There is a construction of symmetry protected topological (SPT) states which roughly goes as follows. We start with a $d$-dimensional system with symmetry $\mathbb{Z}_2 \times G$ in the phase where the $\mathbb{Z}_2$ is spontaneously broken. To the $\mathbb{Z}_2$ domain walls, we attach a $d-1$ dimensional $G$-SPT, and then condense the bound state to restore the $\mathbb{Z}_2$ symmetry. The result is a nontrivial $\mathbb{Z}_2 \times G$-SPT.
However, for $d=1$, the $d-1$ dimensional $G$-SPTs are just objects carrying nonzero $G$-charge. Naively, it seems that attaching $G$-charge to the $\mathbb{Z}_2$ domain wall and condensing the pair ought to spontaneously break the $G$ symmetry. However, the $1d$ $\mathbb{Z}_2 \times \mathbb{Z}_2$ SPT (the cluster state) seems to be constructed in precisely this way, and doesn't break any symmetries. Why doesn't condensing a bound state involving a symmetry charge lead to symmetry breaking?
I believe there's also a similar construction of the bosonic integer quantum hall state in $d=2$. One starts with a bilayer of bosons with $U(1) \times U(1)$ symmetry, i.e. individually conserved boson numbers. Then, we attach a boson from one layer to a vortex of the other layer and condense the pair, and finally break the symmetry to the diagonal subgroup. Why doesn't the condensation break one of the $U(1)$ symmetries? Is the reason the same as the above case?
Finally, I'd like to ask what is different about the above two scenarios from the situation that occurs at the Neel-VBS transition. My understanding is that the system has a $SO(3) \times C_4$ symmetry, and in the VBS phase where $C_4$ is spontaneously broken, the $C_4$ defect carries a projective $SO(3)$ charge. Condensing the $C_4$ defect to restore $C_4$ then necessarily breaks $SO(3)$, which leads to the Neel phase. This seems consistent with the intuition that condensing symmetry charges leads to symmetry breaking, and I'd like to understand why that intuition doesn't apply in the above two scenarios.