\documentclass[numbers=enddot,12pt,final,onecolumn,notitlepage]{scrartcl}% \usepackage[headsepline,footsepline,manualmark]{scrlayer-scrpage} \usepackage[all,cmtip]{xy} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{framed} \usepackage{amsmath} \usepackage{comment} \usepackage{color} \usepackage{hyperref} \usepackage[sc]{mathpazo} \usepackage[T1]{fontenc} \usepackage{amsthm} %TCIDATA{OutputFilter=latex2.dll} %TCIDATA{Version=5.50.0.2960} %TCIDATA{LastRevised=Sunday, June 30, 2019 01:31:04} %TCIDATA{SuppressPackageManagement} %TCIDATA{} %TCIDATA{} %TCIDATA{BibliographyScheme=Manual} %BeginMSIPreambleData \providecommand{\U}{\protect\rule{.1in}{.1in}} %EndMSIPreambleData \theoremstyle{definition} \newtheorem{theo}{Theorem}[section] \newenvironment{theorem}[] {\begin{theo}[#1]\begin{leftbar}} {\end{leftbar}\end{theo}} \newtheorem{lem}[theo]{Lemma} \newenvironment{lemma}[] {\begin{lem}[#1]\begin{leftbar}} {\end{leftbar}\end{lem}} \newtheorem{prop}[theo]{Proposition} \newenvironment{proposition}[] {\begin{prop}[#1]\begin{leftbar}} {\end{leftbar}\end{prop}} \newtheorem{defi}[theo]{Definition} \newenvironment{definition}[] {\begin{defi}[#1]\begin{leftbar}} {\end{leftbar}\end{defi}} \newtheorem{remk}[theo]{Remark} \newenvironment{remark}[] {\begin{remk}[#1]\begin{leftbar}} {\end{leftbar}\end{remk}} \newtheorem{coro}[theo]{Corollary} \newenvironment{corollary}[] {\begin{coro}[#1]\begin{leftbar}} {\end{leftbar}\end{coro}} \newtheorem{conv}[theo]{Convention} \newenvironment{condition}[] {\begin{conv}[#1]\begin{leftbar}} {\end{leftbar}\end{conv}} \newtheorem{quest}[theo]{Question} \newenvironment{algorithm}[] {\begin{quest}[#1]\begin{leftbar}} {\end{leftbar}\end{quest}} \newtheorem{warn}[theo]{Warning} \newenvironment{conclusion}[] {\begin{warn}[#1]\begin{leftbar}} {\end{leftbar}\end{warn}} \newtheorem{conj}[theo]{Conjecture} \newenvironment{conjecture}[] {\begin{conj}[#1]\begin{leftbar}} {\end{leftbar}\end{conj}} \newtheorem{exmp}[theo]{Example} \newenvironment{example}[] {\begin{exmp}[#1]\begin{leftbar}} {\end{leftbar}\end{exmp}} \iffalse \newenvironment{proof}[Proof]{\noindent\textbf{#1.} }{\ \rule{0.5em}{0.5em}} \fi \newenvironment{verlong}{}{} \newenvironment{vershort}{}{} \newenvironment{noncompile}{}{} \excludecomment{verlong} \includecomment{vershort} \excludecomment{noncompile} \newcommand{\kk}{\mathbf{k}} \newcommand{\id}{\operatorname{id}} \newcommand{\ev}{\operatorname{ev}} \newcommand{\Comp}{\operatorname{Comp}} \newcommand{\bk}{\mathbf{k}} \newcommand{\Nplus}{\mathbb{N}_{+}} \newcommand{\NN}{\mathbb{N}} \let\sumnonlimits\sum \let\prodnonlimits\prod \renewcommand{\sum}{\sumnonlimits\limits} \renewcommand{\prod}{\prodnonlimits\limits} \DeclareSymbolFont{bbold}{U}{bbold}{m}{n} \DeclareSymbolFontAlphabet{\mathbbold}{bbold} \setlength\textheight{22.5cm} \setlength\textwidth{15cm} \ihead{Errata to Enumerative Combinatorics on Words''} \ohead{\today} \begin{document} \begin{center} \textbf{Chapter 1: Enumerative Combinatorics on Words} \textit{Dominique Perrin and Antonio Restivo} \url{http://www-igm.univ-mlv.fr/~perrin/Recherche/Publications/BookChapters/enhandbook.pdf} version of 10 July 2014 \textbf{Errata and addenda by Darij Grinberg} \bigskip \end{center} %\setcounter{section}{} \section{Errata} The numbering of pages in the following errata matches the numbering of pages on the paper (not the numbering of pages in the PDF file). I have read this chapter from its beginning to \S 1.5.2 (except for a few tangential examples and remarks). This part of the chapter should be covered (almost) completely by these errata. My comments on the rest of the chapter are spotty and random. \begin{itemize} \item \textbf{page 7:} \textquotedblleft its links with several connexions with\textquotedblright\ should probably just be \textquotedblleft its links with\textquotedblright\ or \textquotedblleft several