Why does the Coleman-Mermin–Wagner–Hohenberg theorem exclude discrete symmetry?

Question

The Mermin–Wagner theorem (also known as Mermin–Wagner–Hohenberg theorem or Coleman theorem) states that continuous symmetries cannot be spontaneously broken at finite temperature in systems with sufficiently short-range interactions in low dimensions. In particular at zero temperature $T=0$, if we have a quantum system, the theorem states:

The continuous symmetries cannot be spontaneously broken in 1+1d quantum systems.

Apparently, discrete finite symmetries can still be spontaneously broken in 1+1d quantum systems.

Why does the reasoning of the theorem work for the continuous symmetry but not for the discrete finite symmetries? 

If we take a  $\mathbb{Z}_N$ discrete symmetry with a large $N$, it is still not a U(1) continuous symmetry. Why does the argument not work in $\mathbb{Z}_N$?

Answer 1

If a continuous symmetry is spontaneously broken, then the system contains a Goldstone boson, which in not possible in 2d because massless scalar fields have IR divergent behaviour in 2d.

If a discrete symmetry is spontaneously broken, there is no Goldstone boson so the previous argument does not apply.

The most famous system with a spontaneously broken discrete symmetry in 2d is the Ising model ($\phi^4$ field theory)(the discrete symmetry being $\mathbb{Z}/2$).

Comments

sorry $\phi^4$ looks like to be compatible with a larger U(1) symmetry? It can be continuous.
 

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$\phi$ is  a real scalar field so there is no $U(1)$ symmetry (even if $\phi$ were complex, $\phi^4$ is not invariant under a general $U(1)$ rotation $\phi \mapsto e^{i \theta} \phi$).
 

Answer 2

To add to 40227's answer, there's also a domain-wall proof of the theorem, of which the basic intuition is to rotate a finite region of spins with arbitrarily small energy cost, achieved by creating a domain wall of finite thickness, and in there carefully massaging a smooth interpolation between the rotated spin region and the unrotated ones.

If the symmetry group were discrete, there's no way to massage a smooth interpolation, and hence always at least a finite cost for creating domain walls.