use the results :
Let $a$ be an element of a regular $\mathcal D-$ class $D$ in a semigroup $S$. Then
If $a'$ is the inverse of $a$, then $a' \in D$ and the two $\mathcal H$ - classes $\mathcal R_a \cap \mathcal L_{a'}$ and $\mathcal L_a \cap \mathcal R_{a'}$ contain respectively the idempotent $aa'$ and $a'a$.
If $b$ in $D$ is such that $\mathcal R_a \cap \mathcal L_{b}$ and $\mathcal L_a \cap \mathcal R_{b}$ contains idempotent $e , f$ respectively , then $\mathcal H_b$ contain an inverse $a'$ of $a$ such that $aa' = e$ and $a'a = f$.
Let $a' , a^*$ are the inverse of $a$ conatined in some class $H_x$ of $\mathcal H$, so $a$ is regular and it is contained in some $\mathcal D-$ class says $D$, so $a' , a^* \in D$. So $aa'$ are the idempotent in $\mathcal R_a \cap \mathcal L_{a'}$ , it follows that $aa' \in D$.
basicly i want to show that $aa'$ and $aa*$ belongs to the same $\mathcal H-$ class.
if $aa'$ and $aa^*$ belongs to the same class , then $aa' = aa^*$ and similarly we can show that $a'a = a*a$ it follows that
$$a^* = a^* aa^* = a^*aa' = a'aa' = a' $$
How to show that $aa'$ and $aa^*$ belongs to the same $\mathcal H-$ class.
Any help would be appreciated. Thank you.
If $\bar a$ is an inverse of $a$, then $\bar aa$ and $a \bar a$ are idempotents. Furthermore, $a$ and $\bar a$ are in the same $\mathcal{D}$-class and the relations $a\bar a \mathrel{\mathcal L} \bar a \mathrel{\mathcal R} \bar aa \mathrel{\mathcal L} a \mathrel{\mathcal R} a\bar a$ hold. Consequently $H(a\bar a) = R(a) \cap L(\bar a)$.
Suppose that an $\mathcal H$-class $H$ contains two inverses $\bar a_1$ and $\bar a_2$ of some element $a$. Then $L(a_1) = L(a_2)$ and thus $R(a) \cap L(\bar a_1) = R(a) \cap L(\bar a_2)$. It follows that $a\bar a_1$ and $a\bar a_2$ are idempotents of the same $\cal H$-class and hence are equal. Similarly, $\bar a_1a$ and $\bar a_2a$ are equal. It follows that $\bar a_1a\bar a_1 = \bar a_2a\bar a_1 = \bar a_2a\bar a_2$, that is, $\bar a_1 = \bar a_2$.