How could the "Riemann mapping group" be a group? Here is the definition of Riemann mapping group:
A set of injective holomorphic functions from a closed unit disk 
$f:D\rightarrow\mathbb{C}$ which is holomorphic on the interior and smooth on boundary, normalized by the condition $f(0)=0$ and $f'(0)=1$.
How to define the group multiplication?
There's another story that this group is isomorphic to the group $\mathrm{Diff}_+(S^1)/\mathrm{Rot}(S^1)$, which is a conformal  transformation group in physics.
For $f:D\rightarrow D$, by Schwarz lemma, $f(z)=az$ for some $a\in\mathbb{C}$ with $|a|=1$. We can then write $f=e^{i\theta}$ which is isomorphic to $\mathrm{Rot}(S^1)$. But I don't know the detail about the general $f$.
I saw it at Bartlett's comment.
And I guess the  Kirillov's paper he mentioned is the origin.
 A: As pointed out in the comments, $\text{Diff}_+(S^1) / \text{Rot}(S^1)$ is not a group. To see this, note that $\text{SL}(2, \mathbb R)$ is a subgroup of $\text{Diff}_+(S^1)$: The homomorphism $\varphi: \text{SL}(2, \mathbb R) \rightarrow \text{Diff}_+(S^1)$ given by
$$\varphi(A)(x,y) = \frac{A(x,y)}{\Vert A(x, y) \Vert}$$
is injective.
Now, the subgroup $\text{Rot}(S^1)$ is actually isomorphic to the subgroup $\text{SO}(2, \mathbb R)$ of $\text{SL}(2, \mathbb R)$ in an obvious way. But $\text{SO}(2, \mathbb R)$ not even normal in $\text{SL}(2, \mathbb R)$ (trial and error will give you a counterexample), so it is also not normal in $\text{Diff}_+(S^1)$. Hence, neither the "Riemann mapping group" nor $\text{Diff}_+(S^1)/\text{Rot}(S^1)$ is a group.
A: It all depends on what the meaning of the word "is" is. As is, the space $S$ of holomorphic maps in your question is only a topological space (it has several natural topologies, I will use the uniform topology, for concreteness). [A side issue is that you did not specify what "smooth" means, I assume that it means $C^\infty$.] 
One can meaningfully ask if this space is homeomorphic to a topological group, since we are not given a binary operation on $S$. This question has positive answer. To prove this, one first observes (using the Riemann mapping theorem plus Caratheodori's extension theorem) that $S$ is homeomorphic to the quotient space $Diff_+(S^1)/PSL(2,R)$, where $Diff_+(S^1)$ is the space of orientation-preserving smooth diffeomorphisms of $S^1$. The group  $PSL(2,R)$ acts simply transitively on the set of positively oriented triples of distinct points on $S^1$. Therefore, fixing three distinct points $t_1, t_2, t_3\in S^1$, one sees that $Diff_+(S^1)/PSL(2,R)$ is homeomorphic to the subspace $G$ of  $Diff_+(S^1)$ fixing the set $\{t_1, t_2, t_3\}$ pointwise.    Lastly, once we equip $G$ with the binary operation of composition, one can easily verify that $G$ is indeed a topological group. Thus, $S$ is homeomorphic to a topological group. Whether this helps you or not, I have no idea. 
