Proof. For any subshift X and any n, define an n-right branch word to be a word in $L(X)$ beginning with a word in $RS_n(X)$ , containing no other word in $RS_n(X)$ , containing no repeated n-letter subwords, and which is maximal with respect to subword inclusion subject to these constraints. Similarly, define an n-left branch word to be a word in $L(X)$ ending with a word in $LS_n(X)$ , containing no other word in $LS_n(X)$ , with no repeated n-letter subwords, and which is maximal with respect to subword inclusion subject to these constraints. An n-branch word is any word that is either an n-left or n-right branch word.

The proof relies on the following three facts about n-branch words.

1. (1) For every n, the number of n-branch words is less than or equal to the quantity $2|A| (c_{n+1}(X) - c_n(X))$ .

2. (2) For every n, each n-branch word has length less than $n + c_n(X)$ .

3. (3) Suppose $\phi _1$ and $\phi _2$ are automorphisms with range $\lfloor (n-1)/2 \rfloor $ induced by block codes $\Phi _{1}, \Phi _{2}$ respectively such that $\phi _{1}, \phi _{2}$ fix all isolated periodic points, and $\Phi _1(w) = \Phi _2(w)$ for all n-branch words w. Then $\phi _1 = \phi _2$ .

Proof of fact (1). Each n-right branch word w is determined completely by its initial word in $RS_n(X)$ and the following letter; then, since w contains no other words in $RS_n(X)$ , each n-letter subword determines the next letter, meaning that all of w is forced. There are obviously at most $|A| |RS_n(X)|$ choices for this initial word and following letter, which is less than or equal to $|A| (c_{n+1}(X) - c_n(X))$ by Lemma 2.9. A similar bound holds for n-left branch words, implying fact (1).

Proof of fact (2). Every n-branch word contains no repeated n-letter subwords, and so contains at most $c_n(X) n$ -letter subwords. This clearly implies that such a word has length less than $n + c_n(X)$ .

Proof of fact (3). We claim that every $w \kern-1pt\in\kern-1pt L_n(X)$ which is not the subword of any n-branch word must be a subword of an isolated periodic point of X. To see this, assume that $w \in L_n(X)$ is not a subword of any n-branch word.

Choose any $x \in X$ with $x([0, n)) = w$ . Define m to be the minimal integer greater than n so that there exists $n \leq i < m$ for which $x([i-n, i)) = x([m-n, m))$ , that is the first place, when moving to the right from $x(0)$ , where an n-letter word appears for the second time. Choose such an i, and suppose $i> n$ . Then $x(i-n-1) \neq x(m-n-1)$ , since otherwise, $x([i-n-1, i-1)) = x([m-n-1, m-1))$ , violating minimality of m. This would imply that $x([i-n, i))$ is a left-special word, and by minimality of m, that $x([0,i))$ is a word ending with a word in $LS_n(X)$ with no repeated n-letter subwords. We could then extend $x([0,i))$ to the left to create a maximal such word $x([j,i))$ , which is an n-left branch word by definition. This n-left branch word would contain $w = x([0,n))$ as a subword, a contradiction.

Therefore $i = n$ , that is $x([0, m))$ begins and ends with w. If $x([0, m))$ contained any words in $LS_n(X)$ , then just as before we could construct an n-left branch word containing w, a contradiction. We have then shown that $x([0,m))$ begins and ends with w and contains no subwords in $LS_n(X)$ . Therefore, the right-most occurrence of w in $x([0,m))$ forces letters to the left until the left-most occurrence of w, and this continues indefinitely. In other words, every $y \in X$ with $y([0,m)) = x([0,m))$ in fact has $y((-\infty , m))$ periodic with period $m-n$ .

A similar argument shows that $x([0,m))$ cannot contain any words in $RS_n(X)$ either; if $j \geq 0$ were minimal so that $x([j, m))$ begins with a word in $RS_n(X)$ , then $x([j,m))$ could be extended to the right to create an n-right branch word containing $x([m-n, m)) = w$ , a contradiction. So $x([0,m))$ contains no words in $RS_n(X)$ , meaning that the left-most occurrence of w forces letters to the right until the right-most occurrence. It follows that if $y \in X$ satisfies $y([0,m)) = x([0,m))$ , then $y([0, \infty ))$ is periodic with period $m-n$ .

Altogether, what we have shown is that every $y \in X$ with $y([0, m)) = x([0, m))$ is a periodic point with period $m-n$ coming from biinfinite repetition of $x([0, m-n))$ . Therefore, x is an isolated periodic point, verifying the claim that every $w \in L_n(X)$ which is not the subword of any n-branch word must be a subword of an isolated periodic point.

Now, choose any $\phi _1,\phi _2 \in \mathrm {Aut}^{(\textit{FIP})}(X, \sigma )$ with range $\lfloor (n-1)/2 \rfloor $ and inducing block maps $\Phi _1$ and $\Phi _2$ , and assume that $\Phi _{1}(v) = \Phi _{2}(v)$ for all n-branch words v. Define $n' = 2 \lfloor (n-1)/2 \rfloor + 1$ , so that $\Phi _1$ and $\Phi _2$ have domain $\mathcal {A}^{n'}$ ; clearly $n' \leq n$ . Since $\phi _1$ and $\phi _2$ fix isolated periodic points, for all $n'$ -letter subwords u of such points, $\Phi _1(u) = \Phi _2(u)$ . Choose any $w \in L_{n'}(X)$ which is not a subword of such a point; since $n' \leq n$ , by the above, it is a subword of an n-branch word v. Now, since $\Phi _{1}(v) = \Phi _{2}(v)$ and w is a subword of v, $\Phi _1(w) = \Phi _2(w)$ . We now know that $\Phi _1$ and $\Phi _2$ agree on all words in $L_{n'}(X)$ , so $\Phi _1 = \Phi _2$ , meaning that $\phi _1 = \phi _2$ .

By fact (3), the number of automorphisms of range $\lfloor (n-1)/2 \rfloor $ which fix isolated periodic points is bounded from above by the number of possible choices for $\Phi (w)$ for all n-branch words w. Each $\Phi (w)$ is determined by the length of w (which is independent of $\phi $ ) and some word of length $|w| - 2\lfloor (n-1)/2 \rfloor \leq |w| - n + 2$ . By fact (2), the number of such words is less than or equal to $c_{1+c_n(X)}(X)$ . By fact (1), the number of w is bounded by $2|A| (c_{n+1}(X) - c_n(X))$ , completing the proof.

