In physics, it’s very common to utilize the group (exceptional) isomorphism $$ \mathrm{Spin}^{+}(1, 3) \approx \mathrm{SL}(2, \mathbb{C}) $$ in problem solving. I’m working with $\mathrm{Pin}$ groups instead and was wondering if the above generalizes to the isomorphism: $$ \mathrm{Pin}^{+}(1, 3) \approx \mathrm{GL}(2, \mathbb{C}) \,? $$
I haven’t been able to find it anywhere, I would think it’s as simple as dividing each $M \in \mathrm{GL}(2, \mathbb{C})$ by it’s determinant via a map $$ M \rightarrow M' = \det(M)^{-n} M \in \mathrm{SL}(2, \mathbb{C}) \,. $$ I don’t know very much about $\mathrm{Pin}$ groups however, and not sure if that would work on the left-hand side. Can anyone show if this does (or doesn’t) work?
I presume you want to show $\mathrm{SL}(2, \mathbb{C})$ is isomorphic to $\mathrm{GL}(2, \mathbb{C})$. The answer is they are not isomorphic as the mapping $M \rightarrow M/\sqrt{\det(M)}$ is not injective since $M$ and $M/\sqrt{\det(M)}$ is mapped to same matrix inside $\mathrm{SL}(2,\mathbb{C})$. But as @runway44 pointed out in the comments there is problem in making this map a homomorphism due to the fact that square root is not unique but we can have the following isomorphism: Let $K = \{ cI \mid c \in \mathbb{C}, c \neq 0 \}$ where $I$ is the identity matrix. We have that $\mathrm{GL}(2,\mathbb{C}) / \mathrm{SL}(2, \mathbb{C}) \cong K$. This can be done using the map $M \rightarrow \det(M)$.
Please define the $\mathrm{Pin}$ group in the question so that I can make my answer better.