Let us consider a short exact sequence: $$ 1\to N\to G\to Q \to 1, $$ where $N$, $Q$, and $G$ can be continuous Lie groups in general (or finite groups).

Suppose I have the data and the computations of the bordism group $$ \Omega_{d}^{(Spin \times Q)/\mathbb{Z}_2}, $$ where ${(Spin \times Q)/\mathbb{Z}_2}$ means the modification of the $Spin$-structure to a new ${(Spin \times Q)/\mathbb{Z}_2}$-structure. Here the $Spin$ and $Q$ shares a normal subgroup $\mathbb{Z}_2$ that was mod out, such that $$ Spin/\mathbb{Z}_2= SO, $$ or more explicitly $$ Spin(d)/\mathbb{Z}_2= SO(d), $$ where we omit the dimension $d$ through this post in the $Spin \equiv Spin(d)$.

I also have the data and the computations of the bordism group $$ \Omega_{d}^{(Spin \times N)/\mathbb{Z}_2}, $$ where ${(Spin \times N)/\mathbb{Z}_2}$ means the modification of the $Spin$-structure to a new ${(Spin \times N)/\mathbb{Z}_2}$-structure. Here the $Spin$ and $N$ shares a normal subgroup $\mathbb{Z}_2$ that was mod out.

QuestionHow can we compute $$\Omega_{d}^{(Spin \times G)/\mathbb{Z}_2}, $$ where the precise extension from $Q$ to $G$ is given (thus the $G$ is determined and chosen), based on the previously known information of $\Omega_{d}^{(Spin \times Q)/\mathbb{Z}_2}$ and $\Omega_{d}^{(Spin \times N)/\mathbb{Z}_2}$?

Is there some simple way to decompose $\Omega_{d}^{(Spin \times G)/\mathbb{Z}_2}$ into $\Omega_{d}^{(Spin \times Q)/\mathbb{Z}_2}$, $\Omega_{d}^{(Spin \times N)/\mathbb{Z}_2}$ and other things?