(1) \(\Rightarrow \) (2): Suppose \(x \cdot y \mathcal{R} x\) and \(x \cdot y \mathcal{L} y\). By Green’s Lemma (12), right multiplication by \(y\) induces a bijection from \([x]_{\mathcal{L}}\) to \([x \cdot y]_{\mathcal{L}} = [y]_{\mathcal{L}}\). In particular, this map is surjective, so there exists \(w \in [x]_{\mathcal{L}} \cap [y]_{\mathcal{R}}\) with \(w \cdot y = y\).

Now we show \(w\) is idempotent. Since \(w \mathcal{R} y\), there exists \(u\) with \(y \cdot u = w\). Then:

Thus \(w\) is an idempotent in \([y]_{\mathcal{R}} \cap [x]_{\mathcal{L}}\).

(2) \(\Rightarrow \) (1): Suppose there exists an idempotent \(e \in [y]_{\mathcal{R}} \cap [x]_{\mathcal{L}}\). By 7, since \(e\) is idempotent and \(y \mathcal{R} e\), we have \(y = e \cdot y\). Similarly, since \(x \mathcal{L} e\), we have \(x = x \cdot e\).

For \(x \cdot y \mathcal{R} x\): we have \(x \cdot y = x \cdot e \cdot y\) (since \(x = x \cdot e\) and \(y = e \cdot y\)). By compatibility of \(\mathcal{R}\) with left multiplication (4), \(x \cdot e \mathcal{R} x \cdot y\) follows from \(e \mathcal{R} y\).

For \(x \cdot y \mathcal{L} y\): similarly, \(x \cdot y = x \cdot e \cdot y\), and by compatibility of \(\mathcal{L}\) with right multiplication, \(x \cdot y \mathcal{L} y\) follows from \(x \mathcal{L} e\).

(1) \(\Leftrightarrow \) (3) in finite semigroups: In a finite semigroup, \(\mathcal{D} = \mathcal{J}\) by 9.

For the forward direction, if \(x \cdot y \mathcal{R} x\) and \(x \cdot y \mathcal{L} y\), then \(x \cdot y \mathcal{D} x\) (via the intermediate element \(x \cdot y\) itself) and \(x \mathcal{D} y\) (via \(x \cdot y\)).

For the reverse direction, suppose \(x \mathcal{D} y\) and \(x \cdot y \mathcal{D} x\). Since \(\mathcal{D} = \mathcal{J}\), we have \(x \cdot y \mathcal{J} x\). Since \(x \cdot y \leq _{\mathcal{R}} x\) trivially, by 10, \(x \cdot y \mathcal{R} x\). Similarly, \(x \cdot y \leq _{\mathcal{L}} y\) and \(x \cdot y \mathcal{J} y\) (by transitivity through \(x\)), so \(x \cdot y \mathcal{L} y\).

(1) Closure under multiplication: Let \(x, y \in [e]_{\mathcal{H}}\). By the Location Theorem (15), \(x \cdot y \in [x]_{\mathcal{R}} \cap [y]_{\mathcal{L}}\) if and only if there exists an idempotent in \([y]_{\mathcal{R}} \cap [x]_{\mathcal{L}}\). Since \(x \mathcal{H} e\) and \(y \mathcal{H} e\), we have \([y]_{\mathcal{R}} = [e]_{\mathcal{R}}\) and \([x]_{\mathcal{L}} = [e]_{\mathcal{L}}\), so \(e \in [y]_{\mathcal{R}} \cap [x]_{\mathcal{L}}\). Thus \(x \cdot y \mathcal{R} x \mathcal{R} e\) and \(x \cdot y \mathcal{L} y \mathcal{L} e\), giving \(x \cdot y \in [e]_{\mathcal{H}}\).

(2) Identity property: For \(x \in [e]_{\mathcal{H}}\), we have \(x \mathcal{R} e\), so \(x \leq _{\mathcal{R}} e\). By 7, since \(e\) is idempotent, \(e \cdot x = x\). Similarly, \(x \mathcal{L} e\) gives \(x \cdot e = x\).

