Which monoids have a faithful irreducible representation?

Question

Let $*$ be a binary operation on a set $M$, with an identity element $e\in M$.

A monoid representation of $(M,*,e)$ is a map $\delta:M\to (S\to S)$ for some set $S$, such that $\delta(e)=\mathrm{id}_S$, and $\delta(a*b)=\delta(a)\circ\delta(b)$ for all $a,b\in M$. (A representation could also be called an action, I suppose?)

• $\delta$ is faithful if $\delta$ is injective as a function from $M$ to $S^S$.

• $\delta$ is irreducible if there is no subset $\emptyset\subsetneq T\subsetneq S$ with $\delta(m)(t)\in T$ for all $t\in T$ and $m\in M$.

Which monoids $(M,*,e)$ have faithful irreducible representations? For example, all groups do have such representations, but the monoid $\{e,a\}$ with $a^2=a\ne e$ does not since we can take $T=\{\delta(a)(t)\}$ for a fixed $t\in S$.

Is there a characterization, or a name for such monoids?

Example: let $M$ be generated by $f,g:\{0,1,2,3\}\to\{0,1,2,3\}$ where $f(0)=1$, $g(0)=2$, $f(1)=g(1)=1$, and $f(2)=3$, $f(3)=2$, $g(2)=3$, $g(3)=3$. The monoid is $$\begin{matrix} & && e &&\\ &&f & &g\\ &f^2 & gf&&fg & g^2\\ & & fgf&&&fg^2\\ \end{matrix}$$ which has the ideals: $$M(gf)=\{fgf,gf\}, M(g^2)=\{g^2,fg^2\}\quad\text{(minimal)}$$ $$ M(f),M(g) \quad\text{(not minimal)}$$

Answer 1

Cleaner rewrite:

I have a bit more time, so here is a cleaner rewrite. This notion is usually called transitive rather than irreducible, although the terms irreducible and minimal are both used.

If $M$ has a minimal left ideal $L$,then $L$ is a transitive $M$-set and every transitive action is a quotient of $L$ and hence $M$ has a faithful transitive action iff it acts faithfully on $L$. In particular, if $M$ is finite then all its minimal left ideals are isomorphic as $M$-sets (since they are quotients of each other and finite) and $M$ has faithful transitive action iff it acts faithfully on one (equals all) of its minimal left ideals.

The proof is trivial. If $S$ is a transitive $M$-set, then $Ls$ is invariant and hence $Ls=S$ for all $s\in S$. Thus, if we fix $s\in S$, then $m\mapsto ms$ is a surjective $M$-set map $L\to S$.

There is some information on Clifford and Preston Volume 2, Chapter 11.5 on the infinite case. For instance they show noncommutative free monoids have a faithful and transitive action and nontrivial free products of semigroups. See also Tully, E. J. (1961). Representation of a Semigroup by Transformations Acting Transitively on a Set. American Journal of Mathematics, 83(3), 533.

Further Remark It is more common to consider transitive actions by partial mappings because many more semigroups have faithful transitive actions in this setting.