On the almost-palindromic width of free groups

We answer a question of Bardakov (Kourovka Notebook, Problem 19.8) which asks for the existence of a pair of natural numbers $(c, m)$ with the property that every element in the free group on the two-element set $\{a, b\}$ can be represented as a concatenation of $c$, or fewer, $m$-almost-palindromes in letters $a^{\pm 1}, b^{\pm 1}$. Here, an $m$-almost-palindrome is a word which can be obtained from a palindrome by changing at most $m$ letters. We show that no such pair $(c, m)$ exists. In fact, we show that the analogous result holds for all non-abelian free groups.


Introduction and statement of the main result
For a group G, an element g ∈ G, and a generating set X ⊆ G, let length(g, X) be the minimal n ≥ 0 with the property that there exist x 1 , . . ., x n ∈ X for which n , that is, the minimal number of elements from the generating set X necessary to generate g.Let us define the width of G with respect to the generating set X as width(G, X) = max g∈G length(g, X) , or width(G, X) = ∞, if the maximum does not exist.If B is a set, we denote by F B the free group on B whose elements are all reduced words with letters in B ±1 = B ∪ {b −1 | b ∈ B}, with the operation given by concatenation and subsequent reduction.We denote by W B the free monoid on the set B ±1 whose elements are now all words with letters in B ±1 , with concatenation as operation.There is the homomorphism of monoids r B : W B → F B which sends a word to its corresponding reduced word, and which is left-inverse to the inclusion of For a word w ∈ W B , let rev(w) be its reverse word, that is, the word given by the letters of w in reverse order.A palindrome is a word p ∈ W B with the property that p = rev(p).An m-almost-palindrome is a word which differs from a palindrome by a change of at most m letters.In other words, p ∈ W B is an m-almost-palindrome if there exists a palindrome p with the property that d(p, p) ≤ m, where d is the Hamming distance on the set of all words in W B whose number of letters equals the number of letters of p, or, equivalently, if d(p, rev(p)) ≤ 2m.Thus, the palindromes are exactly the 0-almost-palindromes.Let us denote by P B,m the set of all malmost-palindromes and observe that the image r B (P B,m ) is a generating set for the free group F B , for every m ≥ 0.
If F is a free group, together with a basis B of F , then there is the canonical isomorphism φ B : F B → F .Furthermore, note that for all m the set X B,m = φ B (r B (P B,m )) is a generating set for F .In fact, we have an increasing sequence X B,0 ⊆ X B,1 ⊆ . . . of generating sets of F .The value of the expression width(F, X B,m ) does not depend on the choice of the basis B of F , since for two bases B, B ′ , the automorphism of F induced by a bijection B → B ′ sends X B,m to X B ′ ,m .We call this value the m-almost-palindromic width of the free group F and write AP-width(F, m) = width(F, X B,m ).
If F is trivial, or a free group of rank one, then it is readily seen that for all m ≥ 0 the m-almost-palindromic width of F is zero, or one.However, Bardakov, Shpilrain, and Tolstykh [1] show that, for a non-abelian free group F (i.e., a free group which is neither trivial nor infinite cyclic), the 0-almost-palindromic width (that is, the palindromic width) of F is infinite.Extending on this, Bardakov [2,Problem 19.8] asks whether there exists a pair of natural numbers (c, m) with the property that every element of F {a,b} can be represented by a concatenation of c, or fewer, m-almost-palindromes in letters a ±1 , b ±1 .In other words, he asks whether there is an m ∈ N for which the m-almost-palindromic width of F is finite, assuming that F is a free group of rank two.
Of course, the same question can be asked without the restriction on the rank of F .In this note, we answer this question negatively for all non-abelian free groups, building on the argument by Bardakov, Shpilrain, and Tolstykh for the case m = 0. Theorem 1.If F is a non-abelian free group, then, for all m ≥ 0, the m-almostpalindromic width of F is infinite.
Throughout this note, whenever u, w ∈ W B , in writing uw we always mean the concatenation of the words u and w, that is, the result under the operation in the monoid W B , in order to avoid ambiguity which may arise from the fact that F B ⊆ W B .For instance, we explicitly write r B (uw) to denote the result under the operation in the group F B , if u and w are reduced words.Moreover, if u, w ∈ W B are words, we mean by u = w that the two are equal as elements of W B , i.e., in a letter-by-letter way.We denote the empty word by ε.If t ∈ B ±1 and k ∈ N, then we denote by t k the word which consists of k-times the letter t.In particular, t 0 = ε, while t 1 denotes the word which consists of the single letter t.If t ∈ B and k < 0, then we write t k = (t −1 ) −k , that is, (−k)-times the letter t −1 ∈ B ±1 .

