# Weyl Group Element $w$ fixing a root, and its presentation as product of simple reflections $w=s_1\dots s_n$

Let $$\Phi$$ be a root system and $$\gamma \in \Phi$$ a root. Let $$W$$ be the Weyl group and $$\Delta$$ a set of simple roots. Let $$w \in W$$ such that $$w(\gamma)=\gamma$$. Is it true that if $$w=s_1\dots s_n$$ with $$s_i$$ simple reflections, then $$s_i(\gamma)=\gamma$$ for $$i=1\dots n$$?

I can't see a reason why this shouldn't be true, but am unable to prove it.

Thanks!

This is definitely not true. For instance, already in $$\Phi=B_2$$, each root has a root orthogonal to it, so for every root there is some nontrivial element (in fact, a reflection) of the Weyl group fixing it. But e.g. a simple root in $$B_2$$ is not fixed by any simple reflection.
However, what you might want to know is the following. If we choose any point $$v$$ in the dominant chamber of our root system, then the stabilizer of $$v$$ is exactly the parabolic subgroup of $$W$$ generated by simple reflections that fix $$v$$. For a proof of this see Lemma 10.3B of Humphreys' "Introduction to Lie Algebras and Representation Theory" (https://www.springer.com/us/book/9780387900537).