Unique idempotents in semigroups

Question

In a finite semigroups every element has a unique idempotent power, just take $s^{n!}$ where $n = |S|, s \in S$.

In an infinite semigroup there are clearly elements without idempotents, just take $(\mathbb N, +)$ for example.

In an arbitrary semigroup if $s \in S$ has an idempotent power, i.e. there exists an $n$ such that $s^{n}$ is idempotent, is it unique? And further if it is unique, I can denote by $s^{\omega}$ the unique idempotent of $s$, then if $st$ and $ts$ has such an unique idempotent, does $(st)^{\omega} s = s(ts)^{\omega}$? (in the finite case this is all satisfied with $\omega := n!$).

Answer 1

If an element $s$ of a semigroup $S$ has an idempotent power, then the subsemigroup generated by $s$ is finite and it contains a unique idempotent.

If both $st$ and $ts$ have an idempotent power, say $(st)^p = (st)^\omega$ and $(ts)^q = (ts)^\omega$, then $$ (st)^{pq} = \bigl((st)^\omega\bigr)^q = (st)^\omega \text{ and } (ts)^{pq} = \bigl((ts)^\omega \bigr)^p = (ts)^\omega. $$ Therefore $(st)^\omega s = (st)^{pq}s= s(ts)^{pq} = s(ts)^\omega$.

Answer 2

1. Let $S$ be a semigroup. Then there exists a semigroup with unit $\tilde S$ such that $S\subset \tilde S$. Indeed, simply add formally a unit $1$ to $S$ and postulate $1\cdot s = s \cdot 1 = s$, $1\cdot 1 = 1$.

2. Let $S$ be semigroup with $1$. Then $S$ imbeds into the monoid of transformations of $S$ (Cayley embedding).

3. Let $S$ be a finite semigroup. Then $S$ embeds into the monoid of transformations of a finite set.

4. Let $S$ be a semigroup, $s \in S$ of finite order, i.e. there exist $1\le a < b$ such that $s^a = s^b$. . Then the semigroup generated by $s$ is finite. Therefore, we can restrict to the case $s$ in a monoid of transformations of a finite set $M$. The transformation $s$ will have several $\rho$-orbits, with tail $t_i$ and cycle length $d_i$. We now see that we have $$s^k = s^{k + m}$$ if and only if $t \ge \max t_i$ and $\operatorname{lcm}( d_i) \mid m$. In particular, $s^d = s^{d+d}$ if and only if $d$ is a multiple $\ge \max t_i$ of $\operatorname{lcm}( d_i)$. For all such $d$ the elements $s^d$ are the same. Therefore, we have a unique idempotent in $\langle s\rangle$.

5. We could prove the uniqueness without the representation of the semigroup on a monoid of transformations. Indeed, consider all such $d$ such that $s^a = s^{a+d}$ for some $a\ge 1$. Then the set of such $d$'s is closed under addition and substraction ( when defined), so it consists of the multiples of some $d$. Now consider $t$ smallest with $s^{t} = s^{t+d}$ &c.