(i) If ${[a_{j_1},\cdots ,a_{j_k}]}_r$ were not left-divisible by an $a_j$ with $j\in J$ then by 3.2 it could be represented by a chain from $a_j$ to an $a_j^{\prime },$ $j\prime \in J$ and thus it would not be right-divisible by $a_j^{\prime }$ in contradiction to its definition. Hence ${[a_{j_1},\cdots ,a_{j_k}]}_r$ is divisible by ${[a_{j_1},\cdots ,a_{j_k}]}_l\text{.}$ Analogously one shows that ${[a_{j_1},\cdots ,a_{j_k}]}_l$ is divisible by ${[a_{j_1},\cdots ,a_{j_k}]}_r$ and hence both these elements of $G^{+}$ are equivalent to one another.

(ii) The assertion (ii) follows trivially from (1) and 4.3.

$\square$

$W{\Delta }_J$ is left-divisible by ${\Delta }_J$ by the same argument as in the proof of 5.1. So by 2.3 there is a uniquely determined ${\sigma }_J\left(W\right)$ such that (ii) holds. From 2.3 it also follows immediately that ${\sigma }_J$ is an automorphism of $G_J^{+}\text{.}$ Since ${\sigma }_J$ preserves lengths it takes letters to letters, and hence arises from a permutation $\sigma$ of $J\text{.}$ From $\sigma \left(a\right){\Delta }_J⩦{\Delta }_J{\sigma }^2\left(a\right)$ it follows by application of rev that ${\sigma }^2\left(a\right){\Delta }_J⩦{\Delta }_J\sigma \left(a\right)\text{.}$ The right hand side is positive equivalent to $a{\Delta }_J$ and hence from 2.3, ${\sigma }^2\left(a\right)⩦a\text{.}$ Thus $\sigma$ is an involution and clearly ${\sigma }_J$ is too. Finally, since for all $i,$ $j,$ $${⟨\sigma \left(a_i\right)\sigma \left(a_j\right)⟩}^{m_{ij}}⩦{⟨\sigma \left(a_j\right)\sigma \left(a_j\right)⟩}^{m_{ij}}$$ it follows that $m_{ij}=m_{\sigma \left(i\right)\sigma \left(j\right)}\text{.}$ Thus (i) is proved.

$\square$

It suffices to show that $\Delta$ is a common multiple of the letters of $J\text{.}$ At least, one letter $a$ from $J$ divides $\Delta \text{.}$ Hence let $\Delta ⩦a\Delta \prime \text{.}$ If $b$ is any letter of $J$ with $m_{ab}>2,$ $ba\Delta ⩦\Delta \sigma \left(b\right)\sigma \left(a\right)⩦a\Delta \prime \sigma \left(b\right)\sigma \left(a\right),$ so by the reduction lemma 2.1, we have that $a\Delta$ is divisible by ${⟨ab⟩}^{m_{ab}-1}$ and thus $b$ is a divisor of $\Delta \text{.}$ Hence $\Delta$ is divisible by all the letters of $J$ since the Coxeter-graph of $M_J$ is assumed connected. By 4.1 the existence of ${\Delta }_J$ then follows.

$\square$

(ii) If $a_j$ neither right-divides $V$ nor left-divides $W$ then, by 3.2, one can represent $V$ by a chain with target $a_j$ and $W$ by a chain with source $a_j\text{.}$ Thus one can represent $VW$ by a chain [Ed: a word in the letters of $J$] which, by 3.1, is not divisible by its source, and hence neither by ${\Delta }_J\text{.}$

$\square$

(i) From 3.5 it follows immediately by induction on the number of elements of $J$ that ${\Delta }_J$ is square free; and consequently so are its divisors. The converse is shown by induction on the length of $U\text{.}$ Let $U⩦Va\text{.}$ By the induction assumption there exists a $W$ with ${\Delta }_J⩦VW\text{.}$ Since $a$ does not right-divide $V,$ it left divides $W$ by 5.3 and hence $U$ is a divisor of ${\Delta }_J\text{.}$

(ii) The assertion (ii) follows trivially from (i).