connexions with\textquotedblright. \item \textbf{page 8:} \textquotedblleft well-kown\textquotedblright% \ $\rightarrow$ \textquotedblleft well-known\textquotedblright. \item \textbf{page 9:} I would put the words \textquotedblleft radix order\textquotedblright\ in italics, since you are just defining this concept here. Also, I would point out that this \textquotedblleft radix order\textquotedblright\ is better known as the length-lexicographic order. \item \textbf{page 9:} I would add the following remark: For any $n\geq0$, the set $A^{n}$ of $n$-tuples of elements of $A$ is identified with a subset of $A^{\ast}$ (indeed, every $n$-tuple $\left( a_{1},a_{2},\ldots,a_{n}\right) \in A^{n}$ is identified with the word $a_{1}a_{2}\cdots a_{n}\in A^{\ast}$). This explains how \textquotedblleft$X\cap A^{n}$\textquotedblright\ is to be understood in \S 1.2.1, and similar expressions elsewhere. \item \textbf{page 10:} \textquotedblleft the the product\textquotedblright% \ $\rightarrow$ \textquotedblleft that the product\textquotedblright. \item \textbf{page 10:} Somewhere here it would be helpful to define the notation $X^{\ast}$. (Namely, if $X$ is a subset of $A^{\ast}$, then $X^{\ast }$ denotes the submonoid of $A^{\ast}$ generated by $X$. This is not always a free monoid, so this is not a particular case of the notation $A^{\ast}$ defined on page 8.) \item \textbf{page 10:} Add a period at the end of the equality (1.2.3). \item \textbf{page 11, Example 2:} \textquotedblleft and $y\in aD_{a}^{\ast}% b$\textquotedblright\ $\rightarrow$ \textquotedblleft and $y\in D_{a}^{\ast}$, so that $d\in aD_{a}^{\ast}b$\textquotedblright. \item \textbf{page 11, Example 2:} Remove the period at the end of equation (1.2.6) (the sentence does not end here). \item \textbf{page 11, Example 2:} Add periods at the end of equation (1.2.7) and at the end of equation (1.2.8). \item \textbf{page 12:} Between (1.2.8) and (1.2.9), the \textquotedblleft% $n4^{n}$\textquotedblright\ should be in parentheses (since both $n$ and $4^{n}$ are meant to be part of the denominator). \item \textbf{page 12, Example 3:} \textquotedblleft The initial edges are indicated with an incoming edge and the terminal ones with with an outgoing edge\textquotedblright\ $\rightarrow$ \textquotedblleft The initial states are indicated with an incoming arrow and the terminal ones with an outgoing arrow\textquotedblright. (There are too many mistakes in this sentence to list.) \item \textbf{page 13, Example 5:} In the equation \textquotedblleft$M=\left[ % \begin{array} [c]{cc}% F_{n+1} & F_{n}\\ F_{n} & F_{n-1}% \end{array} \right]$\textquotedblright, replace \textquotedblleft$M$\textquotedblright% \ by \textquotedblleft$M^{n}$\textquotedblright. \item \textbf{page 14, proof of Proposition 2:} Replace \textquotedblleft% $n/d\geq p/d+q/d$\textquotedblright\ by \textquotedblleft$n/d\geq p/d+q/d-1$\textquotedblright. \item \textbf{page 14, Example 6:} You write \textquotedblleft$x_{n-4}y_{n-3}$ is a prefix of $x_{n-3}$ and thus of $x_{n-1}$\textquotedblright. The first part of this is wrong ($x_{n-4}y_{n-3}$ is usually too long to be a prefix of $x_{n-3}$). The second is correct, but needs to be proven. Here is a sketch of the proof: Forget that we fixed $n$. First, it is clear that% \begin{equation} y_{n+1}=x_{n}y_{n-1}\ \ \ \ \ \ \ \ \ \ \text{for each }n\geq4. \label{p14.fib.yxy.0}% \end{equation} (Indeed, this follows by removing the last two letters from the equality $x_{n+1}=x_{n}x_{n-1}$.) Next, we claim that% \begin{equation} y_{n+1}=x_{n-1}y_{n}\ \ \ \ \ \ \ \ \ \ \text{for each }n\geq3. \label{p14.fib.yxy.1}% \end{equation} (\textit{Proof of (\ref{p14.fib.yxy.1}):} We proceed by induction on $n$. In the case $n=3$, the equality (\ref{p14.fib.yxy.1}) holds because both $y_{4}$ and $x_{2}y_{3}$ equal $a$. This covers the induction base. For the