Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
up vote 3 down vote accepted

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 $x=xb=y$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.