Bordism invariants vanishes in a lifted twisted $Pin^- \times Spin$-structure

Question

It looks to me that the bordism group $$\Omega_3^{SO} (B(O(2) \times SO(3))) \tag{1}$$ (whose Pontryagin dual for the manifold generator) contains at least a nontrivial invariant: $$ w_1(O(2))\big(w_1(O(2))^2 +w_2(SO(3))\big). \tag{2} $$

Question: Is it true that if we lift this invariant respect to a new bordism group $$ \Omega^{SO}_3 (B((Pin^-(2) \times Spin(3))/ \mathbb{Z}_2)), \tag{3} $$ the invariant eq.(2) becomes zero in bordism group (2)?

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My trial/attempt: The lifting is simply that the $O(2)$ bundle in eq(1) is lifted to $Pin^-(2)$ bundle in eq(3), and the $SO(3)$ bundle in eq(1) is lifted to $Spin(3)$ bundle in eq(3). There is an overall constraint: $$ w_2(O(2))+w_1(O(2))^2=w_2(SO(3)). \tag{4} $$

It also looks that, after I look it up, the $w_3(O(2))=w_1(O(2))w_2(O(2))+\frac{d w_2(O(2))}{2}=0$. But I am not sure that $\frac{d w_2(O(2))}{2}$ is well-defined. By using the fact that the O(2) bundle can be lifted to $Pin^-(2)$ bundle, we shall have this constraint $w_2(O(2))+w_1(O(2))^2=0$. But here we may need to apply eq.(4) instead.

Answer 1

Since you have the constraint $$w_1(O(2))^2+w_2(O(2))=w_2(SO(3)),$$ then $$w_1(O(2))(w_1(O(2))^2+w_2(SO(3)))=w_1(O(2))w_2(O(2))=Sq^1(w_2(O(2)))=w_1(TM)w_2(O(2))=0$$ by Wu formula. ($w_1(TM)=0$ since you are considering oriented bordism.)

$\frac{dw_2(O(2))}{2}=Sq^1(w_2(O(2)))$ by the definition of Bockstein homomorphism.