Proof. We first note that the hypothesis immediately implies that there exist infinitely many n where $f(n) - f(n-1) < g(n)$ ; if not, then there would be N where $f(n) - f(n-1) \geq g(n)$ for all $n> N$ , meaning that

$$ \begin{align*} f(n) = f(N) + \sum_{i=N+1}^{n} f(i) - f(i-1) \geq (f(N) - \sum_{i=1}^{N} g(i)) + \sum_{i=1}^n g(i)\ \ \ \mathrm{for\ all }\ n> N, \end{align*} $$

a contradiction to the assumption.

We now break into two cases. First, suppose that there exists N so that $f(n) < \sum _{i=1}^n g(i)$ for $n> N$ . Combining with the previous paragraph then yields the conclusion of the lemma.

Now, suppose that there exist infinitely many n where $f(n) \geq \sum _{i = 1}^{n} g(i)$ . The hypothesis of the lemma implies that there are also infinitely many n where $f(n) < \sum _{i=1}^n g(i)$ . This implies that there are infinitely many n where $f(n-1) \geq \sum _{i=1}^{n-1} g(i)$ and $f(n) < \sum _{i=1}^{n} g(i)$ (that is the sign of the inequality ‘switches infinitely many times’). However, for any such n,

$$ \begin{align*} f(n) - f(n-1) < \sum_{i=1}^{n} g(i) - \sum_{i=1}^{n-1} g(i) = g(n), \end{align*} $$

completing the proof.

Proof of Theorem 1.1

Choose any infinite transitive subshift X with $\liminf (\log (c_n(X)/n) / (\log \log \log n)) = 0$ , and take $\epsilon> 0$ where $5\epsilon $ is less than the constant c from Theorem 2.26. We first claim that

(1) $$ \begin{align} \liminf c_n(X) - \sum_{i = 2}^n \lfloor (\log \log i)^{\epsilon} \rfloor = -\infty. \end{align} $$

To see this, by assumption, there are infinitely many n where $c_n(X) < n (\log \log n)^{\epsilon /2}$ , which is less than $(n/3) (\log \log (n/2))^{\epsilon }$ for large enough n. Also, $\sum _{i = 2}^n \lfloor (\log \log i)^{\epsilon } \rfloor \geq \sum _{i = \lceil n/2 \rceil }^n \lfloor (\log \log i)^{\epsilon } \rfloor \geq (n/2) \lfloor (\log \log (n/2))^{\epsilon } \rfloor $ , and so for infinitely many n, $c_n(X) - \sum _{i = 2}^n \lfloor (\log \log i)^{\epsilon } \rfloor $ is less than

$$ \begin{align*} (n/3) (\log \log (n/2))^{\epsilon} - (n/2) \lfloor (\log \log (n/2))^{\epsilon} \rfloor \leq n/2 - (n/6) (\log \log (n/2))^{\epsilon}, \end{align*} $$

which approaches $-\infty $ , verifying equation (1). We now apply Lemma 3.3, and see that there exist infinitely many n for which

(2) $$ \begin{align} c_n(X) < \sum_{i = 2}^n \lfloor (\log \log i)^{\epsilon} \rfloor < n \lfloor (\log \log n)^{\epsilon} \rfloor \quad\textrm{and}\quad c_n(X) - c_{n-1}(X) < \lfloor (\log \log n)^{\epsilon} \rfloor. \end{align} $$

Now, by Corollary 3.1, for any n satisfying equation (2), $|\mathrm {Aut}_{\lfloor (n-1)/2 \rfloor }(X, \sigma )|$ is bounded from above by

$$ \begin{align*} (c_{1+c_n(X)}(X))^{2|A| (c_{n+1}(X) - c_n(X))} \leq (c_{n\lfloor (\log \log n)^{\epsilon}\rfloor}(X))^{2|A| (\log \log n)^{\epsilon}}. \end{align*} $$

By subadditivity,

$$ \begin{align*} c_{n\lfloor (\log \log n)^{\epsilon}\rfloor}(X) \leq c_n(X)^{\lfloor (\log \log n)^{\epsilon} \rfloor} < ((n\log \log n)^{\epsilon})^{(\log \log n)^{\epsilon}} \end{align*} $$

$$ \begin{align*} \ \, \leq (n^{2})^{(\log \log n)^{\epsilon}}\! = n^{2 (\log \log n)^{\epsilon}}. \end{align*} $$

Therefore,

(3) $$ \begin{align} |\mathrm{Aut}_{\lfloor (n-1)/2 \rfloor}(X, \sigma)| \leq (n^{2 (\log \log n)^{\epsilon}})^{2|A| (\log \log n)^{\epsilon}} \!= n^{4|A| (\log \log n)^{2\epsilon}} \end{align} $$

holds for any of the (infinitely many) n satisfying equation (2).

Now, choose any finite subset S of ${{\mathrm {Aut}(X,\sigma )}/{\langle \sigma \rangle }}$ and let $S^{\prime }$ be a finite set in $\mathrm {Aut}(X,\sigma )$ whose image under the quotient map $\mathrm {Aut}(X,\sigma ) \to {{\mathrm {Aut}(X,\sigma )}/{\langle \sigma \rangle }}$ is the set S. Suppose that k is large enough that all automorphisms in $S'$ and their inverses have range k. Then by additivity of ranges of automorphisms under composition, any composition of m elements of $S'$ is an automorphism of range $km$ .