induction step, we fix an integer $m\geq4$, and we assume that (\ref{p14.fib.yxy.1}) holds for $n=m-1$. We must now prove that (\ref{p14.fib.yxy.1}) holds for $n=m$. We have assumed that (\ref{p14.fib.yxy.1}) holds for $n=m-1$. In other words, we have $y_{m}=x_{m-2}y_{m-1}$. But the recursive definition of the Fibonacci sequence of words yields $x_{m}=x_{m-1}x_{m-2}$. Now, (\ref{p14.fib.yxy.0}) (applied to $n=m$) yields $y_{m+1}=\underbrace{x_{m}}_{=x_{m-1}x_{m-2}}% y_{m-1}=x_{m-1}\underbrace{x_{m-2}y_{m-1}}_{=y_{m}}=x_{m-1}y_{m}$. In other words, (\ref{p14.fib.yxy.1}) holds for $n=m$. This completes the induction step. Thus, (\ref{p14.fib.yxy.1}) is proven by induction.) Now, assume that $n\geq6$. Then, (\ref{p14.fib.yxy.1}) (applied to $n-3$ instead of $n$) yields $y_{n-2}=x_{n-4}y_{n-3}$. Hence, $x_{n-4}y_{n-3}$ is a prefix of $x_{n-2}$ (since $y_{n-2}$ is clearly a prefix of $x_{n-2}$) and thus also a prefix of $x_{n-1}$ (since $x_{n-1}=x_{n-2}x_{n-3}$). This completes the proof of your claim. \item \textbf{page 14, \S 1.3.1:} You write: \textquotedblleft The conjugacy class of a word of length $n$ and period $p$ has $p$ elements if $p$ divides $n$ and has $n$ elements otherwise.\textquotedblright. Here you probably mean \textquotedblleft minimal period\textquotedblright\ rather than \textquotedblleft period\textquotedblright, since otherwise this is clearly false. But I'm not sure how to prove it when $p$ is the minimum period either. (Fortunately you don't actually use this observation; Proposition 3 is clear for simpler reasons.) \item \textbf{page 14, \S 1.3.2:} The notions of a \textit{primitive necklace} (= a necklace consisting of primitive words) and of the \textit{length} of a necklace (= the length of any word in it) should be defined. Also, it should be explained what you mean by \textquotedblleft on $k$ letters\textquotedblright\ (namely, you mean that you are working over a $k$-letter alphabet $A$; you are not requiring that all its $k$ letters are actually being used in the necklace). \item \textbf{page 14, \S 1.3.2:} Add a period after the equality (1.3.11). \item \textbf{page 15:} \textquotedblleft The \textit{M\"{o}bius function} is defined\textquotedblright\ $\rightarrow$ \textquotedblleft The \textit{M\"{o}bius function} is the function $\mu:\mathbb{N}\setminus 0\rightarrow\mathbb{Z}$ defined\textquotedblright. (Otherwise, the \textquotedblleft$\mu$\textquotedblright\ in the equality is out of place.) \item \textbf{page 16:} It makes sense to mention that the convolution product of functions $\mathbb{N}\setminus0\rightarrow R$ is commutative whenever $R$ is commutative. (You use this tacitly.) \item \textbf{page 16, proof of Proposition 4:} At the very end of the proof, replace \textquotedblleft$\sum_{n=de}\mu\left( d\right) k^{e}$% \textquotedblright\ by \textquotedblleft$\sum_{n=de}k^{d}\mu\left( e\right)$\textquotedblright\ (this is equivalent, but closer to what you get from the convolution product, and also closer to the expression you want). \item \textbf{page 17:} \textquotedblleft the set $M_{d}$ of integers $m\leq n$\textquotedblright\ $\rightarrow$ \textquotedblleft the set $M_{d}$ of positive integers $m\leq n$\textquotedblright. \item \textbf{page 17, proof of Proposition 6:} It is not clear what you mean by \textquotedblleft the multiset formed by the $n$ circular shifts of the words of length $n$\textquotedblright. The obvious interpretation would be to consider both the two shifts $12$ and $21$ of the word $12$ and also the two shifts $21$ and $12$ of the word $21$; but this is not what you mean. Instead, the multiset you want is the multiset obtained by picking a representative of each necklace (of length $n$ on $k$ letters), and then writing down all $n$ cyclic shifts of each necklace you have picked. (When the necklace is not primitive, there will be