Symmetry of the 2d anisotropic Heisenberg model? I am getting confused about the symmetries of the 2d anisotropic Heisenberg model. The Hamiltonian is:
$$H=-\sum_{\langle i,j\rangle}(J_x S_i^xS_j^x+J_yS_i^yS_j^y)\tag{1}$$
I have read (source not publicly available) that this has symmetry $\Bbb{Z}_2$. Which although I agree with - I don't think is the full symmetry. Since as far as I can tell we have the following symmetry generators:


*

*Rotation about $z$ axis by $\pi$.

*Reflection about $xz$ plane

*Reflection about $yz$ plane


which does not appear to be simply $\Bbb{Z}_2$. 
My question is therefore: What is the symmetry of the Hamiltonian (1)?
 A: Short Answer
Yes the symmetry group is larger then $\Bbb{Z}_2$ and is $\mathbf{k_4\cong \Bbb{Z}_2 \times \Bbb{Z}_2}$. But the ground states are only related by $\Bbb{Z}_2$ and it is this symmetry the get's broken in spontaneous symmetry breaking.
Long Answer
Let us look at the individual symmetry groups mentioned in the question and there generators:


*

*Rotation about $z$ axis by $\pi$: This has a single generator given by:
$$\pi_z=\begin{pmatrix} -1 &0 \\ 0 &-1\end{pmatrix}$$

*Reflection about  $xz$ plane: This has a single generator given by:
$$ R_y=\begin{pmatrix} 1 &0 \\ 0 &-1\end{pmatrix}$$

*Reflection about  $yz$ plane: This has a single generator given by:
$$ R_x=\begin{pmatrix} -1 &0 \\ 0 &1\end{pmatrix}$$


Note that since $R_x=\pi_z R_y=R_y \pi_z$ the total symmetry group is given by:
$$\langle \pi_z, R_y\rangle\cong k_4\cong \Bbb{Z}_2 \times \Bbb{Z}_2$$
where $k_4$ is the Klein four-group.
Now we note that the ground states (say all spins in the $+x$ or $-x$ direction) are related by the generator $\pi_z$ and thus the group $\langle \pi_z\rangle \cong \Bbb{Z}_2$. Whilst the generator $R_2$ leaves the ground states invariant. Thus it is this $\langle \pi_z\rangle \cong\Bbb{Z}_2$ which the above is likely talking about. We could equally view this group as being generated by the rotations $R_x$ since these are related by a transformation that leaves the ground states invariant - namely $R_y$.
As a side note spontaneous symmetry the generator $\pi_z$ (or $R_x$) is broken leaving the system with a symmetry $\langle R_y \rangle \cong \Bbb{Z}_2$.
