A question on Coxeter groups Let $W$ be a Coxeter group, and $S$ be its set of simple reflections. For any $w \in W$, define $\mathcal L(w)$ 

$\mathcal L(w) = \{s \in S| sw<w \}.$

Is it true:

If $w \in W$, $s_1,s_2 \in S$, the order of $s_1s_2$ is $\alpha$ (an integer $\geq 2$ or $\infty$), and $\mathcal L(w) =\{s_1,s_2\}$, then $\alpha$ is finite and there exists $w' \in W$ such that $w = (s_1s_2)^{\alpha}w'$, $\ell(w)=\ell(w')+\alpha$.

 A: It is not true as written. By definition of $\alpha$, $(s_1 s_2)^\alpha = 1$, so you are requiring $w = w'$,
while $\ell(w) = \ell(w') + \alpha$ where $\alpha \geq 2$; which is absurd.
I suspect that by $(s_1 s_2)^\alpha$ you actually meant the word alternating between the letters $s_1$ and $s_2$ with $\alpha$ letters. I've seen someone call this $(s_1 s_2)^{\alpha/2}$ (maybe Conway?),
so $(s_1 s_2)^{3/2}$ means $s_1 s_2 s_1$ and $(s_1 s_2)^{4/2}$ means $s_1 s_2 s_1 s_2$.
In this case, the claim IS true. In fact, a weaker hypothesis suffices: if $w \in W$ and and $s_1, s_2$ are simple reflections such that $\ell(s_1 w)$ and $\ell(s_2 w)$ are less than $\ell(w)$, then the order $\alpha$ of $(s_1 s_2)$ is finite, and $w$ can be written $w = (s_1 s_2)^{\alpha/2} v$ with $\ell(w) = \ell(v) + \alpha$.
This is Lemma 4.2.9 of Arjeh Cohen's notes on Coxeter groups. I'll reproduce a proof here. We need the exchange condition for Coxeter groups, which is Theorem 4.2.2 in Cohen's notes:

If $s, r_1, \dots, r_q \in S$ satisfy $w = r_1 \dotsm r_q$ and $q = \ell(w) \geq \ell(sw)$, then there is $j \leq q$ such that $s r_1 \dotsm r_{j-1} = r_1 \dotsm r_j$.

First of all, if $w \in \langle s_1, s_2 \rangle$,
then since $s_1 w$ and $s_2 w$ are both shorter, $w$ must be the
longest element of the Coxeter group $\langle s_1, s_2 \rangle$; such an element exists (and is unique) iff the group is finite. Hence $\alpha$ is finite, and the longest element of the group $\langle s_1, s_2 \rangle$ is precisely $(s_1 s_2)^{\alpha/2}$.
Otherwise, let $u = r_1 \dotsm r_k$ be a word in $\langle s_1, s_2 \rangle$ and $v = r_{k+1} \dotsm r_q$ be any word such that $w = uv$ is a minimal-length expression for $w$ and $u$ is as long as possible.
By the exchange condition, there exists $j$ such that
$s_1 r_1 \dotsm r_j = r_1 \dotsm r_{j+1}$.
If $j \geq k$, then $s_1 u v$ is
$$
s_1 r_1 \dotsm r_k r_{k+1} \dotsm r_q =
r_1 \dotsm r_k r_{k+1} \dotsm r_j r_{j+2} \dotsm r_q
= u v'
$$
where $v'$ is shorter than $v$.
But then $w = s_1 u v'$ is a minimal-length expression for $w$
with $(s_1 u) \in \langle s_1, s_2 \rangle$ and $\ell(s_1 u) > \ell(u)$,
contradicting $\ell(u)$'s maximality.
So $j < k$,
and so $s_1 u = r_1 \dotsm r_j r_{j+2} \dotsm r_k$
is shorter than $u$.
Similarly, $s_2 u$ is shorter than $u$.
So by the first case, $u \in \langle s_1, s_2 \rangle$ is the longest element of $\langle s_1, s_2 \rangle$; hence $\alpha$ is finite and $u = (s_1 s_2)^{\alpha/2}$.
