Is there SSB in IQHE? The Wikipedia page on FQHE mentions that the discovery of FQH states is significant partly because it shows the limits of Landau's symmetry breaking theory, since different FQH states all have the same symmetries. But isn't it true that IQHE, which was discovered two years earlier, also doesn't fit within the Landau SB paradigm?
Is the IQHE described by Landau's SB theory, or is it a topological order?
 A: IQHE is a topological phase transition, but it’s non-interacting. FQHE is also a topological phase transition, and it involves interactions.
IQHE is therefore a special (“boring”) case of FQHE.
The Ginzburg-Landau theory of phase transitions relies on interactions. The potential energy usually looks like:
$$ V \propto a\phi^2 + b\phi^4,$$
where $a$ gives you the typical SHM-style potential energy while $b$ is usually an interaction$^\dagger$.
Then you write $a= a_0(T-T_c)$ such that for $T<T_c$ the potential doesn’t have a minimum at $\phi=0$ anymore but rather a ring of minima at $\phi = -a_0/b$. Because the state chooses a particular state out of this degenerate ring, the symmetry is broken:

But the ring-minima result in dependent on having this $b\phi^4$ term in the Hamiltonian. Without it, you’d have no phase transition in the Ginzburg-Landau formalism and hence no symmetry broken.
—————
$^\dagger$: what do I mean by interaction? The equation of motion will go as $ \propto \partial_\phi V \propto a\phi + b |\phi|^2\phi$.
In the specific example of a quantum system, $|\phi|^2$ is related to the number (density of particles. So the interaction controlled by $a$ is independent on the number of particles, “boring”, while $b|\phi|^2$ is a true interaction term in that it changes depending on the number of  participants taking part in it.
