Let $z\in Z(W)\backslash \{\rm{id}\}$ for $W$ the Weyl group of a root system. I would like to show that it has order 2. Here is my attempted solution:
Since $z\in \rm{GL}(E)$ and leaves $\Phi$ invariant, by a certain result and the fact that $z$ is in the center of $W$, we have that $z\sigma_{\alpha}z^{-1} = \sigma_{z(\alpha)} = \sigma_{\alpha}$ for all $\alpha \in \Phi$. Then \begin{align} z\sigma_{\alpha}z^{-1}(\alpha) &= \sigma_{\alpha}(\alpha) = -\alpha \nonumber\\ &= \sigma_{z(\alpha)}(\alpha) = \alpha -<\alpha,z(\alpha)> z(\alpha) \nonumber \end{align} and therefore $2\alpha = <\alpha,z(\alpha)> z(\alpha)$. Since this holds for all $\alpha \in \Phi$ and $<\alpha,z(\alpha)> \in \mathbb{Z}$, it must be that $z(\alpha) = \pm \alpha$. It was assumed that $z\neq \rm{id}$, therefore $z(\alpha) = -\alpha$ and thus has order 2.
It seems I have shown that this particular $z=-\rm{id}$, but the next problem I'm working on asks me to prove that $z=-\rm{id}$ when $\Phi$ is irreducible. Have I made some illogical conclusions?
Just to flag this as answered: My last argument that $z(\alpha) = -\alpha$ is incorrect, since all we know is it can't be the case that $z(\alpha) = \alpha$ for all $\alpha$, but certainly some of them. The element $z$ has, however, order 2.