$\square$

[Ed: In this proof $G_M^{+}$ is written $G^{+}$ for simplicity]

We will show first of all the equivalence of (i) and (ii), where clearly (ii) trivially implies (i). Let $\Lambda ⩦\Delta$ or $\Lambda ⩦{\Delta }^2$ according to whether $\sigma =1$ or not. Then $\Lambda$ is, by 5.2, a central element in $G^{+}$ and for each letter $a_i,$ $i\in I,$ there is by 5.1 a ${\Lambda }_i$ with $\Lambda =a_i{\Lambda }_i\text{.}$ Now, if $A⩦a_{i_1}\dots a_{i_m}$ is an arbitrary element of $G^{+}$ then $${\Lambda }^m⩦a_{i_m}{\Lambda }_{i_m}\cdots a_{i_1}{\Lambda }_{i_1}⩦A{\Lambda }_{i_m}\cdots {\Lambda }_{i_1}\text{.}$$ Hence ${\Lambda }^m$ is divisible by each element $A$ of $G^{+}$ with $L\left(A\right)\le m\text{.}$ In particular, a finite set of elements always has a common multiple and thus by 4.1 a least common multiple. This proves the equivalence of (i) and (ii).

If (ii), then (iv). Since to all $B,C\in G^{+}$ there exists a common multiple, and thus $B\prime ,C\prime ,\in G^{+}$ with $BC\prime ⩦CB\prime \text{.}$ From this and cancellativity, 2.3, it follows by a general theorem of Öre that $G^{+}$ embeds in a group. Thence follows the injectivity of $G^{+}\to G$ and also that each element $A\in G$ can be represented in the form $A=C^{-1}B$ or also $B\prime C^{-1}$ with $B,B\prime ,C,C\prime \in G^{+}\text{.}$ That $C$ can moreover be chosen to be central follows from the fact that — as shown above — to every $C$ with $L\left(C\right)\le m$ there exists $D\in G^{+}$ with ${\Lambda }^m⩦CD$ so that, therefore, $C^{-1}={\Lambda }^{-m}D=D{\Lambda }^{-m}\text{.}$

[Ed: As an alternative to applying Öre’s condition we provide the following proof of the injectivity of $G^{+}\to G$ when there exists a fundamental element $\Delta \text{.}$

By (5.2) it is clear that ${\Delta }^2$ is a central element in $G^{+}\text{.}$ Let $W,W\prime$ be positive words such that $W=W\prime$ in $G\text{.}$ Then there is some sequence $W_1,W_2,\dots ,W_k$ of words in the letters of $I$ and their inverses such that $W\equiv W_1=W_2=\cdots =W_k\equiv W\prime$ where at each step $W_{i+1}$ is obtained from $W_i$ either by a positive transformation (cf 1.3.) or (so-called trivial) insertion or deletion of a subword $a{a}^{-1}$ or $a^{-1}a$ for some letter $a\text{.}$ Note that the number of inverse letters appearing in any word is bounded by $k\text{.}$

Let $C$ denote the central element ${\Delta }^2$ of $G^{+}\text{.}$ Then we may define positive words $V_i$ for $i=1,\dots ,k$ such that $V_i=C^k{W}_i$ as follows. Write $W_i\equiv U{a}^{-1}U\prime$ for $U$ a positive word, $a$ a letter. Then $CU⩦UC,$ so if we let $C_a$ denote the unique element of $G^{+}$ such that $C_{a}a⩦C,$ $C{W}_i=CU{a}^{-1}U\prime =UC{a}^{-1}U\prime =UC{a}^{-1}U\prime =U{C}_{a}U\prime$ where $U{C}_a$ is positive. Repeating this step for successive inverses in $U\prime$ yields a positive word $V_i^{\prime }$ equal in $G$ to $C^r{W}_i$ for some $r\le k\text{.}$ Put $V_i\equiv C^{k-r}{V}_i^{\prime }\text{.}$ Essentially, $V_i$ is obtained from $W_i$ by replacing each occurrence of $a^{-1}$ with the word $C_a,$ and then attaching unused copies of $C$ to the front.

