The Hamiltonian of Heisenberg's XXY model is given by: $$ H=\sum_{j=1}^{N}\left[S_{j}^{x} S_{j+1}^{x}+S_{j}^{y} S_{j+1}^{y}+\Delta S_{j}^{z} S_{j+1}^{z}\right] ,$$ where $S_{j}^{u}=\sigma_{j}^{u} / 2(u=x, y, z), \sigma_{j}^{u}$ are the Pauli spin- $\frac{1}{2}$ operators on site $j, \Delta$ is the anisotropy parameter, and $\sigma_{j+N}^{u}=\sigma_{j}^{u} .$
The Hamiltonian has two symmetries: (i) a discrete parity $\mathbb{Z}_{2}$ symmetry over the plane $x y: \sigma^{z} \rightarrow-\sigma^{z}$, and (ii) a continuous U1 symmetry that rotates the spins in the $x z$ plane by any angle $\theta$. But why does the $\mathbb{Z}_{2}$ symmetry implies that $\left\langle\sigma_{i}^{z}\right\rangle=0$ and $\left\langle\sigma_{i}^{x} \sigma_{j}^{z}\right\rangle=\left\langle\sigma_{i}^{y} \sigma_{j}^{z}\right\rangle=0$? This was described in this paper, while I cannot see how can I only use the symmetry without knowing the specific form of the eigenvector to get such a result.
