This is Exercise 2.6.7 of Howie's "Fundamentals of Semigroup Theory".
The Details:
Let $S$ be a semigroup.
Definition 1: We say $S$ is regular if for all $a\in S$ there exists an $x$ in $S$ such that $a=axa$.
Definition 2: A equivalence relation $R$ on a set $S$ is a right congruence if it is right compatible; that is, $$(\forall s, t, a\in S) (s, t)\in R\implies (sa, ta)\in R.$$
Definition 3: An equivalence $\mathcal{L}$ on $S$ is defined by the rule that $a\mathcal{L}b$ if and only if $S^1a=S^1b$.
Definition 4: Define an equivalence $\mathcal{L}^*$ on $S$ by the rule that $a\mathcal{L}^*b$ if and only if $$(\forall x, y\in S^1) ax=ay\Leftrightarrow bx=by.$$
The Question(s):
(a) Show that $\mathcal{L}\subseteq\mathcal{L}^*$, and that $\mathcal{L}=\mathcal{L}^*$ if $S$ is regular.
My Attempt:
Let $a\mathcal{L}b$. Then there exist $x, y\in S^1$ such that $xb=a$ & $ya=b$. Suppose $u, v\in S^1$ such that $au=av$. Then $bu=(ya)u=y(av)=bv$. Thus $$au=av\implies bu=bv$$ and similarly $$bu=bv\implies au=av.$$ Hence $a\mathcal{L}^*b$, and $\mathcal{L}\subseteq \mathcal{L}^*.$
Suppose that $S$ is regular. Suppose $a\mathcal{L}^*b$. Then what?
(b) Show that $\mathcal{L}^*$ is a right congruence on $S$.
My Attempt:
That $\mathcal{L}^*$ is an equivalence relation follows from $\Leftrightarrow$ being an equivalence relation.
Right compatibility is shown as follows. Let $s\mathcal{L}^*t$. Then for all $x, y\in S^1$, $$sx=sy\Leftrightarrow tx=ty.$$ But then for all $u, v\in S^1$, $$sau=sav\Leftrightarrow tau=tav,$$ so that $sa\mathcal {L}^*ta$.$\square$
(c) Show that, for every idempotent $e$ in $S$, $a\mathcal{L}^* e$ if and only if $ae=a$ and for all $x, y\in S^1$, $$ax=ay\implies ex=ey.$$ In particular, note that $e$ acts as a right identity within its $\mathcal{L}^*$-class.
My Attempt:
Let $e$ be an idempotent in $S$.
Suppose $a\mathcal L^*b$. Then for all $x, y\in S^1$, $ax=ay\implies ex=ey.$ How do I show that $ae=a$?
How do I show that converse implication?
Please help :)
For (a), if $a\mathcal{L}^*b$ then by $a1=axa$ we have $b1=bxa$ and vice versa.
For (c), notice the idempotent satisfies $ee=e1$ thus $ae=a1=a$.
And if $ex=ey$ then $aex=aey$ which implies $ax=ay$ since $ae=a$. By definition, $a\mathcal{L}^*e$.