Now we check that $V_i⩦V_{i+1}\text{.}$

If $W_{i+1}$ differs from $W_i$ by a positive transformation, then $W_{i+1}$ is $W_i$ with some positive subword $U$ switched with a positive subword $U\prime ,$ and so the same transformation applied to $V_i$ gives the word $V_{i+1},$ so they are positive equivalent.

If $W_{i+1}$ is obtained from $W_i$ by insertion of $a{a}^{-1}$ or $a^{-1}a$ Then $V_{i+1}\equiv C^{r}U{C}_{a}aV$ or $C^{r}Ua{C}_{a}V$ for positive words $U,V,$ where $V_i\equiv C^{r+1}UV\text{.}$ By the centrality of $C$ and the fact that $C⩦a{C}_a⩦C_{a}a,$ $V_i$ and $V_{i+1}$ are positive equivalent.

If $W_{i+1}$ is obtained by a trivial deletion, then the proof is identical as above, but with the roles of $W_{i+1}$ and $W_i$ reversed.

Hence we have a sequence of words $V_1,V_2,\dots ,V_k$ such that each is positive equivalent to its predecessor, so that $V_1$ is positive equivalent to $V_k\text{.}$ But $V_1\equiv C_{k}W$ and $V_k\equiv C_{k}W\prime$ so by cancellativity, $W$ is positive equivalent to $W\prime \text{.}$

So $G^{+}$ embeds in $G\text{.}$ ]

Assuming (iv), (iii) follows trivially. And from (iii), (ii) follows easily. Since for $B,C\in G^{+}$ there exist $B\prime ,C\prime \in G^{+}$ with $C^{-1}B=B\prime {C\prime }^{-1},$ and thus $BC\prime =CB\prime$ and consequently $BC\prime ⩦CB\prime$ so by 4.1 $B$ and $C$ have a least common multiple. Thus 5.5 is proved.

$\square$

By Tits, ${\bar{G}}_M$ is finite exactly when $QF{G}_M^{+}$ is finite. By 5.4 and 3.5 this is the case if and only if $\Delta$ exists. Since, if $\Delta$ exists, by 5.4 $QF{G}_M^{+}$ consists of the divisors of $\Delta \text{.}$ And if $\Delta$ does not exist, by 3.5 there exists a sequence of infinitely many distinct square free elements.

$\square$

If an element is not square free, then it is represented by a word $W$ which contains a square. But this is clearly not a reduced word for the Coxeter element $\bar{W},$ and so $l\left(\bar{W}\right)\le L\left(W\right)-2$ (the square cancels).

Conversely, suppose that $W$ represents a square free element of $G_M^{+}\text{.}$ By definition of length there is a $V\in G_M^{+}$ such that $\bar{V}=\bar{W}$ and $L\left(W\right)=l\left(\bar{V}\right)=l\left(\bar{W}\right)\text{.}$ Then by above $V$ is square free. But $\bar{V}=\bar{W}$ and hence by Tits theorem $V⩦W$ and $L\left(W\right)=L\left(V\right)=l\left(\bar{W}\right)\text{.}$

$\square \text{]}$

(i) By 5.4, the elements of $QF{G}_M^{+}$ are the divisors of $\Delta \text{.}$ A proper divisor $W$ of $\Delta$ clearly has $L\left(W\right)<L\left(\Delta \right)\text{.}$ Thus $\Delta$ is the unique square free element of maximal length.

(ii) By the theorem of Tits and (i) there is also in ${\bar{G}}_M$ only one unique element of maximal length, namely $\bar{\Delta }\text{.}$ A positive word with $\bar{W}=\bar{\Delta }$ and $L\left(W\right)=l\left(\bar{\Delta }\right)$ is according to Tits square-free and it has maximal length, so by (i) it represents $\Delta \text{.}$ That only such positive words can represent $\Delta$ is clear.

