# when is the Teichmuller space a group?

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

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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.