Proof. The map $\Psi $ is continuous and injective; hence, it is an homeomorphism onto its image. Moreover, it follows from the definitions that $\Psi \circ T_{\varphi } = S\circ \Psi $ .

Proof. Let $\mathcal {M}$ be the set of invariant measures on $(Z,R)$ and $\mathcal {M}_\mu $ be the subset of those $\zeta \in \mathcal {M}$ such that $\zeta \circ \pi ^{-1}=\mu $ . First, we show that $\mathcal {M}_\mu \neq \emptyset $ . Start by choosing a point $y\in Y$ satisfying

$$ \begin{align*} \lim_{n\to\infty} \frac{1}{n}\sum_{i=0}^{n-1}f(T^iy) = \int_Y f\, d\mu \end{align*} $$

for every continuous function on X; the point y is called a generic point for $\mu $ . It is well known that ergodicity of $\mu $ implies that $\mu $ -almost all points of Y have this property (see for instance [Reference Furstenberg30, Proposition 3.7]); thus, such a point y exists. Take a preimage $z\in \pi ^{-1}(y)$ and let $\zeta _i$ be the point mass measure concentrated on $R^iz$ . Consider the sequence of Cesàro averages $({1}/{n})\sum _{i=0}^{n-1}\zeta _i$ . By the Banach–Alaoglu theorem, we may choose a subsequence that converges for the weak- $*$ topology in the space of all probability measures on $(Z,R)$ , say

$$ \begin{align*} \zeta = \lim_{k\to\infty} \frac{1}{n_k}\sum_{i=0}^{n_k-1}\zeta_i. \end{align*} $$

Observe that $\zeta _i\circ R^{-1} {\kern-1pt}={\kern-1pt} \zeta _{i+1}$ and, thus, $\zeta - \zeta \circ R^{-1} {\kern-1pt}={\kern-1pt} \lim _{k\to \infty }({1}/{n_k})(\zeta _0 {\kern-1pt}-{\kern-1pt} \zeta _{n_k}+1){\kern-1pt}={\kern-1pt}0$ , that is, $\zeta \in \mathcal {M}$ . Moreover, for every measurable subset $B\subseteq Y$ ,

$$ \begin{align*} \zeta\circ\pi^{-1}(B) = \lim_{k\to\infty}\frac{1}{n_k}\sum_{i=0}^{n_k-1}\zeta_i(\pi^{-1}(B)) = \lim_{k\to\infty}\frac{1}{n_k}\sum_{i=0}^{n_k-1} 1_B(T^iy) = \mu(B). \end{align*} $$

Thus, $\zeta \in \mathcal {M}_\mu $ .

Observe that $\mathcal {M}_\mu $ is a closed convex subspace of $\mathcal {M}$ and, since it is non-empty, we can apply the Krein–Milman theorem to conclude that it contains an extreme point. Since the ergodic measures on $(Z,R)$ are precisely the extreme points of $\mathcal {M}$ , it remains only to show that any extreme point $\overline \mu \in \mathcal {M}_\mu $ is also extreme in $\mathcal {M}$ . Assume that $\overline \mu = s\zeta ' + (1-s)\zeta "$ , where $0<s<1$ and $\zeta '$ , $\zeta "\in \mathcal {M}$ . Let us prove that $\overline \mu $ cannot be expressed as a non-trivial convex combination of distinct elements of $\mathcal {M}$ . Then, we have

$$ \begin{align*} \mu = \overline\mu\circ\pi^{-1} = s\zeta'\circ\pi^{-1} + (1-s)\zeta"\circ\pi^{-1}. \end{align*} $$

Since $\mu $ , being ergodic, is an extreme point in the space of invariant measures of $(Y,T)$ , we conclude that $\zeta '\circ \pi ^{-1} = \zeta "\circ \pi ^{-1} = \mu $ . Therefore, $\zeta ',\zeta "\in \mathcal {M}_\mu $ and since $\overline \mu $ is an extreme point in $\mathcal {M}_\mu $ , it follows that $\zeta '=\zeta "$ . Thus, $\overline \mu $ is an extreme point in $\mathcal {M}$ .

Proof. We first assume that $\mu $ is ergodic.

