Action of Coxeter element on mod $2$ root lattice is semisimple Let $\Lambda$ be a simply laced root lattice and $w$ a Coxeter element of the Weyl group of $\Lambda$.
Question: Is it true that the action of $w$ on the $\mathbb{F}_2$-vector space $\Lambda/2\Lambda$ is semisimple?
Motivation: I want to show that, if $F$ denotes the subset of $w$-fixed points on $\Lambda/2\Lambda$, then $w$ has no nonzero fixed points on $\left(\Lambda/2\Lambda\right)/F$. This is more or less equivalent to showing that $w$ acts semisimply on its $1$-eigenspace. A proof of the latter claim would already be great!
 A: The answer seems to be negative.
According to this answer, an operator $X\in\operatorname{M}(n,\mathbb{F}_q)$ is semisimple if and only if $X^{q^m}=X$, where $m=\operatorname{lcm}(2,\ldots,n)$.
Now consider the standard realization of the root system of type $\mathsf{A}_n$ inside $\mathbb{Z}^{n+1}$, where the simple roots are $\alpha_i = e_i-e_{i+1}$ for $i=1,\ldots,n$. Then
$$\Lambda = \{ (x_1,\ldots,x_{n+1})\in\mathbb{Z}^{n+1} \mid x_1+\ldots+x_{n+1}=0 \},$$
and the action of $W(\mathsf{A}_n)\cong S_{n+1}$ is by permutations of the entries.
One possible choice of the Coxeter element $w$ is the long cycle
$$w = (1\ \ 2\ \ \ldots\ \ n+1).$$
For $n=3$ one gets $\dim(\Lambda/2\Lambda)=3$ and $m=\operatorname{lcm}(2,3)=6$. Since $\operatorname{ord}(w)=n+1=4$ and $2^6 = 64 \equiv 0 \pmod{4}$, one has $w^{2^m} = \operatorname{id}$, while $w$ act non-trivially on $\Lambda/2\Lambda$.
The same considerations work for $\mathsf{A}_n$, $n$ odd, and for $\mathsf{D}_n$ (where the Coxeter number equals $2n-2$ and $2^m$ always gives an even remainder modulo $2n-2$).
The conjecture holds for $\mathsf{A}_n$, $n$ even, because in this case $2^m\equiv 1\pmod{n+1}$.