Then for any n for which equation (2) holds, equation (3) implies that the number $B_{\lfloor (n-1)/2 \rfloor /k}(S')$ of compositions of at most $\lfloor (n-1)/2 \rfloor /k$ elements of $S'$ satisfies

$$ \begin{align*} |B_{\lfloor (n-1)/2 \rfloor/k}(S')| \leq n^{4|A| (\log \log n)^{2\epsilon}}. \end{align*} $$

Since equation (2) holds for infinitely many n, we may choose such an n greater than $e^{e^3}$ , $9k^2$ , $e^{2e^{(8|A|/5\epsilon )^{\epsilon ^{-1}}}}$ , and $c^{-2}$ (here c is as in Theorem 2.26). Then $\lfloor (n-1)/2 \rfloor /k> n/3k > \sqrt {n}$ , and $\log \log \sqrt {n} = \log \log n - \log 2> \sqrt {\log \log n}$ since $\log \log n> 3$ , so

$$ \begin{align*} |B_{\sqrt{n}}(S')| \leq n^{4|A| (\log \log n)^{2\epsilon}} \leq \sqrt{n}^{8|A| (\log \log \sqrt{n})^{4\epsilon}}. \end{align*} $$

Since $n> e^{2e^{(8|A|/5\epsilon )^{\epsilon ^{-1}}}}$ , then $8|A| < 5 \epsilon (\log \log \sqrt {n})^{\epsilon }$ , and so

$$ \begin{align*} |B_{\sqrt{n}}(S')| < \sqrt{n}^{5\epsilon (\log \log \sqrt{n})^{5\epsilon}}. \end{align*} $$

Finally, since $\sqrt {n}> c^{-1}$ , by Theorem 2.26, $\langle S' \rangle $ is virtually nilpotent.

Therefore, $\langle S \rangle = {\langle S' \rangle }/{\langle \sigma \rangle }$ is a quotient group of a virtually nilpotent group and so itself virtually nilpotent. Let H be a finite index nilpotent subgroup of $\langle S \rangle $ ; it is finitely generated as it is a finite index subgroup of a finitely generated group. By Theorem 2.21, ${{\mathrm {Aut}(X,\sigma )}/{\langle \sigma \rangle }}$ is periodic, so H is also periodic. Altogether we have that H is finitely generated, periodic, and nilpotent, and therefore finite, implying that $\langle S \rangle $ is finite as well. Since S was an arbitrary finite subset of ${{\mathrm {Aut}(X,\sigma )}/{\langle \sigma \rangle }}$ , we have shown that ${{\mathrm {Aut}(X,\sigma )}/{\langle \sigma \rangle }}$ is locally finite.

Proof of Theorem 1.2

Assume that the gap conjecture holds. We change almost nothing about the proof of Theorem 1.1, but must simply change our estimates for the usage of Lemma 3.3.

Choose any infinite transitive subshift X where $\liminf ({c_n(X)}/{n^{1.25} (\log n)^{-0.5}}) = 0$ . We first claim that for any $\epsilon> 0$ ,

(4) $$ \begin{align} \liminf c_n(X) - \sum_{i = 2}^n \lfloor \epsilon i^{0.25} (\log i)^{-0.5} \rfloor = -\infty. \end{align} $$

To see this, note that by assumption, there are infinitely many n where $c_n(X) < ({\epsilon }/{3}) n^{1.25} (\log n)^{-0.5}$ . Also, $\sum _{i = 2}^n \lfloor \epsilon i^{0.25} (\log i)^{-0.5} \rfloor \geq \sum _{i = \lceil n/2 \rceil }^n \lfloor \epsilon i^{0.25} (\log i)^{-0.5} \rfloor \geq (n/2) \lfloor \epsilon (n/2)^{0.25} (\log n)^{-0.5} \rfloor \geq ({\epsilon }/{2^{1.25}}) n^{1.25} (\log n)^{-0.5} - {n}/{2}$ . So, for infinitely many n,

$$ \begin{align*} c_n(X) - \sum_{i = 2}^n \lfloor \epsilon i^{0.25} (\log i)^{-0.5} \rfloor < \frac{\epsilon }{3} n^{1.25} (\log (n/2))^{-0.5} - \frac{\epsilon }{2^{1.25}} n^{1.25} (\log (n/2))^{-0.5} + \frac{n}{2}. \end{align*} $$

Since this last term approaches $-\infty $ , we have verified equation (4). We now apply Lemma 3.3, and see that there exist infinitely many n for which

(5) $$ \begin{align} \begin{array}{c} c_n(X) < \displaystyle{\sum_{i = 2}^n} \lfloor \epsilon i^{0.25} (\log i)^{-0.5} \rfloor \leq n \lfloor \epsilon n^{0.25} (\log n)^{-0.5} \rfloor \qquad\qquad\qquad\qquad\quad \\ \qquad\qquad\qquad\qquad\qquad\quad \textrm{ and}\quad c_n(X) - c_{n-1}(X) < \lfloor \epsilon n^{0.25} (\log n)^{-0.5} \rfloor. \end{array}\end{align} $$

Now, by Corollary 3.1, if n satisfies equation (5), $|\mathrm {Aut}_{\lfloor (n-1)/2 \rfloor }(X, \sigma )|$ is bounded from above by

$$ \begin{align*} (c_{1+c_n(X)}(X))^{2|A| (c_{n+1}(X) - c_n(X))} \leq (c_{n \lfloor \epsilon n^{0.25} (\log n)^{-0.5} \rfloor}(X))^{2|A| \epsilon n^{0.25} (\log n)^{-0.5}}. \end{align*} $$

By subadditivity,

$$ \begin{align*} c_{n \lfloor \epsilon n^{0.25} (\log n)^{-0.5} \rfloor}(X) &\leq c_n(X)^{\lfloor \epsilon n^{0.25} (\log n)^{-0.5} \rfloor}\\ &\leq (n^{1.25})^{\epsilon n^{0.25} (\log n)^{-0.5}} < n^{2\epsilon n^{0.25} (\log n)^{-0.5}}. \end{align*} $$

Therefore, $|\mathrm {Aut}_{\lfloor (n-1)/2 \rfloor }(X, \sigma )|$ is bounded from above by

$$ \begin{align*} (n^{2\epsilon n^{0.25} (\log n)^{-0.5}})^{2|A|\epsilon n^{0.25} (\log n)^{-0.5}} = n^{4|A| \epsilon^2 n^{0.5} (\log n)^{-1}} = e^{4|A| \epsilon^2 \sqrt{n}}. \end{align*} $$

Now, exactly as in the end of Theorem 1.1, any finitely generated subgroup H of ${{\mathrm {Aut}(X,\sigma )}/{\langle \sigma \rangle }}$ has growth less than $e^{4|A| \epsilon ^2 \sqrt {n}}$ . Since $\epsilon> 0$ was arbitrary, by the gap conjecture, H must be virtually nilpotent. Exactly as in the proof of Theorem 1.1, this implies that ${{\mathrm {Aut}(X,\sigma )}/{\langle \sigma \rangle }}$ is locally finite.