For $C\subseteq G$ , let $U_{C} = C\times X$ ; in the case where $C = \{g\}$ , $g\in G$ , we write simply $U_g$ . Then, we find

$$ \begin{align*} \{g\}\times [L\cap A^i]_X=(\{g\}\times X)\cap T_{\varphi}^{-i}(\{gK\}\times X)=U_g\cap T_{\varphi}^{-i}U_{gK}. \end{align*} $$

Since the projection $G\times X\to X$ is a factor map, by Lemma 3.4, we may take an ergodic measure $\overline \mu $ on $G\rtimes _{\varphi } X$ that projects to $\mu $ . In terms of $\overline \mu $ , we have

$$ \begin{align*} \mu([L\cap A^i]) = \overline\mu(G\times [L\cap A^i]) = \sum_{g\in G} \overline\mu(\{g\}\cap [L\cap A^i]) = \sum_{g\in G} \overline\mu(U_g\cap T_{\varphi}^{-i}U_{gK}). \end{align*} $$

Since $\overline \mu $ is ergodic, we can use (2.1) to obtain

$$ \begin{align*} \delta_\mu(L) &= \lim_{n\to\infty}\frac{1}{n}\sum_{i=0}^{n-1}\mu([L\cap A^i]) \\ &= \sum_{g\in G}\lim_{n\to\infty}\frac{1}{n}\sum_{i=0}^{n-1} \overline\mu (U_g\cap T_{\varphi}^{-i}U_{gK})\\ &= \sum_{g\in G} \overline\mu (U_g)\,\overline\mu(U_{gK}). \end{align*} $$

We now only assume that $\mu $ is a shift-invariant probability measure on X. Let ${\mathcal {E} = \mathcal {E}(X,S)}$ be the set of ergodic invariant probability measures on $(X,S)$ . By e.g. [Reference Phelps49, Ch. 12], there exists a Borel probability measure $\tau $ on $\mathcal E$ such that

$$ \begin{align*} \mu=\int_{\mathcal E}\nu \,d\tau(\nu). \end{align*} $$

Then, by the dominated convergence theorem

$$ \begin{align*} \delta_{\mu}(L) = \lim_{n\to\infty}\frac1n\sum_{i=0}^{n-1}\mu([L\cap A^i]) =\lim_{n\to\infty}\frac1n\sum_{i=0}^{n-1}\!\int_{\mathcal E}\!\nu([L\cap A^i])\,d\tau(\nu) =\!\int_{\mathcal E}\! \delta_{\nu} (L) \,d\tau(\nu), \end{align*} $$

where the densities $\delta _\nu (L)$ exist by the first part of the proof.

Proof. Since $(\nu \times \mu )(U_C)=|C|/|G|$ for all $C\subseteq G$ , the above theorem with $\overline \mu =\nu \times \mu $ yields $\delta _\mu (L) = \sum _{g\in G}|K|/|G|^2=|K|/|G|$ .

Proof. Let $\zeta $ be an invariant measure on $G\rtimes _{\varphi }X$ . Observe that the measure $\zeta \circ P_X^{-1}$ , where $P_X\colon G\rtimes _{\varphi }X\to X$ denotes the projection on the second component, is an invariant measure on X; thus, it must be equal to $\mu $ as X is uniquely ergodic. In other words,

$$ \begin{align*} \zeta(G\times E) = \mu(E)\quad \text{for every measurable set } E\subseteq X. \end{align*} $$

Let $g\in G$ act on the left of $G\rtimes _{\varphi }X$ by $g(h,x) = (gh,x)$ . This action is $T_{\varphi }$ -commuting, as well as measure-preserving as is easily checked on rectangular sets (that is, sets of the form $F\times E$ , where $F \subseteq G$ , and $E\subseteq X$ is measurable). Thus, the measure $g\zeta $ defined by $g\zeta (F) = \zeta (gF)$ is also an invariant measure. We claim that the average measure $\overline \zeta = (\sum _{g\in G}g\zeta )/|G|$ is equal to $\nu \times \mu $ . Indeed, for every measurable set $E\subseteq X$ and $h\in G$ , we have

$$ \begin{align*} \overline\zeta(\{h\}\times E) = \frac{1}{|G|}\sum_{g\in G}\zeta(\{gh\}\times E) = \frac{1}{|G|}\zeta(G\times E) = \frac{1}{|G|}\mu(E). \end{align*} $$

Thus, we conclude that $\nu \times \mu = \overline \zeta $ . If $\nu \times \mu $ is ergodic, then it is an extremal point in the convex set of invariant measures of $G\rtimes _{\varphi } X$ . Since $\nu \times \mu =\overline \zeta $ is a uniform convex combination of the measures $g\zeta $ , it follows that $g\zeta = \nu \times \mu $ for all $g\in G$ .

