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The following question arises in connection with problems in automata theory related to the road problem. Let $f_1, f_2: [N] \to [N]$ be maps such that the transformation semigroup $S = \langle f_1, f_2 \rangle$ is transitive. Let $A$ be the $N \times N$ matrix with $A_{ij} = \mathbf{1}_{\{f_1(i) = j\}} + \mathbf{1}_{\{f_2(i) = j\}}$, and let $p$ be the period of $A$, i.e. the gcd of the set $\{ k - \ell: A^k > 0, A^{\ell} > 0, k, \ell \in \mathbb{N} \}$.

(That is, $A$ is the adjacency matrix of a strongly connected digraph $G$ with period $p$ and out-degree $2$ at every vertex. We label the edges with $\{ 1, 2 \}$ so that every state has each outgoing symbol on one of its edges, and we have $f_k(i) = j$ iff there is an edge from $i$ to $j$ labeled $k$. I have presented the setup more or less backwards, relative to the usual presentation, because I want to emphasize points that I don't often see.)

For a permutation $\sigma \in S_N$, let $\sigma S = \langle f_1 \circ \sigma, f_2 \circ \sigma \rangle$. One way of phrasing the periodic version (due to Béal-Perrin and Budzban-Feinsilver) of Trakhtman's road colouring theorem, for out-degree $2$, is that unless $f_1 \equiv f_2$, in which case $\sigma S = S$ for all $\sigma$, then among the semigroups $\{ \sigma S : \sigma \in S_N \}$ there exist $\sigma$ and $f \in \sigma S$ such that $|f([N]) = p|$. (This is the minimum possible value, given $p$.) When $p = 1$, this means that some $\sigma S$ contains the $N$ constant maps.

Suppose that we didn't know the road colouring theorem. What could we say about the family $\{ \sigma S : \sigma \in S_N \}$? I am basically interested in any response to this vague question.

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    $\begingroup$ Your condition doesn't seem the same as transitive to me. For insurance if f_1 is the idempotent sending n to n-1 and fixing all other inputs and f_2 is the constant map to n you meet your condition but it is not transitive. You can move any element to n and n to n-1 but you can't move n to anything else $\endgroup$ Commented Jul 30, 2021 at 22:24
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    $\begingroup$ To make sure I understand you start with a strongly connected two letter automaton and you allow yourself to relabel the edges and see what semigroups you can get and what you can say about them $\endgroup$ Commented Jul 30, 2021 at 22:31
  • $\begingroup$ I think it is hard to say what happens without road coloring which is already quite hard in some sense but I'd be happy to see an answer that says something concrete $\endgroup$ Commented Jul 30, 2021 at 22:36
  • $\begingroup$ In some sense a similar problem arises in markov chains. Any stochastic matrix is a convex combination of function matrices but in a nonunique way. How do you choose the convex combination so that the semigroup they generate is the most useful to analyze the chain $\endgroup$ Commented Jul 30, 2021 at 22:38
  • $\begingroup$ @BenjaminSteinberg Thanks for these comments! You're right about the transitivity. I have removed the other condition that I stated to prevent confusion. I've been reading your book, but my knowledge of modules is remedial at best, so I thought that while I work on that remediation, I'd see if phrasing the question in a slightly different way could dredge up any ideas that might be latent in the community. What I'm ultimately interested in is a potentially different proof of road colouring that might generalize to other, related problems more readily. $\endgroup$
    – Sophie M
    Commented Jul 31, 2021 at 19:41

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