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Let $M$ be a monoid that acts transitively from the right on a finite set $X$. Assume furthermore that the action of $M$ on $X$ induces for every $m \in M$ a bijection on $X \to X, x \mapsto x.m$. Let $$M_X := \{ m \in M \; | \; \forall x \in X : x.m = x \}$$ be the fixer of $X$.

Question 1: Is there always a finite subset $S$ of $M$ such that $S M_X = M$?

(here $S M_X$ denotes the set $\{ sm \; | \; s \in S, m \in M_X \}$)

Question 2: Assume that there is such a set $S$ as in question 1. Given elements $s \in S$ and $m \in M$ is there always an element $s' \in S$ such that $ms' \in s M_X$?

(here $s M_X$ denotes the set $\{ s m \; | \; m \in M_X \}$)

The answer to both questions is positive if either $M$ is a group and also if $M$ is the monoid of natural numbers with addition. In the case where $M$ is a group the affirmation in question 1 results from the fact that there are only finitely many cosets of $M_X$ and question 2 from the fact that $M$ operates transitively on these cosets.

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The answer is no. Map the free monoid $M$ on $\{a,b\}$ onto $\mathbb Z/2$ by $a$ maps to $1$ and $b$ maps to $0$. Then $M$ acts transitively on the right of $\mathbb Z/2$ via bijections by applying the homomorphism and doing the regular representation.

The fixer is all words with an even number of $a$'s. Consider the words $b^ka$ with $k>0$. No nonempty suffix is in the fixer so all these must belong to any potential $S$.

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  • $\begingroup$ Thank you for your answer. Do you know any class of monoids that includes groups and the natural numbers for which the questions have positive answers? And how about commutative monoids? $\endgroup$
    – skew41
    Nov 2, 2014 at 11:10
  • $\begingroup$ I think this would be very rare. Monoids just don't behave so well. $\endgroup$ Nov 2, 2014 at 11:49

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