Proof of Theorem 1.3

Assume that the gap conjecture holds, and choose any infinite transitive subshift X where ${c_n(X)}/{n^{1.5} (\log n)^{-1}} \rightarrow 0$ . We first claim that for any $\epsilon> 0$ ,

(6) $$ \begin{align} \liminf c_n(X) - \sum_{i = 2}^n \lfloor \epsilon i^{0.5} (\log i)^{-1}\rfloor = -\infty. \end{align} $$

Again, by assumption, there are infinitely many n where $c_n(X) < ({\epsilon }/{3}) n^{1.5} (\log n)^{-1}$ . Also, $\sum _{i = 2}^n \lfloor \epsilon i^{0.5} (\log i)^{-1} \rfloor \geq \sum _{i = \lceil n/2 \rceil }^n \lfloor \epsilon i^{0.5} (\log i)^{-1} \rfloor \geq (n/2) \lfloor \epsilon (n/2)^{0.5} (\log n)^{-1} \rfloor \geq ({\epsilon }/{2^{1.5}}) n^{1.5} (\log n)^{-1} \kern1.2pt{-}\kern1.2pt {n}/{2}$ . So, for infinitely many n,

$$ \begin{align*} c_n(X) - \sum_{i = 2}^n \epsilon \sqrt{n} (\log n)^{-1} < \frac{\epsilon}{3} n^{1.5} (\log n)^{-1} - \frac{\epsilon}{2^{1.5}} n^{1.5} (\log n)^{-1} + \frac{n}{2}. \end{align*} $$

Since this last term approaches $-\infty $ , we have verified equation (6). We now apply Lemma 3.3, and see that there exist infinitely many n for which

(7) $$ \begin{align} c_n(X) - c_{n-1}(X) < \lfloor \epsilon n^{0.5} (\log n)^{-1} \rfloor. \end{align} $$

Rather than using Lemma 3.3 to bound $c_n(X)$ , we simply recall that by assumption, there exists N so that

(8) $$ \begin{align} c_n(X) < \lfloor n^{1.5} \rfloor \end{align} $$

for all $n>N$ . This clearly implies that $c_{1 + c_n(X)}(X) \leq (n^{1.5})^{1.5} = n^{2.25}$ for any $n>N$ . By Corollary 3.1, for any of the infinitely many $n> N$ satisfying equations (7) and (8), $|\mathrm {Aut}_{\lfloor (n-1)/2 \rfloor }(X, \sigma )|$ is bounded from above by

$$ \begin{align*} (c_{1+c_n(X)}(X))^{2|A| (c_{n+1}(X) - c_n(X))} \leq (n^{2.25})^{2|A| \epsilon n^{0.5} (\log n)^{-1}} = e^{4.5 |A| \epsilon \sqrt{n}}. \end{align*} $$

Now, exactly as in the end of Theorem 1.1, any finitely generated subgroup H of ${{\mathrm {Aut}(X,\sigma )}/{\langle \sigma \rangle }}$ has growth less than $e^{4.5 |A| \epsilon \sqrt {n}}$ . Since $\epsilon> 0$ was arbitrary, by the gap conjecture, H must be virtually nilpotent. Exactly as in the proof of Theorem 1.1, this implies that ${{\mathrm {Aut}(X,\sigma )}/{\langle \sigma \rangle }}$ is locally finite.

Proof of Theorem 1.4

Choose any subshift X where ${c_n(X)}/{n^{2} (\log n)^{-1}} \rightarrow 0$ , and any $\epsilon> 0$ . We claim that

(9) $$ \begin{align} \liminf c_n(X) - \sum_{i = 2}^n \lfloor \epsilon i (\log i)^{-1} \rfloor = -\infty. \end{align} $$

Again, by assumption, there are infinitely many n where $c_n(X) < ({\epsilon }/{5}) n^2 (\log n)^{-1}$ . Also,

$$ \begin{align*} \sum_{i = 2}^n \lfloor \epsilon i (\log i)^{-1} \rfloor \geq \!\sum_{i = \lceil n/2 \rceil}^n \lfloor \epsilon i (\log i)^{-1} \rfloor \kern1.2pt{\geq}\kern1.2pt (n/2) \lfloor (\epsilon n/2) (\log n)^{-1} \rfloor \kern1.2pt{\geq}\kern1.2pt \frac{\epsilon}{4} n^2 (\log n)^{-1} \kern1.2pt{-}\kern1.2pt \frac{n}{2}. \end{align*} $$

So, for infinitely many n,

$$ \begin{align*} c_n(X) - \sum_{i = 2}^n \lfloor i (\log i)^{-1} \rfloor < \frac{\epsilon}{5} n^2 (\log n)^{-1} - \frac{\epsilon}{4} n^2 (\log n)^{-1} +\frac{n}{2}. \end{align*} $$

Since this last term approaches $-\infty $ , we have verified equation (9). We now apply Lemma 3.3, and see that there exist infinitely many n for which

(10) $$ \begin{align} c_n(X) - c_{n-1}(X) < \lfloor \epsilon n (\log n)^{-1} \rfloor. \end{align} $$

Rather than using Lemma 3.3 to bound $c_n(X)$ , we simply recall that by assumption, there exists N so that

(11) $$ \begin{align} c_n(X) < n^2 \end{align} $$

for all $n>N$ . This clearly implies that $c_{1 + c_n(X)}(X) \leq (n^{2})^{2} = n^{4}$ for any $n>N$ . Now, by Theorem 3.2, for any of the infinitely many $n> N$ satisfying equations (10) and (11), $|\mathrm {Aut}^{(\textit{FIP})}_{\lfloor (n-1)/2 \rfloor }(X, \sigma )|$ is bounded from above by

$$ \begin{align*} (c_{1+c_n(X)}(X))^{2|A| (c_{n+1}(X) - c_n(X))} \leq (n^{4})^{2|A| \epsilon n (\log n)^{-1}} = e^{8 |A| \epsilon n}. \end{align*} $$