Proof. Let $\overline \mu $ be any $T_{\varphi }$ -invariant measure on the skew product $G\rtimes _{\varphi } X$ that projects to $\mu $ (for instance, the product measure $\nu \times \mu $ , where $\nu $ is the uniform probability measure on G). For $g\in G$ , let $f_g$ be the indicator function of $\{g\}\times X$ . By the ergodic theorem applied to $G\rtimes _{\varphi }X$ , the time average function $\overline f_g$ exists on some subset $E_g \subseteq G \times X$ with $\overline \mu (E_g)=1$ .

Next, note that $\varphi ^{(i)}(x) = g$ if and only if $T_\varphi ^i(1_G,x) \in \{g\}\times X$ , so

$$ \begin{align*} \dot\delta_L(x) = \lim_{n\to\infty}\frac1n\sum_{i=0}^{n-1}{1}_K(\varphi^{(i)}(x)) =\sum_{g\in K}\lim_{n\to\infty}\frac1n\sum_{i=0}^{n-1}f_g(T_\varphi^i(1_G,x)) = \sum_{g\in K}\overline f_g(1_G,x). \end{align*} $$

Let $E = \bigcap _{g \in G} E_g$ and $F = \{ x\in X : (1_G,x)\in E\}$ . Then, $\dot \delta _L(x)$ certainly exists for all $x\in F$ by the above, so it suffices to show that $\mu (F)=1$ . However, observe that $T_\varphi ^i(h,x)\in \{g\}\times X$ is equivalent to $T_\varphi ^i(1_G,x)\in \{h^{-1}g\}\times X$ , so that

$$ \begin{align*} \overline f_g(h,x) = \lim_{n\to\infty}\frac1n\sum_{i=0}^{n-1}f_g(T_\varphi^i(h,x)) = \lim_{n\to\infty}\frac1n\sum_{i=0}^{n-1}f_{h^{-1}g}(T_\varphi^i(1_G,x)) = \overline f_{h^{-1}g}(1_G,x). \end{align*} $$

It follows that if $(h,x)\in E$ , then $\overline f_{g}(1_G,x) = \overline f_{hg}(h,x)$ for all $g\in G$ and, thus, $x\in F$ . Therefore, we have $E\subseteq G\times F$ . Since $\overline \mu $ projects to $\mu $ , we conclude that $\mu (F) = \overline \mu (G\times F) \geq \overline \mu (E)$ . However, E is a finite intersection of sets of $\overline \mu $ -measure 1; thus, $\overline \mu (E)=1$ and $\mu (F)=1$ .

For the last part of the statement, we first note that for $w\in A^i$ , we have $\varphi ^{(i)}(x)\in K$ if and only if $x\in [w]$ for some $w\in L\cap A^i$ , and thus,

$$ \begin{align*} \int_X {1}_K(\varphi^{(i)}(x))\,d\mu = \sum_{w\in L\cap A^i} \int_X{1}_{[w]}(x)\,d\mu = \mu([L\cap A^i]). \end{align*} $$

Hence, we have

$$ \begin{align*} \begin{aligned} \int_X \dot\delta_L(x)\,d\mu &= \int_X \lim_{n\to\infty}\frac1n\sum_{i=0}^{n-1} 1_K(\varphi^{(i)}(x))\,d\mu\\ &= \lim_{n\to\infty}\frac1n\sum_{i=0}^{n-1}\int_X 1_K(\varphi^{(i)}(x))\,d\mu\\ &= \lim_{n\to\infty}\frac1n\sum_{i=0}^{n-1}\mu([L\cap A^i]) = \delta_\mu(L), \end{aligned} \end{align*} $$

where the integral and limit can be swapped by the dominated convergence theorem.

