Positive artin relations preserve homogeneous length

Define a relation $\sim$ on ${\mathrm{\Sigma }}_n^{\ast }$ by $u\sim v$ if and only if $\mathrm{\mid }u\mathrm{\mid }=\mathrm{\mid }v\mathrm{\mid }$. It is reflexive, symmetric and transitive because equality of natural numbers is, so it is an equivalence relation.

For (b): let $W_k:=\{w\in {\mathrm{\Sigma }}_n^{\ast }:\mathrm{\mid }w\mathrm{\mid }=k\}$ be the set of positive words of length $k$. We prove $\mathrm{\mid }{W}_k\mathrm{\mid }=\mathrm{\mid }{\mathrm{\Sigma }}_n{\mathrm{\mid }}^k$ by induction on $k$: $W_0=\{\epsilon \}$ has one element and $\mathrm{\mid }{\mathrm{\Sigma }}_n{\mathrm{\mid }}^0=1$; and each word of length $k+1$ is $ws$ for a unique $w\in W_k$ and a unique letter $s\in {\mathrm{\Sigma }}_n$, so $\mathrm{\mid }{W}_{k+1}\mathrm{\mid }=\mathrm{\mid }{W}_k\mathrm{\mid }\cdot \mathrm{\mid }{\mathrm{\Sigma }}_n\mathrm{\mid }=\mathrm{\mid }{\mathrm{\Sigma }}_n{\mathrm{\mid }}^k\cdot \mathrm{\mid }{\mathrm{\Sigma }}_n\mathrm{\mid }=\mathrm{\mid }{\mathrm{\Sigma }}_n{\mathrm{\mid }}^{k+1}$. Finally $\mathrm{\mid }{\mathrm{\Sigma }}_n\mathrm{\mid }=n-1$ for $n\ge 2$, while for $n\le 1$ the set $\{{\sigma }_1,\dots ,{\sigma }_{n-1}\}$ is empty, which gives the two cases displayed in (b).

The relation $\sim$ is compatible with concatenation: if $\mathrm{\mid }u\mathrm{\mid }=\mathrm{\mid }v\mathrm{\mid }$, then for all words $x,y$ we have $\mathrm{\mid }xuy\mathrm{\mid }=\mathrm{\mid }x\mathrm{\mid }+\mathrm{\mid }u\mathrm{\mid }+\mathrm{\mid }y\mathrm{\mid }=\mathrm{\mid }x\mathrm{\mid }+\mathrm{\mid }v\mathrm{\mid }+\mathrm{\mid }y\mathrm{\mid }=\mathrm{\mid }xvy\mathrm{\mid }$, so $xuy\sim xvy$. Hence $\sim$ is a congruence on ${\mathrm{\Sigma }}_n^{\ast }$.

Every pair of $R_n$ has two sides of equal length: ${\sigma }_i{\sigma }_{i+1}{\sigma }_i$ and ${\sigma }_{i+1}{\sigma }_i{\sigma }_{i+1}$ both have three letters, and ${\sigma }_i{\sigma }_j$ and ${\sigma }_j{\sigma }_i$ both have two letters. Hence each such pair lies in the congruence $\sim$ of step 2.1.

Since ${\equiv }^{+}$ is the smallest congruence containing all pairs in $R_n$ and $\sim$ is one such congruence by steps 2.1--3.1, we have $u{\equiv }^{+}v\Rightarrow u\sim v$, that is, equivalent positive words have equal length. This is (a), first part.

Hence $\mathrm{\ell }(x):=\mathrm{\mid }w\mathrm{\mid }$ for any word $w$ with $x=[w]$ is independent of the chosen representative $w$, and $\mathrm{\ell }$ is a function $B_n^{+}\to \mathbb{N}$; moreover $\mathrm{\ell }([\epsilon ])=\mathrm{\mid }\epsilon \mathrm{\mid }=0$ and $\mathrm{\ell }(xy)=\mathrm{\ell }([uv])=\mathrm{\mid }uv\mathrm{\mid }=\mathrm{\mid }u\mathrm{\mid }+\mathrm{\mid }v\mathrm{\mid }=\mathrm{\ell }(x)+\mathrm{\ell }(y)$ for representatives $u$ of $x$ and $v$ of $y$. Thus $\mathrm{\ell }$ is a monoid homomorphism and (a) is complete.

$\mathrm{\ell }(x)=0$ if and only if $x=1$: if $x=[w]$ with $\mathrm{\mid }w\mathrm{\mid }=0$, then $w=\epsilon$ and $x=[\epsilon ]=1$; conversely $\mathrm{\ell }(1)=0$. If $\mathrm{\ell }(v)=0$ then $v=1$.

The map $W_k\to B_n^{+}$, $w\mapsto [w]$, is a surjection onto the set of elements of length $k$ by step 5.1, and a surjection from a finite set onto a set makes the target finite with cardinality at most that of the source. Hence there are at most $\mathrm{\mid }{\mathrm{\Sigma }}_n{\mathrm{\mid }}^k$ elements of $B_n^{+}$ of length $k$, which is (b).

Let $x_1,\dots ,x_r\in B_n^{+}$ with $x_1\cdots x_r=1$. By step 5.1, $\mathrm{\ell }(x_1)+\cdots +\mathrm{\ell }(x_r)=\mathrm{\ell }(1)=0$, and each $\mathrm{\ell }(x_i)\in \mathbb{N}$; a sum of natural numbers is zero only if every summand is zero, so $\mathrm{\ell }(x_i)=0$ for all $i$, and step 6.1 gives $x_i=1$. Taking $r=2$ shows $xy=1\Rightarrow x=y=1$; hence if $x$ has a two-sided inverse $y$ (so $xy=1$) then $x=1$, and $1$ is the only invertible element. This is (c).

For (d): $\mathrm{\ell }({\bar{\sigma }}_{i}x)=\mathrm{\ell }({\bar{\sigma }}_i)+\mathrm{\ell }(x)=\mathrm{\ell }([{\sigma }_i])+\mathrm{\ell }(x)=1+\mathrm{\ell }(x)$ for every $x$, using step 5.1 and $\mathrm{\mid }{\sigma }_i\mathrm{\mid }=1$. In particular $\mathrm{\ell }({\bar{\sigma }}_i)>0$, so ${\bar{\sigma }}_i\mathrm{\ne }1$ by step 6.1.

Collecting: (a) is steps 4.1--5.1, (b) is steps 1.2 and 6.2, (c) is step 7.1, and (d) is step 7.2. In particular the length function $\mathrm{\ell }$ exists, is additive, takes the value $0$ only on $1$, and satisfies $\mathrm{\ell }({\bar{\sigma }}_{i}x)>\mathrm{\ell }(x)$ for every generator ${\bar{\sigma }}_i$; these are the homogeneity, conicality and strict-increase properties used later on this page. ∎