By compactness, the set of isolated periodic points of X is finite; denote this set by $\mathcal {P}$ . It is straightforward to check that the set $\mathcal {P}$ is invariant under any automorphism of $(X,\sigma )$ , and we may consider the homomorphism

$$ \begin{align*} \begin{gathered} \pi_{\mathcal{P}} \colon \mathrm{Aut}(X,\sigma) \to \textrm{Aut}(\mathcal{P},\sigma|_{\mathcal{P}})\\ \pi_{\mathcal{P}} \colon \phi \mapsto \phi|_{\mathcal{P}}. \end{gathered} \end{align*} $$

By definition, we have $\mathrm {Aut}^{(\textit{FIP})}(X, \sigma ) = \mathrm {ker}(\pi _{\mathcal {P}})$ . Now, exactly as in the end of Theorem 1.1, any finitely generated subgroup H of $\mathrm {Aut}^{(\textit{FIP})}(X, \sigma )$ has growth less than $e^{8 |A| \epsilon n}$ . Since $\epsilon $ was arbitrary, this implies that H has subexponential growth, and so is amenable. Then, every finitely generated subgroup of $\mathrm {Aut}^{(\textit{FIP})}(X, \sigma )$ is amenable, implying that $\mathrm {Aut}^{(\textit{FIP})}(X, \sigma )$ is amenable. Since $\mathcal {P}$ is finite, $\mathrm {Aut}(\mathcal {P},\sigma |_{\mathcal {P}})$ is a finite group, and hence $\mathrm {Aut}^{(\textit{FIP})}(X, \sigma )$ is of finite index in $\mathrm {Aut}(X,\sigma )$ . Since $\mathrm {Aut}^{(\textit{FIP})}(X, \sigma )$ is amenable, this implies $\mathrm {Aut}(X,\sigma )$ is amenable.

Proof. For every $w \in L(X)$ , there exists k so that w is a subword of some $A_k$ -word. However, then w is a subword of every $A_{k+1}$ -word. For every $x \in X$ , x contains an $A_{k+1}$ -word, so contains w. Since x and w were arbitrary, X is minimal.

Proof. We prove by induction on k. The base case $k = 1$ simply says that locations of $A_1$ -words are uniquely determined for such subshifts, which is trivial since we always take $A_1$ to be the alphabet $\mathcal {A}$ . For the inductive step, assume for some k that for every $x \in X$ , x can be uniquely decomposed into $A_k$ -words. Given this decomposition, one simply searches for $2d_{k+1}$ -fold concatenations of the form $v^{d_{k+1}} w^{d_{k+1}}$ (with $v,w \in A_k$ possibly equal), which can only occur with $v^{d_{k+1}}$ at the end of one $A_{k+1}$ -word and $w^{d_{k+1}}$ at the beginning of another. This implies that x can be written in a unique way as a concatenation of $A_{k+1}$ -words, completing the inductive step and the proof.

Proof of Theorem 1.5

Choose any unbounded increasing f and countable locally finite group G; G can be written, by definition, as the union of an increasing chain of finite subgroups, that is there exist finite groups $H_k$ so that $H_k$ is a proper subgroup of $H_{k+1}$ for all n, and G is the union of the $H_k$ . Choose an increasing sequence $(b_k)$ of integers greater than $1$ with the property that $f(b_k)> k(|H_k|^{5|H_{k+1}| + 2|H_k|^2})$ for all k.

Our technique is somewhat similar to that of [Reference Boyle, Lind and Rudolph2], where a subshift X was constructed for which the additive group of rationals embeds into $\mathrm {Aut}(X,\sigma )$ , in that we will construct, for every k, automorphisms defined by their action on the set $A_k$ , and then show that every automorphism of X can be realized in this way. Specifically, for each k, we will define a group of permutations of the words in $A_k$ which is isomorphic to $H_k$ . We will then show that for any k, each $A_k$ -permutation induces an $A_{k+1}$ -permutation by coordinatewise application to $A_k$ -words, in a manner that is compatible with the containment of $H_k$ as a subgroup of $H_{k+1}$ .

We begin with some notation. For every k, fix an ordering $(h_1^{(k)}, \ldots , h_{|H_k|}^{(k)})$ of the elements of $H_k$ , where $h_1^{(k)} = \{\mathrm {id}\}$ . Write $q_k = |H_k|/|H_{k-1}|$ , and choose any set $\{r_i^{(k)}\}_{i = 1}^{q_k}$ of representatives for right cosets of $H_{k-1}$ in $H_{k}$ , that is

$$ \begin{align*} H_{k} = \bigcup_{i=1}^{q_k} H_{k-1} r_i^{(k)}. \end{align*} $$

Without loss of generality, we will always take $r_1^{(k)} = \{\mathrm {id}\}$ , the identity element of G (and of all $H_k$ ). We will now recursively define the sets $A_k$ and lengths $n_k$ . In our construction, $|A_k| = |H_k|$ for every k.

The $k = 1$ case is simple; for every $g \in H_1$ , define a symbol $w_g^{(1)}$ , and define the alphabet $A_1 = \{w_g^{(1)} : g \in H_1\}$ ; clearly $|A_1| = |H_1|$ .

We now define $A_{k+1}$ for $k \geq 1$ , assuming that $A_k = \{w_g^{(k)} : g \in H_k\}$ has been defined already. Informally, the idea is that we will define $w_g^{(k+1)}$ for every $g \in H_{k+1}$ by assigning a different ‘template’ concatenation of $A_k$ -words to each coset representative $r_i^{(k+1)}$ , and then permuting the $A_k$ -words $\{w_g^{(k)}\}_{g \in H_k}$ in those templates by multiplying on the left by a properly chosen $g'$ in the subscripts.