A map with a useful property
If B is a set, and w ∈ F B , then, for n ≥ 0, we may write w = t k1 1 • • • t kn n , where t i ∈ B and k i ∈ Z \ {0} for all i ∈ {1, . . ., n}, and where t i = t i+1 for all i ∈ {1, . . ., n − 1}.In this way, the number n, as well as the k i and t i are uniquely determined.If n = 0, this means that w = ε.We call the words t i ki , for i ∈ {1, . . ., n}, the syllables of the reduced word w.
Let the map ∆ B : W B → Z be defined in two steps as follows: In particular, ∆ B (ε) = 0. Otherwise, w ∈ F B has at least two syllables, that is, it is of the form w = t k1 1 • • • t kn n , where the t ki i are the syllables of w, with t i ∈ B for all i ∈ {1, . . ., n}, for n ≥ 2. In this case, we set Here, sign : Z → Z is the map which takes the value −1 on all negative integers, the value 1 on all positive integers, and for which sign(0) = 0.This concludes the definition of ∆ B on F B .In order to extend this definition to all of W B , we define for a not necessarily reduced word w ∈ W B .
The map ∆ B has the following useful property which resembles [1, Lemma 1.3].
Lemma 1.Let n ≥ 0 be an integer and w 1 , . . ., w n ∈ W B .Then as well as n i=1 ∆ B (w i ) = 0. Thus, for n = 0 the inequality holds.If n = 1, the inequality is obviously true.
For n = 2, we consider two cases: First, if r B (w 2 ) = r B (w 1 ) −1 , then ∆ B (w be the decompositions of r B (w 1 ) and r B (w 2 ) into syllables, where t i ∈ B, k i ∈ Z \ {0} for all i ∈ {1, . . ., n}, and s i ∈ B, l i ∈ Z \ {0} for all i ∈ {1, . . ., m}.Let j be the largest index with the property that 0 ≤ j ≤ min{n, m}, and that t 1 where t ∈ B is a letter, k, l ∈ Z with k + l = 0, where neither u 1 ends with the letter t or t −1 , nor u 2 begins with t or t −1 , and where u 1 and z −1 each consist of whole syllables of r B (w 1 ) (i.e., the sum of the numbers of syllables of the reduced words u 1 , t k , and z −1 is equal to the number of syllables of r B (w 1 )), and where z and u 2 each consist of whole syllables of r B (w 2 ).Here, it may happen that u 1 , u 2 , or z, of course, are empty.Also, one of the numbers k and l may be zero.Hence, we observe that where where, similarly, ǫ 3 ∈ {−2, −1, 0, 1, 2}.Now, Finally, for n ≥ 3, we proceed by induction: Assuming that the assertion holds for n − 1, and applying the considerations for the case n = 2, we see that = 6n , completing the proof.

Bounds for some values of the map
Throughout this section, let B be a fixed set.We simplify notation and put F = F B , W = W B , P m = P B,m , r = r B , and ∆ = ∆ B .In this context, the following propositions hold.Proof.Let p be a palindrome from which p is obtained by changing m or fewer letters.Let t 1 , . . ., t α ∈ B ±1 be all the letters of p which are subject to the change, in the order in which they appear within p, so that , where the w i are (potentially empty) words.Here, α ≤ m.