$\square$

By Bourbaki, V §4 number 8 corollary to proposition 8, or from the classification of finite Coxeter groups we know that if $(W,S)$ is irreducible and finite then its graph is a tree. So the statement about decompositions boils down to the following statement about trees: if $\Gamma$ is a tree with a finite set $S$ of vertices then there exists a unique partition $(S\prime ,S″)$ (up to interchange of $S\prime$ and $S″\text{),}$ of $S$ into two sets such that no two elements of $S\prime$ and no two elements of $S″$ are joined by an edge.

We prove this by induction on the number of vertices of $\Gamma \text{.}$ For a graph on one vertex $a$ it is clear that the only suitable partitions are $(\left\{a\right\},\varphi )$ and $(\varphi ,\left\{a\right\})\text{.}$ Now let $\Gamma$ be an arbitrary tree with a finite set of vertices and let $a$ be a terminal vertex. Then applying the assumption to the subgraph of $\Gamma$ whose vertices are those vertices $n\ne a$ of $\Gamma$ we see that there exists a unique partition $(S_1^{\prime },S_l^{″})$ (up to interchange of $S_1^{\prime },S_1^{″}\text{)}$ of $S\backslash \left\{a\right\}$ such that no two elements of $S_1^{\prime }$ and no two elements of $S_1^{″}$ are joined by an edge of $\Gamma \prime \text{.}$ Now, by definition of a tree, $a$ is joined to exactly one vertex $b$ of $\Gamma \prime \text{.}$ Without loss of generality let $b\in S_1^{\prime }\text{.}$ Then it is easy to see that $(S_1^{\prime },S_1^{″}\cup \left\{a\right\})$ is a partition of $S$ satisfying the above conditions and that it is unique up to interchanging $S_1^{\prime }$ and $S_1^{″}\cup \left\{a\right\}\text{.}$

$\square \text{]}$

According to Bourbaki, V §6 ex 2 (6) the corresponding equations for $\bar{\Delta },$ $\bar{\Pi },$ $\bar{\Pi }\prime ,$ $\bar{\Pi }″$ hold. Since, in addition, the elements on the right hand sides of the equations have length $nh/2$ the statement follows from Proposition 5.7 (ii).

$\square$

By [Bou1968] V §6.1 Lemma 1, all products $P$ of generators in $G_M$ are conjugate to one another. Thus $P^h$ is conjugate to ${\Pi }^h$ and $P^{h/2}$ is conjugate to ${\Pi }^{h/2}$ if $h$ is even. If $\Delta$ is central and $h$ is even then $\Delta ={\Pi }^{h/2}$ is central and hence $P^{h/2}={\Pi }^{h/2}$ in $G_M$ and thus $P^{h/2}⩦{\Pi }^{h/2}⩦\Delta$ in $G_M^{+}\text{.}$ Likewise it follows immediately that $P^h⩦{\Pi }^h⩦{\Delta }^2$ since ${\Delta }^2⩦{\Pi }^h$ is always central. Hence (i) and (ii) are shown. [Note: we are using the fact that $G_M^{+}⟶G_M$ is injective here.]

(iii) Suppose $\Delta$ is not central, i.e. $\sigma \ne \text{id}$ and let $h$ be even. If for all products $P⩦a_{i_1}\cdots a_{i_n}$ we were to have the equation $P^{h/2}⩦\Delta$ then this also would be true for the product $a_{i_n}P{a}_{i_n}^{-1}$ which arises from it by cyclic permutation of the factors. Now, in if $P$ were such a product with $\sigma \left(a_{i_n}\right)\ne a_{i_n}$ then we would have $a_{i_n}{P}^{h/2}{a}_{i_n}^{-1}⩦\Delta$ and thus $a_{i_n}\Delta ⩦\Delta a_{i_n}$ in contradiction to $a_{i_n}\Delta ⩦\Delta \sigma \left(a_{i_n}\right)\text{.}$

$\square$