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The Coxeter graph has vertex set $I$, joins $i$ and $j$ when $m_{ij}\ge 3$, and labels the edge by $m_{ij}$ when $m_{ij}>3$. A Coxeter system is irreducible exactly when this graph is connected.

A Coxeter group is a group with a presentationwhere $m_{ii}=1$ and $m_{ij}=m_{ji}\in \{2,3,\dots ,\infty \}$ for $i\ne j$. The pair consisting of $W$ and its distinguished generating reflections is a Coxeter system.

The Coxeter matrix of a Coxeter system is the symmetric matrix whose entry $m_{ij}$ is the order of $x_i{x}_j$.

For a finite-rank Coxeter system, let $V$ have basis $(e_i{)}_{i\in I}$ and symmetric bilinear formIts geometric representation sends the generator $x_i$ to the reflection

First suppose $(W,S)$ is a Coxeter system. The usual exchange condition implies that multiplication by a simple generator changes length by exactly one. Let $w=t_1\cdots t_n$ be reduced and suppose both $s_{i}w$ and $w{s}_j$ have length $n+1$. The length of $s_{i}w{s}_j$ is therefore either $n+2$ or $n$. In the latter case, apply exchange to the reduced word $s_i{t}_1\cdots t_n$ followed by $s_j$. If exchange deleted one of the $t_k$, multiplying the resulting equality on the left by $s_i$ would express the length-$n+1$ element $w{s}_j$ using only $n-1$ generators. Hence exchange must delete the initial $s_i$, giving $s_{i}w{s}_j=w$. This is exactly the folding condition.

Conversely, suppose the folding condition holds, and let $W_M$ be the abstract Coxeter group with generators $S$ and matrix $(m_{ij})$. The defining relations hold in $W$, so there is a surjective homomorphismIt remains to prove injectivity. Take any word in the kernel. If its image word in $W$ is not reduced, choose its shortest nonreduced prefix $ut$, where $u$ is reduced and $t$ is its last generator. Part b gives $ut=v$, where $v$ is obtained by deleting one letter from $u$. Hence $u=vt$, and $u$ and $vt$ are two reduced expressions for the same element. By the assumed braid-equivalence theorem they are related by braid moves. Those moves are defining relations in $W_M$, after which the end of the prefix becomes $vtt$ and $t^2=1$ shortens the original word by two.

Repeating this process turns the kernel word, using only Coxeter relations, into a word that is reduced in $W$. Since its image is the identity, that reduced word is empty. The original word is therefore already the identity in $W_M$, so $\ker \pi =1$. Hence $W\cong W_M$ is the Coxeter group with generators $S$.

Solved by gpt-5.6-sol high.

The Coxeter graph has vertex set $I$. Distinct vertices $i,j$ are joined precisely when $m_{ij}\ge 3$, and the edge is labelled $m_{ij}$ when $m_{ij}>3$; the customary unlabelled edge therefore means $m_{ij}=3$. The Coxeter system is an Irreducible Coxeter system precisely when this graph is connected.

Solved by gpt-5.6-sol high.