Subshifts with a free semigroup

Let $X$ be a subshift on a finite alphabet. I'm interested in the following property: there exist words $s,t\in\mathcal L(X)$ (the language of $X$) such that $\{s,t\}^*\subset \mathcal L(X)$. That is, $s^{k_1}t^{m_1}\dots s^{k_n}t^{m_n}\in \mathcal L(X)$ for all $n\ge1$ and all non-negative $k_j$ and $m_j$.

Question. What natural property of $X$ will guarantee the existence of such a free semigroup? (Apart from positive topological entropy, which $X$ must have, of course.)

For instance, if $X$ is an irreducible sofic subshift, does it always hold?

• Are you perhaps missing a condition? If $X$ contains a periodic points with periodic word $u$, then just set $s=u$ and $t=uu$. I guess you want something like $s^\infty \neq t^\infty$? (some kind of 'independence' condition on $s$ and $t$) – Dan Rust Sep 27 '17 at 0:33
• Even still, for irreducible sofic subshifts, it's enough to find two distinct loops in the corresponding right resolved graph which have the same 'root'. I think all you need for that is to have two distinct periodic orbits and irreducibility? – Dan Rust Sep 27 '17 at 0:37
• The precise characterization for sofic shifts to have the property is that they contain uncountably many points, or equivalently have positive entropy. – Ville Salo Jun 8 '18 at 14:53
• If you want that all the words obtained by concatenating are distinct, that is, e.g. s and t are distinct and have the same length. Otherwise take a periodic point w^\omega and pick s = t = w. – Ville Salo Jun 8 '18 at 15:12

For an irreducible sofic shift which is not periodic you will have this property. The Fischer cover gives a strongly connected deterministic partial automaton with all states initial and final recognizing the $\mathcal L(X)$. For any vertex v, the set of words labeling a loop at v is a free monoid on the words labeling a loop at v that do not visit v except at the beginning and end. If the strongly connected graph is not a cycle you have at least two such words.