Tu к I a, Homeomorphic conjugates of Fuchsian groups

53

So we will prove that T/H is compact.

Suppose first that H is not isomorphic to a Fuchsian group. Then H can have neither hyperboHc simple axes nor parabolic elements since these would imply the existence of such an isomorphism by (c) and (e). By Theorem 4 D, the 3-manifold T/H must be compact, otherwise Я would have either parabolics or hyperbolic simple axes.

If H is isomorphic to a Fuchsian group, then Я is conjugate to a Fuchsian group by (i) and so we can assume that Я is Fuchsian. If T/H is non-compact, so is D/H (see 4B). However, the definition of a semitriangle group precludes the possibility that D/H is non-compact. This follows since Я is a free product of cyclic groups if D/H is non- compact (in the orientation preserving case in which we are) and in this situation the product of two elements of finite order is of finite order only if the elements are in a finite cyclic subgroup. We include this as

Lemma 6B. Let H be the free product of cyclic groups. If u,veH\{\d} have finite order and are not in a cyclic subgroup of Я, then и v has infinite order.

Proof . Let Я be the free product of Яр / e /, where each Я, is cyclic. By the Kurosh subgroup theorem [M], p. 219, any element of finite order of Я is conjugate to an element of some Я,. Suppose that uv has finite order. We can assume that и e Я^-, V G hHjh~^, and uv e gHf^g~^ for some i, j, ke I and g, he H. It is easy to check by the cancellation rule that if Я, Ф йЯ^.^г"\ then this is impossible. П

6C . In view of Theorem 6 В it would be very interesting to know whether there are such semitriangle groups which are not conjugate to Fuchsian groups. This we do not know although such a group would have some interesting properties: it could not have torsionless subgroups of finite index nor could the manifold T/G be virtually Haken, and any finite-index subgroup contains another semitriangle subgroup. theless we have

Corollary 6 C. Let H be a semitriangle convergence group. Then T/H is compact and if H is a subgroup of another convergence group G, then H is of finite index in G.

Proof We must prove only that T/H is compact. The proof is exactly the same as the one for the compactness of the manifold T/H in the conclusion of the proof of Theorem 6B. D

6D . The Nielsen realization problem. The Nielson realization problem would be a consequence of our theorem if we could prove it for all groups. The reason is that if G is a Fuchsian group such that D/G is compact, and if Я is any finite extension of G, then there is a unique way to realize Я as a group of homeomorphisms of S in such a way that G remains unchanged, see [Z], 41. 2. It is easy to see that Я is a convergence group (since G is and Я is of finite index in G). Suppose that f=/Я/~^ is Fuchsianjor some homeomorphism / of S. Actually, one can extend / to a homeomorphism of D in such a way that G'=fGf~^ is still Fuchsian [DE]. Then G' is geometrically equivalent to G and the extension f of G' corresponds to the extension Я of G which means the solution of Nielsen's problem.

That is why we have been careful not to use the solution of the Nielsen problem by Kerckhoff [K] whenever possible. (Therefore we included separately cases (b) and (i) even if they are subcases of (j).) In Theorem 6B, we use the Nielsen realization problem in cases 0) and (k), cases (a)—(i) being independent of it, and even in (k) it is not needed if the manifold T/Gq is already Haken, as a sHght modification of our proof shows. If we

Зз Journal fur Mathematik Band 391