TITLE: Why doesn't this group have a name?
QUESTION [9 upvotes]: $$\{A\in GL_n(\mathbb{C}) : |det(A)|=1\}$$  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.

REPLY [7 votes]: 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 $\lbrace\text{diag}(z,z^{-1},1,...,1) \mid z \neq 0\rbrace$.