repetitions among its shifts, whence you get a multiset, not a set.) \item \textbf{page 17, proof of Proposition 6:} When you defined periods of a word, you required them to be positive integers. Here, you are using $0$ as a period of a word. Either this or the definition of a period needs to be modified (although it shouldn't be difficult to do so). \item \textbf{page 18, \S 1.3.3:} \textquotedblleft has a unique factorization\textquotedblright\ $\rightarrow$ \textquotedblleft has at most one factorization\textquotedblright. I would furthermore illustrate this notion of a \textquotedblleft factorization\textquotedblright\ of a necklace by a picture: If we draw out the necklace on a circle, then a factorization means a way of partitioning this circle into arcs (with each arc starting and ending in the middle between two consecutive letters) such that each arc gives a word in $X$ (when you read the letters on this arc in clockwise order). Such factorizations are considered equal when the corresponding arcs are the same (or, more precisely: the points where one arc ends and the next one begins are the same). Thus, $\left\{ aa\right\}$ is not a circular code (since the necklace corresponding to the word $aaaa$ has two different factorizations; these factorizations are different even though the factors are all the same!), and $\left\{ aba,aabaa\right\}$ is not a circular code either (since the necklace $baaabaaa$ has two factorizations: either of the two \textquotedblleft$b$\textquotedblright s can be part of an \textquotedblleft% $aba$\textquotedblright\ factor or part of an \textquotedblleft$aabaa$% \textquotedblright\ factor). \item \textbf{page 18, \S 1.3.3:} I think you want to require circular codes to consist of \textbf{nonempty} words. Otherwise, the \textquotedblleft Formally\textquotedblright\ version of the definition of a circular code would allow $X=\left\{ \varnothing\right\}$ (since the condition is vacuously true: $s$ can never be nonempty if $\varnothing=ps$), and then the denominator $1-f_{X}\left( z\right)$ in (1.3.16) would vanish. \item \textbf{page 19:} On the line directly after (1.3.15), replace \textquotedblleft in such a way that\textquotedblright\ by \textquotedblleft, so that\textquotedblright. See, e.g., \href{http://scg.unibe.ch/wiki/howtos/commonwritingerrors/SuchThatVsSoThat}{this page} about \textquotedblleft so that\textquotedblright\ vs. \textquotedblleft such that\textquotedblright. \item \textbf{page 19, Theorem 1.3.1:} \textquotedblleft let $S$ be the set of words\textquotedblright\ $\rightarrow$ \textquotedblleft let $S$ be the set of nonempty words\textquotedblright. It is not quite clear whether the empty word counts as being conjugate to itself, but let's avoid this issue altogether by forbidding it. (If you allowed the empty word to be in $S$, then the left hand side of (1.3.17) would have constant term $1$ while the right hand side has constant term $0$.) \item \textbf{page 19, Theorem 1.3.1:} The period at the end of (1.3.16) should be a comma. \item \textbf{page 19, proof of Theorem 1.3.1:} \textquotedblleft is of this form for some $x\in X$\textquotedblright\ $\rightarrow$ \textquotedblleft is of this form for some unique $x\in X$\textquotedblright. \item \textbf{page 19, proof of Theorem 1.3.1:} \textquotedblleft Thus $g_{x,n}$\textquotedblright\ $\rightarrow$ \textquotedblleft Thus $g_{n,x}%$\textquotedblright. \item \textbf{page 19, proof of Theorem 1.3.1:} \textquotedblleft Formula (1.3.16) is obtained from (1.3.17) by taking the derivative of the logarithm of each side\textquotedblright\ $\rightarrow$ \textquotedblleft Formula (1.3.16) is obtained from (1.3.17) by dividing both sides by $z$, then taking antiderivatives (with constant term $0$) and exponentiating\textquotedblright. \item \textbf{page 20:} \textquotedblleft in such a way