More formally, for every $g \in H_{k+1}$ , we first write $g = g' r_i^{(k+1)}$ for some $g' \in H_k$ and $1 \leq i \leq q_{k+1}$ . We then define

$$ \begin{align*} w_{g}^{(k+1)} = w_{g' r_i^{(k+1)}}^{(k+1)} &= \Big(w_{g' h_2^{(k)}}^{(k)}\Big)^{2b_{k+1} q_{k+1}}\\ &\ \, \quad\Big( \Big(w_{g' h_1^{(k)}}^{(k)}\Big)^{b_{k+1}} \Big(w_{g' h_2^{(k)}}^{(k)}\Big)^{b_{k+1}} \Big| \cdots \Big| \Big(w_{g' h_{|H_k|-1}^{(k)}}^{(k)}\Big)^{b_{k+1}} \Big(w_{g' h_{|H_k|}^{(1)}}^{(1)}\Big)^{b_{k+1}}\Big)\\ &\ \, \quad\Big(w_{g' h_{1}^{(k)}}^{(k)}\Big)^{ib_{k+1}} \Big(w_{g' h_{2}^{(k)}}^{(k)}\Big)^{b_{k+1}(3q_{k+1}-i)}. \end{align*} $$

Here, the concatenation inside the largest parentheses is a list of all pairs of the form $\Big (w_{g' h_a^{(k)}}^{(k)}\Big )^{b_{k+1}} \Big (w_{g' h_b^{(k)}}^{(k)}\Big )^{b_{k+1}}$ , with pairs $(a,b)$ listed in lexicographic order. We then define $A_{k+1} = \{w_g^{(k+1)} : g \in H_{k+1}\}$ , and note that all $A_{k+1}$ -words begin and end with $2b_{k+1} q_{k+1}$ -fold repetitions of an $A_k$ -word. We also note that every $A_{k+1}$ -word contains no other such repetitions. To see this, note that in the central line of the definition of $w_g^{(k+1)}$ , no $\Big (w_{g' h_i^{(k)}}^{(k)}\Big )^{b_{k+1}}$ can repeat four times, since the pair $(i, i)$ is used only once and it is not possible to have consecutive lexicographic pairs of the form $(j, i) (i, i) (i, k)$ . Therefore, the largest number of times that an $A_k$ -word can consecutively repeat aside from the beginning and end of $w_g^{(k+1)}$ is $\max (3b_{k+1}, ib_{k+1}) < 2b_{k+1} q_{k+1}$ . The length of all $A_{k+1}$ -words is

(13) $$ \begin{align} n_{k+1} = b_{k+1} n_k(2|A_k|^2 + 5q_{k+1})> b_{k+1} q_{k+1}. \end{align} $$

Recursively, this defines $n_k$ and $A_k$ for all k, and so an associated block concatenation subshift X. We here note a few properties of X which will be useful later.

• • X is minimal by Lemma 4.1.

• • X is uniquely decomposable by Lemma 4.2 (with $d_k = 2b_k q_k$ ).

• • Concatenations of the form $uvw$ with $u,v,w \in A_k$ , $u \neq v$ , and $v \neq w$ never appear in points of X, and that every other concatenation $uvw$ with $u = v$ or $v = w$ appears within every $A_{k+1}$ -word.

We now wish to bound the complexity of X from above to show that ${c_n(X)}/{nf(n)} \rightarrow 0$ . Choose any length n; there exists k so that $n_k \leq n < n_{k+1}$ . We first treat the case where n belongs to $[n_k, b_{k+1} n_k)$ . All points of X are concatenations of $A_k$ -words, and by definition of $A_{k+1}$ , each $A_k$ -word is repeated some number of times which is a multiple of $b_{k+1}$ . Therefore, since $n < b_{k+1} n_k$ , any word $w \in L_n(X)$ is of the form $s qq \ldots qq rr \ldots rr p$ , where $q,r$ are $A_k$ -words, s is a suffix of q, and p is a prefix of r. Then w is determined completely by the location at which the transition from q to r occurs, and the choices of q and r. (The case where w has no transition is still included here by just taking $q = r$ .) So, $c_n(X) \leq n |A_k|^2 = n|H_k|^2$ . Since $|H_k|^2 < f(b_k)/k < f(n_k)/k$ and $n \geq n_k$ , we have $c_n(X) \leq n f(n_k)/k \leq n f(n)/k$ .

Now consider the case where n belongs to $[b_{k+1} n_k, n_{k+1})$ . Any word $w \in L_n(X)$ must be of the form $s u_1^{b_{k+1}} u_2^{b_{k+1}} \ldots u_{j-1}^{b_{k+1}} p$ , where $u_1, \ldots , u_{j-1} \in A_k$ , s is a suffix of some word $u_0^{b_{k+1}}$ for $u_0 \in A_k$ , and p is a prefix of some word $u_j^{b_{k+1}}$ for $u_j \in A_k$ . By equation (13), ${n_{k+1}}/{b_{k+1} n_k} < 5q_{k+1} + 2|A_k|^2$ , and so $j < 5|H_{k+1}| + 2|H_k|^2$ . Clearly, w is determined by the words $u_0, \ldots , u_j$ and the length of s, so $c_n(X) \leq b_{k+1} n_k |A_k|^{j+1} = b_{k+1} n_k |H_k|^{j+1}$ . Since $j < 5|H_{k+1}| + 2|H_k|^2$ ,

$$ \begin{align*} |H_k|^{j+1} \leq |H_k|^{5|H_{k+1}| + 2|H_k|^2} < f(b_{k})/k < f(b_{k+1} n_k)/k. \end{align*} $$

Since $n \geq b_{k+1} n_k$ , this implies that $c_n(X) \leq n f(b_{k+1} n_k)/k \leq n f(n)/k$ .

We have shown that for all $n \geq n_k$ , $c_n(X) \leq nf(n)/k$ , and so ${c_n(X)}/{nf(n)} \rightarrow 0$ . It remains only to show that ${{\mathrm {Aut}(X,\sigma )}/{\langle \sigma \rangle }}$ is isomorphic to G.

For any k and $h \in H_k$ , define the permutation $\pi _{k,h}$ of the set $\{w_g^{(k)}\}$ of $A_k$ -words by left multiplication of the subscript, that is $\pi _{k,h}(w_g^{(k)}) = w_{hg}^{(k)}$ . In a slight abuse of notation, we also define $\pi _{k,h}$ to act on concatenations of $A_k$ -words by ‘coordinatewise’ application, that is if $w_1, \ldots , w_m \in A_k$ , $\pi _{k,h}(w_1 \ldots w_m) := \pi _{k,h}(w_1) \ldots \pi _{k,h}(w_m)$ . (We note that this definition is well defined because X is uniquely decomposable.)

