S3 bruhat order and inversion sets

Direct composition gives $1=123$, $a=213$, $b=132$, $ab=231$, $ba=312$, and $aba=bab=321$. No word of length at most two represents $321$: after canceling $aa=bb=1$, the list of such words is $1,a,b,ab,ba$. Thus these six words have lengths $0,1,1,2,2,3$ respectively. At lengths one and two the indicated reduced words are unique. At length three any adjacent repetition cancels, so the only reduced words are $aba$ and $bab$, both for $321$. This enumerates every reduced expression of every element.

The reduced subwords of $ab$ give $\{1,a,b,ab\}$ and those of $ba$ give $\{1,a,b,ba\}$. The reduced subwords of each of $aba$ and $bab$ give all six elements: their one-letter choices give $a,b$, their adjacent two-letter choices give $ab,ba$, their empty subword gives $1$, and the full word gives $321$; the nonadjacent equal-letter choice is not reduced. For $1,a,b$ the lower sets are respectively $\{1\},\{1,a\},\{1,b\}$. Hence the subword relation is independent of the reduced expression in this group. These explicitly nested lower sets also show reflexivity, antisymmetry (distinct same-rank elements are incomparable), and transitivity, so they define a partial order.

Its covers are $1\to a,b$; $a\to ab,ba$ and $b\to ab,ba$; and $ab,ba\to 321$. Each is left multiplication by a transposition. For the middle four covers, compute $(ab)a^{-1}=aba=(13)$, $(ba)a^{-1}=b$, $(ab)b^{-1}=a$, and $(ba)b^{-1}=bab=(13)$. For the last two use $321=bab$ to get $321(ab{)}^{-1}=b$, and $321=aba$ to get $321(ba{)}^{-1}=a$. The first two use $a,b$ themselves. Each cover raises length by one. Conversely any transposition multiplication raising length by one must connect adjacent ranks, and the displayed list includes every possible pair of adjacent ranks. Thus chains of such multiplications give exactly the subword order computed in 2.1; both usual strong Bruhat conventions agree here.

Direct inequalities between the three entries give the inversion sets: $123$ has $\mathrm{\varnothing }$, $213$ has $\{(1,2)\}$, $132$ has $\{(2,3)\}$, $231$ has $\{(1,3),(2,3)\}$, $312$ has $\{(1,2),(1,3)\}$, and $321$ has all three pairs. Their sizes match the lengths in 1.1. Yet $213\le 231$ by 2.1 while $(1,2)$ belongs only to the former inversion set. This is the required failed containment witness. The empty word is the unique minimum and either longest reduced word gives the same unique maximum. Every computation is finite and explicit, so no choice or general exchange theorem is assumed.