It is known that the universal Teichmuller space $T(1)=\{quasisymmetric \ homeomorphisms \ of \ S^1 \}/ SL (2, \mathbb R)$ is a group. My question is, under what conditions does the Teichmuller space $T(G)$ of a Fuchsian goup $G$ which is finitely generated and of the first kind a group. Or, basically, e.g., (under what conditions does it ture that:) if a quasiconformal homeomorphism $f: \mathbb H \to \mathbb H$ is compatible with $G$ then is $f^{-1}$ also compatible with $G.$ 


I have checked the mathoverflow, and found the following question which is related to the above question: [Conjugate Groups of (quasi) Fuchsian Groups](https://mathoverflow.net/questions/47998/conjugate-groups-of-quasi-fuchsian-groups) 



when is the Teichmuller space a group?
Since you do not say what group operation you have in mind, your question is rather difficult to answer. But what you seem to be proposing in your "Or, basically..." sentence does not work.


For $f : \mathbb{H} \to \mathbb{H}$ to be compatible with $G$ means that the Fuchsian groups $G$ and $f G f^{-1}$ are conjugate under some automorphism of $G$, which means that the points of $T(G)$ represented by those two Fuchsian groups are in the same orbit under the action on $T(G)$ of the mapping class group of the quotient surface $\mathbb{H} / G$ (I am assuming implicitly that $G$ has no torsion and so the quotient is indeed a surface as opposed to an orbifold). The mapping class group of $\mathbb{H} / G$, aka the Teichmuller modular group, is a finitely generated group acting properly discontinuously on $T(G)$, in particular the mapping class group orbit of any point of $T(G)$ is a discrete set. 
You might wish to say that the effect is to identify the orbit of a point of $T(G)$ with the mapping class group itself, and so you might wish to conclude that this puts a group structure on the orbit (this is itself problematical because orbits of group actions need not correspond bijectively to the group; but that is beside the point of your question). The real point is that this identification misses every point of $T(G)$ which is not on that orbit. 