Proof. This comes from the definition of $A_{k+1}$ . Informally, it is due to the fact that a left multiplication by any $h \in H_k$ in the subscript of an $A_{k+1}$ -word $w_{g' r_i^{(k+1)}}^{(k+1)}$ passes through to left multiplications of all subscripts of the component $A_k$ -words.

More formally, choose any $k \ge 1$ , $h \in H_k$ and $g \in H_{k+1}$ ; we can write $g = g' r_i^{(k+1)}$ for some $g^{\prime } \in H_k$ and $1 \leq i \leq q_{k+1}$ . Then,

$$ \begin{align*} \pi_{k,h}\Big(w_{g}^{(k+1)}\Big) &= \pi_{k,h}\Big(w_{g' r_i^{(k+1)}}^{(k+1)}\Big)\\& = \pi_{k,h}\Big( \Big(w_{g' h_2^{(k)}}^{(k)}\Big)^{2b_{k+1} q_{k+1}}\\ &\quad\Big( \Big(w_{g' h_1^{(k)}}^{(k)}\Big)^{b_{k+1}} \Big(w_{g' h_2^{(k)}}^{(k)}\Big)^{b_{k+1}} \Big| \cdots \Big| \Big(w_{g' h_{s_k-1}^{(k)}}^{(k)}\Big)^{b_{k+1}} \Big(w_{g' h_{s_k}^{(1)}}^{(1)}\Big)^{b_{k+1}}\Big)\\ &\quad\Big(w_{g' h_{1}^{(k)}}^{(k)}\Big)^{ib_{k+1}} \Big(w_{g' h_{2}^{(k)}}^{(k)}\Big)^{b_{k+1}(3q_{k+1}-i)} \Big)\\ &= \Big(w_{hg' h_2^{(k)}}^{(k)}\Big)^{2b_{k+1} q_{k+1}}\\ &\quad\Big( \Big(w_{hg' h_1^{(k)}}^{(k)}\Big)^{b_{k+1}} \Big(w_{hg' h_2^{(k)}}^{(k)}\Big)^{b_{k+1}} \Big| \cdots \Big| \Big(w_{hg' h_{s_k-1}^{(k)}}^{(k)}\Big)^{b_{k+1}} \Big(w_{hg' h_{s_k}^{(1)}}^{(1)}\Big)^{b_{k+1}}\Big)\\ &\quad\Big(w_{hg' h_{1}^{(k)}}^{(k)}\Big)^{ib_{k+1}} \Big(w_{hg' h_{2}^{(k)}}^{(k)}\Big)^{b_{k+1}(3q_{k+1}-i)}\\ &= w_{hg' r_i^{(k+1)}}^{(k+1)} = \pi_{k+1,h}\Big(w_{g' r_i^{(k+1)}}^{(k+1)}\Big) = \pi_{k+1,h}\Big(w_{g}^{(k+1)}\Big).\\[-42pt] \end{align*} $$

In particular, by induction, Lemma 4.3 implies that for any $k < m$ , $\pi _{k,h}$ induces the permutation $\pi _{m,h}$ of $A_m$ -words by coordinatewise action. By passing to limits, we see that in fact $\pi _{k,h}$ induces a self-bijection of X, which we denote by $\phi _{k,h}$ . Since X is uniquely decomposable, $\phi _{k,h}$ is continuous and shift-commuting (in fact, it is one of the maps $\alpha _{\tau }$ defined in equation (12)), and so is in $\mathrm {Aut}(X, \sigma )$ .

For every k, the group $\{\phi _{k,h}\}_{h \in H_k}$ is clearly isomorphic to $H_k$ itself (since $\phi _{k,h} \circ \phi _{k,h'} = \phi _{k,hh'}$ ). The collection $\{\phi _{k,h}\}_{k \in \mathbb {N}, h \in H_k}$ then forms a subgroup of $\mathrm {Aut}(X,\sigma )$ , which we denote by $G_{\phi }$ , and it follows from the above that $G_{\phi }$ is isomorphic to G. We now claim that even after quotienting out by the subgroup generated by the shift, this is still true.

Finally, we must show that under the quotient map $\rho \colon \mathrm {Aut}(X,\sigma ) \to {{\mathrm {Aut}(X,\sigma )}/{\langle \sigma \rangle }}$ , the image of $G_{\phi }$ is all of ${{\mathrm {Aut}(X,\sigma )}/{\langle \sigma \rangle }}$ ; in other words, that every automorphism of X can be written as $\sigma ^j \phi _{k,h}$ for some $j \in \mathbb {Z}$ , $k \in \mathbb {N}$ , and $h \in H_k$ .

We need a technical definition; we say that $\phi \in \mathrm {Aut}(X, \sigma )$ preserves locations of $A_k$ -words if, for all $x \in X$ and $m < n$ , if $x([m,m+n_k))$ is an $A_k$ -word, then $(\phi x)([m,m+n_k))$ is an $A_k$ -word. (Note that preserving locations of $A_k$ -words clearly implies preserving locations of $A_j$ -words for any $j < k$ .) We say that $\phi $ simply preserves locations if it preserves locations of $A_k$ -words for all k.

It is clear that for every $\phi \in \mathrm {Aut}(X,\sigma )$ of range $n_{k}$ and every $x \in X$ , there exists a shift i with $|i| \leq n_k/2$ so that $(\phi \circ \sigma ^i)(x)$ has $A_k$ -words at the same locations as x. In theory though, this is weaker than preserving locations of $A_k$ -words, as i could depend on x. For our examples, we can show that this is not possible as long as k is large enough.

We now wish to show that all automorphisms of X preserve locations up to a shift.

Proof. Choose any $\phi \in \mathrm {Aut}(X, \sigma )$ , and choose k so that both $\phi $ and $\phi ^{-1}$ have range $n_k/2$ . Choose any $x \in X$ with $x([0, n_{k+1}))$ an $A_{k+1}$ -word. There clearly exists i with $|i| < n_k/2$ so that for all m, $(\phi \circ \sigma ^i x)([mn_k, (m+1)n_k))$ is an $A_k$ -word. Write $\phi ' := \phi \circ \sigma ^i$ . By additivity of ranges under composition, $\phi '$ and $\phi ^{\prime -1}$ have range $n_k$ , and by Lemma 4.5, $\phi '$ preserves locations of $A_k$ -words.