Proof. Take the substitution $\overline \sigma $ defined as follows on the alphabet $\mathbb {Z}/2\mathbb {Z}\times A$ , whose letters are denoted $a_0, a_1, b_0, b_1$ for conciseness,

$$ \begin{align*} \begin{array}{lll} \overline\sigma\colon & a_0\mapsto a_0b_1a_0a_1b_1,\quad & a_1\mapsto a_1b_0a_0a_1b_0, \\ & b_0 \mapsto a_0b_1a_1,\quad & b_1 \mapsto a_1b_0a_0. \end{array} \end{align*} $$

This substitution is primitive and satisfies $\pi \circ \overline \sigma =\sigma ^3$ , where $\pi \colon (\mathbb {Z}/2\mathbb {Z}\times A)^*\to A^*$ is the natural projection (mapping $a_i$ to a and $b_i$ to b). Moreover, $\varphi \circ \sigma ^3 = \varphi $ and $\overline \sigma (\mathcal {L}(Y))\subseteq \mathcal {L}(Y)$ , where $Y = \Psi (G\rtimes _{\varphi } X)$ is the skew product viewed as a shift space on $\mathbb {Z}/2\mathbb {Z}\times A$ via the map $\Psi $ of Lemma 3.1. It follows that Y is the shift space generated by $\overline \sigma $ . Therefore, by Michel’s theorem, Y is uniquely ergodic; hence, so is $G\rtimes _{\varphi } X$ .

Proof. (i) As u is left special, then $au$ and $bu$ are in $\mathcal {L}(X)$ . Let c and d be right extensions of respectively $au$ and $bu$ ; let $c'$ and $d'$ be such that $\sigma ^2(c) = abc'$ and $\sigma ^2(d) = abd'$ ( $c'$ and $d'$ are either a or $\varepsilon $ ). Then, the following words also belong to $\mathcal {L}(X)$ :

$$ \begin{align*} \sigma^2(auc) &= aba\sigma^2(u)abc' = abau'c',\\ \sigma^2(bud) &= ab\sigma^2(u)abd' = abu'd'. \end{align*} $$

In particular, $au'$ and $bu'$ belong to $\mathcal {L}(X)$ .

(ii) The proof of the second part follows similar lines. Let c and d be right extensions of respectively $va$ , $vb$ (in fact, $d=a$ since $bb$ does not occur in X); let $c'$ be such that $\sigma ^2(c) = ac'$ . Then, we find that the following words belong to $\mathcal {L}(X)$ :

$$ \begin{align*} \sigma^2(vac) = v'ac',\quad \sigma^2(vba) = \sigma^2(v)ababa = v'ba. \end{align*} $$

Hence, both $v'a$ , $v'b\in \mathcal {L}(X)$ .

Proof. Let us first prove that $bu_n = v_nb$ when $n=F(2k+2)-1$ , $k\geq 0$ . Consider the following recursively defined sequence of words:

$$ \begin{align*} w_0 = \varepsilon,\quad w_{k+1} = \sigma^2(w_kb). \end{align*} $$

In other words, this is the sequence of words starting with

$$ \begin{align*} \varepsilon,\quad \sigma^2(b),\quad \sigma^4(b)\sigma^2(b),\quad \sigma^6(b)\sigma^4(b)\sigma^2(b),\quad\ldots. \end{align*} $$

It is clear from Lemma 3.19 that this is a sequence of left special factors of X. We claim that $|w_k| = F(2k+2) - 1$ for every $k\geq 0$ . Indeed, observe that, for every $i\in \mathbb {N}$ , $|\sigma ^{2i}(b)| = F(2i+1)$ (a fact easily established by induction) and thus,

$$ \begin{align*} 1+|w_k| = 1+\sum_{i=1}^k|\sigma^{2i}(b)| = \sum_{i=0}^kF(2i+1) = F(2k+2). \end{align*} $$

Thus, we are reduced to show that $bw_k = v_nb$ , where $n = F(2k+2)-1$ ; or in other words that removing the last letter from $bw_k$ yields a right special factor. We do so by induction on k. The basis $k=0$ is trivial since $w_0 = \varepsilon $ . Assume that the equality ${bw_k = v_nb}$ holds for some $k\geq 0$ . Then, we have

$$ \begin{align*} abw_{k+1}a = ab\sigma^2(w_kb)a = \sigma^2(bw_kb)a = \sigma^2(v_nbb)a = \sigma^2(v_na)ba. \end{align*} $$

By Lemma 3.19, the word $\sigma ^2(v_na)$ is right special and, thus, so is the word $a^{-1}\sigma ^2(v_na)$ obtained by removing its leading letter a. Finally, we observe that $a^{-1}\sigma ^2(v_na)b = bw_{k+1}$ and, thus, $bw_{k+1}=v_{m}b$ , where $m=F(2k+4)-1$ .

