In a given coxeter group $(W,S)$, a reflection is an element of $W$ that can be written with a symmetric word in the generators $S$.


In multiple sources, I found the following formula:
$$
 \mathrm{dp}(\alpha) = \frac{1}{2}(l(t\_\alpha) + 1)
$$
where $\alpha$ is a positive root, $t\_\alpha$ the corresponding reflection and the depth $\mathrm{dp}(\alpha)$ is the length of a shortest word $w$ such that $w\cdot \alpha$ is a negative root.


Assuming that any reflection has length achievable by a symmetric word, this formula is rather easy to check, but I couldn't find a proof for just this fact.


In [X Fu's](http://www.maths.usyd.edu.au/u/PG/Theses/X-Fu-2010_thesis.pdf) thesis, Lemma 1.3.19, the formula is proven but I'm looking for a more elementary proof of this fact:


**Question:** Is the length of a reflection in a Coxeter group achievable by a symmetric word?



Reference Request: Length of a reflection in a Coxeter group can be achieved by symmetric word
Let $\beta$ be a positive root. Let $u$ be an element of length $\mathrm{dp}(\beta)$ such that $u(\beta) <0$. Then $u(\beta)=-\alpha$ for some simple root $\alpha$ (because otherwise we could multiply $u$ by a left descent to get an element of shorter length inverting $\beta$), so $u^{-1}(\alpha)=-\beta$ and hence $\ell(s\_\alpha u) <\ell(u)$.


Now since $u^{-1}s\_\alpha(\alpha) = \beta$, we have that $(s\_\alpha u)^{-1}s\_\alpha (s\_\alpha u)=s\_\beta$. Thus $s\_\beta$ has a symmetric word of length $2\mathrm{dp}(\beta)-1$.


This proves that $\ell(s\_\beta) \leq 2\mathrm{dp}(\beta)-1$. For the opposite inequality, let $(s\_1,\ldots,s\_k)$ be a reduced word for $s\_\beta$. Let $k+1-i$ be the maximal index such that $s\_{k+1-i}\cdots s\_k(\beta) =-\alpha<0$. Then $k+1-(k+1-i)= i\geq\mathrm{dp}(\beta)$. Now $s\_1\cdots s\_{k-i }s\_{k+1-i}(\alpha)=-\beta$, so $s\_{k+1-i}\cdots s\_{2}s\_{1}(\beta)=-\alpha$, so $k+1-i\geq\mathrm{dp}(\beta)$. Hence $k+1\geq 2\mathrm{dp}(\beta)$, and the result follows. 


