Avoiding conjugacy classes on the 5-letter alphabet
aa r X i v : . [ m a t h . C O ] N ov Avoiding Conjugacy Classes on the 5-letter Alphabet
Golnaz Badkobeh a,1 , Pascal Ochem b,2, ∗ a Goldsmiths, University of London b LIRMM, Université de Montpellier, CNRS, Montpellier, France
Abstract
We construct an infinite word w over the -letter alphabet such that for everyfactor f of w of length at least two, there exists a cyclic permutation of f thatis not a factor of w . In other words, w does not contain a non-trivial conjugacyclass. This proves the conjecture in Gamard et al. [TCS 2018] Keywords:
Combinatorics on words, Conjugacy classes
1. Introduction
We consider infinite words over a finite alphabet that do not contain all theconjugates of the same word w , with the necessary condition that | w | > . Arecent interest in such words appeared in the context of pattern avoidance [1]and of iterative algebras [2]. Bell and Madill [2] obtained a pure morphic wordwith this property (and some additional properties) over the -letter alphabet.Gamard et al. [1] independently obtained a morphic word over the -letter al-phabet. They also conjectured that the alphabet size can be lowered to , whichis best possible. In this paper, we prove this conjecture using a morphic word.Together with the construction of a morphic binary word avoiding everyconjugacy class of length at least and a morphic ternary word avoiding everyconjugacy class of length at least [1], this settles the topic of the smallestalphabet needed to avoid every conjugacy class of length at least k .
2. Main result
Let ε denote the empty word. We consider the morphic word w = G ( F ω ( )) defined by the following morphisms. ∗ Corresponding author
Email addresses: [email protected] (Golnaz Badkobeh), [email protected] (PascalOchem) Golnaz Badkobeh is supported by the Leverhulme Trust on the Leverhulme Early CareerScheme. This work is supported by the ANR project CoCoGro (ANR-16-CE40-0005).
Preprint submitted to Elsevier November 21, 2018 ( ) = ,F ( ) = ,F ( ) = ,F ( ) = ,F ( ) = . G ( ) = abcd ,G ( ) = ε,G ( ) = eacd ,G ( ) = becd ,G ( ) = be . Theorem 2.1.
The morphic word w ∈ Σ ∗ avoids every conjugacy class oflength at least 2. In order to prove this theorem, it is convenient to express w with the largermorphisms f = F and g = G ◦ F given below. Clearly, w = g ( f ω ( )) . f ( ) = ,f ( ) = ,f ( ) = ,f ( ) = ,f ( ) = . g ( ) = abcdeacd ,g ( ) = abcdbecd ,g ( ) = abcdeacdbe ,g ( ) = abcdbecdeacdbecd ,g ( ) = abcdbecdeacdbe . F ω ( ) Here we study the pure morphic word and the conjugacy classes it contains.
Lemma 2.2.
The infinite word F ω ( ) contains only the conjugacy classes listedin C = (cid:8) F ( ) , F ( ) , F d ( ) , f d ( ) (cid:9) , for all d > .Proof. Notice that the factor only occurs as the prefix of the f -image ofevery letter in F ω ( ) . Moreover, every letter only occurs in F ω ( ) as thesuffix of the factor . Let us say that the index of a conjugacy class is thenumber of occurrences of in any of its elements. An easy computation showsthat the set of complete conjugacy classes in F ω ( ) with index at most one is C = (cid:8) F ( ) , F ( ) , F ( ) , F ( ) , f ( ) , f ( ) (cid:9) . Let us assume that F ω ( ) containsa conjugacy class c with index at least two. Let w ∈ c be such that is a prefixof w . We write w = ps such that the leftmost occurrence of in w is the prefixof s . Then the conjugate sp of w also belongs to c and thus is a factor of F ω ( ) .This implies that the pre-image v = f − ( w ) is a factor of F ω ( ) , and so doesevery conjugate of v . Thus, F ω ( ) contains a conjugacy class c ′ such that theelements of c with prefix are the f -images of the elements of c ′ . Moreover,the index of c ′ is strictly smaller than the index of c .Using this argument recursively, we conclude that every complete conjugacyclass in F ω ( ) has a member of the form f i ( x ) such that x is an element of aconjugacy class in C .Now we show that F ( ) does not generate larger conjugacy classes in F ω ( ) .We thus have to exhibit a conjugate of f ( F ( )) = F ( ) = that is not a factor of F ω ( ) . A computer check shows that the conjugate is not a factor of F ω ( ) . Similarly, F ( ) does not generatelarger conjugacy classes in F ω ( ) since the conjugate of f ( F ( )) = F ( ) = is not a factor of F ω ( ) .2 .2. Avoiding conjugacy classes in w We are ready to prove Theorem 2.1. Notice that ab only appears in w asthe prefix of the g -image of every letter. A computer check shows that w avoidsevery conjugacy class of length at most . Consider a word w with length atleast whose conjugacy class is complete in w . So w contains at least twooccurrences of ab .Let us assume that w contains a conjugacy class c . Let w ∈ c be such that ab is a prefix of w . Since | w | > , w contains at least occurrences of ab and we write w = ps such that the leftmost occurrence of ab in w is the prefixof s . Then the conjugate sp of w also belongs to c and thus is a factor of w .This implies that the pre-image v = g − ( w ) is a factor of F ω ( ) , and so doesevery conjugate of v . Thus, F ω ( ) contains a conjugacy class c ′ such that theelements of c with prefix ab are the f -images of the elements of c ′ .To finish the proof, it is thus sufficient to show that for every c ′ ∈ C ,there exists a conjugate of g ( c ′ ) that is not a factor of w . Recall that C = (cid:8) F ( ) , F ( ) , F d ( ) , f d ( ) (cid:9) for all d > . The computer check mentioned abovesettles the case of F ( ) and F ( ) since | g ( F ( )) | < | g ( F ( )) | = 40 .The next four lemmas handle the remaining cases: • g ( f d ( F ( ))) = g ( f d ( )) • g ( f d ( F ( ))) = g ( f d ( )) • g ( f d +1 ( )) = g ( f d ( )) • g ( f d +1 ( )) = g ( f d ( )) Lemma 2.3.
