If $ar \mathcal R a$ , then the map $x \mapsto xr$ is a bijection from $\mathcal H_a$ onto $\mathcal H_{ar}$.
I know that
If D is an arbitrary $\mathcal D - $ class in a Semigroup $S$ and if $a, b \in D$ are such that $a \mathcal R b$ , then by the definition of $\mathcal R$ there exist $s , s' \in S^1 $ such that
$$ as = b , \ \ \ \ bs' = a\ .$$
The right translation map $\rho_s : S \rightarrow S$ such that $(x)\rho_s = xs$ thus maps $a$ to $b$ and maps $\mathcal L_a$ into $\mathcal L_b$.
Similarly $\rho_{s'}$ maps $\mathcal L_b$ into $\mathcal L_a$, the composition $\rho_s \rho_{s'}$ is the identity map on $\mathcal L_a$ and $\rho_{s'} \rho_{s}$ is the identity map on $\mathcal L_b$
We deduce that $\rho_s|\mathcal L_a$ and $\rho_{s'}|\mathcal L_b$ are mutually inverse bijections from $\mathcal L_a$ onto $\mathcal L_b$ and $\mathcal L_b$ onto $\mathcal L_a$.
We can say even more about these maps; if $x \in \mathcal L_a$, then the element $ y = x\rho_s$ of $\mathcal L_b$ has the property that
$$ y = xs , \ \ \ \ x = ys'\ .$$
Thus $y \mathcal R x $, and so the map $\rho_s$ is $\mathcal R$ - class preserving. It maps each $\mathcal H -$ class in $\mathcal L_a$ in a one-one manner onto the corresponding $\mathcal H-$ class in $\mathcal L_b$
I want to prove the required result by using this theory. Any help would be appreciated. Thank you