$H_{x}$ denotes the class of $x$ for the Green relation $\mathcal{H}$.
Let $S$ be a finite semigroup where all elements can be written as a product of idempotents, that is, $x=e_1 e_2\dots e_n,$ for idempotents $e_1,e_2,\dots, e_n \in S$ for any $x \in S.$
Assume also that $H_{ef}$ is a group for all idempotents $e,f \in S$ Show that $H_x$ is a group for all $x \in S .$
$ \newcommand{\oH}{\mathop{\cal H}\nolimits} \newcommand{\oL}{\mathop{\cal L}\nolimits} \newcommand{\oR}{\mathop{\cal R}\nolimits} $
Since $S$ is finite, there exists a positive integer $\omega$ such that, for all $s \in S$, $s^\omega$ is idempotent. Let us prove by induction on $n$ that if $x$ is a product of $n$ idempotents, then $H_x$ is a group, that is, $x \oH x^\omega$, or, equivalently $x^{\omega+1} = x$.
For $n = 1$ et $n = 2$, this follows from the hypothesis (for $n= 1$, it suffices to take $e = f$). Let $n \geqslant 2$ and let $x = e_1e_2 \dotsm e_ne_{n+1}$ be a product of $n+1$ idempotents. By the induction hypothesis, $e_1e_2 \dotsm e_n \oH\, (e_1e_2 \dotsm e_n)^\omega$ and $e_ne_{n+1} \oH\, (e_ne_{n+1})^\omega$. I claim that the following diagram is part of the $\cal D$-class of $x$ (as usual, a star means that the corresponding $\cal H$-class is a group). \begin{array}{|c|c|} \hline x = e_1e_2 \dotsm e_{n-1}e_ne_{n+1} &{}^*\ u = e_1e_2 \dotsm e_{n-1}(e_ne_{n+1})^\omega \\ \hline {}^*\ v = (e_1e_2 \dotsm e_n)^\omega e_{n+1} &{}^*\ w = (e_1e_2 \dotsm e_n)^\omega (e_ne_{n+1})^\omega \\ \hline \end{array} Indeed, by the induction hypothesis, $e_1e_2 \dotsm e_n \oH\ (e_1e_2 \dotsm e_n)^\omega$ and thus $e_1e_2 \dotsm e_ne_{n+1} \oL\ (e_1e_2 \dotsm e_n)^\omega e_{n+1}$. Since $n \geqslant 2$, the induction hypothesis also implies that $e_ne_{n+1} \oH\ (e_ne_{n+1})^\omega$, whence $e_1e_2 \dotsm e_{n-1}e_ne_{n+1} \oR e_1e_2 \dotsm e_{n-1}(e_ne_{n+1})^\omega$ and $(e_1e_2 \dotsm e_n)^\omega e_{n+1} = (e_1e_2 \dotsm e_n)^\omega e_ne_{n+1} \oR (e_1e_2 \dotsm e_n)^\omega (e_ne_{n+1})^\omega$. The "stars" occur because the elements $v$ and $w$ are the product of two idempotents and $u$ is the product of $n$ idempotents. Now, since $R(v) \cap L(u)$ contains an idempotent, the product $s = u^\omega v^\omega$ belongs to $H_x$. Thus $H_x = H_s$, but since $s$ is the product of two idempotents, $H_s$ is a group. Thus $H_x$ is a group.
