Every finite reflection-invariant set of vectors is crystallographic

Let $\mathrm{\Phi }=\{\pm (\cos (2\pi k\mathrm{/}5),\sin (2\pi k\mathrm{/}5)):k=0,1,2,3,4\}\subset {\mathbb{R}}^2$; it is a finite subset of ${\mathbb{R}}^2\setminus \{0\}$ spanning ${\mathbb{R}}^2$, and its ten elements lie on five distinct lines, so $\mathbb{R}\alpha \cap \mathrm{\Phi }=\{\pm \alpha \}$ for each $\alpha \in \mathrm{\Phi }$.

The set $\mathrm{\Phi }$ consists exactly of the unit vectors whose angles are $m\pi \mathrm{/}5$, $m\in \mathbb{Z}\mathrm{/}10\mathbb{Z}$. If $\alpha$ has angle $m\pi \mathrm{/}5$, then $s_{\alpha }$ is reflection in the perpendicular line at angle $m\pi \mathrm{/}5+\pi \mathrm{/}2$. It therefore sends the vector at angle $n\pi \mathrm{/}5$ to the vector at angle $2(\frac{m\pi }{5}+\frac{\pi }{2})-\frac{n\pi }{5}=\frac{(2m+5-n)\pi }{5},$ which again belongs to $\mathrm{\Phi }$. Hence $s_{\alpha }(\mathrm{\Phi })=\mathrm{\Phi }$ for every $\alpha \in \mathrm{\Phi }$.

The integrality axiom fails. Take $\alpha =(1,0)$ and $\beta =(\cos (2\pi \mathrm{/}5),\sin (2\pi \mathrm{/}5))$; then $2(\beta ,\alpha )\mathrm{/}(\alpha ,\alpha )=2\cos (2\pi \mathrm{/}5)=:y$. With $\zeta =e^{2\pi i\mathrm{/}5}$ one has $y=\zeta +{\zeta }^{-1}$ and $1+\zeta +{\zeta }^2+{\zeta }^3+{\zeta }^4=0$, so dividing by ${\zeta }^2$ gives $0={\zeta }^2+\zeta +1+{\zeta }^{-1}+{\zeta }^{-2}=(y^2-2)+y+1$, that is $y^2+y-1=0$ and $y(y+1)=1$. If $y$ were an integer, the integer factor pair $(y,y+1)$ would have to be $(1,1)$ or $(-1,-1)$, neither of which consists of consecutive integers. Hence $y\mathrm{\notin }\mathbb{Z}$.

Thus $\mathrm{\Phi }$ is a finite, spanning, reduced, reflection-invariant set of nonzero vectors whose Cartan integer $y$ is not an integer, so $\mathrm{\Phi }$ is not a crystallographic root system by [L1]; this refutes the claim that reflection invariance alone suffices.