Now, consider any $x \in X$ for which $x([-n_{k+1}, 0))$ and $x([0, n_{k+1}))$ are both $A_{k+1}$ -words, and $x([-n_{k+1}, 0))$ ends with the same $2b_{k+1} c_{k+1}$ -fold repetition of an $A_k$ -word that $x([0, n_{k+1}))$ begins with. Then $x([-2b_{k+1} c_{k+1} n_k, 2b_{k+1} c_{k+1} n_k)) = w^{4b_{k+1} c_{k+1}}$ for some $w \in A_k$ . Since $\phi $ preserves locations of $A_k$ -words and has range less than $n_k$ , $(\phi ' x)([(-2b_{k+1} c_{k+1} + 1)n_k, (2b_{k+1} c_{k+1} - 1)n_k)) = v^{4b_{k+1} c_{k+1} - 2}$ for some $v \in A_k$ . However, the only $(4b_{k+1} c_{k+1} - 2)$ -fold repetitions of $A_k$ -words within points of x are within the $(4b_{k+1} c_{k+1})$ -fold repetitions occurring across the boundary of some pairs of $A_{k+1}$ -words. Therefore, there exists some $j \in \{0, \pm n_k\}$ for which $((\phi ' \circ \sigma ^j) x)([-2b_{k+1} c_{k+1} n_k, 2b_{k+1} c_{k+1} n_k]) = v^{4b_{k+1} c_{k+1}}$ , which implies that $((\phi ' \circ \sigma ^j) x)([-n_{k+1}, 0))$ and $((\phi ' \circ \sigma ^j) x)([0, n_{k+1}))$ are $A_{k+1}$ -words, and so that $(\phi ' \circ \sigma ^j) x$ has $A_{k+1}$ -words in the same locations as x. Define $\phi '' := \phi ' \circ \sigma ^j$ ; since $\phi ', \phi ^{\prime -1}$ had range $n_k$ and $|j| \leq n_k$ , $\phi ''$ and $\phi ^{\prime \prime -1}$ have range $2n_k$ by additivity of ranges under composition. Since $2n_k < n_{k+1}$ , Lemma 4.5 implies that $\phi ''$ preserves locations of $A_{k+1}$ -words. We claim that in fact $\phi ''$ preserves locations for all $A_m$ -words for $m> k$ , which will complete the proof since $\phi '' = \phi \circ \sigma ^{i+j}$ . We prove by induction on m. The base case $m = k+1$ is completed, so we assume that $m> k$ and that $\phi ''$ preserves locations of $A_m$ -words.

Choose two $A_{m}$ -words $y, z$ which are not equal, but agree on their first and last $2n_m$ letters (for instance, $w_{\mathrm {id}}^{(m)}$ and $w_{r_2^{(m)}}^{(m)}$ would work). Choose $x \in X$ so that $x([-n_{m+1}, 0))$ and $x([0, n_{m+1}))$ are both in $A_{m+1}$ , $x([-n_{m+1}, 0))$ ends with $y^{2b_{m+1} c_{m+1}}$ , and $x([0, n_{m+1}))$ begins with $z^{2b_{m+1} c_{m+1}}$ .

Since $\phi ''$ preserves locations of $A_{m}$ -words, all words $(\phi '' x)([in_{m}, (i+1)n_{m}))$ for $-2b_{m+1} c_{m+1} \leq i < 2b_{m+1} c_{m+1}$ are $A_{m}$ -words. Since $\phi ''$ has range $2n_k$ , each $(\phi '' x)([in_{m}, (i+1)n_{m}))$ depends only on $x(in_{m} - 2n_k, (i+1)n_{m} + 2n_k))$ . Since y and z agree on their first and last $2n_m \geq 2n_k$ letters and

$$ \begin{align*} x([-2b_{m+1} c_{m+1} n_{m}, 2b_{m+1} c_{m+1} n_{m})) = y^{2b_{m+1} c_{m+1}} z^{2b_{m+1} c_{m+1}}, \end{align*} $$

$(\phi '' x)([in_{m}, (i+1)n_{m}))$ is the same for $-2b_{m+1} c_{m+1} < i < 0$ (call this $A_{m}$ -word a) and for $0 \leq i < 2b_{m+1} c_{m+1} - 1$ (call this $A_{m}$ -word b).

If $a = b$ , then since $\phi ^{\prime \prime -1}$ has range $2n_k < n_{m}$ , it would have to be the case that $y = z$ (formally, if $\Psi $ is an inducing block map for $\phi ^{\prime \prime -1}$ , then since $a = (\phi '' x)([-2n_{m} - 2n_k, -n_{m} + 2n_k) = (\phi '' x)([n_{m} - 2n_k, 2n_{m} + 2n_k) = b$ , it must be the case that $y = \Psi (a) = x([-2n_{m}, -n_{m}) = x([n_{m}, 2n_{m}) = \Psi (b) = z$ .) This is a contradiction, so $a \neq b$ . Now, we know that $(\phi '' x)([(-2b_{m+1} c_{m+1} + 1)n_{m}, (2b_{m+1} c_{m+1} - 1)n_{m})) = a^{2b_{m+1} c_{m+1} - 1} b^{2b_{m+1} c_{m+1} - 1}$ for $a \neq b \in A_{m}$ , and the only occurrence of such a word is with midpoint at the border between $A_{m+1}$ -words. Therefore, $(\phi '' x)([-n_{m+1}, 0))$ and $(\phi '' x)([0, n_{m+1}))$ are $A_{m+1}$ -words, and so $\phi '' x$ has $A_{m+1}$ -words at the same locations as x. Since $\phi ''$ has range $2n_k < n_{m+1}$ , $\phi ''$ preserves locations of $A_{m+1}$ -words by Lemma 4.5. This completes the induction and the proof.

The following lemma now completes the proof of Theorem 1.5.

By Lemmas 4.6 and 4.7, every $\phi \in \mathrm {Aut}(X, \sigma )$ can be written as $\sigma ^j \phi _{k,h}$ for some $j \in \mathbb {Z}$ , $k \in \mathbb {N}$ , and $h \in H_k$ , and so ${{\mathrm {Aut}(X,\sigma )}/{\langle \sigma \rangle }}$ is isomorphic to G, completing the proof.