that\textquotedblright% \ $\rightarrow$ \textquotedblleft, so that\textquotedblright. \item \textbf{page 20:} Between the first two paragraphs of this page, I would add the following observation: The set $S$ is not a semigroup, but (due to (1.3.15)) it is closed under taking powers and radicals (i.e., for a word $w\in X^{\ast}$ and a positive integer $m$, we have $w\in S$ if and only if $w^{m}\in S$). Furthermore, $S$ is closed under cyclic shifting, whence $S$ is a disjoint union of necklaces. \item \textbf{page 20:} Replace \textquotedblleft and let $p_{n}% =\operatorname*{Card}\left( P\cap A^{n}\right)$\textquotedblright\ by \textquotedblleft and let $p_{n}$ be the number of necklaces of length $n$ contained in $P$\textquotedblright. (The expression \textquotedblleft$P\cap A^{n}$\textquotedblright\ is meaningless, since the elements of $P$ are necklaces, not words.) \item \textbf{page 20:} Add a period after \textquotedblleft Thus $c_{n}% =\sum_{d\mid n}p_{d}$\textquotedblright. \item \textbf{page 22, Proposition 7:} \textquotedblleft of its proper suffixes\textquotedblright\ $\rightarrow$ \textquotedblleft of its nonempty proper suffixes\textquotedblright. \item \textbf{page 22, proof of Proposition 7:} After \textquotedblleft But $uv\left\vert v^{\prime}\right\vert$)\textquotedblright. \item \textbf{page 23:} You write: \textquotedblleft It motivated Knuth to call Lyndon words \textit{prime} words in \textquotedblright. This is correct, but it is worth mentioning that David Radford already called them \textquotedblleft primes\textquotedblright\ in his 1977 paper \textit{A Natural Ring Basis for the Shuffle Algebra and an Application to Group Schemes} (Journal of Algebra \textbf{58} (1979), pp. 432--454), except that he works with the reverse of the lexicographic order on $A^{\ast}$. \item \textbf{page 23, proof of Lemma 1:} \textquotedblleft let $v$ be the minimal suffix\textquotedblright\ $\rightarrow$ \textquotedblleft let $v$ be the minimal nonempty suffix\textquotedblright. \item \textbf{page 23, Proposition 9:} This proposition also easily follows from Exercise 6.1.32 in \href{https://arxiv.org/src/1409.8356v5/anc/HopfComb-v73-with-solutions.pdf}{Darij Grinberg and Victor Reiner, \textit{Hopf Algebras in Combinatorics}, arXiv:1409.8356v5}. (Indeed, combining parts (c) and (b) of this exercise yields the implication $\mathcal{F}^{\prime}\Longrightarrow\mathcal{G}% ^{\prime}$, which shows that each word in $P$ is a sesquipower of a Lyndon word; conversely, if $w$ is a sesquipower of a Lyndon word distinct from the maximal letter, then parts (f) and (b) of the exercise yield the relation $\mathcal{G}^{\prime}\Longrightarrow\mathcal{F}^{\prime}$, which entails that $w\in P$.) I am mentioning this because the exercise gives a few more equivalent characterizations of $P$. \item \textbf{page 24, proof of Lemma 2:} After the first sentence of this proof, add: \textquotedblleft Pick $n$ minimal with this property\textquotedblright. \item \textbf{page 24, proof of Lemma 2:} How exactly does $x1$. [Note: I still don't understand this proof, so I am not completely sure that this change is the right thing to do.] \item \textbf{page 34:} You write: \textquotedblleft we denote by $X_{s}$ the set of words such that no factor has a conjugate in $\left\{ \ell_{1}% ,\ldots,\ell_{s}\right\}$\textquotedblright. This seems to be a wrong definition of $X_{s}$, as it would imply that (for $s\geq1$) the words in $X_{s}$ cannot contain the letter $a$ (since such a letter would be a factor conjugate to $a=\ell_{1}$). Or are you talking of factors in the factorization of Theorem 1.4.1? (Even in this case, I guess you mean \textquotedblleft conjugate in $\left\{ \ell_{1},\ldots,\ell_{s-1}\right\}$\textquotedblright% \ rather than \textquotedblleft conjugate in $\left\{ \ell_{1},\ldots ,\ell_{s}\right\}$\textquotedblright, though. I