Why doesn't this group have a name? $$\{A\in GL_n(\mathbb{C}) : |det(A)|=1\}$$  This seems to me to be a perfectly natural group to study; it is easy to define and contains $U(n), SL_n$, and all the torsion.    Is there any good reason why this group isn't among the usual classical groups that are so well-understood and thoroughly discussed?  I understand that most of those are studied/defined by looking at groups preserving particular inner products, but it still surprises me that I've never heard of any interesting results/properties of this group.  The only guess I currently have is it's not compact.
I could ask the same question with $\mathbb{R}$ but then "morally" the group is just two copies of $SL_n(\mathbb{R})$  so I understand why it's less interesting.  
A perfectly acceptable answer is that I'm totally misinformed and this group is perfectly understood, classical, named, etc., in which case any reference would be appreciated.  
 A: Denote the group in question by $G$. Then there is a split extension 
$$1 \to SL_n(\mathbb{C}) \to G \xrightarrow{\text{det}} S^1 \to 1$$
where the splitting is given by $S^1 \to G,\; z \mapsto \text{diag}(z,1,...,1)$. Hence, from the group theoretical point of view $G$ is just the semi-direct product 
$$G = SL_n(\mathbb{C}) \ltimes S^1$$
Added: Your guess that $G$ isn't compact (in Euclidean topology) if $n>1$ is correct. For, suppose $G$ is compact. Then, the closed subgroup $SL_n(\mathbb{C}) = \text{det}^{-1}(1)$ is also compact, in contradiction to the fact that it contains the unbounded subset $\lbrace\text{diag}(z,z^{-1},1,...,1) \mid z \neq 0\rbrace$. 
A: This is just a guess: If you're going to sacrifice complex analyticity by allowing complex conjugation in the defining equations, then you should take full advantage of that by using lots of complex conjugates and gaining something important in return, like compactness (as in the case of the unitary group).
A: I guess one answer is there's an isomorphism between your group and 
$$SL_n \mathbb C \times_{\mathbb Z /n\mathbb Z} SO_2$$
My notation means take the product and mod out by the diagonally embedded copy of $\mathbb Z/n \mathbb{Z}$. 
The embedding of $\mathbb Z/n\mathbb Z$ in $SO_2$ is as the cyclic subgroup of order $n$, and the embedding in $SL_n \mathbb C$ is the matrices of the form $\lambda I$ where $\lambda \in S^1$ is an $n$-th root of unity and $I$ is the identity matrix in $SL_n \mathbb C$. 
So it's almost a direct product of two fairly nice groups.    
