The simple braids and divisibility lattice for b three

The six simple braids and their permutations. By [F2] there are exactly $3!=6$ simple braids. Applying [F3] to the six displayed elements gives the images $\mathrm{i}\mathrm{d}$, $s_1=(12)$, $s_2=(23)$, $s_1{s}_2=(123)$, $s_2{s}_1=(132)$ and $s_1{s}_2{s}_1=(13)$ in this order; these are the six elements of $S_3$, so they are pairwise distinct, and since $\pi$ is a function the six braids $1,{\sigma }_1,{\sigma }_2,{\sigma }_1{\sigma }_2,{\sigma }_2{\sigma }_1,\mathrm{\Delta }$ are pairwise distinct and exhaust the left divisors of $\mathrm{\Delta }$. In particular ${\sigma }_1{\sigma }_2$ and ${\sigma }_2{\sigma }_1$ are proper simple braids, because ${\sigma }_1{\sigma }_2\cdot {\sigma }_1=\mathrm{\Delta }$ and ${\sigma }_2{\sigma }_1\cdot {\sigma }_2=\mathrm{\Delta }$ by [F2].

The atom pair $({\sigma }_1,{\sigma }_2)$. The left divisors of ${\sigma }_1$ are $1$ and ${\sigma }_1$, since a divisor $d$ of ${\sigma }_1$ has $\mathrm{\ell }(d)\le 1$, so either $d=1$ or $d$ is an atom equal to ${\sigma }_1$; the same holds for ${\sigma }_2$, and the distinct atoms are incomparable, so the common left divisors of ${\sigma }_1$ and ${\sigma }_2$ are just $1$: thus ${\sigma }_1{\wedge }_L{\sigma }_2=1$. Their least common left multiple is ${\sigma }_1{\sigma }_2{\sigma }_1$ by [F4], which is $\mathrm{\Delta }$ by [F2], so ${\sigma }_1{\vee }_L{\sigma }_2=\mathrm{\Delta }$.

A pair with different left and right meets. For $a={\sigma }_1{\sigma }_2$ and $b={\sigma }_1$: ${\sigma }_1$ left-divides $b$ trivially and left-divides $a={\sigma }_1{\sigma }_2$, so ${\sigma }_1$ is a common left divisor; every common left divisor $d$ of $a$ and $b$ divides $b={\sigma }_1$, hence is $1$ or ${\sigma }_1$, so $d{\preccurlyeq }_L{\sigma }_1$: therefore $a{\wedge }_{L}b={\sigma }_1$. On the right: the right divisors of $b={\sigma }_1$ are $1$ and ${\sigma }_1$ by [F1], while the right divisors of $a={\sigma }_1{\sigma }_2$ are $1,{\sigma }_2,{\sigma }_1{\sigma }_2$ — indeed if $a=dc$ then $\mathrm{\ell }(c)\le 2$; for $\mathrm{\ell }(c)=2$ one has $\mathrm{\ell }(d)=0$, so $d=1$ and $c=a$; for $\mathrm{\ell }(c)=1$ the element $c$ is an atom, and $d$ then has length $1$ and is an atom as well, so $a$ must be a product of two atoms, the four candidates being ${\sigma }_1{\sigma }_1,{\sigma }_1{\sigma }_2,{\sigma }_2{\sigma }_1,{\sigma }_2{\sigma }_2$, which are distinct because every defining relation has length three; only ${\sigma }_1{\sigma }_2=a$ occurs, so the only one-letter right divisor is $c={\sigma }_2$. Hence the common right divisors of $a$ and $b$ are $\{1,{\sigma }_2,{\sigma }_1{\sigma }_2\}\cap \{1,{\sigma }_1\}=\{1\}$, and $a{\wedge }_{R}b=1$.

Conclusion. Steps 1.1--1.3 give the six simple braids, ${\sigma }_1{\wedge }_L{\sigma }_2=1$, ${\sigma }_1{\vee }_L{\sigma }_2=\mathrm{\Delta }$, and the pair $a,b$ with $a{\wedge }_{L}b={\sigma }_1\mathrm{\ne }1=a{\wedge }_{R}b$. So already on the four-element subfamily $\{1,{\sigma }_1,{\sigma }_2,{\sigma }_1{\sigma }_2\}$ the two orders have different meets, even though on $\mathrm{\Delta }$ itself the left and right divisor sets coincide by [F2]; the balancedness established in Delta is the lcm of the artin atoms and has the same left and right divisors is therefore a property of $\mathrm{\Delta }$ and does not identify the two orders. No choice principle is used. ∎