# Semigroup which satisfied $(xy)^{\pi} = (xy)^{\pi}x$ and principal right ideals

Let $(S,\cdot)$ be a finite semigroup, then every $s \in S$ has an unique idempotent power, i.e. there exists a smallest $i \in \mathbb N$ such that $s^i$ is idempotent. It is the unique idempotent in the the subsemigroup generated by $s$. Let's denote this element by $s^{\pi}$. Now suppose in an finite semigroup the following identiy holds $$(xy)^{\pi} = (xy)^{\pi}x$$ then show that two principal right ideals coincide iff there generators are the same, i.e. $$xS^1 = yS^1 \quad \textrm{ iff } \quad x = y$$ where $S^1 = S \cup \{ 1 \}$, i.e. $S$ adjoined with a unity.

I have no idea how to solve this, any hints?

Your assumption on ideals is equivalent to the fact that there are $a,b$ in $S$ such that $x=ya$ and $y=xb$.
From this, you get $x=xba$, so $x=x(ba)^n$, where $n$ is the idempotent power of $(ba)$.
By assumption on idempotent powers, we obtain $x=x(ba)^n b$, but we can replace $x(ba)^n$ by $x$, so finally $x=xb=y$.