Next, let us fix $k\geq 0$ and let $n=F(2k+2)-1$ . Then, by the first part of the proof, we know that $w_k = u_n$ , $bu_n=v_nb$ , and then

$$ \begin{align*} \varphi(u_n) = \varphi(bu_n) = \varphi(v_nb) = \varphi(v_n). \end{align*} $$

We need to show that $\varphi (w_k) \equiv k \bmod 2$ . We do so by induction on k. Since $w_0 = \varepsilon $ and $w_1 = \sigma ^2(b)=ab$ , the result is clear for $k=0,1$ . Assume then that the result holds for some $k\geq 0$ . By definition, $w_{k+2} = \sigma ^4(w_{k}b)\sigma ^2(b) =\sigma ^4(w_kb)ab$ . However, observe that

$$ \begin{align*} \sigma^4(a) = abaababa \quad\text{and}\quad \sigma^4(b)=abaab. \end{align*} $$

Thus, we have the relation $\varphi (\sigma ^4(x)) \equiv |x| \bmod 2$ for all $x\in A^*$ . By induction, we then have

$$ \begin{align*} \varphi(w_{k+2}) {\kern-1pt}={\kern-1pt}\varphi(\sigma^4(w_kb)ab) {\kern-1pt}={\kern-1pt} \varphi(\sigma^4(w_kb))+\varphi(ab) {\kern-1pt}\equiv{\kern-1pt} |w_kb| {\kern-1pt}+{\kern-1pt}1 {\kern-1pt}\equiv{\kern-1pt} |w_k|{\kern-1pt}+{\kern-1pt}2 {\kern-1pt}\equiv{\kern-1pt} k{\kern-1pt}+{\kern-1pt}2\bmod 2, \end{align*} $$

which concludes the proof.

Proof of Proposition 3.18.

Recall that $L = \{w\in \{a,b\}^* : |w|_a\equiv 0 \bmod 2\} = \varphi ^{-1}(0)$ , where $\varphi \colon A^*\to \mathbb {Z}/2\mathbb {Z}$ is the morphism defined by $\varphi (a)=1$ , $\varphi (b)=0$ . Let $\leq _{\mathrm {lex}}$ denote the lexicographic order on words. It follows immediately from the description of $\leq _{\mathrm {lex}}$ on $\mathcal {L}(X)\cap A^n$ by Perrin and Restivo [Reference Perrin and Restivo47, Theorem 2] that for every $w\in \mathcal {L}(X)$ with $|w|=n+1$ ,

$$ \begin{align*} \varphi(w) = \begin{cases} \varphi(v_{n}a) & \text{if}\ w\leq_{\mathrm{lex}} v_{n}a,\\ \varphi(v_{n}b) & \text{if}\ w\geq_{\mathrm{lex}} v_{n}b. \end{cases} \end{align*} $$

Moreover, [Reference Perrin and Restivo47, Proposition 2] states that the lexicographically maximal element of $\mathcal {L}(X)\cap A^{n+1}$ is $bu_{n}$ . In particular, if $bu_n = v_nb$ , then $v_nb$ is maximal in $\mathcal {L}(X)\cap A^{n+1}$ and for such values of n,

$$ \begin{align*} [L\cap A^{n+1}] = \begin{cases} [v_nb] & \text{if }\varphi(v_nb)=0, \\ X\setminus[v_nb] & \text{if }\varphi(v_nb)=1. \end{cases} \end{align*} $$

However, by Lemma 3.20, we know that $\varphi (v_nb) = \varphi (v_n)\equiv k \bmod 2$ , where $n = F(2k+2)-1$ , and

$$ \begin{align*} \mu([L\cap A^{n+1}]) = \begin{cases} \mu([v_nb]) & \text{if } k \text{ is even,}\\ 1-\mu([v_nb]) & \text{if } k \text{ is odd.} \end{cases} \end{align*} $$

Finally, k is even precisely when $n+1 = F(4\ell +2)$ for some $\ell $ and k is odd precisely when $n+1 = F(4\ell )$ for some $\ell $ . We conclude the proof by observing that $\lim _{n\to \infty }\mu ([v_n]) = 0$ , which is a straightforward consequence of [Reference Durand24, Proposition 13].

