We prove the result for \(\mathcal{R}\)-equivalent elements; the dual statement for \(\mathcal{L}\)-equivalent elements follows by a symmetric argument.

Suppose \(x \mathcal{R} y\) with \(x \cdot v = y\) and \(y \cdot u = x\). We show that \(\rho _v : w \mapsto w \cdot v\) is a bijection from \([x]_{\mathcal{L}}\) to \([y]_{\mathcal{L}}\) by establishing that it maps into the correct set, is injective, is surjective, and has \(\rho _u\) as its inverse.

Translation Identity. The key observation is that for any \(w \mathcal{L} x\), right translation by \(v\) followed by \(u\) returns \(w\) to itself: \(w \cdot v \cdot u = w\). To see this, since \(w \mathcal{L} x\), there exists \(z\) such that \(w = z \cdot x\). Then:

This identity is central to all parts of the proof.

Maps To (\(\rho _v\) maps \([x]_{\mathcal{L}}\) to \([y]_{\mathcal{L}}\)). Let \(w \mathcal{L} x\). By 4, right multiplication preserves \(\mathcal{L}\)-equivalence, so \(w \cdot v \mathcal{L} x \cdot v = y\). Thus \(\rho _v\) maps the \(\mathcal{L}\)-class of \(x\) into the \(\mathcal{L}\)-class of \(y\).

Injectivity. Suppose \(w, z \in [x]_{\mathcal{L}}\) and \(w \cdot v = z \cdot v\). By the translation identity, \(w = w \cdot v \cdot u = z \cdot v \cdot u = z\). Thus \(\rho _v\) is injective on \([x]_{\mathcal{L}}\).

Surjectivity. Let \(z \in [y]_{\mathcal{L}}\). We claim that \(z \cdot u \in [x]_{\mathcal{L}}\) and \(\rho _v(z \cdot u) = z\). First, since \(z \mathcal{L} y\) and \(y \cdot u = x\), by 4 we have \(z \cdot u \mathcal{L} y \cdot u = x\). Second, by the translation identity (applied with the roles of \(u\) and \(v\) swapped, using \(z \mathcal{L} y\)), we have \((z \cdot u) \cdot v = z\). Thus \(\rho _v\) is surjective.

Inverse (\(\rho _u\) is the inverse of \(\rho _v\)). The translation identity shows that for \(w \in [x]_{\mathcal{L}}\), we have \(\rho _u(\rho _v(w)) = w \cdot v \cdot u = w\). Symmetrically, for \(z \in [y]_{\mathcal{L}}\), we have \(\rho _v(\rho _u(z)) = z \cdot u \cdot v = z\). Thus \(\rho _u\) and \(\rho _v\) are inverses on the respective \(\mathcal{L}\)-classes.

Preservation of \(\mathcal{H}\)-classes. Let \(a, b \in [x]_{\mathcal{L}}\). We show \(a \mathcal{H} b \Leftrightarrow a \cdot v \mathcal{H} b \cdot v\). Recall that \(\mathcal{H} = \mathcal{R} \cap \mathcal{L}\).

For the \(\mathcal{L}\)-part: by 4, \(a \mathcal{L} b\) implies \(a \cdot v \mathcal{L} b \cdot v\), and conversely, \(a \cdot v \mathcal{L} b \cdot v\) implies \(a \cdot v \cdot u \mathcal{L} b \cdot v \cdot u\), i.e., \(a \mathcal{L} b\).

For the \(\mathcal{R}\)-part: since \(a, b \in [x]_{\mathcal{L}}\), we have \(a \cdot v \cdot u = a\) and \(b \cdot v \cdot u = b\). This means \(a \cdot v \mathcal{R} a\) (witnessed by \(u\)) and \(b \mathcal{R} b \cdot v\) (witnessed by \(u\)). Thus \(a \mathcal{R} b\) if and only if \(a \cdot v \mathcal{R} b \cdot v\), by transitivity through these intermediate equivalences.

Dual Version. For \(x \mathcal{L} y\) with \(v \cdot x = y\) and \(u \cdot y = x\), the proof is entirely symmetric: left translation \(\lambda _v : w \mapsto v \cdot w\) is a bijection from \([x]_{\mathcal{R}}\) to \([y]_{\mathcal{R}}\), with inverse \(\lambda _u\), and preserves \(\mathcal{H}\)-classes.