(3) Existence of inverses: Let \(x \in [e]_{\mathcal{H}}\). Since \(x \mathcal{R} e\), there exists \(y\) with \(x \cdot y = e\). By Green’s Lemma, right multiplication by \(x\) is a bijection on \([e]_{\mathcal{L}}\) (since \(e \cdot x = x\) from part (2)). This bijection maps \([e]_{\mathcal{H}}\) to itself. Since \(e \in [e]_{\mathcal{L}}\), there exists \(z \in [e]_{\mathcal{L}}\) with \(z \cdot x = e\). We can show \(z \in [e]_{\mathcal{H}}\) and that the same element works for both left and right inverse.

(4) Subgroup structure: Parts (1)-(3) establish that \([e]_{\mathcal{H}}\) satisfies all the axioms of a subgroup with identity \(e\).

(5) Maximal subgroups are \(\mathcal{H}\)-classes: Let \(H\) be a maximal subgroup with identity element \(e_H\). The identity \(e_H\) is idempotent (since \(e_H \cdot e_H = e_H\)). All elements of \(H\) are \(\mathcal{H}\)-equivalent to \(e_H\): for \(x, y \in H\), we have \(x = x \cdot e_H\) and \(y = e_H \cdot y\), giving the required \(\mathcal{R}\) and \(\mathcal{L}\) relations. Thus \(H \subseteq [e_H]_{\mathcal{H}}\). Since \([e_H]_{\mathcal{H}}\) is itself a subgroup containing \(H\), maximality of \(H\) implies \(H = [e_H]_{\mathcal{H}}\).

By 16, \(H = [e]_{\mathcal{H}}\) and \(K = [f]_{\mathcal{H}}\) for some idempotents \(e, f \in S\). Since \(x \in H\) and \(y \in K\), we have \(x \mathcal{H} e\) and \(y \mathcal{H} f\). From \(x \mathcal{D} y\), we get \(e \mathcal{D} f\) by transitivity.

Since \(e \mathcal{D} f\), there exists \(s \in S\) with \(e \mathcal{R} s\) and \(s \mathcal{L} f\). Let \(t\) be a witness of \(f \leq _{\mathcal{L}} s\), so \(t \cdot s = f\).

We claim the map \(\phi : [e]_{\mathcal{H}} \to [f]_{\mathcal{H}}\) defined by \(\phi (x) = t \cdot x \cdot s\) is a group isomorphism.

Well-defined and bijective: By Green’s Lemma (12), right multiplication by \(s\) gives a bijection from \([e]_{\mathcal{H}}\) to \([s]_{\mathcal{H}}\) (since \(e \mathcal{R} s\)). Similarly, left multiplication by \(t\) gives a bijection from \([s]_{\mathcal{H}}\) to \([f]_{\mathcal{H}}\) (since \(s \mathcal{L} f\)). The composition \(x \mapsto t \cdot x \cdot s\) is therefore a bijection from \([e]_{\mathcal{H}}\) to \([f]_{\mathcal{H}}\).

Preserves multiplication: We need to show \(\phi (x) \cdot \phi (y) = \phi (x \cdot y)\) for \(x, y \in [e]_{\mathcal{H}}\). That is:

The key observation is that \(s \cdot t\) is idempotent: since \(s \mathcal{L} f\) and \(t \cdot s = f\), we have \(s \cdot t \cdot s \cdot t = s \cdot f \cdot t = s \cdot t\) (using \(s \cdot f = s\) which follows from \(s \mathcal{L} f\) and \(f\) being idempotent).

Furthermore, for \(y \in [e]_{\mathcal{H}}\), we have \(y \mathcal{R} e \mathcal{R} s\), so \(s \cdot t \cdot y = y\) (since \(s \cdot t\) is an idempotent in the \(\mathcal{R}\)-class of \(y\)). Therefore:

Thus \(\phi \) is a group isomorphism.