don't know for sure since I haven't read Moreno's paper.) \item \textbf{page 34 and further:} \textquotedblleft$\left( a_{n}\right) _{n\in Z}$\textquotedblright\ should be \textquotedblleft$\left( a_{n}\right) _{n\in\mathbb{Z}}$\textquotedblright. \item \textbf{page 34, \S 1.6:} \textquotedblleft It is of course equivalent to ask\textquotedblright: Why? \item \textbf{page 35, proof of Proposition 11 and further:} \textquotedblleft% $\left( a_{n}\right) _{n\in\mathcal{Z}}$\textquotedblright\ should be \textquotedblleft$\left( a_{n}\right) _{n\in\mathbb{Z}}$\textquotedblright. \item \textbf{page 36, Figure 1.6.9:} This figure collides with the header of the page. \item \textbf{page 39, Theorem 1.6.2:} Here you seem to be repeating the last paragraph before Example 18 from page 38. \item \textbf{page 39, Proposition 16:} \textquotedblleft a proper suffix of $m$\textquotedblright\ $\rightarrow$ \textquotedblleft a proper nonempty suffix of $m$\textquotedblright. \item \textbf{page 39, proof of Proposition 16:} \textquotedblleft Let $t$ be a proper suffix\textquotedblright\ $\rightarrow$ \textquotedblleft Let $t$ be a proper nonempty suffix\textquotedblright. \item \textbf{page 39, proof of Proposition 16:} In Case 1, replace \textquotedblleft$\ell^{n}s$\textquotedblright\ by \textquotedblleft$\ell ^{i}s$\textquotedblright. \item \textbf{page 39, proof of Proposition 16:} In Case 2, why do we have \textquotedblleft$t>\ell^{i}s$\textquotedblright? \item \textbf{page 39, proof of Proposition 16:} In Case 3, replace \textquotedblleft proper suffix\textquotedblright\ by \textquotedblleft proper nonempty suffix\textquotedblright. \item \textbf{page 42, \S 1.6:} \textquotedblleft are build in this way\textquotedblright\ $\rightarrow$ \textquotedblleft are built in this way\textquotedblright. \item \textbf{page 42, \S 1.7:} \textquotedblleft Suppose $w$ is\textquotedblright\ $\rightarrow$ \textquotedblleft Suppose $w=a_{1}% a_{2}\cdots a_{n}$ is\textquotedblright. (You later use the notation $a_{i}$ for the letters of $w$.) \item \textbf{page 43, (1.7.25):} Add a period at the end of this equality. \item \textbf{page 43, (1.7.26):} Add a period at the end of this equality. \item \textbf{page 43:} \textquotedblleft Indeed, $b_{\sigma\left( i\right) }$ is the last letter of $w_{\sigma\left( i\right) }$\textquotedblright% \ $\rightarrow$ \textquotedblleft Indeed, $b_{\sigma\left( j\right) }$ is the last letter of $w_{\sigma\left( j\right) }$\textquotedblright. \item \textbf{page 43:} Remove the indentation after (1.7.28). \item \textbf{page 44, Remark 1.7.1:} Replace \textquotedblleft$\sigma\left( i\right) =\pi^{i-1}\left( 1\right)$\textquotedblright\ by \textquotedblleft$\sigma\left( i\right) =\pi^{i-1}\left( \sigma\left( 1\right) \right)$\textquotedblright. \item \textbf{page 47:} When you speak of \textquotedblleft the order $\curlyeqprec_{\omega}$\textquotedblright, it is worth pointing out that it is a (total) pre-order, not a (total) order. Indeed, if two words $u$ and $v$ are powers of one and the same word, then $u\curlyeqprec_{\omega}v$ and $v\curlyeqprec_{\omega}u$. \item \textbf{page 47:} \textquotedblleft multiset of necklaces\textquotedblright\ $\rightarrow$ \textquotedblleft multiset of primitive necklaces\textquotedblright. \item \textbf{page 47:} \textquotedblleft$u^{L\setminus\left\vert u\right\vert }$\textquotedblright\ $\rightarrow$ \textquotedblleft$u^{L/\left\vert u\right\vert }$\textquotedblright. \item \textbf{page 48, Theorem 1.8.1:} Have you really shown that $\Phi$ is a bijection? I see at most a proof of the injectivity of $\Phi$ here. \item \textbf{page 55, Theorem 1.9.4:} Add a period at the end of this equation. \item \textbf{page 56:} \textquotedblleft By Lemma4\textquotedblright% \ $\rightarrow$ \textquotedblleft By Lemma 4\textquotedblright. \end{itemize} \end{document}