Partition function of a $1+1$ D bosonic SPT with $D_4$ symmetry

Question

There should be a nontrivial $1+1$ D bosonic SPT with $D_4$ symmetry, which can be obtained by breaking the $SO(3)$ symmetry of a Haldane chain to $D_4$.

Concretely, write $D_4=Z_4\rtimes Z_2$, denote the generator of $Z_4$ by $r$, and denote the generator of $Z_2$ by $s$. Then $srs=r^{-1}$. Suppose this SPT is put on a triangulated manifold $X$. Denote the gauge connection corresponding to the $Z_4$ by $a\in C^1(X, Z_4)$, and the gauge connection corresponding to $Z_2$ by $b\in C^1(X, Z_2)$. Here $a$ and $b$ are thought of as 1-cochains that represent the flat $Z_4$ and $Z_2$ gauge connections, respectively.

I have a couple of related questions:

1. Is the partition function of this SPT something like $\exp(i\pi\int_X a\cup b)$? I think it somehow makes sense, but I do not fully understand it. In fact, now that $Z_4$ and $Z_2$ do not commute, what does such a cup product mean?
2. If the above (or something like it) is indeed the partition function, how should I obtain it from the known partition function of the parent $SO(3)$ SPT, $\exp(i\pi\int_X w_2(SO(3)))$, where $w_2(SO(3))$ is the second Stiefel-Whitney class of the $SO(3)$ gauge bundle that the parent Haldane chain is coupled to? I understand perhaps I need to first embed this $D_4$ group into $SO(3)$, and then pullback $w_2(SO(3))$. But how should this be done properly?
3. Now that the $Z_4$ and $Z_2$ do not commute, I think there should be a twist if we perform a coboundary operation on $a$. But besides this, is there any other constraint or relation between $a$ and $b$? For example, does $a\cup b$ have any special property?