Proof. Let $r\geq 1$ such that X is an r-step SFT. Take $g\in G$ . Since $\varphi $ is onto, we may find letters $a_1,\ldots , a_k\in A$ such that $\varphi (a_1\cdots a_k) = g$ . For $i=1,\ldots ,k$ , let $u_i$ be a word of length $r+1$ in $\mathcal {L}(X)$ starting with $a_i$ . Let $t_i$ and $v_i$ be the suffix and prefix of length r of $u_i$ . By assumption, there exists, for each $i = 1,\ldots k-1$ , a word $w_i$ such that $t_iw_iv_{i+1}\in \mathcal {L}(X)$ and $\varphi (t_iw_i)=1_G$ . Since X is an r-step shift, it follows that the word $z = u_1w_1\cdots w_{k-1}u_kw_k$ belongs to $\mathcal {L}(X)$ and

$$ \begin{align*} \varphi(z) = \varphi(a_1)\varphi(t_1w_1)\cdots\varphi(a_k)\varphi(t_kw_k) = \varphi(a_1\cdots a_k) = g, \end{align*} $$

which proves that X is fiber ergodic.

Proof of Theorem 5.3.

For $g\in G$ , define a relation $\prec _g$ on $\mathcal {L}(X)$ by $u\prec _gv$ if there exists $w\in \mathcal {L}(X)$ such that $uwv\in \mathcal {L}(X)$ and $\varphi (uw)=g$ . Observe that $u\prec _gv$ precisely when $T^m(\{1_G\}\times [u]_X)$ intersects $\{g\}\times [v]_X$ for some $m\geq |u|$ . Therefore, $G\rtimes _{\varphi } X$ is topologically transitive precisely when all relations $\prec _g$ , $g\in G$ , are total. In particular, whenever this is the case, $\prec _{1_G}$ must contain all pairs $u,v\in \mathcal {L}(X)$ , which is precisely the definition of $\varphi $ -irreducibility. Thus, topological transitivity of $G\rtimes _{\varphi } X$ implies $\varphi $ -irreducibility of X. It remains to prove the converse.

Assume that X is $\varphi $ -irreducible, so that ${\prec } = {\prec _{1_G}}$ contains all pairs of words in $\mathcal {L}(X)$ . Take $u,v\in \mathcal {L}(X)$ and $g\in G$ ; we need to show that $u\prec _g v$ .

By fiber ergodicity (which holds thanks to Lemma 5.5), there is $u' \in \mathcal {L}(X)$ such that $\varphi (u')=g$ . Since $u\prec u'$ , there is $z\in \mathcal {L}(X)$ such that $uzu'\in \mathcal {L}(X)$ and $\varphi (uz)=1_G$ . Then, $w_0=zu'$ satisfies $uw_0 \in \mathcal {L}(X)$ and $\varphi (uw_0)=g$ .

Extend $uw_0$ to a word $wu_0v' \in \mathcal {L}(X)$ with $|v'| \geq r+1$ . Since $v'\prec v$ , there is a word $v"$ such that $v'v"v \in \mathcal {L}(X)$ and $\varphi (v'v")=1_G$ . Then, all subwords of length $r+1$ of $uw_0v'v"v$ are in $\mathcal {L}(X)$ and, hence, $uw_0v'v"v \in \mathcal {L}(X)$ . Finally, letting $w=w_0v'v"$ , we have $uwv \in \mathcal {L}(X)$ and $\varphi (uw)=g$ . This shows that $u\prec _g v$ , which concludes the proof.

Proof. Let $u,v\in \mathcal {L}(X)$ . If $|v|<r$ , then we may simply replace v by one of its extensions in $\mathcal {L}(X)$ of length r. If $|v|>r$ , then we may likewise replace it by its prefix of length r. Thus, we may assume moving forward that $|v|=r$ .

First, we assume that $|u|>r$ . Let $u = pq = p'q'$ , where $|q| = |p'| = r$ . By assumption, there are words $w,w'$ such that $qw'p', qwv\in \mathcal {L}(X)$ and $\varphi (qw) = \varphi (qw') = 1_G$ . Observe that, for every $n\in \mathbb {N}$ , $u(w'u)^nwv$ has all of its factors of length r in $\mathcal {L}(X)$ . Thus, the word $z_n = (w'u)^nw$ is such that $uz_nv\in \mathcal {L}(X)$ , while $\varphi (uz_n) = \varphi (p)^n$ . Taking $n = |G|$ , we get $\varphi (uz_n) = 1_G$ , as needed.

It remains to handle the case where $|u|<r$ . Take a word p such that $|p|=r-|u|$ and $pu\in \mathcal {L}(X)$ . Then, we may find words $w, w'$ such that $puw'uwv\in \mathcal {L}(X)$ with ${\varphi (puw') = 1_G}$ and $\varphi (puw'uw) = 1_G$ ; thus, $\varphi (uw) = 1_G$ .