Let p = e .g ( f ( ) . . . f d − ( ) .f d ( )) and s = g ( f d − ( ) .f d − ( ) . . . f ( ) ) . abcdeacdb . For every d > , the word T = p s is a conjugate of g ( f d ( )) that is not a fac-tor of w .Proof. Let us assume that T appears in w .The letter in f ω ( ) appears after either or . However e is a suffix of g ( ) and not of g ( ) . Therefore, e .g ( ) is a suffix of g ( ) only. Since is a suffixof f ( ) and not of f ( ) , then g ( f ( )) is a suffix of g ( f ( )) only. Using thisargument recursively, p is a suffix of g ( f d ( )) only.Now, the letter in f ω ( ) appears before either or , however abcdeacdb is a prefix of g ( ) and not of g ( ) . Thus g ( ) . abcdeacdb is a prefixof g ( ) only. Since is a prefix of f ( ) and not of f ( ) , then g ( f ( ) ) is a prefix of g ( f ( )) only. Using this argument recur-sively, s is a prefix of g ( f d − ( )) only. Thus T is a factor of g ( f d ( )) but is not a factor of f ω ( ) , contradiction. Lemma 2.4.
Let p = acdbecd .g ( f ( ) . . . f d − ( )) .f d ( )) and s = g ( f d ( ) .f d − ( ) . . . f ( ) . ) . abcdbecde . For every d > , the word T = p g ( f d ( )) s is a conjugate of g ( f d ( )) that is not a factor of w . roof. Let us assume that T appears in w .The letter in f ω ( ) appears after either or . However acdbecd is a suffix of g ( ) and not of g ( ) . Therefore acdbecd .g ( ) is a suffix of g ( ) only. Since is a suffix of f ( ) and not of f ( ) , then g ( f ( )) is a suffix of g ( f ( )) only. Using this argument recursively, p is a suffix of g ( f d ( )) only.Now, the letter in f ω ( ) appears before either or . However abcdbecde is a prefix of g ( ) and not of g ( ) . Thus g ( ) . abcdbecde is a prefixof g ( ) only. Since is a prefix of f ( ) and not of f ( ) , then g ( f ( ) ) is a prefix of g ( f ( )) only. Using this argument re-cursively, s is a prefix of g ( f d − ( )) only. Thus T is a factor of g ( f d ( )) but is not a factor of f ω ( ) , contradiction. Lemma 2.5.
Let p = ecdeacdbe .g ( f ( ) · · · f d − ( ) .f d ( )) and s = g ( f d ( ) f d − ( ) . · · · f ( ) ) . abcdb . For every d > , theword T = p s is a conjugate of g ( f d ( ) that is not a fac-tor of w .Proof. Let us assume that T appears in w .The factor in f ω ( ) appears after either or . However ecdeacdbe is a suffixof g ( ) and not of g ( ) . Therefore ecdeacdbe .g ( ) is a suffix of g ( ) only. Since is a suffix of f ( ) and not of f ( ) , then g ( f ( )) isa suffix of g ( f ( )) , using this argument recursively, p is a suffix of g ( f d ( )) only.Now, the factor in f ω ( ) appears before either or . However abcdb isa prefix of g ( ) and not of g ( ) . Thus g ( ) . abcdb must only be a prefix of g ( ) and since is a prefix of f ( ) and not of f ( ) then g ( f ( ) ) is a prefix of g ( f ( )) only. Using this argument recursively, s is aprefix of g ( f d ( )) only. Thus T is a factor of g ( f d ( )) but is not a factor of f ω ( ) , contradiction. Lemma 2.6.
Let p = d .g ( f ( ) . . . f d − ( ) .f d ( )) and s = g ( f d ( ) f d − ( ) .f d − ( ) . . . f ( ) ) . abcdeac . For every d > , the word T = p s is a conjugate of g ( f d ( )) that is not afactor of w .Proof. Let us assume that T appears in w .The letter in f ω ( ) appears after either or . however d is a suffix of g ( ) and not of g ( ) . Therefore d .g ( ) is a suffix of g ( ) only. Since is a suffixof f ( ) and not of f ( ) , then g ( f ( )) is a suffix of g ( f ( )) only. Using thisargument recursively, p is a suffix of g ( f d ( )) only.Now, in f ω ( ) appears before either or , however abcdeac is onlya prefix of g ( ) and not of g ( ) . Thus g ( ) . abcdeac is a prefix of g ( ) only. Since is a prefix of f ( ) and not of f ( ) , then g ( f ( ) ) is aprefix of g ( f ( )) only. Using this argument recursively, s is a prefix of g ( f d ( )) . Thus T is a factor of g ( f d ( )) but is not a factorof f ω ( ) , contradiction. 4 eferences [1] G. Gamard, P. Ochem, G. Richomme, and P. Séébold. Avoidability of cir-cular formulas. Theoret. Comput. Sci (2018), 1–4.[2] J.P. Bell and B.W. Madill. Iterative Algebras.
Al-gebr. Represent. Theor. (2015), 1533–1546. https://doi.org/10.1007/s10468-015-9550-yhttps://doi.org/10.1007/s10468-015-9550-y