L} y\).

Proof

Suppose \(x \mathcal{J} y\) and \(x \leq _{\mathcal{R}} y\). Since we are in a finite semigroup, \(x \mathcal{J} y\) implies \(x \mathcal{D} y\) by 9. So there exists a \(z\) such that \(x \mathcal{R} z\) and \(z \mathcal{L} y\). From \(x \leq _{\mathcal{R}} y\), we can show that \(y \leq _{\mathcal{R}} x\), which gives \(x \mathcal{R} y\). The argument for \(\mathcal{L}\) is analogous.

Lemma 11 H-Equivalence from Sandwiching

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 \(xv\).

Proof

The condition \(x = uxv\) implies \(x \leq _{\mathcal{J}} ux\) and \(x \leq _{\mathcal{J}} xv\). It also implies \(ux \leq _{\mathcal{R}} x\) and \(xv \leq _{\mathcal{L}} x\). Using the property that \(\mathcal{J}\)-equivalence strengthens preorders to equivalences in finite semigroups (10), we can establish the \(\mathcal{R}\) and \(\mathcal{L}\) equivalences needed to show \(x \mathcal{H} ux\) and \(x \mathcal{H} xv\).