Proof. (i) Take a word $\omega \in (G\times A)^*$ with $\omega _i = (g_i,w_i)$ and $m=|\omega |\geq r+1$ . Observe that $\omega $ belongs to the language of $\Psi (G\rtimes _{\varphi } X)$ as long as $w = w_0\cdots w_{m-1}$ belongs to $\mathcal {L}(X)$ and $g_i\varphi (w_i) = g_{i+1}$ . Both of those conditions only need to be verified on factors of length $r+1$ of $\omega $ ; hence, $\Psi (G\rtimes _{\varphi } X)$ is indeed an r-step SFT.

(ii) Fix an r-step Markov measure on X and let $\pi = (\nu \times \mu )\circ \Psi ^{-1}$ . Fix a word $\omega \in (G\times A)^*$ , with $\omega _i = (g_i,w_i)$ , $m=|\omega |\geq r$ , and $w = w_0\cdots w_{m-1}$ . Assume that $\omega $ belongs to the language of $\Psi (G\rtimes _{\varphi } X)$ , which means that $w\in \mathcal {L}(X)$ and $g_{i+1} = g_i\varphi (w_i)$ , $i=0,\ldots ,m-2$ . Take a letter $\alpha = (g,a)\in G\times A$ . In case $g\neq g_{m-1}\varphi (a)$ , then it is clear that

$$ \begin{align*} \pi(\xi_m=\alpha \mid \xi_{[0,m)} = \omega) = \pi(\xi_m=\alpha \mid \xi_{[m-r,m)} = \omega_{[m-r,m)}) = 0. \end{align*} $$

Thus, we may suppose from now on that $g=g_{m-1}\varphi (a)$ . Then, we have

$$ \begin{align*} \pi( \xi_m = \alpha \mid \xi_{[0, m)} = \omega ) &= \frac{\pi( \xi_m = \alpha,\ \xi_{[0, m)} = \omega )}{\pi( \xi_{[0, m)} = \omega )} \\ &= \frac{(\nu\times\mu)(\{g_0\}\times [wa])}{(\nu\times\mu)(\{g_0\}\times [w])} \\ &= \frac{\mu(x_m = a,\ x_{[0, m)} = w_{[0, m)})}{\mu(x_{[0,m)} = w)} \\ &= \mu( x_m = a \mid x_{[0,m)}=w ). \end{align*} $$

Using a similar argument together with the invariance of $\pi $ , we find

$$ \begin{align*} \pi( \xi_m = \alpha \mid \xi_{[m-r,m)} = \omega_{[m-r,m)}) = \mu( x_m = a \mid x_{[m-r,m)}=w_{[m-r,m)}). \end{align*} $$

The fact that $\pi $ is r-step Markov now follows directly from the fact that $\mu $ is.

Proof of Theorem 1.2.

In light of Lemma 5.10, the measure $(\nu \times \mu )\circ \Psi ^{-1}$ is an r-step Markov measure fully supported on a SFT, which is topologically transitive by Theorem 5.3. Therefore, $(\nu \times \mu )\circ \Psi ^{-1}$ must be ergodic and since $\Psi $ is a topological conjugacy, we deduce that the measure $\nu \times \mu $ is ergodic as well.

The last part of the statement is then an immediate consequence of Corollary 3.6.

Proof. Let X be an irreducible r-step SFT. We start by showing that given $u,v \in \mathcal {L}(X)$ with $u \sim _r v$ , there is a word w such that $|w|\geq r$ , $uwv \in \mathcal {L}(X)$ , and $\varphi (uw)=1_G$ . By definition of $\sim _r$ , there is a word z such that $|z|=r$ and $zu,zv \in \mathcal {L}(X)$ . Let $t\in A^*$ be a word such that $zut\in \mathcal {L}(X)$ and $|ut|\geq r$ . Since X is irreducible, there exists a word $w_0\in A^*$ such that $utw_0zut \in \mathcal {L}(X)$ . Then, $(utw_0z)^nu, z(utw_0z)^nv \in \mathcal {L}(X)$ for all $n \geq 1$ , since every factor of length r in either of those words is a factor of either $zv$ , $zut$ , or $utw_0z$ , all of which are in $\mathcal {L}(X)$ . Taking $n=|G|$ , we find $\varphi ((utw_0z)^n) = 1_G$ . The word $w=(tw_0zu)^{n-1}tw_0z$ then satisfies $uw = (tw_0zu)^{n}$ , and, thus, we get $uwv\in \mathcal {L}(X)$ and $\varphi (uw)=1_G$ , as required.

