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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 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 $LS=S$. 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.

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.

I would call your action minimal or transitive instead of irreducible. I know the answer for finite monoids but I'd have to think a bit about the general case and it might be a bit tricky in that level of generality.

If $M$ is a finite monoid, or more generally has both a minimal left and a minimal right ideal, then all minimal left ideals of $M$ are isomorphic as left $M$-sets and are transitive (irreducible in the sense of the OP). Conversely, every transitive action factors through this action. Thus $M$ has a faithful transitive action if and only if it acts faithfully on some (equals any of its) minimal left ideals.

To see, this first note that a minimal left ideal is generated by some idempotent $e$, that is, it can be written in the form $Me$ (this is obvious if $M$ is finite and follows from structure theory if there is both a left and a right minimal ideal). Then if $M$ acts transitively on $S$, we have that $MeS=S$ by transitivity. Fix $s\in eS$. Then $m\mapsto ms$ is a surjective $M$-set homomorphism from $Me$ to $S$.

Rhodes calls finite semigroups with this property left mapping semigroups with respect to their minimal ideal.

Added. In fact on further reflection if a monoid has a minimal left ideal then it acts faithfully and transitively on some set iff it does on its minimal left ideal by the above argument. The existence of idempotents is irrelevant. I'm just used to the finite case. The above argument does not need $e$ to be an idempotent.

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.

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 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 $LS=S$. 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.

I would call your action minimal or transitive instead of irreducible. I know the answer for finite monoids but I'd have to think a bit about the general case and it might be a bit tricky in that level of generality.

If $M$ is a finite monoid, or more generally has both a minimal left and a minimal right ideal, then all minimal left ideals of $M$ are isomorphic as left $M$-sets and are transitive (irreducible in the sense of the OP). Conversely, every transitive action factors through this action. Thus $M$ has a faithful transitive action if and only if it acts faithfully on some (equals any of its) minimal left ideals.

To see, this first note that a minimal left ideal is generated by some idempotent $e$, that is, it can be written in the form $Me$ (this is obvious if $M$ is finite and follows from structure theory if there is both a left and a right minimal ideal). Then if $M$ acts transitively on $S$, we have that $MeS=S$ by transitivity. Fix $s\in eS$. Then $m\mapsto ms$ is a surjective $M$-set homomorphism from $Me$ to $S$.

Rhodes calls finite semigroups with this property left mapping semigroups with respect to their minimal ideal.

Added. In fact on further reflection if a monoid has a minimal left ideal then it acts faithfully and transitively on some set iff it does on its minimal left ideal by the above argument. The existence of idempotents is irrelevant. I'm just used to the finite case. The above argument does not need $e$ to be an idempotent.

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.

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

I would call your action minimal or transitive instead of irreducible. I know the answer for finite monoids but I'd have to think a bit about the general case and it might be a bit tricky in that level of generality.

If $M$ is a finite monoid, or more generally has both a minimal left and a minimal right ideal, then all minimal left ideals of $M$ are isomorphic as left $M$-sets and are transitive (irreducible in the sense of the OP). Conversely, every transitive action factors through this action. Thus $M$ has a faithful transitive action if and only if it acts faithfully on some (equals any of its) minimal left ideals.

To see, this first note that a minimal left ideal is generated by some idempotent $e$, that is, it can be written in the form $Me$ (this is obvious if $M$ is finite and follows from structure theory if there is both a left and a right minimal ideal). Then if $M$ acts transitively on $S$, we have that $MeS=S$ by transitivity. Fix $s\in eS$. Then $m\mapsto ms$ is a surjective $M$-set homomorphism from $Me$ to $S$.

Rhodes calls finite semigroups with this property left mapping semigroups with respect to their minimal ideal.

Added. In fact on further reflection if a monoid has a minimal left ideal then it acts faithfully and transitively on some set iff it does on its minimal left ideal by the above argument. The existence of idempotents is irrelevant. I'm just used to the finite case. The above argument does not need $e$ to be an idempotent.

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.