Suppose next $u = u_0 \sim _r u_1 \sim _r \cdots \sim _r u_{n-1} \sim _r u_n = v$ . We want to find a word w such that $uwv \in \mathcal {L}(X)$ and $\varphi (uw)=1_G$ . Using the above claim, there are words $w_i$ such that $u_iw_iu_{i+1} \in \mathcal {L}(X)$ , $\varphi (u_iw_i)=1_G$ , and $|w_i| \geq r$ . Then, defining $w=w_0u_1w_1\cdots u_{n-1}w_{n-1}$ produces the requisite word such that $uwv\in \mathcal {L}(X)$ and ${\varphi (uw)=1_G}$ .

Proof. For the first part of the statement, note that when X is strongly irreducible, then by Proposition 5.14, the relation ${\prec } = {\prec _{1_G}}$ must be total for every morphism $\varphi \colon A^*\to G$ onto a finite group G, which is precisely the definition of X being $\varphi $ -irreducible.

We now suppose that $r=1$ . We prove the contrapositive: assuming that X is not strongly irreducible, then it is not $\varphi $ -irreducible for some $\varphi $ . The construction presented is essentially the same as that used in the proof of [Reference Lummerzheim, Pogorzelski and Zimmermann40, Theorem 4.3].

Fix an equivalence class C of $\simeq _1$ restricted to $A\times A$ . For each $a\in A$ , observe that the set of right extensions of a in X,

$$ \begin{align*} \operatorname{R}(a) = \{b\in A \mid ab\in\mathcal{L}(X)\}, \end{align*} $$

is contained in a class of $\simeq _1$ ; thus, either $\operatorname {R}(a)\subseteq C$ or $\operatorname {R}(a)\subseteq A\setminus C$ . Let

$$ \begin{align*} B = \{ a\in C \mid \operatorname{R}(a) \subseteq A\setminus C \} \cup \{ a\in A\setminus C \mid \operatorname{R}(a) \subseteq C \}. \end{align*} $$

Define a morphism $\varphi \colon A^*\to \mathbb {Z}/2\mathbb {Z}$ by $\varphi (a) = 1$ if $a\in B$ and $\varphi (a) = 0$ otherwise. Let $a\in C$ and $b\in A$ ; take a word w such that $awb\in \mathcal {L}(X)$ . We claim that $\varphi (aw) = 0$ if and only if $b\in C$ . Observe that this claim finishes the proof, as it implies that $a\not \prec b$ whenever $b\in A\setminus C$ .

To establish the claim, we argue by induction on $|w|$ . We first consider the case $|w|=0$ . We have to check that $\varphi (a)=0$ if and only if $b\in C$ . However, $\varphi (a)=0$ if and only if ${a \not \in B}$ , which is in turn equivalent to $\operatorname {R}(a) \subseteq C$ , that is, $b \in C$ . For the induction step, suppose that $w = w'c$ , $c\in A$ , with $aw'b \in \mathcal {L}(X)$ , and that the claim holds for $w'$ . First, assume that $\varphi (aw)=0$ . If $c\in C$ , then by induction, $\varphi (aw')=0$ and $\varphi (aw) = \varphi (c) = 0$ , and as ${b\in \operatorname {R}(c)}$ , we get $b\in C$ . If, however, $c\in A\setminus C$ , then $\varphi (aw') = 1$ , since the claim holds for $w'$ and $ c \in A \setminus C$ ; hence, $\varphi (aw) = 1+\varphi (c) = 0$ , and so $\varphi (c) = 1$ and $c \not \in B$ . As $b\in \operatorname {R}(c)$ , it again follows that $b\in C$ . Conversely, assume that $b\in C$ . If $c\in C$ , then as $b\in \operatorname {R}(c)$ , we get $c \not \in B$ and $\varphi (c) = 0$ , while by induction, $\varphi (aw') = 0$ ; hence, $\varphi (w) = 0+0 = 0$ . Likewise, if $c\in A\setminus C$ , then $c\in B$ , and $\varphi (aw') = 1 = \varphi (c)$ and $\varphi (aw) = 1+1 = 0$ .